Answer:
object would move but it could be difficult to slow down or stop.
The Lunar Module could make a safe landing if its vertical velocity at impact is 2.8 m/s or less. Use conservation of energy to determine h in each case. The acceleration due to gravity at the surface of the Moon is 1.62 m/s2.
Suppose that you want to determine the greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is zero.
Express your answer to two significant figures and include the appropriate units.
Suppose that you want to determine the greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s downward.
Express your answer to two significant figures and include the appropriate units.
Suppose that you want to determine the greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s upward.
Express your answer to two significant figures and include the appropriate units.
(1) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is zero is 0.4 m.
(2) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s downward is 0.25 m.
(3) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s upward is 0.25 m.
Maximum height
v² = u² - 2gh
where;
v is final velocityu is initial velocitywhen the lander's velocity = 0
0 = u² - 2gh
u² = 2gh
h = u²/2g
h = (2.8²)/(2 x 9.8)
h = 0.4 m
when the velocity of the lander is 1.7 m/s downwardh = (u² - v²)/2g
h = (2.8² - 1.7²)/(2 x 9.8)
h = 0.25 m
when the velocity of the lander is 1.7 m/s upwardh = (u² - v²)/2g
h = (2.8² - 1.7²)/(2 x 9.8)
h = 0.25 m
Thus, the greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is zero is 0.4 m.
The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s downward is 0.25 m.
The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s upward is 0.25 m.
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6. Determine the number of significant figure of:
a. 0.2001 b. 2.0000 c. 243 d. 0.00010300
Answer:
b
Explanation:
the answer is b because if you do the math it equals to b
The Earth’s diameter is about 8,000 miles; our Moon’s diameter is about 2,000 miles; how
many Moon’s would fit inside of the volume of the Sun?
Three moons can fit inside the volume of the sun.
What is the moon?The moon is a non luminous body found in the space. It could cause a solar eclipse when it comes between the sun and the earth.
Since the Earth’s diameter is about 8,000 miles and the Moon’s diameter is about 2,000 miles, to obtain the number of moons that could fit inside the sun we have;
8,000 miles/ 2,000 miles = 3
Hence, three moons can fit inside the volume of the sun.
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Figure 21.62 shows a box on whose surfaces the electric field is measured to be horizontal and to the right. On the left face (3 cm by 2 cm) the magnitude of the electric field is 400 V/m, and on the right face the magnitude of the electric field is 1000 V/m. On the other faces only the direction is known (horizontal). Calculate the electric flux on every face of the box, the total flux, and the total amount of charge that is inside the box.
(a) The electric flux on left face is 0.24 Vm and on the right face is 1 Vm.
(b) The total flux is 1.24 Vm.
(c) The total amount of charge that is inside the box is 1.1 x 10⁻¹¹ C.
Area of the left face
The area of the left face is calculated as follows;
A1 = 0.03 m x 0.02 m = 0.0006 m²
Electric flux on the left faceФ1 = EA1
Ф1 = (400 V/m)( 0.0006 m²) = 0.24 Vm
Let the dimension of the right face = 5 cm by 2 cm
Area of the right faceA2 = 0.05 m x 0.02 m = 0.001 m²
Electric flux on the right faceФ2 = EA2
Ф2 = (1000 V/m)( 0.001 m²) = 1 Vm
Total fluxФ = Ф1 + Ф2
Ф = 0.24 Vm + 1 Vm = 1.24 Vm
Total charge inside the boxФ = Q/ε
Q = εФ
Q = (8.85 x 10⁻¹²)(1.24)
Q = 1.1 x 10⁻¹¹ C
Thus, the electric flux on left face is 0.24 Vm and on the right face is 1 Vm.
The total flux is 1.24 Vm and the total amount of charge that is inside the box is 1.1 x 10⁻¹¹ C.
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On planet Zog, Mr. Spock measures that it takes 1.38 s for a mass of 0.5 kg to hit the ground when released from rest from a height of 2.85 m.
1. Calculate the size of the acceleration of gravity on that planet.
2. He decides to repeat the experiment. Calculate the work he must do to move the mass from the ground back up to its initial height.
The acceleration due to gravity is 3 m/s^2 and the work done is -4.3 J.
What is the acceleration due to gravity?Now we must use the formula;
h = ut + 1/2gt^2
Since it was dropped from a height u = 0 m/s
h = height
u = initial velocity
g = acceleration due to gravity
t = time
h = 1/2gt^2
g = 2h/t^2
g = 2 * 2.85 /(1.38)^2
g = 5.7/1.9
g = 3 m/s^2
The work that must be done is against gravity hence;
W = -(mgh)
W = - (0.5 kg * 3 m/s^2 * 2.85 m)
W = - 4.3 J
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The following table lists the speed of sound in various materials. Use this table to answer the question.
