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puteri [66]
3 years ago
8

A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the g

round to rise 1.15 m above the floor in an attempt to get the ball
Physics
1 answer:
Anastaziya [24]3 years ago
3 0

Answer:

the basketball player must leave the ground with a velocity of 4.748 m/s

Explanation:

Given that data in the question;

From the third equation of motion;

v² - u² = 2as

such that;

u² = v² - 2as

where u is the initial velocity, v is the final velocity, s is the displacement and a is acceleration

so initial velocity of the basket ball player will be;

u = √( v² - 2as )

so from the question; s is 1.15 m and a = - 9.8 m/s² { player is under negative acceleration to get to the ball } and final velocity of the player will be 0.

so we substitute

u = √( (0)² - (2 × -9.8 × 1.15 )

u = √ -( - 22.54 )

u = √ ( 22.54 )

u = 4.748 m/s

Therefore, the basketball player must leave the ground with a velocity of 4.748 m/s

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A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic frictio
Diano4ka-milaya [45]

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

F - External force exerted on the sled, measured in newtons.

f - Friction force, measured in newtons.

N - Normal force from the ground on the mass, measured in newtons.

W - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

W = m\cdot g (1)

Where:

m - Mass, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

If we know that m = 50\,kg and g = 9.807\,\frac{m}{s^{2}}, the weight of the sled is:

W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

W = 490.35\,N

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

F_{min,s} = \mu_{s}\cdot W (2)

Where:

\mu_{s} - Static coefficient of friction, dimensionless.

W - Weight of the sled, measured in newtons.

If we know that \mu_{s} = 0.3 and W = 490.35\,N, then the force needed to start the sled moving is:

F_{min,s} = 0.3\cdot (490.35\,N)

F_{min,s} = 147.105\,N

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

F_{min,k} = \mu_{k}\cdot W (3)

Where \mu_{k} is the kinetic coefficient of friction, dimensionless.

If we know that \mu_{k} = 0.1 and W = 490.35\,N, then the force needed to keep the sled moving at a constant velocity is:

F_{min,k} = 0.1\cdot (490.35\,N)

F_{min,k} = 49.035\,N

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

8 0
3 years ago
Please help with this and explain it,if you can.
vladimir2022 [97]

Answer:

displacement at 45 s  =  30

                           65 s  = 50

So the average speed over the interval from 45 s to 65 s is

(50 - 30) cm / 20 s = 1 cm / sec

As a check an average speed of 1 cm / sec for 20 sec will produce a

displacement of 1 cm / sec * 20 sec = 20 cm  or from 30 to 50 cm

4 0
3 years ago
.<br><img src="https://tex.z-dn.net/?f=25%20%7B%3F%7D%5E%7B%3F%7D%20%20%5Ctimes%20%5Cfrac%7B%3F%7D%7B%3F%7D%20" id="TexFormula1"
mr_godi [17]

Answer:

what is your exact question.

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2 years ago
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D. There are two phosphate ions in a molecule of magnesium phosphate
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A digital signal differs from an analog signal because it a.consists of a current that changes smoothly. b. consists of a curren
zzz [600]

Answer:

d.it is used in electronic devices

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