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andriy [413]
3 years ago
6

Potential energy can be converted into kinetic energy, a good example of this is when a pole-vaulter bends the pole during a lea

p. When the pole is bent the most, does it store elastic or gravitational potential energy? Question 7 options: Only gravitational potential energy because that is more powerful than elastic potential energy. It has neither elastic or gravitational potential energy. The pole stores elastic potential energy when the pole is bent because its shape is change from its natural shape and it will want to go back to its original form, just like a spring or stretched elastic material. The pole stores gravitational potential energy because it is bent from its natural shape when off the ground.
Physics
2 answers:
lesantik [10]3 years ago
6 0

Answer:

The pole stores elastic potential energy when the pole is bent because its shape is change from its natural shape and it will want to go back to its original form, just like a spring or stretched elastic material.

Question:

Why did you lie about being in college?

balandron [24]3 years ago
3 0

Answer:

The pole stores gravitational potential energy because it is bent from its natural shape when off the ground.

Explanation: i took the test

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32. A ball is going to fall downward through a vacuum chamber. If it has maximum potential energy at the top and maximum kinetic
QveST [7]

Half maximum velocity occurs at the point of half maximum kinetic energy which is exactly halfway down.

<h3>Conservation of energy</h3>

The principle of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to another.

M.A = K.E + P.E

At the maximum height, the kinetic energy of the ball while potential energy is maximum.

At the half-way down, the ball kinetic energy of the ball is equal to the potential energy.

Thus, half maximum velocity occurs at the point of half maximum kinetic energy which is exactly halfway down.

Learn more about kinetic energy here: brainly.com/question/25959744

8 0
1 year ago
A car traveling at 37m/s starts to decelerate steadily. It comes to a complete stop in 15 seconds. What is it’s acceleration
LuckyWell [14K]

<u>Answer</u>

The acceleration is

a=-2.5ms^{-2} to the nearest tenth

<u>Explanation</u>

Since the car was travelling at 37ms^{-1} before it starts to decelerate, the initial velocity is

u=37ms^{-1}.

The final velocity is v=0ms^{-1}, because the car came to a stop.

The time taken is t=15s.

Using the Newton's equation of linear motion,

v=u +at, we find the acceleration by substituting the known values.


This implies that,

0=37 +a(15)

This gives us,

0-37=15a


\Rightarrow -37=15a


We divide both sides by 15 to get,

a=-\frac{37}{15}ms^{-2}

or

a=-2.46667ms^{-2}




6 0
3 years ago
A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height
AleksandrR [38]

a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

E_i=\frac{1}{2}m_1v_1^2

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

v_3 = \frac{m_1 v_2}{m_1 + m_2}=\frac{(60 kg)(5.0 m/s)}{60 kg+15 kg}=4 m/s

c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J

While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:

E_f = (m_1 +m_2)gh_{max}

where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find

E_i = (m_1+m_2)gh_{max}\\h_{max}=\frac{E_i}{(m_1+m_2)g}=\frac{2070 J}{(75 kg)(9.8 m/s^2)}=2.8 m

4 0
3 years ago
To do your homework correctly you should (5 points)
pychu [463]

Answer:

C according to me

Explanation:

ITS HARD

4 0
3 years ago
Read 2 more answers
A string is wrapped around a pulley with a radius of 2.0 cm. The pulley is initially at rest. A constant force of 50 N is applie
Ray Of Light [21]

Answer:

0.20kg-m^2

Explanation:

Let the linear velocity of the rope(=of pulley) is v m/s

Using kinematic equation

=> v = u + at

=>v = 0 + 4.9a

=>v = 4.9a ------------ eq1

By v^2 = u^2 + 2as

=>v^2 = 0 + 2 x v/4.9 x 1.2

=>4.9v^2 - 2.4v = 0

=>v(4.9v - 2.4) = 0

=>v = 2.4/4.9 = 0.49 m/s

Thus by v = r x omega

=>omega = v/r = 0.49/0.02 = 24.49 rad/sec

BY W = F x s = 50 x 1.2 = 60 J

=>KE(rotational) = W = 1/2 x I x omega^2

=>60 = 1/2 x I x (24.49)^2

=>I = 0.20 kg-m^2

5 0
3 years ago
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