Substance Speed (m/s)
Glass 5,200
Aluminum 5,100
Iron 4,500
Copper 3,500
Salt water 1,530
Fresh water 1,500
Mercury 1,400
Hydrogen at 0°C 1,284
Ethyl Alcohol 1,125
Helium at 0°C 965
Air at 100°C 387
Air at 0°C 331
Oxygen at 0°C 316
Sound will travel fastest in air at _____.
-5°C
0°C
10°C
15°C
Sound will travel fastest in air at 15°C.
Speed of sound in air
The speed of sound in air, given in the range of 100 degrees Celsius and 0 degree Celsius include;
Air at 100°C 387 m/s
Air at 0°C 331 m/s
From the date above, the speed of sound in air increases with increases in temperature. Thus, Sound will travel fastest in air at 15°C.
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A certain force gives mass m1, an acceleration of 12 m/s² and mass m2 an acceleration of 3.3. m/s². What acceleration will this force give to the combined mass?
Acceleration is the rate of change of velocity. Usually, acceleration means the speed is transforming, but not always. When an object moves in a circular path at a steady speed, it is still accelerating, because the focus of its velocity is changing.
a = 4,552 m / s², b) a = 2,588 m / s²
Newton's second law is
F = ma
a = F / m
in this case the force remains constant
indicate us
* for a mass m₁
a₁ = F/m₁
a₁ = 12, m/ s²
* for a mass m₂
a₂= 3.3 m / s²
acceleration m = m₂-m₁
substitute
[tex]$\begin{aligned}&a=\frac{F}{m_{2}-m_{1}} \\&1 / a=\frac{m_{2}}{F}-\frac{m_{1}}{F}\end{aligned}$[/tex]
let's calculate
1/a=1/3.3 - 1/12 = 0.21969
a = 4,552 m / s²
a= 4,552m/s²
What is speed and acceleration?Speed estimates the rate of movement of an object, that is, the distance traveled per unit of time. Acceleration calculates the rate of change of velocity, that is, the change in velocity between two different moments
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Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km.
1. What is the First Cosmic Speed i.e. the speed of a satellite on a low lying circular orbit around this planet? (Planet-X doesn't have any atmosphere.)
2. What is the Second Cosmic Speed i.e. the minimum speed required for a satellite in order to break free permanently from the planet?
3. If the period of rotation of the planet is 16.6 hours, then what is the radius of the synchronous orbit of a satellite?
Solutions for the three problems are is mathematically given as
v=7338.9349[tex]V=14677.86986 \mathrm{~m} / \mathrm{s}[/tex][tex]&r=30581248.06 \times 10^{4} \mathrm{~m} / \mathrm{s}[/tex]What is the First Cosmic Speed i.e. the speed of a satellite on a low-lying circular orbit around this planet?(a) First cosmic speed (arbitral velocily)
[tex]v=\sqrt{\frac{G M}{r}}\\v=\sqrt{\frac{6.67 \times 10^{-11} \times 4.74 \times 10^{24}}{5870 \times 10^{3}}}[/tex]
v=7338.9349
(b) Second cosmic speed (escape velo.)
$$
\begin{gathered}
[tex]V=\sqrt{\frac{2 G M}{r}}\\\\V=\sqrt{2} \sqrt{\frac{G M}{r}}\\\\=V\sqrt{2} \times 7338.9349 \\[/tex]
[tex]V=14677.86986 \mathrm{~m} / \mathrm{s}[/tex]
(c) In conclusion, in a circular orbit, the gravitational force is gets balanced by centripetal force
[tex]&m_{q \omega \omega^{2}}=\frac{G M M}{r^{2}} \\&r^{3}=\frac{G M}{\omega^{2}}=\frac{G M}{4 \pi^{2}} T^{2} \\&r^{3}=\frac{6.67 \times 10^{-11} \times 4.74 \times 10^{24} \times(16.6 \times 3600)^{7}}{4 \pi^{2}} \\[/tex]
[tex]&r=30581248.06 \times 10^{4} \mathrm{~m} / \mathrm{s}[/tex]
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A candle 4.75 cm tall is 40.0 cm to the left of a plane mirror.
a) Where is the image formed by the mirror?
- to the right of the mirror
- to the left of the mirror
b) What is the height of this image?
(Express your answer in centimeters.)
Answer: A candle 4.75 cm tall is 40.0 cm to the left of a plane mirror. Then, the image formed by the mirror will be to the right of the mirror and the height of this image is same as that of the object.
Explanation: To find the answer, we need to know about the reflection at a plane mirror.
What is the reflection at a plane mirror?There are some points about the reflection at a plane surface.
Distance of image from mirror = distance of object from mirror.Size of the image = size of the object.If object moves with certain velocity, then the image also moves with same velocity in opposite direction.Image formed by a plane mirror will be on the opposite side of the object and which is erect, virtual and laterally inverted.Thus, we can conclude that, a candle 4.75 cm tall is 40.0 cm to the left of a plane mirror. Then, the image formed by the mirror will be to the right of the mirror and the height of this image is same as that of the object.
