The small piston of a hydraulic lift has a crosssectional area of 2.87 cm2 and the large piston 314 cm2 . What force must be applied to the small piston for the lift to raise a load of 2.4 kN
Answer:
P1 = P2 pressure is uniform
F1 / A1 = F2 / A2
F1 = F2 (A1 / A2) = 2,400 N * (2.87 / 314) = 21.9 N
How does the gravitational force that the Earth exerts on the Moon which has a much smaller mass compare to the gravitational force that the Moon exerts on the Earth?
a. The gravitational force the Moon exerts on the Earth is larger
b. The gravitational force the Earth exerts on the Moon is larger
C. Both the Earth and the Moon exerts equal gravitational force on each other.
I
A crate (100 kg) is in an elevator traveling upward and slowing down at 6 m/s2. Find the normal force exerted on the crate by the elevator. Assume g
Answer:
380 N
Explanation:
Forces: gravity mg (downward), Normal force N (upward)
Acceleration: Note velocity is up, but slowing down, so acceleration is opposite to velocity, or downward
Newton's 2nd law:
mg - N = ma
N = m(g - a) = 100(9.8 - 6) = 380 N
pls tell the answer asap
For the following distance vs time graph
Answer:
3
5
4
Explanation:
x = (8=(9+9)
(9+9) = 4, 5, 3
it's raining in 60 degree and man is walking in an inclined plain of base angle 30 degree find angle with vertical at which the man must hold his umbrella to be safe
A 100V battery is connected to two oppositely charged plates that are 10cm apart.
i. What is the magnitude of the electric field?
ii. Calculate the Electric Force exerted on a +200 C point charge.
iii. What is the electric potential energy of the charge when it’s 8cm and 2cm from the negatively
charged plate?
iv. How much work was required to move the charge from 8cm to 2cm?
v. If the 10gram point charge was at rest at point A, what is the final speed at point B?
Answer:
A 100V battery is connected to two oppositely charged plates that are 10cm apart.
i. what is the magnitude of the electric field?
ii. Calculate the electric force exerted on a +200uC point charge.
Explanation:
a.
13. A planet orbiting the sun
Newton's 1st Law
b. Newton's 2nd Law
c. Newton's 3rd Law
Newtons first law of motion.
What is the reactance of a 60.0- mH inductor when connected to an AC current source that has a frequency of 180 Hz
Answer:
67.9 Ω
Explanation:
L = 60 mH = 60 x 10⁻³ H = 0.060 H
f = 180 Hz, ω = 2πf = 1131 rad/s
reactance = ωL = 1131 x 0.060 = 67.9 Ω
1.which of the following terms is the measure of the distance between two neighboring Crest or trough?
A.wavelength
B.amplitude
C.hertz
D.Frequency
2.What wave can travel space?
A.mechanical
B.transverse
C.longitude
D.elctromagnet
3.what is being transferred by a wave?
A.energy
B.matter
C.particles
D.light
4.what is the speed of a wave with a frequency of sound?
A.645 m/s
B.576 m/s
C.875 m/s
D.975 m/s
Paki Sagot po
Answer:
A
D
A
Is all the information here?
If so than I do not know
The first 3 are correct though
Explanation:
This interaction is not possible in nature because it violates a conservation law. Which quantity is not conserved?
p+ e-+ --> π+n+ve
Answer:
its D i just did it on the app
Two students are discussing how the speed of the car compares to the speed of the truck when both vehicles are in front of the house. Student 1 says, "The distance traveled by the car and the truck is the same, and the time is the same, so they must have the same speed." Student 2 says, "I don’t see how that can be. The car catches up to the truck, so the car has to be going faster."
a. Which aspects of Student 1’s reasoning, if any, are correct? Support your answer in terms of relevant features of your graphs in part (a).
b. Which aspects of Student 2’s reasoning, if any, are correct? Support your answer in terms of relevant features of your graphs in part (a).
c. Derive an expression for the acceleration of the car. Express your answer in terms of D and vt
d. Determine the time at which the speed of the car is equal to the speed vt of the truck. Express your answer in terms of tD. Justify your answer.
