Answer:
Projectile motion formula or equations derived (In Tabular format)Motion Path equation:
y = (tanθ) x – (1/2) g . x2/(V0 cosθ)2
Explanation:
Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radius of Earth = 6.37×103 km, mass of Earth = 5.98×1024 kg, G = 6.67×10-11 Nm2/kg2.)
The orbiting velocity of the satellite is 4.2km/s.
To find the answer, we need to know about the orbital velocity of a satellite.
What's the expression of orbital velocity of a satellite?Mathematically, orbital velocity= √(GM/r)r = radius of the orbital, M = mass of earthWhat's the orbital velocity of a satellite orbiting earth with a radius 3.57 times the earth radius?M= 5.98×10²⁴ kg, r= 3.57× 6.37×10³ km = 22.7×10⁶mOrbital velocity= √(6.67×10^(-11)×5.98×10²⁴/22.7×10⁶)=4.2km/s
Thus, we can conclude that the orbiting velocity of the satellite is 4.2km/s.
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Piston 1 in the figure has a diameter of 1.87 cm. Piston 2 has a diameter of 9.46 cm.
In the absence of friction, determine the force F, necessary to support an object with a mass of 991 kg placed on piston 2. (Neglect the height difference between the bottom of the two pistons, and assume that the pistons are massless).
Force necessary to support the object on piston 2 is 24× 10⁴ N.
To find the answer, we need to know about the force and pressure on piston 1 and piston 2.
What's the pressure on piston 2?The force on piston 2= mass × acceleration due to gravity= 991 Kg × 9.8 = 9414.5N
Mathematically, force= pressure/areaPressure= force × area of piston= 9414.5N × π(9.46² cm² /4)
= 9414.5N × π(9.46²× 10^(-4)m²/4)
= 66.2 N/m²
What's the force needed to held the mass on piston 2?Pressure on piston 2 = pressure on piston 1Force on piston 1= pressure on piston 1/area of piston 1= 66.2/ π(1.87² cm² /4)
= 66.2/ π(1.87²×10^(-4)m² /4)
= 24× 10⁴ N
Thus, we can conclude that force necessary to support the object on piston 2 is 24× 10⁴ N.
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1. A block of mass 0.4kg resting on the top of an inclined plane of height 20m starts to slide down on the surface of the incline to its foot, and then continues its slide horizontally. At a distance of 5m from the foot of the incline there is another block of the same mass resting on the horizontal surface to undergo an elastic collision. Next to the second block, there is a light spring of constant k = 4000N/m fixed freely against a wall. The spring is supposed to make a head-on collision with the second block. See the arrangements as in 1. Assuming all surfaces being frictionless, (a) calculate the kinetic energy of the firs block just at the foot of the incline; (b) calculate the kinetic and gravitational potential energies of the first block halfway down the incline; (c) calculate the speeds of the two blocks just after their collision; (d) compute the maximum compression of the spring resulted from its collision with the second block; (e) determine the maximum work done by the spring on the second block.
The kinetic energy of the first block just at the foot of the incline is 78.4J, the kinetic and gravitational potential energies of the first block halfway down the incline are same, and which is equal to 39.2J. The speeds of the two blocks just after their collision interchange with the values before collision.
To find the answer, we need to know about the concept of collision and kinetic energy.
