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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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An astronomer observes that the wavelength of light from a distant star is shifted toward the red part of the visible spectrum.
balandron [24]

Answer:

The distance between the earth and the star is increasing.

Explanation:

When we observe an object and its electromagnetic radiation has been displaced to blue, it means that it is getting closer to us, causing the light waves it emits to get closer together and its wavelength to decrease towards blue, this is knowm as blueshift.

On the contrary, when an object is rapidly moving away from us, the light waves or electromagnetic radiation it emits have been stretched from their normal wavelength to a longer wavelength, towards the red part of the spectrum. This is known as redshift.

This phenomenon of changes in wavelength and frequency due to movement (whether the source approaches or moves away) is described by the Doppler effect.

So for this case because the light we perceive from the star has moved to the red part of the visible spectrum, we can conclude that it is moving away from the earth, and that the distance between the star and the earth is increasing.

7 0
2 years ago
The density of ice is 0.93 g/cm3. what is the volume, in cm3, of a block of ice whose mass is 5.00 kg? remember to select an ans
serious [3.7K]

The Volume of the ice block is 5376.344 cm^3.

The density of a material is define as the mass per unit volume.

Here, the density of ice given is 0.93 g/cm^3

Mass of the ice block  given is 5 kg or 5000 g

Now calculate the volume of the ice block

density=mass/volume

0.93=5000/Volume

Volume =5376.344 cm^3

Therefore the volume of  ice block is 5376.344 cm^3

7 0
2 years ago
An object experiences an acceleration of -6.8 m/s​2.​ As a result, it accelerates from 54 m/s to a complete stop. How much dista
inessss [21]

Answer:

The distance traveled during its acceleration, d = 214.38 m

Explanation:

Given,

The object's acceleration, a = -6.8 m/s²

The initial speed of the object, u = 54 m/s

The final speed of the object, v = 0

The acceleration of the object is given by the formula,

                                      a = (v - u) / t   m/s²

       ∴                              t = (v - u) / a

                                         = (0 - 54) / (-6.8)

                                         = 7.94 s

The average velocity of the object,

                                       V = (54 + 0)/2

                                           = 27 m/s

The displacement of the object,

                                 d = V x t   meter

                                    = 27 x 7.94

                                    = 214.38 m

Hence, the distance the object traveled during that acceleration is, a = 214.38 m

3 0
2 years ago
It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you
Juli2301 [7.4K]

Answer:

0.0984

Explanation:

From the first diagram attached below; a free flow diagram shows the interpretation of this question which will be used  to solve this question.

From the diagram, the horizontal component of the force is:

F_X = F_{cos \ \theta}

Replacing 42°  for θ and 87.0° for F

F_X =87.0 \ N \ *cos \ 42 ^\circ

F_X =64.65 \ N

On the other hand, the vertical component  is ;

F_Y = Fsin \ \theta

Replacing 42°  for θ and 87.0° for F

F_Y =87.0 \ N \ *sin \ 42 ^\circ

F_Y =58.21  \ N

However, resolving the vector, let A be the be the component of the mutually perpendicular directions.

The magnitude of the two components is shown in the second attached diagram below and is now be written as A cos θ and A sin θ

The expression for the frictional force is expressed as follows:

f = \mu \ N

Where;

\mu is said to be the coefficient of the friction

N = the  normal force

Similarly the normal reaction (N) = mg - F sin θ

Replacing F_Y \ for \ F_{sin \  \theta}. The normal reaction can now be:

N = mg \ - \ F_Y

By balancing the forces, the horizontal component of the force equals to frictional force.

The horizontal component of the force is given as follows:

F_X = \mu \ ( mg - \ F_Y)

Making \mu the subject of the formular in the above equation; we have the following:

\mu \ = \ \frac{F_X}{mg - F_Y}

Replacing the following values: i.e

F_X \ = \ 64.65 \  N

m = 73 Kh

g  = 9.8 m/s²

F_Y = \ 58.21 N

Then:

\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}

\mu = 0.0984

Thus, the coefficient of friction is = 0.0984

5 0
2 years ago
Two blocks, 1 and 2, are connected by a rope R1 of negligible mass. A second rope R2, also of negligible mass, is tied to block
alekssr [168]

Answer:

Explanation:

Given

Two block are connected by rope R_1

R_2 rope is attached to block 2

suppose F_2 is a force applied to Rope R_2

Applied force F_2=Tension in Rope 2

F_2=(m_1+m_2)a---1

where a=acceleration of system

Tension in rope R_1 is denoted by F_1

F_1=m_1a---2

divide 1 and 2 we get

\frac{F_2}{F_1}=\frac{(m_1+m_2)a}{m_1a}

also m_1=2.11\cdot m_2

\frac{F_2}{F_1}=\frac{2.11m_2+m_2}{2.11m_2}

\frac{F_2}{F_1}=\frac{3.11}{2.11}

\frac{F_1}{F_2}=\frac{2.11}{3.11}

               

3 0
2 years ago
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