Answer:
a) Student 1 is right in the way he calculates the speed cid d of the truck, since going at constant speed
b) Student 2 is right that for the car to reach the truck it must have some relationship,
c) t = √d/a
Explanation:
In this exercise you are asked to analyze the movement in a mention using kinematics.
a) Student 1 is right in the way he calculates the speed cid d of the truck, since going at constant speed he can use the equation
v = d / t
b) Student 2 is right that for the car to reach the truck it must have some relationship, which is given by
v = v₀ + a t
x = v₀ t + ½ a t²
c) let's use the equations
v = d / t
v = v₀ + at
d / t = vo + at
If we can assume that the car starts from rest, so v₀ = 0
d / t = a t
t² = d / a
t = √d / a
if the speed of the car is not zero the result is a little more complicated
d = vo t + a t²
at² + vo t - d = 0
let's solve the quadratic equation
t = [-v₀ ± √ (v₀² + 4a d)] / 2a
Theweight ofof body is 420N .Calculate its mass
Answer:
42.87
Explanation:
OR
420/9.8 which equals 42.87.
If the orbit of a rocket is a circle around the earth with a radius
= 8 x 106m, and the mass of the earth is 5.97 x 1024 kg while
that of the rocket is 0.5 x 106 kg, then what is the speed of the
rocket under uniform circular motion (in m/s, G = 6.67 * 10-11 N
(m/kg) ?)?
Speed of the rocket under uniform circular motion is 4.97 × [tex]10^7[/tex] m/s.
What is centrifugal force?A fictitious force that moves in a circle and is directed away from the center of the circle is called centrifugal force. When measurements are taken in an inertial frame of reference, the force does not exist. It only becomes relevant when we switch from a ground/inertial reference frame to a rotating reference frame.
As the rocket is in uniform circular motion, the gravitational force due to earth and centrifugal force must be equal in magnitude.
The magnitude of gravitational force due to earth, [tex]F_G = \frac{G M m}{R^2}[/tex]
And, the magnitude of centrifugal force, [tex]F_C = \frac{mv^2}{R}[/tex]
Where,
G = universal gravitational constant = 6.67 * 10-11 N (m/kg)².
M = the mass of the earth = 5.97 x [tex]10^{24[/tex] Kg.
m = the mass of the rocket = 0.5 x [tex]10^6[/tex]kg.
R = the orbit of the rocket = [tex]{8*10^6}[/tex]m.
For uniform circular motion,
[tex]\frac{G M m}{R^2}[/tex] = [tex]\frac{mv^2}{R}[/tex]
⇒ v² = [tex]\frac{GM}{R}[/tex]
⇒v² =[tex]\frac{6.67 * 10^{-11 } *5.97 x 10^{24 } }{8*10^6}[/tex]
⇒ v = 4.97 × [tex]10^7[/tex] m/s.
Hence, the speed of the rocket is 4.97 × [tex]10^7[/tex] m/s.
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A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge density λ = 3.5 nC/m. The point P is located on the postive y-axis at a distance y0 = 15 cm from the origin. The z-axis points out of the screen. 25% Part (a) By symmetry the electric field at point P has no component in the _____________. 25% Part (b) Choose the correct expression for the y-component of the electric field at P due to a thin slice of the rod of thickness dx located at point x. 25% Part (c) Integrate your correct choice in part (b) over the length of the rod and choose the correct expression for the y-component of the electric field at point P. 25% Part (d) Calculate the magnitude of the electric field at point P, in newtons per coulomb.
Answer:
a) The electric field at point P has no component in the x and z directions.
b) dEy = kλdxy₀ / (√( x² + y₀²))^3/2
c)Ey = = 2kλd / y₀( d² + 4y₀²))^1/2
d) Ey = 411.84 N/C
Explanation:
a)
from the uploaded image;
the electric field will only be in y-direction.