How to find the kinetic energy of the first block just at the foot of the incline?Given that, the block of mass 0.4kg resting on the top of an inclined plane of height 20m.Thus, at the top of the incline it has a potential energy, and the kinetic energy will be equal to zero, or we can say that the total energy of the system is equal to the potential energy at topmost point.[tex]TE=KE+PE\\T=PE=m_1gh=(0.4*9.8*20)=78.4 J[/tex]
We have to find the kinetic energy of the first block just at the foot of the incline, and at the bottom point the PE=0, or we can say that the total energy or the potential energy is converted into kinetic energy.[tex]TE=KE=78.4J[/tex]
What is the kinetic and gravitational potential energies of the first block halfway down the incline?At the halfway, the PE will be,[tex]U'=m_1gh'=mg\frac{h}{2} \\U'=39.2J[/tex]
As we know that, the energy is conserved at each point of the motion.[tex]TE=78.4 J\\KE'+PE'=78.4J\\KE'=78-U'=78.4-39.2=39.2J[/tex]
How to find the speeds of the two blocks just after their collision?We have the KE at bottom point as, 78.4J. Thus, the velocity of first block at the bottom before collision will be,[tex]KE=\frac{1}{2} mv^2=78.4J\\v=\sqrt{\frac{2KE}{m} } =4m/s[/tex]
This is the velocity of the block 1 of mass m1 before collision, we can say, u1.As we know that, the 2 nd block of mass m2 is at rest, thus, u2=0.Given that, the collision is elastic. Thus, both the KE and the momentum will be conserved.[tex]\frac{1}{2}m_1u_1^2+ \frac{1}{2}m_2u_2^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2[/tex]
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
We have,[tex]m_1=m_2=m\\u_1=4m/s\\u_2=0\\v_1=?\\v_2=?[/tex]
Substituting this in both the equations, we get,[tex]\frac{1}{2}m*4^2=\frac{1}{2}m(v_1^2+v_2^2)\\(v_1^2+v_2^2)=16[/tex] from resolving KE equation.
[tex]4m=m(v_2+v_1)\\4=v_2+v_1\\v_1=4-v_2[/tex] From resolving momentum conservation.
solving both, we get,[tex]v_2=4m/s\\v_1=0[/tex]
Thus, we can conclude that, the kinetic energy of the first block just at the foot of the incline is 78.4J, the kinetic and gravitational potential energies of the first block halfway down the incline are same, and which is equal to 39.2J. The speeds of the two blocks just after their collision interchange with the values before collision.
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A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at the other end as shown. The cable is attached at a height H=1.70 m above the pivot and the plank's CM is located a distance d=0.700 m from the pivot.
Calculate the tension in the cable.
Hello!
Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.
[tex]\Sigma \tau = 0[/tex]
We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)
- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)
Both of these act in opposite directions. Let's use the equation for torque:
[tex]\tau = r \times F[/tex]
Doing the summation using their respective lever arms:
[tex]0 = L Tsin\theta - dF_g[/tex]
[tex]dF_g = LTsin\theta[/tex]
Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:
[tex]tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o[/tex]
Now, let's solve for 'T'.
[tex]T = \frac{dMg}{Lsin\theta}[/tex]
Plugging in the values:
[tex]T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}[/tex]
At t=0s a small "upward" (positive y) pulse centered at x = 4.0 m is moving to the right on a string with fixed ends at x=0.0m and x = 13.0 m . The wave speed on the string is 3.5 m/s . At what time will the string next have the same appearance that it did at t=0s?
The next time the string will have the same appearance that it did at t=0s is 2.29 s.
Frequency of the wave
v = fλ
f = v/λ
where;
λ is wavelengthhalf of the upward pulse is a quarter of wavelength = ¹/₄ x 4 m = 1 m
f = 3.5/1
f = 3.5 Hz
Time of motion when the pulse is at 4 mt1 = 4/3.5 = 1.143 s
The next time the string will have the same appearance that it did at t=0s.
d = 4 m x 2 = 8 m
t2 = 8/3.5
t2 = 2.29 s
Thus, the next time the string will have the same appearance that it did at t=0s is 2.29 s.
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Part B
A roller coaster ride starts with the roller coaster car being pulled to the top of the first hill with pulley system. The car is
released from the top with an initial velocity close to zero, then accelerates downward. From that first hill, the roller coaster just
coasts; there is no driving force, other than gravity, to keep It going. Assuming no friction, what can you say about the height of
the other hills in the roller coaster ride?
The highest point of a roller coaster is almost always the first hill. In the majority of roller coasters, the hills get smaller as the train travels down the track.
To find the answer, we have to know more about the mechanical energy of a system.
