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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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Practice questions, will mark brainliest!
andrew-mc [135]

Answer:

266.67Watts

Explanation:

Time = 2.5hr to seconds

3600s = 1hr

2.5hrs = 3600×2.5= 9000s

Force = 32N

Distance = 75km  to m

1000m = 1km

75km = 1000×75 = 75000m

Power = workdone / time

Work = force × distance

Therefore work = 32N × 75000m

Work = 2400000Nm

Power = work ➗ time

Power = 2400000Nm ➗ 9000s

Power = 266.67Watts

Watts is the S. i unit of power

I hope this was helpful, please mark as brainliest

4 0
3 years ago
The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9
kirill [66]

Answer:

Speed of the car 1 =V_1=8.98m/s

Speed of the car 2 =V_2=17.96m/s

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

KE =\frac{1}{2}mv^2

where,

m = mass of the body

v = velocity of the body

thus,

\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

2M_2V_1^2=0.5\times M_2V_2^2

or

2V_1^2=0.5\times V_2^2

or

4V_1^2= V_2^2

or

2V_1= V_2  .................(1)

also,

\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2

or

\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2

or

2(V_1+9.0)^2=(2V_1+9.0)^2

or

\sqrt{2}(V_1+9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)

or

(2V_1-\sqrt{2}V_1)=(12.72-9.0)

or

(0.404V_1)=(3.72)

or

V_1=8.98m/s

and, from equation (1)

V_2=2\times 8.98m/s = 17.96m/s

Hence,

Speed of car 1 =V_1=8.98m/s

Speed of car 2 =V_2=17.96m/s

8 0
3 years ago
Read 2 more answers
A box of weight 74 N is sliding on a horizontal surface at a constant velocity due to an external force F-> of magnitude 4.8
n200080 [17]

Answer

4.8 N

If the box is moving with a constant velocity, then we can say that the system is in equilibrium. This is because if the external force (F->) was greater than other forces the box would be accelerating. This tells us that this force (F->) is just enough to overcome friction and so it must be equal to 4.8 N.

The normal force has no effect to the horizontal velocities or forces. It is equal to -Weight. That is -74 N. The negative sign shows that the force is in opposite direction.

6 0
3 years ago
If you move 50 meters in 10 seconds, what is your speed ​
iogann1982 [59]

Explanation:

My answer didn't save :(

8 0
3 years ago
Two students push on a 5-kg cart from opposite sides
sweet-ann [11.9K]

Answer:

nothing will happen the cart will be broken or as it is

4 0
3 years ago
Read 2 more answers
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