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drek231 [11]
3 years ago
8

Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 25.0 m above water wit

h an initial speed of 20.0 m/s strikes the water with a final speed of what, independent of the direction thrown.
Physics
1 answer:
Rasek [7]3 years ago
7 0

Complete question is;

Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 25.0 m above water with an initial speed of 20.0 m/s strikes the water with a final speed of 31.1 m/s, independent of the direction thrown

Answer:

It is proved that the final speed is truly 31.1 m/s

Explanation:

From energy - conservation principle;

E_i = Initial potential energy + Initial Kinetic Energy

Or

E_i = U_i + K_i

Similarly, for final energy

E_f = U_f + K_f

So, expressing the formulas for potential and kinetic energies, we now have;

E_i = (m × g × y_i) + (½ × m × v_i²)

Similarly,

E_f = (m × g × y_f) + (½ × m × v_f²)

We are given;

y_i = 25 m

y_f = 0 m

v_i = 20 m/s

v_f = 31.1 m/s

So, plugging in relevant values;

E_i = m((9.8 × 25) + (½ × 20²))

E_i = 485m

Similarly;

E_f = m((9.8 × 0) + (½ × v_f²)

E_f ≈ ½m•v_f²

From energy conservation principle, E_i = E_f.

Thus;

485m = ½m•v_f²

m will cancel out to give;

½v_f² = 485

v_f² = 485 × 2

v_f² = 970

v_f = √970

v_f ≈ 31.1 m/s

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3 years ago
A car (mass = 1090 kg) is traveling at 30.4 m/s when it collides head-on with a sport utility vehicle (mass = 2880 kg) traveling
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Answer:

The sport utility vehicle was traveling at V2= 11.5 m/s.

Explanation:

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7 0
3 years ago
A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1 2 and t = 1 s? Use G
Romashka-Z-Leto [24]

There is one mistake in the question.The Correct question is here

A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.

Answer:

y(1s) - y(1/2s) =  - 3.675 m  

The cat falls 3.675 m between time 1/2 s and 1 s.

Explanation:

Given data

time=1/2 sec to 1 sec

v(t)=-9.8t m/s

To find

Distance

Solution

As the acceleration as first derivative of velocity with respect to time  

So

acceleration(-g)=  dv/dt

Solve it

dv  =  a dt

dv =  -g dt

v - v₀  =  -gt

v=  dy/dt

dy  =  v dt

dy =  ( v₀ - gt ) dt

y(1s) - y(1/2s)  =  ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]

y(1s) - y(1/2s)  = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]

y1s - y1/2s  = ( - 4.9 m/s² ) ( 3/4 s² )

y(1s) - y(1/2s) =  - 3.675 m  

The cat falls 3.675 m between time 1/2 s and 1 s.

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Which of the following includes an example of a chemical property of an element?
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Answer: D

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