How to find the answer?Since it influences the mechanical energy of the system, the first hill must be the highest.One of the fundamental tenets of physics is that, in the absence of friction, mechanical energy must be conserved. Mechanical energy is the product of kinetic energy and potential energy.When the vehicles ascend the first hill on the roller coaster, mechanical energy is provided to the system because the speed is zero at this point.Mechanical energy = U = mgh
Where m represents the car mass, g represents gravity, and h represents height
If the system is to continue moving, the other hills on the mountain must be lower than the first hill. When the vehicles are released, this energy is converted into kinetic and potential energy when it lowers and ascends, but the sum of these two cannot be larger than the starting energy.Finally, by applying the principle of energy conservation, we may determine that, the initial hill must be the highest.
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The escape velocity of a bullet from the surface of planet Y is 1695.0 m/s. Calculate the escape velocity from the surface of the planet X if the mass of planet X is 1.59 times that of Y, and its radius is 0.903 times the radius of Y.
The escape velocity from the surface of the planet X is 2,249.2 m/s.
Escape velocity of planet X[tex]v = \sqrt{\frac{2GM}{r} } \\\\v^2 = \frac{2GM}{r}\\\\v^2r = 2GM\\\\G = \frac{v^2r}{2M}[/tex]
where;
M is mass of the planetr is radius of the planetG is universal gravitation constant[tex]\frac{v_x^2 \ r_x}{2M_x} = \frac{v_y^2 \ r_y}{2M_y} \\\\\frac{v_x^2 \ r_x}{M_x} = \frac{v_y^2 \ r_y}{M_y} \\\\v_x^2 = \frac{v_y^2 \ r_yM_x}{M_yr_x}\\\\v_x^2 = \frac{(1695)^2 (r_y)(1.59M_y)}{M_y(0.903r_y)} \\\\v_x^2 = 5,058,814.78\\\\v_x = \sqrt{5,058,814.78} \ \ = 2,249.2 \ m/s[/tex]
Thus, the escape velocity from the surface of the planet X is 2,249.2 m/s.
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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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Currents 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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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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Hi I have a question it’s not about the subject but is at the same time what is Physics?
Answer:
the branch of science that is concerned with nature and properties of matter and energy.
Explanation:
a study of the basis of what does what in science.
What is grandfather Paradox?
A grandfather paradox is a situation where individual travels to the past and then introduces a change which affects or contradicts the present.
What is a grandfather paradox?A paradox is a situation or statement which involves two contradictions.
A grandfather paradox is a situation which is defined by the ability of an individual to travel to a time in the past usually before the birth of their grandfather and then introduces a change which affects or contradicts the present. For example, killing the grandfather to prevent their birth.
In conclusion, a grandfather paradox is is an event which contradicts the present as a result of a change done to the past.
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An object 10mm in height is located 20cm from a crown glass spherical surface whose power is +10.00DS. Locate the image
The image is present at 20cm from the crown glass spherical surface.
To find the answer, we need to know about the lens formula.
What's the lens formula?It's (1/V)-(1/U)= (1/f)V= image distance from the lens, U= object distance, f= focal length of the lensWhat's the image distance, if object is present at 20cm from crown glass of power 10DS?Focal length (f)= 1/ power = 1/10 = 0.1 m U= -20cm = -0.2m (-ve sign due to sign convection)(1/V)-(-1/0.2)= (1/0.1)=> (1/V)+5=10
=> 1/V= 5
=> V=0.2m = 20cm
Thus, we can conclude that the image is present at 20cm.
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A 500 N force accelerates an object at 20 m s-2. What is its mass?
Answer: The mass of the object is 25kg.
The given question deals with Newton's second law of motion and its applications.
Explanation: Given force, F=500N
acceleration, a=20 m/[tex]s^{2}[/tex]
From Newton's 2nd law of motion , we have
F=ma where m=mass of the object
⇒500=m×20
⇒m=500/20=25
∴ Mass of the object is 25 kg .
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a body has a mass of 2kg.it accelerats from 20m/s to 40m/s in 4 seconds.the resultant force is
The resultant force is 8N
Given that mass is 2kg , v= 40m/s, u =20m/s and we need to calculate resultant force
F=ma
m is given
so for a
v-u/t=a { first equation of motion }
40-20/4= 4
so a=4
F = ma =2*4 = 8N
The difference between the forces that are acting on an object as part of a system is known as the resultant force.
v = u + at is the first equation of motion. Here, v denotes the end speed, u the starting speed, an acceleration, and t the passage of time. The first equation of motion is provided by the velocity-time relation, which may be used to calculate acceleration.
