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goblinko [34]
2 years ago
7

At the beginning of a basketball game, the referee tosses the ball straight up with a speed of 4.6m/s. A player cannot touch the

basketball until it has reached maximum height and begins to fall back down. What is the minimum time a player must wait before touching the ball?
Physics
1 answer:
maria [59]2 years ago
8 0

Answer:

t=0.47s

Explanation:

the ball has uniformly accelerated movement due to gravity

Vo=initial speed=4.6m/s

g=gravity=-9.8m/s^2

Vf=final speed=0, the player must wait for the ball to stop. so the final speed will be 0

we can use the following ecuation

T=(Vf-Vo)/g

T=(0-4.6)/-9.8m/s^2

T=0.47s

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A feather, a marble, and a cannonball are dropped from the same height at
Vika [28.1K]

The correct answer is that they would all hit the ground at the same time. If no air resistance is present, the rate of descent depends only on how far the object has fallen, no matter how heavy the object is. This means that two objects will reach the ground at the same time if they are dropped simultaneously from the same height. This statement follows from the law of conservation of energy and has been demonstrated experimentally by dropping a feather and a lead ball in an airless tube.

7 0
2 years ago
The temperature of 1 m^3 of water is decreased by 10°C. If this thermal energy is used to lift the water vertically against grav
Rama09 [41]

Answer:

h = 4271.43 m

Explanation:

given,

Volume of the water = 1 m³

temperature decrease by = 10°C

heat removed from water

Q = m c ΔT                            

Q = ρ V c ΔT                            

   = 1000 × 1 × 4186 × 10

   = 4.186 × 10⁷ J

energy is used to do work to move the water against its weight

Q = force  × displacement

4.186 × 10⁷ J =  m g × h                    

4.186 × 10⁷ J =  1000 × 1 × 9.8 × h                

h = 4271.43 m                                

hence, the change in height of is equal to h = 4271.43 m

8 0
3 years ago
3. How are fossil fuels used to generate electrical energy?
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5 0
3 years ago
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7 0
3 years ago
While taking off from an aircraft carrier, a jet starting from rest accelerates uniformly to a final speed of 40. meters per sec
mariarad [96]

The acceleration of the jet is 11.4 m/s^2

Explanation:

Since the jet motion is a uniformly accelerated motion (=constant acceleration), we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the jet in this problem, we have

u = 0 (it starts from rest)

v = 40 m/s (final velocity)

s = 70 m (length of the runway)

Solving for a, we find the acceleration:

a=\frac{v-u}{t}=\frac{40^2-0}{2(70)}=11.4 m/s^2

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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