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Katarina [22]
3 years ago
11

A student sets up an experiment to determine the inertial mass of a cart. The student has access to the following measurement eq

uipment: a spring scale, a meterstick, and a stopwatch. The student uses the spring scale to pull the cart starting from rest along a horizontal surface such that the reading on the spring scale is always constant. All frictional forces are negligible. In addition to the spring-scale reading, which two of the following quantities could the student measure with the available equipment and then use to determine the inertial mass of the cart? Select two answers.
Physics
2 answers:
sergiy2304 [10]3 years ago
7 0

Answer:

The answer is choice A and choice D

Explanation:

The answer below is just a joke, don't listen to them.  

:)

Vikki [24]3 years ago
5 0

Answer:

i. The total distance traveled by the cart after it has been in motion.

ii. The average speed of the cart while in motion.

Explanation:

Inertial mass is a measure of an object's resistance to acceleration when a force is applied. So in determining the inertial mass, force and acceleration are important.

From Newton's second law of motion, the force on an object is equal to the objects mass multiplied by it's acceleration;

                       F = m × a

The spring scales determines or measures the value of the force.

The acceleration of the cart can be determined as it moves by the use of the meter stick and stop watch.

With the known values of displacement of the spring and acceleration, then the mass can be calculated.

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Answer:

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where:

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a = acceleration = 5 [m/s^2]

F = (1000*5)

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8 0
4 years ago
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Two identical 0.200kg mass are pressed against opposite ends of a light spring of force constant 1.75N/cm compressing the spring
arlik [135]

This type of a problem can be solved by considering energy transformations. Initially, the spring is compressed, thus having stored something called an elastic potential energy. This energy is proportional to the square of the spring displacement d from its normal (neutral position) and the spring constant k:

E_p=\frac{1}{2}kd^2= \frac{1}{2}175\frac{N}{m}\cdot 0.37^2m^2=11.98J

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The kinetic energy of a mass m moving with a velocity v is given by:

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And we know that the energies are conserved, so the two kinetic energies will equal the elastic potential one:

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E_p =mv^2\implies v=\pm \sqrt{\frac{E_p}{m}}=\pm\sqrt{\frac{11.98J}{0.2kg}}=\pm 7.74\frac{m}{s}

The speed will be 7.74m/s in in one direction (+), and same magnitude in the opposite direction (-).

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3 years ago
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