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In response, a plane mirror is 40.0 cm to the left of a candle that is 4.75 cm tall. The mirror's generated image will then be to the right of the mirror and have the same height as the object.
In order to determine the solution, we must understand how a plane mirror reflects light.
What appears in a plane mirror's reflection?There are a few things to note concerning reflections on flat surfaces.The distance of an object from a plane mirror is equal to the distance of its image.Size of the object equals size of the image.If an object moves at a specific speed, the picture will follow suit and move in the opposite direction at the same speed.The other side of the object will have an erect, virtual, and laterally inverted image created by a plane mirror.As a result, we may say that a flat mirror is 40.0 cm to the left of a candle that is 4.75 cm tall. The mirror's generated image will then be to the right of the mirror and have the same height as the object.
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What force pulls a falling apple down to the ground?
A. Spring force
B. Tension
C. Normal force
D. Gravity
Answer:
d
Explanation:
Gravitational force is a type of force that pulls objects to the Earth's surface.
A golf ball is hit so that it leaves the club face at a velocity of 45m/s at an angle of 40° to the horizontal. by ignoring the effect of air resistance and spin on the ball, use the information to answer the question (a) to (d).
a) Calculate the horizontal component of the velocity
b) Determine the vertical component of the velocity.
c) Find the time taken for the ball to reach its maximum height.
d) Calculate the horizontal distance travelled when the ball is at its maximum height.
Answer:a)45cos40 b)45sin40 c) 2.95 d)102
Explanation:
a) [tex]v_{x} = 45 cos40[/tex]
b) [tex]v_{y}=45sin40[/tex]
c) Since air resistance is negligible, the only force acting on the golf ball is gravity. Thus, its vertical acceleration is -g. We know the final velocity must be 0 m/s, because this will be when the golf ball reaches the maximum height and starts to change direction (it falls back to the ground). We also know initial velocity in the vertical direction (see part b). Thus, we can use this equation: [tex]v_{f} = v_{i} + at[/tex].
[tex]0 m/s = 45sin40 + (-9.8m/s^2)t\\t = 2.95s[/tex]
d) The horizontal distance traveled is dependent on (1) how long the ball is in the air and (2) what the horizontal velocity is. (1) was found in part c, and (2) was found in part a.
[tex]x=vt\\x=45cos40(2.95) =102m[/tex]
A thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (vertical axes) which are a distance of 30.0 cm apart horizontally (x-axis), as shown in the figure below which is drawn to scale.
Evaluate the magnitude of the force on the left hand pole.
The magnitude of the force on the left-hand pole is mathematically given as
f'=0.167N
Q=45 degrees
What is the magnitude of the force on the left-hand pole.?Generally, the equation for Force is mathematically given as
F=mg/sinФ
Therefore
F(17.1*10^{-3})*9.8/sin 45
F=0.237N
Considering horizontal axis or plane
f'-fcos=0
Therefore
f'=0.237*cos45
f'=0.167N
In conclusion, the slope
tanФ=30/30
tanФ=1
Ф=45
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The three displacement vectors have magnitude of A=5.00m,B=5.00m and C=4.00m.Find the resultant (magnitude and directional angle) of the three vectors?
3.00 m is the magnitude of the resultant vector
137.1° is the directional angle of the resultant vector.
1) To find the magnitude of the resultant
Discover each vector's individual components first.
We must consider the signs of the components because the angles in the figure are measured in various ways.
In this case, both the y component of vector C and the x component of vector A are negative.
The vectors' components are as follows, using a little trigonometry:
Magnitude of A = 5 m
[tex]A_{x}[/tex] = - (5.00m) cos20° = -5 ×0.408 = -4.698 m
[tex]A_{y}[/tex] = + (5.00m)sin20° = +5 × 0.342 = +1.710 m
Magnitude of B = 5m
[tex]B_{x}[/tex] = +(5.00m)cos60° = 5 × 0.5 = +2.5m
[tex]B_{y}[/tex] = +(5.00m)sin60° = 5 × √3/2 = +4.33 m
Cx = 0
Cy = -4.00m
The sum of all three vectors, which we refer to as R, produces components.
Rₓ = Aₓ + Bₓ + Cₓ
= -4.698 + 2.5 + 0
= -2.198 m
[tex]R_{y}[/tex] = [tex]A_{y} + B_{y} + C_{y}[/tex]
= +1.71 +4.33 - 4.00
= 2.040 m
R = [tex]\sqrt{R_{x^{2} }+R_{y^{2} }[/tex]
= [tex]\sqrt{(-2.198)^{2 } + (2.040)^{2} }[/tex]
= 3.00 m
2) To find the directional angle of resultant
tan θ = 2.040/-2.198 = -0.928
θ = -42.9°
Such a vector would be in the so-called "fourth quadrant," as it is well known. However, we discovered that R has a negative x component and a positive y component, indicating that such a vector must reside in the "second quadrant.
