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3 years ago

A 33-kg girl is bouncing on a trampoline. During a certain interval after she leaves the

Physics

1 answer:

Mandarinka [93]3 years ago

6 0

The way to do this is to times 33 x 5 then that answer will be times by 2 boom thats your answer!!

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Given the nuclear equation:

Leokris [45]

This is a nuclear fission reaction, in which a larger nucleus is bombarded with a neutron to make it break down into two smaller nuclei and release energy.

6 0

3 years ago

a boy looks at the reflection of his digital watch in a plane mirror and thinks the time is 10:11. what is the correct time?

Gekata [30.6K]

Answer:

11:10 will be the time. reflection causes the object to be flipped when you see its image at the mirror

6 0

3 years ago

Se coloca una piedra de 600 g en una Honda de 50 cm y se la hace girar a una velocidad de 4 m/s. Dibuja la fuerza que ejerce la

Dima020 [189]

Answer:

Fc = 19.2 N

Explanation:

In this case, the force of the Honda over the rock, is a centripetal force. Then, you have:

$F_c=m\frac{v^2}{r}$

m: mass of the rock = 600g = 0.6 kg

v: tangential velocity of the Honda = 4m/s

r: radius of the Honda = 50cm = 0.5m

You replace the values of m, r and v in the equation for Fc:

$F_c=(0.6kg)\frac{(4m/s)^2}{0.5m}\\\\F_c=19.2N$

hence, the force has a magnitude of 19.2 N

If the rock would have more mass the centripetal force would be higher

4 0

3 years ago

A 244.0 N block is at rest on a flat, frictionless table. A hooked cable applies an upward force of 24.0 N on the block. What is

blagie [28]

Answer:

268N

Explanation:

The upward force acting on the block are the reaction and the hooked table..

The total normal force acting = normal reaction + 24N

Note that the normal reaction is always equal the weight of the table

Hence the normal force acting in the block is 244.0+24 = 268.0N

4 0

3 years ago

A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar

notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2> $\omega = 0.23 rad/s$ </h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

$L_i = L_f$

now we can say

$m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega$

so we will have

$0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega$

$\omega = 0.23 rad/s$

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0

3 years ago