Therefore the electric field at P have no component in the x and z directions.
b)
dEy = dEsin θ
dE = kdq / r²
from figure
sinθ = y₀ / √( x² + y₀²)
r = √( x² + y₀²)
dq = λdx
dEy = kdq sinθ / r²
dEy = kλdxy₀ / (√( x² + y₀²))^3/2
c)
the net electric field at p is ,
Ey = ∫^d/2_-d/2 kλdxy₀ / (√( x² + y₀²))^3/2
= kλy₀ ∫^d/2_-d/2 dx / (√( x² + y₀²))^3/2
= 2kλd / y₀( d² + 4y₀²))^1/2
d)
let y₀ = 15 × 10⁻²m, λ = 3.5 nC/m , d = 1.5m
Ey = 2kλd / y₀( d² + 4y₀²))^1/2
Ey = 2(9×10⁹ N.m²/C²)(3.5×10⁻⁹m)(1.5m) / 0.15×(1.5)² + 4(0.15)²))^1/2
Ey = 411.84 N/C
A boat and a trailer are being pulled over an undulating road at a velocity v. The contour of the road is such that it can be approximated by a sine wave having a wavelength of l = 10 ft and an amplitude of Y = 0.5 in. The total static deflection of the springs and tires of the trailer due to the weight of the boat and trailer has been measured as 1.5 in. Assume that the damping inherent in the system is viscous and the damping ratio is 0.05. Determine: (a) The speed v at which the amplitude of the boat and the trailer will be a maximum; (b) The value of this maximum amplitude referred to in part (a); (c) The amplitude when the boat and trailer are traveling at the speed of 55 mph
Answer:
a) v = 25.54 ft/s
b) Xmax = 0.4180
c) Xmax = 0.0048 ft
Explanation:
a) determine the speed v at which the amplitude of the boat and the trailer will be a maximum;
to calculate the distance travelled
S = vt
given an expression of wavelength
l = vt
l = 2πv / ω
ω = 2πv / l
equation for counter of the road
y = Y sin ωt
= Y sin ( (2πv / l)t)
next we calculate the angular frequency
ω = √ ( k/m)
= √ ( g / Sst )
= √( 32.2 / ( 1.5/12))
= 16.05 rad/s
Now we calculate the speed v at which amplitude of the boat will be maximum
ω = 2πv / l
16.05 = (2 × π × v ) / 10
160.5 = 2πv
v = 160.5 / 2π
v = 25.54 ft/s
b)
we calculate the maximum amplitude using the following expression;
X/Y = [ √( 1 + ( 2Sr)²)] / [ √(( 1 - r²)² + ( 2Sr)²))]
Xmax / (0.5/12) = [ √( 1 + ( 2×0.05×1)²)] / [ √(( 1 - 1²)² + ( 2×0.05×r)²))]
Xmax / 0.0416 = 1.004987 / 0.1
Xmax 0.1 = 0.0418
Xmax = 0.0418 / 0.1
Xmax = 0.4180
c)
we convert speed of the boat from mph velocity m/s
v = 55 mph × 1.46667ft/s / 1mph
v = 80.66 ft/s
next we calculate angular velocity
ω = 2πv / l
= 2π × 80.66 / 10
= 50.68 rad/s
next is the frequency ratio
r = 50.68 / 16.05
= 3.16
finally we find the amplitude of the boat
X/Y = [ √( 1 + ( 2Sr)²)] / [ √(( 1 - r²)² + ( 2Sr)²))]
Xmax / (0.5/12) = [ √( 1 + ( 2×0.05×3.16)²)] / [ √(( 1 - 3.16²)² + ( 2×0.05×3.16)²))]
Xmax / 0.0416 = √1.0998 / √ ( 80.74 + 0.0998)
Xmax / 0.0416 = 1.0487 / 8.9910
Xmax 8.9910 = 0.0436
Xmax = 0.0436 / 8.9910
Xmax = 0.0048 ft
What is the atomic number of this element?
A student conducts an investigation on electricity and magnetism. Which relationship will the student discover between the current and the magnetic field strength in a coiled wire?
Answer:
As current increases, the magnetic field strength increases.
So you're correct! :)
As the strength of the magnetic field increases, the current also increases.
From classical electromagnetism, we know that current is induced on conductor owing to relative motion between the conductor and the magnet. This is the principle upon which electromagnetic induction is based.
In the experiment, the student will find that the strength of the magnetic field is directly proportional to the induced current. Hence, as the strength of the magnetic field increases, the current also increases.