The calculator accidentally returned an angle that is 180 degrees off, thus we must add 180 degrees to the naïve angle in order to get the right angle.
Therefore, R's actual direction is determined by-
θ = -42.9° + 180° = 137.1°
Hence, the magnitude of resultant vector is 3.00 m and directional angle is 137.1°
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A small object with mass 3.95 kg moves counterclockwise with constant speed 1.65 rad/s in a circle of radius 2.95 m centered at the origin. It starts at the point with position vector 2.95î m. Then it undergoes an angular displacement of 9.10 rad.
(a) What is its new position vector? _____ m
(b) In what quadrant is the particle located and what angle does its position vector make with the positive x-axis? ______ quadrant at _______°
(c) What is its velocity? _____ m/s
(d) In what direction is it moving? _____ ° from the +x direction
(e) What is its acceleration? _____ m/s2
(f) Make a sketch of its position, velocity, and acceleration vectors.
a)New position vector in vector form= r = 0.233404 î + 2.94056j m
b) Lies in second quadrant at 161.391°
c)Velocity =4.8675 m/s
d)It is moving in a direction making 161.391° with positive x-direction.
e)Acceleration will be centripetal acceleration (8.031 m/s²).
Given:
Mass of the object m = 3.95 kg
ω=1.65 rad/s
Radius of the circle = 2.95 m
a)
new position vector in vector form
=R cos1.65 î + R sin 1.65 j
= 2.95 cos1.65 î +2.95 sin1.65 j
= 2.95 x 0.07912 î + 2.95 x 0.9968 j
r = 0.233404 î + 2.94056j
b)
Angular Displacement = θ₀ = 9.10 rad
9.10 radian = 180/π× 9/10 degree
= 521.391°
=521.391°- 360°
=161.391°
This will lie in second quadrant.
Angle made with positive x-axis
=161.391°
c)
Velocity
v = ω R
= 1.65 x 2.95
=4.8675 m/s
d)
It is moving in a direction making 161.391° with positive x-direction.
e)
Acceleration will be centripetal acceleration.
= v²/R
=(4.8675)² / 2.95
=23.6925562 / 2.95
=8.031 m/s²
f) Position, Velocity and Acceleration graph:
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a) New position vector is r = 0.233404 î + 2.94056j m
b) Lies in second quadrant at 161.391°
c) Velocity =4.8675 m/s
d) It is moving in a direction making 161.391° with positive x-direction.
e) Acceleration will be centripetal acceleration (8.031 m/s²).
Given:
Mass of the object m = 3.95 kg
ω=1.65 rad/s
The radius of the circle = 2.95 m
a) new position vector in vector form
r =R cos1.65 î + R sin 1.65 j
r = 2.95 cos1.65 î +2.95 sin1.65 j
r = 2.95 x 0.07912 î + 2.95 x 0.9968 j
r = 0.233404 î + 2.94056j
b) Angular Displacement = θ₀ = 9.10 rad
9.10 radian = 180/π× 9/10 degree
= 521.391°
= 521.391°- 360°
=161.391°
This will lie in the second quadrant.
Angle made with the positive x-axis =161.391°
c) Velocity
v = ω R
v = 1.65 x 2.95
v = 4.8675 m/s
d) It is moving in a direction making 161.391° with positive x-direction.
e) Acceleration will be centripetal acceleration.= v²/R
=(4.8675)² / 2.95
=23.6925562 / 2.95
=8.031 m/s²
f) Position, Velocity, and Acceleration graph:
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A 850-kg sports car accelerates from rest to 95 km/h in 6.8 s .
What is the average power delivered by the engine?
Express your answer to two significant figures and include the appropriate units.
From the calculations, the power expended is 43650 W.
What is the power expended?Now we can find the acceleration from;
v = u + at
u = 0 m/s
v = 95 km/h or 26.4 m/s
t = 6.8 s
a = ?
Now
v = at
a = v/t
a = 26.4 m/s/ 6.8 s
a = 3.88 m/s^2
Force = ma = 850-kg * 3.88 m/s^2 = 3298 N
The distance covered is obtained from;
v^2 = u^2 + 2as
v^2 = 2as
s = v^2/2a
s = (26.4)^2/2 * 3.88
s = 696.96/7.76
s = 90 m
Now;
Work = Fs
Work = 3298 N * 90 m = 296820 J
Power = 296820 J/ 6.8 s
= 43650 W
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A crate with a mass of 175.5 kg is suspended from the end of a uniform boom with a mass of 94.7 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall.
Calculate the tension in the cable. (You'll need to get the various positions from the graph. Many are exactly on one of the tic marks.)
323.5 N is the tension in the cable.
Given
Mass of crate(M) = 175.5 kg
Mass of boom(m) = 94.7 kg
The tension, T in the cable can be calculated by taking moments of force about the central point of marked X.
