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3 years ago

A basketball rolls to a stop along a gym floor without anyone touching it. What answer choice BEST explains why this occurred?

Physics

2 answers:

horsena [70]3 years ago

5 0

Answer: The correct option is B.

Explanation:

In the given problem, a basketball rolls to a stop along a gym floor without anyone touching it.

The basketball stops own its own because of the force of friction. Friction is a force which opposes the motion of the object. The direction of the force of friction is in an opposite direction to the motion of the object.

The frictional force will act between the basket ball and the floor.

Therefore, the correct option is B.

Alex777 [14]3 years ago

4 0

<h3><em>Frictional forces are exerting force against the basketball, causing it to stop.</em></h3>
<h3>If USATestPrep, I think answer choice B is correct </h3>

<u><em>Tell me if i'm wrong </em></u>

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A block of mass 3.6 kg, sliding on a horizontal plane, is released with a velocity of 1.7 m/s. The block slides and stops at a d

pentagon [3]

Answer:

The block would have slid <u>12.6 m</u> if its initial velocity were increased by a factor of 2.8.

Explanation:

Given:

Mass of the block (m) = 3.6 kg

Initial velocity (u) = 1.7 m/s

Final velocity (v) = 0 m/s

Displacement (S) = 1.6 m

First we will find the acceleration of the block.

Using the equation of motion, we have:

$v^2=u^2+2aS\\\\a=\dfrac{v^2-u^2}{2S}$

Now, plug in the given values and solve for 'a'. This gives,

$a=\frac{0-1.7^2}{2\times 1.6}\\\\a=\frac{-2.89}{3.2}=0.903\ m/s^2$

The acceleration is negative as it is resisting the motion.

Now, the initial velocity is increased by a factor of 2.8. So,

New initial velocity = 2.8 × 1.7 = 4.76 m/s

Again using the same equation of motion and expressing the result in terms of 'S'. This gives,

$v^2=u^2+2aS\\\\S=\dfrac{v^2-u^2}{2a}$

Now, plug in the given values and solve for 'S'. This gives,

$S=\frac{0-4.76^2}{2\times -0.903}\\\\S=\frac{-22.6576}{-1.806}=12.6\ m$

Therefore, the block would have slid 12.6 m if its initial velocity were increased by a factor of 2.8

6 0

3 years ago

Please help..!

iren2701 [21]

Answer:

idk ha ha ha

Explanation:

7 0

2 years ago

camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Usin

son4ous [18]

The question is incomplete. Here is the complete question.

The image below was taken with a camera that can shoot anywhere between one and two frames per second. A continuous series of photos was combined for this image, so the cars you see are in fact the same car, but photographed at differene times.

Let's assume that the camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Using this information and the photo itself, approximately how fast did the car drive?

Answer: v = 6.5 m/s

Explanation: The question asks for velocity of the car. Velocity is given by:

$v=\frac{\Delta x}{\Delta t}$

The camera took 7 pictures of the car and knowing its length is 5.3, the car's displacement was:

Δx = 7(5.3)

Δx = 37.1 m

The camera delivers 1.3 frames per second and it was taken 7 photos, so time the car drove was:

1.3 frames = 1 s

7 frames = Δt

Δt = 5.4 s

Then, the car was driving:

$v=\frac{37.1}{5.4}$

v = 6.87 m/s

The car drove at, approximately, a velocity of 6.87 m/s

7 0

3 years ago

Write a hypothesis about the effect of temperature and surface area on the rate of chemical reactions using this format: “If . .

grandymaker [24]

Effect of temperature.

"If the temperature of the substance is increased then the rate of chemical reaction is also increased because the kinetic energy is greater."

Effect of surface area.

"If the surface area is increased then the rate of reaction is increased because there will be more active sites for the reaction to occur.

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3 years ago

Read 2 more answers

Two trains are traveling on the same track and in the same direction. The first train, which is behind the second train, blows a

Andrews [41]

Answer:

37.545 m/s

Explanation:

f' = Actual frequency of horn = 269 Hz

f = Observed frequency of horn = 290 Hz

v = Speed of sound in air = 343 m/s

$v_0$ = Speed of second train = 13.7 m/s