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andre
1 month ago
12

Darryl throws a basketball at the gym floor.The ball bounces once on the floor comes to rest in his coach's hands.At which point

are all the forces on the basketball balanced?
a.the moment the ball leaves Darryl's handB.the moment the ball touches the floorC.when the ball is in the airD.when the ball starts falling downward before coming to restE.the moment the ball comes to complete rest
Physics
2 answers:
serg [3.5K]1 month ago
6 0
B. When the ball makes contact with the floor.
Keith_Richards [3.2K]1 month ago
6 0

Response:

E. When the ball comes to complete rest

Explanation:

A basketball is in a balanced state when the net force acting on it equals zero and its velocity remains constant over time.

Therefore, when the ball begins to fall, its velocity increases which indicates an unbalanced situation.

When the ball hits the ground, it rebounds in the opposite direction meaning that the velocity is changing, which is also unbalanced.

When the ball is traveling through the air, gravity causes its speed to vary, so this is yet another unbalanced state.

Thus, the point where the basketball is balanced is when it is at rest in the coach's hands since at that moment its speed is steady.

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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
Maru [3345]

Answer:

0.0984

Explanation:

The first diagram below illustrates a free body diagram that will aid in resolving this problem.

According to the diagram, the force's horizontal component can be expressed as:

F_X = F_{cos \ \theta}

Substituting 42° for θ and 87.0° for F

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

F_X =64.65 \ N

Meanwhile, the vertical component is:

F_Y = Fsin \ \theta

Again substituting 42° for θ and 87.0° for F

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

F_Y =58.21 \ N

In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

The frictional force can be expressed as:

f = \mu \ N

Where;

\mu is the coefficient of friction

N = the normal force

Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting F_Y \ for \ F_{sin \ \theta}. Normal reaction becomes:

N = mg \ - \ F_Y

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

The horizontal component is described as follows:

F_X = \mu \ ( mg - \ F_Y)

Rearranging the equation above to isolate \mu leads to:

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

Substituting in the following values:

F_X \ = \ 64.65 \ N

m = 73 kg

g = 9.8 m/s²

F_Y = \ 58.21 N

Thus:

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

\mu = 0.0984

Therefore, the coefficient of friction is = 0.0984

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Result:

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