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Firdavs [7]
2 years ago
12

Yo yo was good I know its Friday and schools over but I need this, due today. I give BRAINLIEST. I need this in atleast 20 minut

es.

Physics
2 answers:
hoa [83]2 years ago
8 0
Did you get the answer to this?
12345 [234]2 years ago
7 0
I’ve seen this before
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Learning Goal: To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circ
andreev551 [17]

Answer:

v = 15.8 m/s

Explanation:

Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement.  So the variation of mechanical energy is equal to the work of the fictional force

    W_{fr} = ΔEm = Em_{f} -Em₀

Let's write the mechanical energy at each point

Initial

    Em₀ = Ke = ½ k x²

Final

   Em_{f} = K + U = ½ m v² + mg y

Let's use Hooke's law to find compression

    F = - k x

    x = -F / k

    x = 4400/1100

    x = - 4 m

Let's write the energy equation

    fr d = ½ m v² + mgy - ½ k x²

Let's clear the speed

   v² = (fr d + ½ kx² - mg y) 2 / m

   v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50)   2/60.0

   v² = (160 + 8800 - 1470) / 30

   v = √ (229.66)

   v = 15.8 m/s

5 0
3 years ago
Read 2 more answers
On a playground slide, a child has potential energy that decreases by 1000 J while her kinetic energy increases by 900 J. what o
mel-nik [20]

Answer:

Friction     100 j

Explanation:

Friction causes the inequality between PE and KE

100 J

8 0
1 year ago
What human disease can result from exposure to ultraviolet radiation from the sun?
aev [14]
Melanoma skin cancer
8 0
3 years ago
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Who was the second man to walk on the moon?
klemol [59]

Answer:

Buzz Aldrin

Explanation:

6 0
3 years ago
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A ball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the i
Darya [45]

The position of the first ball is

y_1=h-\dfrac g2t^2

while the position of the second ball, thrown with initial velocity v, is

y_2=vt-\dfrac g2t^2

The time it takes for the first ball to reach the halfway point satisfies

\dfrac h2=h-\dfrac g2t^2

\implies\dfrac h2=\dfrac g2t^2

\implies t=\sqrt{\dfrac hg}

We want the second ball to reach the same height at the same time, so that

\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2

\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)

\implies h=v\sqrt{\dfrac hg}

\implies v=\sqrt{hg}

8 0
3 years ago
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