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3 years ago

what type of energy appears when a gymnast jumps on to a spring board? Question is on energy stores!Thanks

1 answer:

5 0

Hello There!

It is Spring potential energy. Also called Elastic potential energy.

Hope This Helps You!
Good Luck :)

- Hannah ❤

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Can any juniors who go to Texas connections academy help me out with physics?

zimovet [89]

I'm not from that school but I can help you.

3 0

2 years ago

A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.

Nesterboy [21]

Answer:

<em>The velocity after the collision is 2.82 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2$

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

The common velocity after this situation is:

$\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

$\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}$

$\displaystyle v'=\frac{22.287}{7.91}$

$\boxed{v' = 2.82\ m/s}$

The velocity after the collision is 2.82 m/s

6 0

2 years ago

Look at a photograph of a fault. notice how the right side appears lower than the left side. this happens when pieces of crust a

dybincka [34]

I believe it’s divergent boundary but I might be wrong

5 0

3 years ago

Read 2 more answers

A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.460 Hz. The pendulum ha

zlopas [31]

Answer:

The moment of inertia is $I =1.0697 \ kg m^2$

Explanation:

From the question we are told that

The frequency is $f = 0.460 \ Hz$

The mass of the pendulum is $m = 2.40 \ kg$

The location of the pivot from the center is $d = 0.380 \ m$

Generally the period of the simple harmonic motion is mathematically represented as

$T = 2 \pi * \sqrt{ \frac{I}{ m * g * d } }$

Where I is the moment of inertia about the pivot point , so making I the subject of the formula it

=> $I = [ \frac{T}{2 \pi } ]^2 * m* g * d$

But the period of this simple harmonic motion can also be represented mathematically as

$T = \frac{1}{f}$

substituting values

$T = \frac{1}{0.460}$

$T = 2.174 \ s$

$I = [ \frac{2.174}{2 * 3.142 } ]^2 * 2.40* 9.8 * 0.380$

$I =1.0697 \ kg m^2$

4 0

3 years ago

6000 kg train moving 5 m/sec to east collides with 5000 kg train moving 3 m/sec to west. What is their velocity

Svetlanka [38]

The final velocity is 1.37 m/s east

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two trains must be conserved before and after the collision.

So we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$