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If a→=3i^-2j^-k^ and b→=i^+4j^+k^, find a unit vector n^ normal to the plane containing a→ and b→ such that a→, b→ and n^, in this order, form a right-handed system.
Since vector a = 3i - 2j - k and vector b = i + 4j + k, the unit vector, n, normal to the plane is (7i - j + 7k)/3√11 where a, b and n, in this order, form a right-handed system.
The normal vector to the planeThe normal vector to the plane is c = a × b
Since a = 3i - 2j - k and b = i + 4j + k,
c = a × b
c = (3i - 2j - k) × (i + 4j + k)
= 3i × i + (- 2j) × i + (-k) × i + 3i × 4j + (- 2j) × 4j + (-k) × 4j + 3i × k + (- 2j) × k + (-k) × k
= 3i × i - 2j × i - k × i + 12i × j - 8j × j - 4k × 4j + 3i × k - 2j × k - k × k
Given that
i × j = k, k × i = j, j × k = i, j × i = -k, i × k = -jk × j = -i, i × i = 0, j × j = 0 and k × k = 0. we havec = 3(0) - 2(-k) - (-j) + 12k - 8(0) - 16(-i) + 3(-j) - 2i - 0
c = 0 + 2k + j + 12k - 0 + 16i - 3j - 2i
c = 16i - 2i + j - 3j + 2k + 12k
c = 14i -2j + 14k
Unit vector normal to the plane nThe unit vector normal to the plane n = c/|c| where |c| is the magnitude of c = √(14² + (-2)² + 14²) = √(196 + 4 + 196)
= √396
= √(36 × 11)
= 6√11
So, n = c/|c|
= (14i -2j + 14k)/6√11
= 2(7i - j + 7k)/6√11
= (7i - j + 7k)/3√11
So, the unit vector normal to the plane is (7i - j + 7k)/3√11
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A small object with mass m, charge q, and initial speed v0 = 5.00 * 103 m>s is projected into a uniform electric field between two parallel metal plates of length 26.0 cm (Fig. P21.78). The electric field between the plates is directed downward and has magnitude E = 800 N>C. Assume that the field is zero outside the region between the plates. The separation between the plates is large enough for the object to pass between the plates without hitting the lower plate. After passing through the field region, the object is deflected downward a vertical distance d = 1.25 cm from its original direction of motion and reaches a collecting plate that is 56.0 cm from the edge of the parallel plates. Ignore gravity and air resistance. Calculate the object’s charge-to-mass ratio, q>m
Answer:
q/m = 2177.4 C/kg
Explanation:
We are given;
Initial speed v_o = 5 × 10³ m/s = 5000 m/s
Now, time of travel in electric field is given by;
t_1 = D_1/v_o
Also, deflection down is given by;
d_1 = ½at²
Now,we know that in electric field;
F = ma = qE
Thus, a = qE/m
So;
d_1 = ½ × (qE/m) × (D_1/v_o)²
Velocity gained is;
V_y = (a × t_1) = (qE/m) × (D_1/v_o)
Now, time of flight out of field is given by;
t_2 = D_2/v_o
The deflection due to this is;
d_2 = V_y × t_2
Thus, d_2 = (qE/m) × (D_1/v_o) × (D_2/v_o)
d_2 = (qE/m) × (D_1•D_2/(v_o)²)
Total deflection down is;
d = d_1 + d_2
d = [½ × (qE/m) × (D_1/v_o)²] + [(qE/m) × (D_1•D_2/(v_o)²)]
d = (qE/m•v_o²)[½(D_1)² + D_1•D_2]
Making q/m the subject, we have;
q/m = (d•v_o²)/[E((D_1²/2) + (D_1•D_2))]
We have;
E = 800 N/C
d = 1.25 cm = 0.0125 m
D_1 = 26.0 cm = 0.26 m
D_2 = 56 cm = 0.56 m
Thus;
q/m = (0.0125 × 5000²)/[800((0.26²/2) + (0.26 × 0.56))]
q/m = 312500/143.52
q/m = 2177.4 C/kg
The ratio of the charge to mass of the object will be equal to:
[tex]\dfrac{q}{m}=2177.4\ \frac{C}{kg}[/tex]
what is electric field?
