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3 years ago

What type of energy does the box have after it is done being pulled?

1 answer:

3 0

Assuming that the box is moving when it is being pulled, Work is done on the box.

So work is the Force times the distance

W=Fd

But what is work actually ? When something moves due to force over some change in distance, it have energy.

But where does this energy come from ? Does it magically appear ? The energy comes from the applied force onto the box.

So the energy have been transferred. And it’s like that throughout the universe

Now to save time, I’ll just tell you the answer: kinetic energy

:)

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The jumping gait of the kangaroo is efficient because energy is stored in the stretch of stout tendons in the legs; the kangaroo

Anna [14]

Answer:

the period T of whole motion should be twice the value for half at he bottom so T is 0.2sec.

w is angular frequency

formula:2π/T

now k is spring constant

F/R-->mw²

putting values:70*(2π/0.2)²

=4.9x10⁶

so we can say that SHM is not affected by the amplitude of the bounce.

6 0

3 years ago

Once out in space, the space shuttle needs to ________________ to keep going in the same direction at the same speed.

MissTica

Accelerate.
Hope it helps<span />

3 0

3 years ago

Read 2 more answers

A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh

alekssr [168]

(a) 392 N/m

Hook's law states that:

$F=k\Delta x$ (1)

where

F is the force exerted on the spring

k is the spring constant

$\Delta x$ is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

$F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N$

- The stretching of the spring is

$\Delta x=15 cm-10 cm=5 cm=0.05 m$

Solving eq.(1) for k, we find the spring constant:

$k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m$

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

$F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N$

And so, the stretch of the spring is

$\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm$

And since the initial lenght of the spring is

$x_0 = 10 cm$

The final length will be

$x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm$

8 0

3 years ago

Read 2 more answers

The density of aluminum is 2.7 × 103 kg/m3 . the speed of longitudinal waves in an aluminum rod is measured to be 5.1 × 103 m/s.

andrey2020 [161]

<span>The speed of longitudinal waves, S, in a thin rod = âšYoung modulus / density , where Y is in N/m^2. So, S = âšYoung modulus/ density. Squaring both sides, we have, S^2 = Young Modulus/ density. So, Young Modulus = S^2 * density; where S is the speed of the longitudinal wave. Then Substiting into the eqn we have (5.1 *10^3)^2 * 2.7 * 10^3 = 26.01 * 10^6 * 2.7 *10^6 = 26.01 * 2.7 * 10^ (6+3) = 70.227 * 10 ^9</span>

5 0

3 years ago

A stone is dropped from from rest at the top of a mine shaft. It takes 95 seconds for the stone to fall to the bottom of the min

Marianna [84]

Distance of fall from rest,
without air resistance = (1/2) (gravity) (time)²

   = (1/2) (9.8 m/s²) (95 sec)²

   = (4.9 m/s²) (9,025 sec²)

   = 44,222.5 meters .

The depth of the mine shaft is five times the height of Mt. Everest !

6 0

2 years ago