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Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radius of Earth = 6.37×103 km, mass of Earth = 5.98×1024 kg, G = 6.67×10-11 Nm2/kg2.)
The speed of a satellite in a circular orbit around the Earth is 4,188 m/s.
Speed of the satelliteThe speed of the satellite is calculated as follows;
v = √GM/r
where;
M is mass of Earthr is radius of satellitev = √[(6.67 x 10⁻¹¹ x 5.98 x 10²⁴) / (3.57 x 6.37 x 10⁶)]
v = 4,188 m/s
Thus, the speed of a satellite in a circular orbit around the Earth is 4,188 m/s.
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The thermal emission of the human body has maximum intensity at a wavelength of approximately 9.5 μm.What photon energy corresponds to this wavelength?
Answer:
Explanation:
2.1 x 10^2 - 20J
A constant force F is applied on a body for a time interval of delta t.This force changes the velocity of the body from V1 to V2.then change in momentum in time interval will be
The change in momentum in time interval, given the data will be F × Δt
What is momentum?Momentum is defined as the product of mass and velocity. It is expressed as
Momentum = mass × velocity
What is impulse?This is defined as the change in momentum of an object.
Impulse = change in momentum
But
Impulse = force × time
Therefore
Force × time = change in momentum
How to determine the change in momentumInitial velocity = v₁ Finalvelocity = v₂Force = FChange in time = ΔtChange in momentum = ?Force × time = change in momentum
F × Δt = change in momentum
Change in momentum = F × Δt
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6.
A swimmer bounces straight up from a diving board and falls feet first into a pool. She
starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. (3pt)
a) How long are her feet in the air?
b) What is her highest point above the board?
c) What is her velocity when her feet hit the water?
a) Her feet are in the air for 0.73+0.41 = 1.14 seconds
b) Her highest height above the board is 0.82 m
c) Her velocity when her feet hit the water is 7.16 m/s
Given,t = Time taken
u = Initial velocity = 4 m/s
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
a) Her feet are in the air for 0.73+0.41 = 1.14 seconds
s = ut + [tex]\frac{1}{2}[/tex]at²
2.62 = 0t + [tex]\frac{1}{2}[/tex] ₓ 9.8 ₓ [tex]t^{2}[/tex]
t = 0.73 s
b) Her highest height above the board is 0.82 m
The total height she would fall is 0.82+1.8 = 2.62 m
v = u + at
0 = 4 ₋ 9.8 ₓ t
t = 0.41 s
s = ut +[tex]\frac{1}{2}[/tex] at²
s = 4 ₓ 0.41 ₊ [tex]\frac{1}{2}[/tex] ₓ ₋9.8 ₓ 0.41 [tex]t^{2}[/tex]
c) Her velocity when her feet hit the water is 7.16 m/s
[tex]v = u + at \\v = 0 + 9.8[/tex] ₓ [tex]0.73[/tex]
v = 7.16 m/s
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a)Calculate the effective value of g, the acceleration of gravity, at 7900 m above the Earth's surface.
b)Calculate the effective value of g, the acceleration of gravity, at 7900 km above the Earth's surface.
The solution for the acceleration of gravity is given as
[tex]g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$[/tex][tex]g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$[/tex]This is further explained below.
