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r-ruslan [8.4K]
2 years ago
12

The force sensor measures the force on the sensor due to the bumper, but the cart's momentum change arises from the force on the

cart due to the bumper. Which of the following facts are needed to assert that the magnitude of these two forces are nearly equal at all times. 1. Their magnitudes differ by the magnitude of the net force on the bumper. 2. The force of friction is small. 3. The net force on the bumper is small. 4. The cart's weight is canceled by the the normal force exerted by the track.
Physics
1 answer:
Irina-Kira [14]2 years ago
5 0

Answer:

Answered

Explanation:

1 and 3 are necessary

Every bit of force applied to the bumper will be transmitted to the cart EXCEPT for the force needed to accelerate the bumper. This is the net force on the bumper.

If the bumper was heavy then a significant amount of force might be needed to accelerate the bumper so the amount transmitted to the cart would be substantially reduced.

If the net force on the bumper is small then the amount transmitted to the cart is almost the entire force applied.

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A spring with a spring constant of 0.70 N/m is stretched 1.5 m. What was the force?
Talja [164]

Answer:

1.05 N

Explanation:

K = 0.7 N/m

e = 1.5 m

F = ?

from Hooke's law of elasticity

F = Ke

= 0.7×1.5

= 1.05 N

5 0
2 years ago
An object of mass 24kg is accelerated up a frictionless place incline at an angle of 37° with horizontal by a constant force, st
RoseWind [281]

a) Average power: 1425 W

b) Instantaneous power at 3.0 sec: 2850 W

Explanation:

a)

The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation

s=ut+\frac{1}{2}at^2

where

s = 18 m is the displacement along the ramp

u = 0 is the initial velocity

t = 3.0 s is the time taken

a is the acceleration of the object along the ramp

Solving for a,

a=\frac{2s}{t^2}=\frac{2(18)}{(3.0)^2}=4 m/s^2

Now we can apply Newton's second law to find the net force on the object:

F=ma=(24 kg)(4 m/s^2)=96 N

This net force is the resultant of the applied force forward (F_a) and the component of the weight acting backward (mg sin \theta), so we can find what is the applied force:

F=F_a - mg sin \theta\\F_a = F+mg sin \theta = 96+(24)(9.8)(sin 37^{\circ})=237.5 N

where

m = 24 kg is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity

Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:

W=F_a s = (237.6)(18)=4276 J

where s = 18 m is the displacement.

Therefore the average power needed is:

P=\frac{W}{t}=\frac{4276}{3}=1425 W

b)

The instantaneous power at any point of the motion is given by

P=F_av

where

F_a is the force applied

v is the velocity of the object

We already calculated the applied force:

F_a=237.5 N

While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:

v=u+at=0+(4)(3.0)=12.0 m/s

And therefore, the instantaeous power at 3.0 sec is:

P=Fv=(237.5)(12)=2850 W

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0
2 years ago
Write a hypothesis about how the height of the cylinder affects the temperature of the water. Use the "if . . . then . . . becau
IgorLugansk [536]

If the mass of the cylinder increases, the temperature of the water increases, because a greater mass means the cylinder has more potential energy that can be converted to thermal energy, increasing the temperature of the water.


4 0
2 years ago
If the period of a clock signal is 500 ps, the frequency is:_____
NNADVOKAT [17]

Answer:

The frequency of the signal is 2 GHz

Explanation:

Given;

period of the clock signal, T = 500 ps = 500 x 10⁻¹² s

the frequency of the signal is given by;

F= \frac{1}{T}\\\\F = \frac{1}{500*10^{-12}}\\\\F = 2*10^{9} \ Hz

F = 2 GHz

Therefore, the frequency of the signal is 2 x 10⁹ Hz or 2 GHz

4 0
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A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .
nevsk [136]

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

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v is image distance

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm

Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3

So, the magnification of the lens is 3.

3 0
2 years ago
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