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4 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:

Minchanka [31]4 years ago

7 0

Frictional forces are exerting force against the basketball, causing it to stop.

salantis [7]4 years ago

5 0

its B,The opposing force of friction stopped the ball.

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What is the sequence of energy transformations from the moment the ball is dropped to the moment the board is bent to its maximu

Mila [183]

Answer:

Explanation:

Transformation of energy involves conversion of energy from one form to another for example our movement around involves the conversion of chemical energy stored in the food we eat to other forms of energy such as kinetic energy for the movement, electrical energy in the neurons for impulses and others

The ball posses gravitational potential energy since it is held at a displacement to the ground ( zero point) and when released, the gravitational potential energy is converted to kinetic energy which leads to the fall of the ball until it is at zero displacement to the earth. The board likewise when bent to its maximum extent stored elastic potential energy as a result of the partial displacement of its constituent particle provided it is not stretch beyond its elastic limit which can lead to deformation of the board and the elastic potential energy lost.

5 0

3 years ago

Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes

pantera1 [17]

Answer:

The magnitude of F1 is

$|F1|=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

$\displaystyle -2F\ sin21^o+F\ cos\alpha =0$

The sum of the vertical components of F1 and F2 must equal the total force Ft

$\displaystyle -2F\ cos21^o+F\ sin\alpha =460$

Solving for $\alpha$ in the first equation

$\displaystyle cos\alpha =\frac{2F\ sin21^o}{F}=2sin21^o$

$\displaystyle cos\alpha =0.717=>\alpha =44.214^o$

$\displaystyle F(2cos21^o+sin\alpha)=460$

$\displaystyle F=\frac{460}{2cos21^o+sin\alpha}$

$\displaystyle F=\frac{460}{2cos21^o+sin44.214^o}$

$\displaystyle F=179.37\ N$

The magnitude of F1 is

$|F1|=2*F=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

4 0

3 years ago

During a demonstration of the gravitational force on falling objects to her class, Sarah drops an 11 lb. bowling ball from the t

tia_tia [17]

1.A) 4.9 m

AL2006 Ace

The instant it was dropped, the ball had zero speed.

After falling for 1 second, its speed was 9.8 m/s straight down (gravity).

Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.

Falling for 1 second at an average speed of 4.9 m/s, is covered 4.9 meters.

ANYTHING you drop does that, if air resistance doesn't hold it back.

Read more on Brainly.com - brainly.com/question/11776597#readmore

2 idk sorry

5 0

3 years ago

Read 2 more answers

How do noise and vibration affect you when operating a boat?

kogti [31]

The noise and vibration could affect your body and mood in many ways while operating a boat. The noise and vibration could make you moody or uncomfortable, because it will also distract you in doing what you're supposed to do. Your body will also be affected, making your body tired because of the things surrounding you and the noise and vibration, it could affect your body in many ways.

3 0

3 years ago

Read 2 more answers

A World War II bomber flies horizontally over level terrain, with a speed of 287 m/s relative to the ground and at an altitude o

Scorpion4ik [409]

Answer: 7.38 km

Explanation: The attachment shows the illustration diagram for the question.

The range of the bomb's motion as obtained from the equations of motion,

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2

R = 287 √(2×3240/9.8) = 7380 m = 7.38 km

6 0

3 years ago