The Angle of the boom with the horizontal can be calculated by
tanθ = 5/10
θ = tan⁻¹(5/10) = 26.56°
Angle of the boom with horizontal is 26.56°
The angle of cable with horizontal can be calculated by
tan B = 4/10
B = tan⁻¹(4/10) = 21.80°
Angle of cable with horizontal is 21.80°
Taking moments of force about the point X
(Mcosθ + mcosθ) 0.5 = T(sin(θ +B)1
(175.5 × cos 26.56 + 94.7 × cos 26.56 )× 0.5 = T (sin(26.56 + 21.80) X 1
By calculating, we get
Tension(T) = 241.68/0.747
Tension(T) = 323.5 N
Hence, 323.5 N is the tension in the cable.
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A thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (vertical axes) which are a distance of 30.0 cm apart horizontally (x-axis), as shown in the figure below which is drawn to scale.
Evaluate the magnitude of the force on the left hand pole.
The magnitude of the force on the left-hand pole of the thin flexible gold chain of uniform linear density with a mass of 17.1 and, hangs between two 30.0 cm long vertical sticks, which are a distance of 30.0 cm apart will be, 0.167N.
To find the correct answer, we have to know more about the Basic forces that acts upon a body.
What is force and which are the basic forces that acts upon a body?A push or a pull which changes or tends to change the state or rest, or motion of a body is called Force.Force is a polar vector as it has a point of application.Positive force represents repulsion and the negative force represented attraction.There are 3 main forces acting on a body, such as, weight mg, normal reaction N, and the Tension or pulling force.How to solve the problem?We have given that, the gold chain hangs between the vertical sticks of 30cm and the horizontal distance between then is 30cm.From the given data, we can find the angle (in the free body diagram, it is given as θ).[tex]tan\alpha =\frac{30}{30}\\ \alpha =45^0[/tex]
From the free body diagram given, we can write the balanced equations of total force along y direction as,[tex]y-direction\\T_2sin\alpha =mg\\\\T_2=\frac{mg}{sin\alpha } =\frac{17.1*10^{-3}*9.8}{sin45} \\\\T_2=0.236N[/tex]
From the free body diagram given, we can write the balanced equations of total force along x direction as,[tex]x-direction\\T_1-T_2cos\alpha =0\\T_1=T_2cos\alpha \\T_1=0.236*cos45=0.167N[/tex]
Thus, we can conclude that, the magnitude of force on the left-hand pole will be 0.167N.
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The force acting on the left-hand pole of a thin, flexible gold chain of uniform linear density weighing 17.1 grams that is suspended between two vertical sticks that are 30.0 cm apart will be 0.167 N in magnitude.
We need to learn more about the fundamental forces that affect a body in order to arrive at the right answer.
What exactly is force, and what are the fundamental forces that affect a body?Force is a push or a pull that modifies or tends to modify the condition, rest, or motion of a body.Given that it has a point of application, force is a polar vector.Repulsion is represented by positive force, and attraction by negative force.A body is subject to three main forces: weight mg, normal response N, and tension or pulling force.How can the issue be resolved?We have given that, the gold chain hangs between the vertical sticks of 30cm and the horizontal space between then is 30cm.We can determine the angle (shown as in the free body figure) using the information provided.[tex]\alpha =tan^{-1}(\frac{30}{30})=45^0[/tex]
We may get the balanced equations for the total force in the y direction from the provided free body diagram as follows:[tex]T_2sin\alpha =mg\\T_2=\frac{mg}{sin\alpha } =0.236N[/tex]
We can create the balanced equations for the total force in the x direction using the provided free body diagram:[tex]T_1=T_2cos\alpha \\T_1=0.167N[/tex]
Thus, we may infer that the force acting on the left-hand pole will have a value of 0.167N.
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If you were in a spaceship watching a ball hover at rest (inside the spaceship) in mid-air, and the spaceship suddenly began rapidly accelerating, what would you see happen to the ball? Why?
Answer:
It will still hover until the spaceship "hits" or exerts a force on it.
Explanation:
Remember, if there is no net force, there is no acceleration or movement.
In this case, our ball is hovering in the spaceship, and in space, we can assume there is no [tex]F_g[/tex], and we can assume there is no [tex]F_N[/tex], nor no forces acting against it.
So, the ball would not move.
However, once the spaceship starts accelerating, the ball would still hover until the spaceship exerts a force on it.
This is because of the same thing as explained above, no forces acting on it, therefore, no acceleration.
Think about it this way.
Imagine you jumped up, then someone threw a ball at you. Now let's imagine you can't move until you hit the floor, meaning that in an ideal situation only [tex]F_g[/tex] is acting on you. Now again, let's imagine time slows really down for you, but not the ball. Before the ball comes and hits you, you are "hovering" like a ball. But after the ball hits you, you move a little because the ball exerted a force on you.
