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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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An automobile tire having a temperature of
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Answer:

Psm = 30.66 [Psig]

Explanation:

To solve this problem we will use the ideal gas equation, recall that the ideal gas state equation is always worked with absolute values.

P * v = R * T

where:

P = pressure [Pa]

v = specific volume [m^3/kg]

R = gas constant for air = 0.287 [kJ/kg*K]

T = temperature [K]

<u>For the initial state</u>

<u />

P1 = 24 [Psi] + 14.7 = 165.47[kPa] + 101.325 = 266.8 [kPa] (absolute pressure)

T1 = -2.6 [°C] = - 2.6 + 273 = 270.4 [K] (absolute Temperature)

Therefore we can calculate the specific volume:

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v1 = (0.287 * 270.4) / 266.8

v1 = 0.29 [m^3/kg]

As there are no leaks, the mass and volume are conserved, so the volume in the initial state is equal to the volume in the final state.

V2 = 0.29 [m^3/kg], with this volume and the new temperature, we can calculate the new pressure.

T2 = 43 + 273 = 316 [K]

P2 = R*T2 / V2

P2 = (0.287 * 316) / 0.29

P2 = 312.73 [kPa]

Now calculating the manometric pressure

Psm = 312.73 -101.325 = 211.4 [kPa]

And converting this value to Psig

Psm = 30.66 [Psig]

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