What is the effective value of g, the acceleration of gravity, at 7900 km above the Earth's surface.?Generally,
Mass of earth [tex]$M=5.97 \times 10^{24} \mathrm{~kg}$[/tex]
Radius of earth [tex]$R=6371 \mathrm{~km}$[/tex]
Gravitational const. [tex]$G=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$[/tex]
height [tex]$h_{1}=7900 \mathrm{~m}=7.9 \mathrm{~km}$$$[/tex]
[tex]R+h_{1}=6371+7.9 &\\\\R+h_{1}=6378.9 \mathrm{~km} \\\\&R+h_{1}=6378.9 \times 10^{3} \mathrm{~m}\end{aligned}[/tex]
In conclusion, acceleration due to gravity at this point will be
[tex]g_{1}=\frac{G M}{\left(\bar{R}+\overline{h_{1}}\right)^{2}}$\\\\$g_{1}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(6378.9 \times 10^{3}\right)^{2}}$\\\\$g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$[/tex]
for [tex]$h_{2}=7900 \mathrm{~km}$[/tex]
[tex]R+h_{2}=6371+7900\\\\R+h_{2}=14271 \mathrm{~km}\\\\$g_{2}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(14271 \times 10^{3}\right)^{2}}$\\\\$g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$[/tex]
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A 45-kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.
_______m
The altitude or height of the pole vaulter as she crosses the bar is 4.04 m.
What is the height of the pole vaulter?The height of the pole vaulter is determined from the change in kinetic energy which is equal to the potential energy at that height.
Potential energy = Change in kinetic energymgh = m(v - u)²/2h = (v - u)²/2g
h = (10 - 1.1)²/2 * 9.8
h = 4.04 m.
In conclusion, the height is determined from the potential energy at that height.
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Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.17 times a second. A tack is stuck in the tire at a distance of 0.351 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed
The tangential speed of the wheel is determined as 4.786 m/s.
Tangential speed of the wheel
The tangential speed of the wheel is calculated as follows;
v = ωr
where;
ω is angular speed in rad/sr is radius of the circular pathv = (2.17 x 2π rad)/s x 0.351 m
v = 4.786 m/s
Thus, the tangential speed of the wheel is determined as 4.786 m/s.
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Consider a message signal m(t) with the spectrum shown in the following Figure. The message signal bandwidth W=1KHz, This signal is applied to a product modulator, together with a carrier Accos(2πfc t)wave producing the DSB-SC modulated wave S(t). This modulated wave is next applied to a coherent detector. Assuming perfect synchronism between the carrier waves in the modulator and detector, determine the spectrum of the detector output when: (a) The carrier frequency fc=1.25 KHz. And (b) The carrier frequency fc=0.75 KHz. What is the lowest carrier frequency for which each component of the modulated wave S(t) is uniquely determined by m(t)?
The lowest carrier frequency for which each component of the modulated wave S(t) is uniquely determined by m(t) is
[tex]-Fc,-Fc+W,-Fc-wW = > -1.5k,-0.5k,-2.5kHz[/tex]"-w,w -1kHz,1kHz" are the frequency components of the detector output in this case.
[tex]-Fc,-Fc+w,-Fc-W = > -0.75k,0.25k,-1.75kHz[/tex]This is further explained below.
What is the lowest carrier frequency for which each component of the modulated wave S(t) is uniquely determined by m(t)?Generally, the equation for the modulated wave containing frequency components is mathematically given as
[tex]Fc,Fc+w,Fc-W = > 1.5k,2.5k,0.5kHz[/tex]
[tex]-Fc,-Fc+W,-Fc-wW = > -1.5k,-0.5k,-2.5kHz[/tex]
"-w,w -1kHz,1kHz" are the frequency components of the detector output in this case.
b)
In conclusion,, then the modulated wave contains frequency components as
[tex]Fc,Fc+W,Fc-W = > 0.75k,1.75k,-0.25kHz[/tex]
[tex]-Fc,-Fc+w,-Fc-W = > -0.75k,0.25k,-1.75kHz[/tex]
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A solid cylinder of uniform density of 0.85 g/cm3 floats in a glass of water tinted light blue by food coloring.
Its circular surfaces are horizontal. What effect will the following changes, each made to the initial system, have on X, the height of the upper surface above the water? The liquids added do not mix with the water, and the cylinder never hits the bottom.