If you did not understand what I meant above, just forget about it, and think about the fact that if there is a Net force (all the force values added up), then there is acceleration and movement.
In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.430 m (1.41 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 15.00 millibar. The atmospheric pressure outside is 940 millibar.
1. Calculate the force required to pull the two hemispheres apart. [Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
2. Two equal teams of horses, are attached to the hemispheres to pull it apart. If each horse can pull with a force of 1450N (i.e., about 326 lbs), what is the minimum number of horses required?
The force required to pull the two hemispheres apart 53696.25N and the minimum number of horses required is 37 .
( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 )
(1)The contact area between the hemispheres is (π x 0.430²) = 0.5805m²
Pressure difference = (940 - 15) = 925millibars.
(925 x 100) = 92,500N/m^2.
(92500 x 0.5805) = 53696.25N. is the force required to part the hemispheres.
(2)[tex]\frac{53696.25N}{1450N}[/tex] = 37 horses for each side .
37 + 37 = 74 horses will be required.
Force is something which can change the motion of an object, stop it or move it, change its shape or size with it. There are two types of forces, contact forces and non-contact forces. Here, it is a contact force at first, then when the horses come it becomes non-contact force.
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a) Calculate the magnitude of the magnetic field at point P due to the current in the semicircular section of wire shown in the figure (Figure 1). (Hint: Does the current in the long, straight section of the wire produce any field at P?)
Express your answer in terms of the variables I , R , and magnetic constant μ0 .
b) Find the direction of the magnetic field at point P.
Hello!
a)
To begin, let me first clarify that no section of the wire along the axis of the point 'P' will contribute to the magnetic field (Aka, the straight part of the wire) because the radius vector and current vectors would be parallel.
Now, let's use Biot-Savart's Law:
[tex]dB = \frac{\mu_0}{4\pi }\frac{id\vec{l}\times \hat{r}}{r^2}[/tex]
B = Magnetic field strength (? T)
μ₀ = Permeability of free space (Tm/A)
i = Current in wire ('I' A)
dl = differential length element
r = distance from wire to point P ('R' m, remains constant!)
We can rewrite Biot-Savart as:
[tex]B = \frac{\mu_0}{4\pi } \int \frac{id\vec{l}\times \hat{r}}{r^2}[/tex]
First, let's mind the cross-product.
The angle between the radius vector (Along 'R') and the current vector is ALWAYS 90° since the two vectors are perpendicular along the arc. At a certain point, think about the current as being "tangential" to the differential length and thus perpendicular to the radius.
Therefore, we can disregard the cross-product since sin(90) = 1.
Let's plug in what we already know, replacing 'dl' with 'ds' since this is an arc:
[tex]B = \frac{\mu_0}{4\pi } \int \frac{ids}{R^2}[/tex]
We have a semi-circle, so we are integrating from 0 to π radians.
[tex]B = \frac{\mu_0}{4\pi } \int\limits^{\pi}_{0} {\frac{ids}{R^2}}[/tex]
Simplifying to make the integral easier, we can take constants out of the integral.
[tex]B = \frac{\mu_0i}{4\pi R^2 } \int\limits^{\pi R}_{0} {} \, ds[/tex]
Evaluating:
[tex]B = \frac{\mu_0i}{4\pi R^2} (\pi R- 0) = \boxed{\frac{\mu_0 i}{4R}}[/tex]
b)
Using the current Right-Hand-Rule at the top of the arc, point your thumb to the right. Curl your fingers as if you are gripping the wire over the top and all the way over to the other side of the wire (Where point 'P' would be). Your fingers should point INTO THE PAGE.
two particles P and Q are shot vertically up. Particle P is first up with a velocity of 40m/s. After 4 seconds, particle Q is also shot up. Find a)where the two particles meet,if at the point of meeting,P has a velocity 15m/s. b ) the velocity with which Q is shot. take g=10m/s^2
The velocity with which the particle Q is fired is 15m/s upwards.
What is velocity?The vector quantity velocity (v), represented by the equation v = s/t, quantifies displacement (or change in position, s) over the change in time (t). Speed (or rate, r) is a scalar number, denoted by the equation r = d/t, that quantifies the distance traveled (d) over the change in time (t).
Assume the upward direction is positive and the downward direction is negative.
Velocity P, = 40m/s
Distance traveled by P,
Using the first equation of motion for particle P,
v = u + at
⇒ 0 = 40 + (-10)t
⇒ t = 4s
This is the time it takes for P to rise
Now, the maximum height(s) reached by the particle P is,
Using the second equation of motion,
s = ut + 1/2at²
⇒ s = 40×4 + 1/2 × (-10) × 4²
⇒ s = 80m
a) A particle P falls when Q is shot after 4 seconds from the initial time:
V² = U² + 2aH₁
⇒ H₁ = 15² - 0/ 2(-10)
⇒ H₁ = 11.25m
Particles P and Q meet at a distance from the ground (H₂).