The electric field is defined as the whenever an atom carries a charge whether positive or negative the their influence of force is spread around the charge at a particular distance this effect of force is called as the electric field.
We are given;
Initial speed [tex]V_o[/tex] = 5 × 10³ m/s = 5000 m/s
Now, time of travel in electric field is given by;
[tex]t_1=\dfrac{D_1}{v_o}[/tex]
Also, deflection down is given by;
[tex]d_1=\dfrac{1}{2}at^2[/tex]
Now,we know that in electric field;
F = ma = qE
Thus, [tex]a=\dfrac{qE}{m}[/tex]
So;
[tex]d_1=\dfrac{1}{2}\times(\dfrac{qE}{m})(\dfrac{D_1}{v_o})^2[/tex]
Velocity gained is;
[tex]V_y=(a\timest_1)=(\dfrac{qE}{m})\times (\dfrac{D_1}{v_o})[/tex]
Now, time of flight out of field is given by;
[tex]t_2=\dfrac{D_2}{v_o}[/tex]
The deflection due to this is;
[tex]d_2=V_y\times t_2[/tex]
Thus, [tex]d_2=(\dfrac{qE}{m}\times (\dfrac{D_1}{v_o})\times(\dfrac{D_2}{v_o})[/tex]
[tex]d_2=(\dfrac{qE}{m})\times (\dfrac{D_1\times D_2}{(v_o)^2}[/tex]
Total deflection down is;
[tex]d=d_1+d_2[/tex]
[tex]d=[\dfrac{1}{2}\times(\dfrac {qE}{m}) \times (\dfrac{D_1}{v_o^2})]+[(\dfrac{qE}{m})\times (\dfrac{D_1\times D_2}{v_o^2}})][/tex]
Making q/m the subject, we have;
[tex]\dfrac{q}{m}=\dfrac{(d\times v_o^2)}{[E(\dfrac{D_1^2}{2})+(D_1\times D_2))]}[/tex]
We have;
E = 800 N/C
d = 1.25 cm = 0.0125 m
D_1 = 26.0 cm = 0.26 m
D_2 = 56 cm = 0.56 m
Thus;
[tex]\dfrac{q}{m}=\dfrac{(0.0125\times 5000^2)}{[800((\dfrac{0.26^2}{2})+(0.26\times 0.56))]}[/tex]
[tex]\dfrac{q}{m}= 2177.4[/tex]
Hence the ratio of the charge to mass of the object will be equal to:
[tex]\dfrac{q}{m}=2177.4\ \frac{C}{kg}[/tex]
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How is the speed of an object related to the amount of
kinetic energy the object has?
Answer:
directly proportional to the square of its speed.
Explanation:
;)
A raging bull of mass 700 kg runs at 36 km/h .How much kinetic energy does it have ?
Answer:
Use equation for kinetic energy: Ek=mV²/2
m=700 kg
V=10m/s
Ek=700kg*100m²7s²/2
Ek=35000 J=35kJ
Explanation:
Hope this helps you
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Your friend has been hired to design the interior of a special executive express elevator for a new office building. This elevator has all the latest safety features and will stop with an acceleration of g/3 in the case of an emergency. The management would like a decorative lamp hanging from the unusually high ceiling of the elevator. He designs a lamp which has three sections which hang one directly below the other. Each section is attached to the previous one by a single thin wire, which also carries the electric current. The lamp is also attached to the ceiling by a single wire. Each section of the lamp weighs 7.0 N. Because the idea is to make each section appear that it is floating on air without support, he wants to use the thinnest wire possible. Unfortunately the thinner the wire, the weaker it is. Since he knows that you have taken a course in physics, he asks you to calculate the force on each wire in case of an emergency stop.
Required:
Calculate the force.
Answer:
Sorry, don't understand the question
Explanation:
“Relative” is an important word. Block L of mass mL = 1.90 kg and block R of mass mR = 0.460 kg are held in place with a compressed spring between them. When the blocks are released, the spring sends them sliding across a frictionless floor. (The spring has negligible mass and falls to the floor after the blocks leave it.) (a) If the spring gives block L a release speed of 1.10 m/s relative to the floor, how far does block R travel in the next 0.740 s? (b) If, instead, the spring gives block L a release speed of 1.10 m/s relative to the velocity that the spring gives block R, how far does block R travel in the next 0.740 s?
