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mariarad [96]
3 years ago
5

A teacher does a demonstration in front of a class. For the demonstration, a motion sensor is placed at the end of a long, horiz

ontal track of negligible friction. A force sensor is used to pull a cart of known mass along the track, and the motion sensor tracks the motion of the cart. For several positions of the cart, a computer records the reading on the force sensor and the reading on the motion sensor for both the speed of the cart and the distance the cart moves away from the motion sensor. Based on only these data, which of the following scientific hypotheses could a student test?
a. The force exerted on the cart is directly proportional to the time the cart has moved.
b. The gravitational acceleration for an object on Earth is constant near Earth's surface.
с. The net work done on the cart is equal to the change in kinetic energy of the cart.
d. As the force the sensor exerts on the cart increases, so does the acceleration of the cart.
e. Earth's gravitational force is a conservative force.
Physics
1 answer:
pogonyaev3 years ago
8 0

Answer:

с. The net work done on the cart is equal to the change in kinetic energy of the cart.

Explanation:

In the context, a motion sensor detects the motion of the cart where a cart is used to move using a force sensor that is attached to the cart of certain mass. As the cart moves with a speed at a particular time, the computer records both the readings on the force sensor as well as the motion sensor. From these readings the is can be concluded that the work done on the cart is same as the change in the kinetic energy in the cart.

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An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea
melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, v=1.37\times10^7\ \rm m/s

Explanation:

<u>Given:</u>

  • Linear charge density of the ring=0.1\ \rm \mu C/m
  • Radius of the ring R=0.2 m
  • Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C

Potential due the ring at a distance x from the centre of the rings is given by

V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V

Let\Delta U be the change in potential Energy given by

\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s

So the electron will be moving with v=1.37\times10^7\ \rm m/s

5 0
3 years ago
A 50.0 kg woman climbs a flight of stairs 6.00 m high in 15.0 s. How much power does she use.
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Her weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)

Work = (weight) · (height) = (50kg) · (9.8 m/s²) · (6 m)

Power = (work) / (time) = (50kg) · (9.8 m/s²) · (6 m) / (15 s)

Power = (50 · 9.8 · 6 / 15) · (kg · m² / s³)

Power = 196 (kg · m / s²) · (m) / s

Power = 196 Newton-meter/second

<em>Power = 196 watts</em>

6 0
4 years ago
A crowbar makes our work easier by multiplying effort true or false​
OLEGan [10]

Answer:

True

Explanation:

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And first class lever makes work easier by multiplying the effort

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The light coming out of a concave lens will never meet.

So, the answer is A. will never meet.

Happy Studying! ^^
4 0
3 years ago
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