Height, H₂ = s - H
⇒ H₂ = 80 - 11.25
= 68.75m
Particles P and Q meet at a distance of 68.75m from the ground.
v = u + at₁
⇒ 15 = 0 + (-10) t₁
⇒ t₁ = 1.5s
It is equal to P's fall time and Q's rise time.
b) For particle Q
H₂ = u₂t₁ + 1/2at₁
⇒ 11.25 = u₂ × 1.5 + 1/2 (-10) × 1.5
⇒ u₂ = 15 m/s
Therefore,
The velocity with which the particle Q is fired is 15m/s upwards.
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An object of height 5 cm is kept in front of a
lens. An inverted image of height 5 is formed. identity the lens and position of the
object and image.
Answer:
The type of lens involved is a convex lens and the object is positioned at the center of the curvature of the convex lens.
According to the question, a real, inverted, and same-sized image of the object is formed.
A concave lens is a diverging lens and always forms virtual images. But convex lenses are converging lenses and are the only type of lens that produce real and inverted images of the corresponding objects.
When an object is placed at the center of the curvature of a convex lens, its corresponding image is formed on the opposite side of the convex lens. The image formed is real and inverted.
The distance of the image from the lens is equal to the distance of the object from the lens.
For a convex lens, the distance of the center of curvature from the lens is double the focal length of the lens.
That's why the convex lens and center of curvature are the correct answer to this question.
Explanation:
(These are multiple choice questions)
1) The magnitude of the charge of five electrons is
A. 1.0×10-19 C.
B. 1.6×10-19 C.
C. 6.0×10-19 C.
D. 8.0×10-19 C.
2) Total resistance (in ohms-Ω) between the points A and B of the following circuit is
A. 70 Ω
B. 30 Ω
C. 25 Ω
D. 20 Ω
3) Given the electron configuration of a neutral atom 1s22s22p63s23p5. The atom
A. can become a negative ion easily.
B. is a halogen.
C. has 7 electrons in the outer most shell.
D. all of the above.
4) Identify the number of protons, neutrons and electrons in an atom of
A. 136 protons, 92 neutrons and 92 electrons
B. 92 protons, 136 neutrons and 92 electrons
C. 92 protons, 138 neutrons and 92 electrons
D. 230 protons, 92 neutrons and 92 electrons
5)Which of the following is not an application of Total Internal Reflection?
A. mirage
B. optical fiber
C. prismatic binocular
D. hologram
The application of total Internal Reflection occurs in a mirage. Option A
What is the charge?1) We know that a charge can be positive or negative. If the charge is negative, we call it an electron. If the charge is positive, we call it a proton. Now we know that the magnitude of charge on each electron is 1.6×10-19 C. Hence, the magnitude of charge on five electrons = 5 * 1.6×10-19 C = 8.0×10-19 C. Option D
2) The details of question 2 are not shown hence the question can not be answered.
3) Looking at the electronic configuration of the element; 1s22s22p63s23p5, it is clear to see that it has the ns2 np5 outermost configuration that is common to halogens thus it;
can become a negative ion easily. is a halogen.has 7 electrons in the outer most shell.4) The atom is composed of the protons, neutrons and electrons. Given the nuclide identified as 230/92U we have 92 protons, 138 neutrons and 92 electrons. Option C
5) The application of total Internal Reflection occurs in a mirage. Option A
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In our solar system, which celestial object is known as the dwarf planet?
Answer:
unfournatletly
Explanation:
i have no clue sorry to waste ur time ill rather not say a answer that will be incorrect.
Answer:
Pluto
Explanation:
Pluto was part of our solar system till 2006In 2006 international scientific committee removed it from planets listIt's known as dwarf planet nowCurrents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers.
a) For a line that has current 150 A and a height of 8.0 m above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas.
b) What magnetic field does the line produce at ground level as a percent of the earth's magnetic field, which is 0.50 G .
c) Is this value of magnetic field cause for worry?
Yes. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.
No. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.
Yes. Since this field is much greater than the earth's magnetic field, it would be expected to have more effect than the earth's field.
No. Since this field is much smaller than the earth's magnetic field, it would be expected to have less effect than the earth's field.
(a) The magnetic field the line produced at ground level is 3.75 x 10⁻⁶ T.
(b) The lines magnetic field as a percent of the earth's magnetic field is 7.5%.
(c) No, since this field is much smaller than the earth's magnetic field, it would be expected to have less effect than the earth's field.
Magnetic field the line produced at ground levelB = (μ x I) / (2πr)
B = (4π x 10⁻⁷ x 150) / (2π x 8)
B = 3.75 x 10⁻⁶ T
Percent of the earth's magnetic fieldx = 3.75 x 10⁻⁶ /0.5G
x = (3.75 x 10⁻⁶ ) / (0.5 x 10⁻⁴)
x = 0.075
x = 0.075 x 100% = 7.5%
Thus, we can conclude that, since this field is much smaller than the earth's magnetic field, it would be expected to have less effect than the earth's field.
