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suter
4 months ago
14

A cart with an unknown mass is at rest on one side of a track. A student must find the mass of the cart by using Newton’s second

law. The student attaches a force probe to the cart and pulls it while keeping the force constant. A motion detector rests on the opposite end of the track to record the acceleration of the cart as it is pulled. The student uses the measured force and acceleration values and determines that the cart’s mass is 0.4kg . When placed on a balance, the cart’s mass is found to be 0.5kg . Which of the following could explain the difference in mass?
Answer choices:

A) The track was not level and was tilted slightly downward.

B) The student did not pull the cart with a force parallel to the track.

C) The wheels contain bearings that were rough and caused a significant amount of friction.

D) The motion sensor setting was incorrect. The student set it up so that motion away from the sensor would be the negative direction.
Physics
1 answer:
Maru [3.3K]4 months ago
5 0

Explanation:

D) The motion sensor was improperly configured. The student arranged it to consider motion away from the sensor as moving in a negative direction.

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help please!! A runner at a speed of 3.5 m/s comes to a stop in 0.15 seconds. What is the deceleration?
Maru [3345]

Answer:

The runner's deceleration is -23.33 \frac{m}{s^{2} }

Given:

Initial speed = 3.5 \frac{m}{s}

Final speed = 0 \frac{m}{s}

Time taken = 0.15 s

To determine:

Deceleration of the runner =?

Used Formula:

Using the first equation of motion,

v = u + at

Where, v = final speed

u = initial speed

a = deceleration

t = duration

Solution:

<pusing the="" first="" equation="" of="" motion="">

v = u + at

Where, v = final speed

u = initial speed

a = deceleration

t = duration

0 = 3.5 + a (0.15)

-3.5 = 0.15 (a)

a = \frac{-3.5}{0.15}

a = -23.33 \frac{m}{s^{2} }

The negative sign indicates that this represents deceleration.

Hence, the deceleration of the runner is -23.33 \frac{m}{s^{2} }

</pusing>
7 0
3 months ago
A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the cen
Yuliya22 [3333]
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Ostrovityanka [3204]

Answer:

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