1. The cylinder is replaced with one that has the same density and diameter, but with half the height.
2. Some of the water is removed from the glass.
3. A liquid with a density of 1.06 g/cm3 is poured into the glass.
4. The cylinder is replaced with one that has the same height and diameter, but with density of 0.83 g/cm3.
5. A liquid with a density of 0.76 g/cm3 is poured into the glass.
6. The cylinder is replaced with one that has the same density and height, but 1.5× the diameter.
Options are: Increase, Decrease, No change
The buoyant force acting on the cylinder is, [tex]Fb = \rho Ahg[/tex]. Here A is the cross-sectional area of the cylinder, h is the height of the cylinder, ρ is the density of the cylinder, and g is the acceleration due to gravity.
This buoyant force is also equal to the volume of the fluid displaced. [tex]Fb = \sigma h(A-x)g[/tex]. Here σ is the density of the fluid.
Equate the above two equations and solve for x.
[tex]\rho Ahg = \sigma A(h-x)g\\\rho h = \sigma h - \sigma x\\x = \frac{(\sigma - \rho)h}{\sigma}[/tex]
So, the distance x depends on the density of the fluid, density of the cylinder and the height of the cylinder.
1. The density of the cylinder is same and distance x is independent of the diameter of the cylinder. Therefor, there will be no change in the distance x. Hence, the correct answer is No change.
2. Now the height is changing keeping the density same. As the distance x is directly proportional to the height, the distance x will increase.
3. The density of the added liquid is greater than of the water and it does not mix with the water. So, the liquid will settle down and there will be no change in the distance x.
4. The density of the added liquid is less than that of the water and it does not mix with the water. So, the liquid will not settle down and the distance x will change. The change in distance x can be determined as follow:
[tex]\rho Agh = \sigma' Axg + \sigma A(h-x)g\\\rho h=\sigma' x + \sigma h - \sigma x\\x=(\frac{\sigma - \rho}{\sigma - \sigma'})h[/tex]
Here, σ' is the density of the added liquid.
From the above relation it is clear, that on adding the liquid of the density less than that of water, the denominator term become small ando so the value of x will increase.
5. On removing some of the water inside the glass, the height of the water column will decrease, but the value of x does not depend on the height of the water column. So, there will be no change in the distance x.
6. The density of the new cylinder is smaller than that of the earlier one. So, the numerator term will increase. Therefore, the value of x will increase.
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1.A virtual image is formed 20.5 cm from a concave mirror having a radius of curvature of 31.5 cm.
(a) Find the position of the object.
(b) What is the magnification of the mirror?
2.A contact lens is made of plastic with an index of refraction of 1.45. The lens has an outer radius of curvature of +2.04 cm and an inner radius of curvature of +2.48 cm. What is the focal length of the lens?
The position of the object is = -68cm
The magnification of the mirror= 0.3
Calculation of object distanceThe image distance = 20.5cm
The focal length= R/2 = 31.5/2= 15.75
The object distance= ?
Using the lens formula,1/f = 1/v-1/u
1/u = 1/v- 1/f
1/u = 1/20.5 - 1/15.75
1/u = 0.0489- 0.0635
1/u = -0.0146
u = -68cm
The magnification of the mirror is image size/object size
= 20.5cm/-68cm
= 0.3
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Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radius of Earth = 6.37×10^3 km, mass of Earth = 5.98×10^24 kg, G = 6.67×10^-11 Nm^2/kg^2.)
Answer: The speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth is 4188.11 m/s.
Explanation: To find the answer, we need to know about the equation of motion of a satellite around earth.
What is the equation of motion of a satellite around earth?We have gravitational force of attraction between the satellite of mass m and earth of mass M as,[tex]F_g=\frac{GMm}{r^2}[/tex]
The expression for centripetal force of,[tex]F_c=\frac{mv^2}{r} \\[/tex]
These two forces are equal for a satellite around earth.[tex]\frac{GMm}{r^2} =\frac{mv^2}{r} \\thus,\\v=\sqrt{\frac{GM}{r} }[/tex]
How to solve the problem?Given that,[tex]r=3.57 R_E=3.57*6.37*10^3=22.74*10^3 km\\M=5.98*10^24kg\\G=6.67*10^{-11}Nm^2/kg^2[/tex]
Thus, the speed of the satellite will be,[tex]v=\sqrt{\frac{6.67*10^{-11}*5.98*10^{24}}{22.74*10^6m} } =4188.11 m/s[/tex]
Thus, we can conclude that the speed of satellite will be 4188.11 m/s.