Answer:m
R
ν
R
+m
L
ν
L
=0 ⇒ (0.500kg)ν
R
+(1.00kg)(−1.20m/s)=0
which yields ν
R
=2.40m/s . Thus , Δx=ν
R
t=(2.40m/s)(0.800s)=1.92m .
(b) Now we have m
R
ν
R
+m
L
(ν
R
−1.20m/s)=0 which yields
ν
R
=
m
L
+m
R
(1.2m/s)m
L
=
1.00kg+0.500kg
(1.20m/s)(1.00kg)
=0.800m/s .
Consequently , Δx=ν
R
t=0.640m .
Explanation:
A 5-kilogram bowling ball is traveling down a lane with a velocity of 40ms. What is the kinetic energy and momentum of the bowling ball
The Kinetic energy of the bowling ball is 4000 J while the momentum of the ball is 200 kg m/s
what is kinetic energy?Kinetic energy, KE is the energy exerted by a body in motion. This is represented mathematically as half the product of mass, m and velocity, v squared.
Kinetic energy
KE = 1 / 2 m v^2
m = 5 kg
v = 40 m/s
KE = 0.5 * 5 *40 ^2
KE = 4000 J
Momentum, M
M = mass * velocity
M = 5 * 40
M = 200 kg m/s
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Which of the following experiments demonstrates Newton's Third Law? Select two answers
a rocket goes forward by burning fuel behind it
A student jumps 2 feet by using his muscles to exert a force on the ground
Matter cannot be created or destroyed
A person claps because they enjoy Physical Science
For the circuit shown in (Figure 1), find the potential difference between points a and b. Each resistor has
R = 160 N and each battery is 1.5 V
The potential difference between points a and b is zero.
Total emf of the series circuit
The total emf in the circuit is the sum of all the emf in the circuit.
emf(total) = 1.5 + 1.5 = 3.0 V
Potential differenceThe potential difference between two points, a and b is calculated as follows;
V(ab) = Va - Vb
V(ab) = 1.5 - 1.5
V(ab) = 0
Thus, the potential difference between points a and b is zero.
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Think of a person standing still at the end of a diving board, getting ready to make a dive into a pool. When the diver is ready, the diver will make 3 small jumps on the diving board, and then dive into the water. Where is the amount of potential energy increasing? Where is the amount of potential energy decreasing and the amount of kinetic energy increasing? Where are the points of maximum kinetic energy? Where are the points of maximum potential energy?
Answer:
The amount of potential energy is increasing when they jump on the board. The amount of potential energy is decreasing and the amount of kinetic energy is increasing when they are falling towards the water. The point of maximum kinetic energy is right before they splash into the water. The point of maximum potential energy is when they jump the highest into the air.
Explanation:
The points of maximum potential energy when they jump highest into the air, their potential energy is at its highest.
What is potential energy?Potential energy develops in systems where the configuration, or relative position of the pieces, determines the amount of the forces they exert on one another. Potential energy directly proportional to height of object higher the object higher the potential energy with respect to same reference point.
When they jump onto the board, potential energy is building. When they are descending toward the sea, their potential energy is dwindling and their kinetic energy is rising. Just before they splash into the water is when they have the most kinetic energy. When they jump highest into the air, their potential energy is at its highest.
Potential energy is highest at its highest point of jump.
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What is the symbol that is used to indicate the fulcrum in a lever?
A triangle is the symbol used to indicate a fulcrum in a lever, a triangle has three sides and three vertices.
What is a fulcrum?A fulcrum is a pivot that has a base and a point that serves as a balance for a lever.
What is a lever?Basically a lever is a simple machine that is a beam with two sides pivoted to fulcrum.
A lever has Load, Effort and the last part is fulcrum
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Which scenario is an example of a scientific way of thinking ?
A example of a scientific way of thinking is making observations, forming questions, making hypotheses, doing an experiment, analyzing the data, and forming a conclusion.