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Alja [10]
2 years ago
15

A ball is dropped from rest at the top of a 6.10 m

Physics
1 answer:
natita [175]2 years ago
3 0

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

\frac{v_1}{v} = e ( coefficient of restitution ) = \frac{1}{\sqrt{10} }

and

\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

e = \sqrt{\frac{h_1}{6.1} }

e = \sqrt{\frac{h_2}{h_1} }

So on

e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }

(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}= .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

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4- by eating food only will you get energy?​
Serga [27]

Answer:

yes

Explanation:

All parts of the body (muscles, brain, heart, and liver) need energy to work. This energy comes from the food we eat. Our bodies digest the food we eat by mixing it with fluids (acids and enzymes) in the stomach.

4 0
3 years ago
What is the angular momentum of a 0.25 kg mass rotating on the end of a piece of
finlep [7]

L = r x p = rmv = mr²ω

L = 0.25 x 0.75² x 12.5 = 1.758

4 0
2 years ago
A freight car moves along a friction less level railroad track at constant speed. The car is open on top. A large load of coal i
maxonik [38]

Answer:

The velocity of the freight car decreases.

Explanation:

This question is answered by the conservation of momentum principle.

When the freight car is moving at a certain speed, it has a constant momentum.

We will call this M1.

The equation for M1 will be:

M1 = Mass * Speed

Now when the coal is dumped into the freight car, the Mass increases.

Since conservation of momentum states that the momentum will remain the same. We have:

M1 = (Mass of freight + Mass of coal) * Speed

Since M1 is constant, if the mass increases, the speed had to decrease to keep the equation true.

8 0
3 years ago
An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 43.8 N,
maw [93]

A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant <em>k</em> is

43.8 N = <em>k</em> (0.155 m)   ==>   <em>k</em> = (43.8 N) / (0.155 m) ≈ 283 N/m

The total work done on the spring to stretch it to 15.5 cm from equilibrium is

1/2 (283 N/m) (0.155 m)² ≈ 3.39 J

The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be

1/2 (283 N/m) (0.259 m)² ≈ 9.48 J

Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.

5 0
3 years ago
A drop
exis [7]

Answer:

27.5\  m

Explanation:

As we know that volume of cylinder is

v=\pi r^{2} *h

Where v=volume , h= height or thickness and r= radius

Here,

v= 10 m ,\  diameter= 10, \ r=\frac{diameter}{2} \ r=\frac{10}{2}\\ r=5

Putting these values in the previous equation , we get

10\ = \frac{22}{7} *5 *5*h\\ 14\ =\ 110*h\\h=\frac{110}{14} \\h=\frac{55}{2} \\\\h=27.5\  m

Therefore thickness is 27.5 m

7 0
3 years ago
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