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expeople1 [14]
3 years ago
14

A basketball has a coefficient of restitution of 0.821 in collisions with the wood floor of a basketball court. The ball is drop

ped on such a floor from a height of 2.07 m . How high is its fourth bounce?
Physics
1 answer:
Tanya [424]3 years ago
7 0

Answer: The height of its fourth bounce = 0.43m

Explanation:

The coefficient of restitution denoted by (e), is the ratio that shows the  final velocity to initial relative velocity between two objects after collision

IT is given by the formula in terms of height as

Coefficient of Restitution, e  = √(2gh))/√(2gH) = √(h/H)

Where

Coefficient of Restitution, e= 0.821

H = 2.07 m

At fourth bounce ,   we have that

Coefficient of Restitution, e⁴  =√(h₄/H)  

Putting the given values and solving , we have,

e⁴  =√(h₄/H)  

= 0.821⁴ = √(h₄/2.07)

 (0.821⁴ )² =h₄/2.07

0.2064 x 2.07 = 0.427 = 0.43

At  fourth bounce,  h₄ height = 0.43m

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Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface
FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

\tau=I\alpha     (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

\tau=Fd        (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

Fd=I\alpha\\\\I=\frac{Fd}{\alpha}

Finally, you replace the values of all parameters in the previous equation for I:

I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2

The moment of inertia of the door around the hinges is 2 kgm^2

3 0
3 years ago
A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle
tatiyna

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

-h= vsinA\times t-gt^2/2

putting values h=15 m, v=0.8

-15 = 0.8vt - 4.9t^2  ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

-15 = 0.8\times40/0.6 - 4.9t^2

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) =  17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = -H =v sinA t - gt^2/2

⇒ 4.9t^2 - 8.9\times0.8t - 100 = 0

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

3 0
3 years ago
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