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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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Consider a thin rod of mass 3.2 kg, length 1.2 m and uniform density. The rod is pivoted at one end on a frictionless horizontal
notsponge [240]

Answer:

the angular acceleration is 9.7 rad/s^{2}

Explanation:

given information:

mass of thin rod, m = 3.2 kg

the length of the rod, L = 1.2

angle, θ = 38

to find the acceleration of the rod, we can use the torque's formula as below,

τ = Iα

where

τ = torque

I = inertia

σ = acceleration

moment inertia of this rod, I

I = \frac{1}{3} mL^{2}

τ = F d, d = \frac{L}{2}cosθ

τ = m g \frac{L}{2}cosθ

now we can substitute the both equation,

τ = Iα

α = τ/I

  = (m g \frac{L}{2}cosθ)/(\frac{1}{3} mL^{2})

  = 3gcosθ/2L

  = 3 (9.8)cos 38°/(2 x 1.2)

  = 9.7 rad/s^{2}

 

3 0
3 years ago
PLZ hurry! Thank you
Alenkinab [10]

Answer:do it again

Explanation:

It will work

3 0
3 years ago
A _______ is a repeating disturbance or vibration that transfers or moves energy from place to place without transporting mass.
cupoosta [38]

the answer is a wave


7 0
3 years ago
Read 2 more answers
Umar has two copper pans, each containing 500cm3 of water. Pan A has a mass of 750g and pan B has a mass of 1.5kg. Which pan wil
Olin [163]

Answer:

heat required in pan B is more than pan A

Explanation:

Heat required to raise the temperature of the substance is given by the formula

Q = ms\Delta T

now we know that both pan contains same volume of water while the mass of pan is different

So here heat required to raise the temperature of water in Pan A is given as

Q_1 = (m_w s_w + m_ps_p)\delta T

Q_1 = (0.5(4186) + 0.750(s))\Delta T

Now similarly for other pan we have

Q_2 = (m_w s_w + m_ps_p)\delta T

Q_2 = (0.5(4186) + 1.50(s))\Delta T

So here by comparing the two equations we can say that heat required in pan B is more than pan A

3 0
3 years ago
A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a
sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

5 0
3 years ago
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