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In a circular orbit with a radius that is 3.57 times the mean radius of the Earth, a satellite moves at a speed of 132.43km/s.
In order to get the solution, we must understand the satellite's planetary motion equation.
What is the satellite's orbital motion equation?The earth's mass M and the satellite's mass M are attracted to one another by gravity.[tex]F_g=\frac{GMm}{r^2}[/tex]
The term used to describe centripetal force of,[tex]F_c=\frac{MV^2}{r}[/tex]
When a satellite orbits the earth, these two forces are equivalent. Thus, the velocity will be,[tex]\frac{GMm}{r^2}=\frac{mV^2}{r}\\V=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10{-11}*5.98*10^{24}}{22.74*10^3} } \\V=132.43*10^3m/s[/tex]
As a result, we may estimate that the satellite will move at a speed of 132.43 km/s.
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Question 2
A photon of green light has a wavelength of 520 nm. Find the green photon's frequency in Hz?
Hints: C=fa ; this will give you the frequency in Hz; 1 nm = 1x10-⁹ nm
5.77 ×[tex]10^1^4[/tex] Hz is the green photon's frequency .
The distance between similar points (adjacent crests) in adjacent cycles of a waveform signal that is propagated in space is known as the wavelength. A wave's wavelength is often measured in meters (m), centimeters (cm), or millimeters (mm) (mm). The relationship between frequency and wavelength is inverse.
Given:Wavelength of green light = 520 nm
f = c / λ
where, f = Frequency
c = Speed of light = 3 × [tex]10^8[/tex] m/s
λ = Wavelength of light
∴ f = c / λ
f = [tex]\frac{3*10^8}{520 * 10^-^9}[/tex]
= 5.77 ×[tex]10^1^4[/tex] Hz
Therefore, 5.77 ×[tex]10^1^4[/tex] Hz is the green photon's frequency .
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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:
A man pushing a crate of mass
m = 92.0 kg
at a speed of
v = 0.845 m/s
encounters a rough horizontal surface of length
ℓ = 0.65 m
as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.351 and he exerts a constant horizontal force of 280 N on the crate.
(a) Find the magnitude and direction of the net force on the crate while it is on the rough surface.
magnitude_____N
What is the direction?
1. Opposite as the motion of the crate
2. Same as the motion of the crate
(b) Find the net work done on the crate while it is on the rough surface.
______J
(c) Find the speed of the crate when it reaches the end of the rough surface.
_______m/s
(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
(b) The net work done on the crate while it is on the rough surface is 23.7 J.
(c) The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
Magnitude of net force on the crateF(net) = F - μFf
F(net) = 280 - 0.351(92 x 9.8)
F(net) = -36.46 N
Net work done on the crateW = F(net) x L
W = -36.46 x 0.65
W = - 23.7 J
Acceleration of the cratea = F(net)/m
a = -36.46/92
a = - 0.396 m/s²
Speed of the cratev² = u² + 2as
v² = 0.845² + 2(-0.396)(0.65)
v² = 0.199
v = √0.199
v = 0.45 m/s
Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
The net work done on the crate while it is on the rough surface is 23.7 J.
The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
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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.
The initial speed of the rock as it left the astronaut's hand is 168 m/s.
The speed of the satellite is 60.2 m/s.
Acceleration due to gravity of the satellite
g = GM/R²
where;
M is mass of the satelliteR is radius of the satelliteg = (6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(6.32 x 10⁴)²
g = 0.132 m/s²
initial speed of the rock when it reaches maximum heightv² = u² - 2gh
0 = u² - 2gh
u² = 2gh
u = √2gh
u = √(2 x 9.8 x 1440)
u = 168 m/s
Speed of the satellitev = √GM/r
v = √[(6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(1.45 x 10⁵)]
v = 60.2 m/s
Thus, the initial speed of the rock as it left the astronaut's hand is 168 m/s.
The speed of the satellite is 60.2 m/s.
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