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Alina [70]
2 years ago
8

A student must design an experiment to determine the relationship between the mass of an object and the resulting acceleration w

hen the object is under the influence of a net force. Which of the following experiments should the student conduct in order to determine the relationship between all three quantities?
Answer choices:

A) Drop objects of different masses from a known height above the ground for multiple trials such that they reach their respective terminal speeds. Use a stopwatch to measure the time it takes each object to reach the ground, and record the mass of each object by using a mass scale.

B) Slide objects of different masses across the same rough surface so that each object travels at a constant speed while under the influence of the force of kinetic friction. Then measure the force required to keep each object at a constant speed by using a force sensor, and record the mass of each object by using a mass scale. Perform this experiment multiple times with objects of different masses.

C) Place an object on a rough surface so that the object is at rest. Use a force sensor to exert a force on the object until just after the object overcomes the force of static friction. Record this force. Repeat the experiment for objects of different masses.

D) Slide an object of known mass across a rough surface, using a constant applied force that can be measured by a force sensor. Place a motion detector behind the object so that its speed can be measured as it slides across the surface. Repeat the experiment for different applied forces.
Physics
1 answer:
SSSSS [86.1K]2 years ago
4 0

Answer:

D) Slide an object of known mass across a rough surface, using a constant applied force that can be measured by a force sensor. Place a motion detector behind the object so that its speed can be measured as it slides across the surface. Repeat the experiment for different applied forces.

Explanation:

"The motion detector will provide information about the object’s speed as a function of time as it slides as a result of the applied force. The information about the object’s speed as a function of time can be used to determine the acceleration of the object. The force sensor measures the applied force exerted on the object, and the mass of the object is known. Therefore, this experiment can be used to determine how an object’s mass is related to the net force exerted on the object and the acceleration of the object."

It cannot be A because we need an acceleration will be determined by gravity.

It cannot be B because the term constant speed means that there is no net force, which is required by the initial question.

It cannot be C because the experiment is good for determining the coefficient of friction but not for determining how the mass relates to the acceleration.

It must be D because the object is moving and we have a motion detector, we can graph the acceleration vs time graph. So D allows you to have a lot of the different acceleration values which helps with determining the relationship between acceleration and the mass.

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A child is riding a bike at a speed of 6m/s with a total kinetic energy of 1224J. If the mass of the child is 30kg, what is the
UkoKoshka [18]

Answer:

 Mass of bike = 38 kg.

Explanation:

Kinetic energy is given by the expression, KE = \frac{1}{2} mv^2, where m is the mass and v is the velocity.

Here speed of child riding bike = 6 m/s

Mass of child = 30 kg

Total kinetic energy = 1224 J

Let the mass of bike be, m kg

So, total mass of child and bike = (m + 30) kg

Substituting,

  1224 = \frac{1}{2}* (m+30)*6^2\\ \\ m+30=68\\ \\ m=38kg

So, mass of bike = 38 kg.

3 0
2 years ago
Now that we have a feel for the state of the circuit in its steady state, let us obtain the expression for the current in the ci
vesna_86 [32]

Answer:

i(t) = (E/R)[1 - exp(-Rt/L)]

Explanation:

E−vR−vL=0

E− iR− Ldi/dt = 0

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Separating te variables,

dt/L = di/(E - iR)

Let x = E - iR, so dx = -Rdi and di = -dx/R substituting for x and di we have

dt/L = -dx/Rx

-Rdt/L = dx/x

interating both sides, we have

∫-Rdt/L = ∫dx/x

-Rt/L + C = ㏑x

x = exp(-Rt/L + C)

x = exp(-Rt/L)exp(C)     A = exp(C) we have

x = Aexp(-Rt/L) Substituting x = E - iR we have

E - iR = Aexp(-Rt/L) when t = 0, i(0) = 0. So

E - i(0)R = Aexp(-R×0/L)

E - 0 = Aexp(0) = A × 1

E = A

So,

E - i(t)R = Eexp(-Rt/L)

i(t)R = E - Eexp(-Rt/L)

i(t)R = E(1 - exp(-Rt/L))

i(t) = (E/R)(1 - exp(-Rt/L))

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2 years ago
You drop your keys in a high-speed elevator going up at a constant speed. Part APart complete Do the keys accelerate faster towa
anzhelika [568]

Answer:

Explained

Explanation:

a) No, the keys were initially moving upward in the elevator only effects the initial velocity of the key and not the rate of change of velocity that is acceleration. So, the keys accelerate with the same acceleration as before.

b)Yes, keys will accelerate towards the floor faster if it is a constant speed than it is moving downward because if the elevator is accelerating downward, the downward change in velocity of the keys is at least partially matched by a downward change in the velocity of the of the elevator.

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Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib
Alexus [3.1K]

Answer:

<em>a) 318.2 W/m^2</em>

<em>b) 2.5 x 10^-4 J</em>

<em>c) 1.55 x 10^-8 v/m</em>

<em></em>

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = \pi d^{2} /4 = (3.142 x (2*10^{-3} )^{2})/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>

<em></em>

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>

<em></em>

c) The peak electric field is given as

E = \sqrt{2I/ce_{0} }

where I is the intensity of the beam

E is the electric field

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e_{0} = 8.85 x 10^9 m kg s^-2 A^-2

E = \sqrt{2*318.2/3*10^8*8.85*10^9}  = <em>1.55 x 10^-8 v/m</em>

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Explanation:

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C. Rope strength value

B. Your weight

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