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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 uniform Rectangular Parallelepiped of mass m and edges a, b, and c is rotating with the constant angular velocity ω around an
Sonbull [250]

Answer:

(a) k = \frac{Mw^{2} }{6} (a^{2} +b^{2} )

(b)  τ = \frac{M}{3} (a^{2} +b^{2} ) ∝

Explanation:

The moment of parallel pipe rotating about it's axis is given by the formula;

I = \frac{M}{3} (a^{2} +b^{2} )   ---------------------------------1

(a) The kinetic energy of a parallel pipe is also given as;

k =\frac{1}{2} Iw^{2} --------------------------------2

Putting equation 1 into equation 2, we have;

k = \frac{M}{6} (a^{2} +b^{2} )w^{2}

k = \frac{Mw^{2} }{6} (a^{2} +b^{2} )

(b) The angular momentum is given by the formula;

τ = Iw -----------------------3

Putting equation 1 into equation 3, we have

τ = \frac{Mw}{3} (a^{2} +b^{2} )

But

τ = dτ/dt = \frac{M}{3} (a^{2} +b^{2} )\frac{dw}{dt}   ------------------4

where

dw/dt = angular acceleration =∝

Equation 4 becomes;

τ = \frac{M}{3} (a^{2} +b^{2} ) ∝

8 0
2 years ago
A sample of a gas occupies a volume of 90 mL at 298 K and a pressure of 702 mm Hg. What is the correct expression for calculatin
aleksandr82 [10.1K]

Answer:

Explanation:

Given

volume of sample V_1=90\ mL

Temperature T_1=298\ K

Pressure P_1=702\ mm\ of\ Hg

for different conditions

Temperature T_2=273\ K

Pressure P_2=760\ mm\ of\ Hg

suppose V_2 is the volume of sample

Using ideal gas equation

PV=nRT

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

\frac{702\times 90}{298}=\frac{760\times V_2}{273}

V_2=76.15\ mL

             

8 0
2 years ago
A 0.110 kg cube of ice (frozen water) is floating in glycerine. The glycerine is in a tall cylinder that has inside radius 3.70
Sonbull [250]

Answer:

the distance by which the height of the  liquid in the cylinder change after the ice gets melted = 0.528 cm

Explanation:

The change in volume of glycerin when the ice cube is placed on the surface of the glycerin can be represented as:

V = \frac{m}{ \rho}

Given that ;

the mass of the ice cube (m) = 0.11 kg = 0.110 × 10³ g

density of the glycerine (\rho) = 1.260 kg/L = 1.260 g/cm³

Then:

V = \frac{0.110*10^3 \ g}{1.260 \ g/cm^3}

V = 0.0873*10^3 \ cm^3 (\frac{1L}{10^3 cm^3})

V = 0.0873 L

Now;Initially the volume of the glycerin before the ice cube starts to melt is:

V_1 = V_i + V\\\\V_1 = V_i+ 0.0873 \ L

However; the volume of the water produced by the 0.11 kg ice cube = 0.11*10^3 \ cm^3

The expression for change in the volume of glycerin after the ice cube starts to melt is as follows:

V_2 = V_i + V"

replacing V" with 0.11*10^3 \ cm^3 ; we have:

V_2 = V_i (0.11*10^3 \ cm^3 )(\frac{1 \ L }{10^3 \ cm^3})

V_2 = V_i + 0.11 \ L

The overall total change in the volume of the glycerin is illustrated as:

V_f = V_2 - V_1

Now; from the foregoing ; lets replace the respective value of V_2 and V_1 in the above equation ; we have;

V_f = (V_i + 0.11 \ L) - (V_i + \ 00873 \ L)\\ \\V_f = 0.11 L - 0.0873 \ L\\\\V_f = 0.0227 \ L

The formula usually known to be the volume of a cylinder is :

V = \pi r ^2 h

For the question ; we will have:

V_f = \pi r ^2 h

making h the subject of the  formula ; we have:

h = \frac{V_f}{\pi r^2}

replacing 0.0227 L for V_fand the given value of radius which is = 3.70 cm; we have:

h = \frac{0.0227 \ L ( \frac{10^3 \ cm^3}{1\ L})}{\pi * (3.70 cm)^2}

h = \frac{22.7 \ cm^3}{\pi * (3.70 cm)^2}\\\\h = 0.528 \ \ cm

Thus ; the distance by which the height of the  liquid in the cylinder change after the ice gets melted = 0.528 cm

8 0
2 years ago
7. A local sign company needs to install a new billboard. The signpost is 30 m tall, and the ladder truck is parked 24 m away fr
wolverine [178]
<h2>Solution :</h2>

Here ,

• Height of sign post = 30 m

• Distance between signpost and truck = 24 m

Let the

• Top of signpost = A

• Bottom of signpost = B

• The end of truck facing sign post be = C

Now as we can clearly imagine that the ladder will act as an hypotenuse to the Triangle ABC .

Where

• AB = Height of signpost = 30 m

• BC = distance between both = 24 m

• AC = Minimum length of ladder

→ AC² = AB² + BC² ( As we can see AB is perpendicular to BC )

→ AC² = (30)² + (24)²

→ AC² = 900 + 576

→ AC² = 1476

→ AC = 38.41875

or AC apx = 38.42

So minimum height of ladder = 38.42

6 0
2 years ago
For an object starting from rest and accelerating with constant acceleration, distance traveled is proportional to the square of
natali 33 [55]

The problem states that the distance travelled (d) is directly proportional to the square of time (t^2), therefore we can write this in the form of:

d = k t^2

where k is the constant of proportionality in furlongs / s^2

 

<span>Using the 1st condition where d = 2 furlongs, t = 2 s, we calculate for the value of k:</span>

2 = k (2)^2

k = 2 / 4

k = 0.5 furlongs / s^2

The equation becomes:

d = 0.5 t^2

 

Now solving for d when t = 4:

d = 0.5 (4)^2

d = 0.5 * 16

<span>d = 8 furlongs</span>

<span>
</span>

<span>It traveled 8 furlongs for the first 4.0 seconds.</span>

8 0
2 years ago
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