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what is fundamental quantity
Answer:
The Fundamental Quantity is a physical quantity that exists independently and cannot be expressed by any other physical quantity.
Explanation:
Examples of these could be : Length, electric current and mass
Hope this helps! :)
Can momentum be negative?
A. No.
B. Yes, if it has a very small mass.
C. Yes, if it is moving very slow.
D. Yes, if it is moving backward.
Answer:
D
Explanation:
momentum can be negative as it's a vector quantity. positive momentum indicates that the object is traveling in the positive direction as defined by the coordinate system. negative momentum indicates that the object is moving in the opposite direction (backwards). the momentum equation itself is p = mass x velocity.
and only if you consider velocity a a directed vector, can you have a negative momentum. this equation shows that the momentum is in the same direction as velocity.
so, the sign is only for the direction.
the force is the same (going with the absolute value). a negative momentum is not smaller than a positive momentum.
It is possible that momentum can be negative if an object is moving backward. The correct option is D.
A vector quantity called momentum depends on an object's mass and velocity. It is described as the result of the mass and the velocity of an object.
When an item is moving against a selected positive direction, its velocity will be negative since velocity is a vector variable that comprises both magnitude and direction. As a result, the negative velocity will yield a negative value for momentum when it is calculated.
Thus the correct option is D.
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The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.
1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10-11 Nm2/kg2.)
2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Calculate the speed of the satellite.
Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
Explanation: To find the answer, we need to know about the different equations of planetary motion.
How to find the initial speed of the rock as it left the astronaut's hand?We have given with the following values,[tex]m_g=7.88*10^{18}kg\\r_g=6.32*10^4 m.\\h_{max}=1.44*10^3m.\\[/tex]
We have the expression for the initial velocity as,[tex]v=\sqrt{2gh}[/tex]
Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,[tex]g_g=\frac{GM_g}{r_g^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2} =0.132m/s^2.[/tex]
Now, the velocity will become,[tex]v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s.[/tex]
How to find the speed of the satellite?As we know that, by equating both centripetal force and the gravitational force, we get the equation of speed of a satellite as,[tex]v=\sqrt{ \frac{GM}{r}}=\sqrt{\frac{7.88*10^{18}*6.67*10^{-11}}{1.45*10^5 }} =3.624km/s.[/tex]
Thus, we can conclude that,
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
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1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.
2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.
Understanding the planetary motion equations is necessary in order to determine the solution.
How to determine the rock's original speed when it left the astronaut's hand?The starting velocity's expression is as follows:[tex]v=\sqrt{2gh}[/tex]
So, in order to determine v, we must determine the acceleration of glob caused by gravity. We already have,[tex]a=\frac{GM}{R^2} =0.132m/s^2[/tex]
The velocity will now change to,[tex]v=\sqrt{2*9.8*0.132} =19.46m/s[/tex]
How can I determine the satellite's speed?As we are aware, the centripetal force and gravitational force are equivalent, and thus leads to the following satellite speed equation:[tex]v=\sqrt{\frac{GM}{R} } =3.624km/s\\where, M=7.88*10^{18}kg[/tex]
Consequently, we can say that
1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.
2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.
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A space craft is moving relative to the earth , an observer on the earth finds that, between 1pm and 2pm according to her clock, 3601 seconds elapse on the space craft clock . What is the space craft speed relative to the earth?c=2.998×10^8ms
The speed of the space craft relative to the earth is given as: 0.024c. This is solved using the the equation for time dilation.
What is time dilation?
Time dilation is the "slowing down" of a clock as determined by an observer in relative motion with regard to that clock under the theory of special relativity.
The formula is given as :
Δt = [Δr]/ √ 1 - (v²/c²)
Thus,
v = c√1 - (Δr/Δt)²
= c √(1 - (3600/3601)²
v = 0.024c
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A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 4.05 m/s. If the roof is pitched at 34.0° below the horizon and the roof edge is 2.10 m above the ground, find the time the baseball spends in the air and the horizontal distance from the roof edge to the point where the baseball lands on the ground.
(a) the time the baseball spends in the air (in s)
(b) the horizontal distance from the roof edge to the point where the baseball lands on the ground (in m)
(a) The time the baseball spends in the air is 0.92 s.
(b) The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.
Time spent in air by the baseball
h = vt - ¹/₂gt²
-2.1 = (4.05 x sin 34)t - ¹/₂(9.8)(t²)
-2.1 = 2.26t - 4.9t²
4.9t² - 2.26t - 2.1 = 0
t = 0.92 s
Horizontal distance traveled by the baseballR = Vx(t)
R = (4.05 x cos 34)(0.92)
R = 3.1 m
Thus, the time the baseball spends in the air is 0.92 s.
The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.
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