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BabaBlast [244]

2 years ago

9

If a frictional force of 4 N is acting against a push of 53 N and the box that is being pushed is 4 kg, what is its acceleration

?

1 answer:

kifflom [539]2 years ago

3 0

Answer:

gggggggggggg

Explanation:

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telescopes on the earth suffer from the fact that the light coming from the stars has to pass through air or space

Alexeev081 [22]

True. as the light is diffracted it is diffused and must be corrected for

3 0

3 years ago

A 25N force is applied to a bar that can pivot around its end. The force is r=0.75 m away from the end of an angle at 0= 45. wha

andrew-mc [135]

Answer:

The answer is B. I had a question like this last week in my physics practice exam. dont have alot of time so I cant explain right now

Explanation:

8 0

2 years ago

A squirrel is trying to locate some nuts he buried for the winter. He moves 4.3 m to the right of a stone and dogs unsuccessfull

krek1111 [17]

Answer:

The total displacement from the starting point is 1.5 m.

Explanation:

You need to sum and substract, depending on the movement (to the right, sum; to the left, substract).

First, it moves 4.3 m right and return 1.1 m. So the new distance from the starting point is 3.2 m.

Second, it moves 6.3 m right, so the new distance is 9.5 m.

Finally it moves 8 m to the left, so 9.5 m - 8 m= 1.5 m.

Summarizing, at the end the squirrel is 1.5 m from its starting point.

8 0

3 years ago

A 6.00 kg ball is dropped from a height of 12.0 m above one end of a uniform bar that pivots at its center. The bar has mass 5.0

Andre45 [30]

Answer:

The height of the other ball go after the collision is 2.304 m.

Explanation:

Given that,

Mass of ball = 6.00 kg

Height = 12.0 m

Mass of bar =5.00 kg

Length = 4.00 m

Suppose we need to calculate how high will the other ball go after the collision

We need to calculate the velocity of ball

Using formula of velocity

$v=\sqrt{2gh}$

$v=\sqrt{2\times9.8\times12.0}$

$v=15.33\ m/s$

We need to calculate the angular momentum

Using formula of angular momentum

$l_{before}=mvr$

Put the value into the formula

$l_{before}=6.00\times15.33\times2.0$

$l_{before}=183.96\ kgm^2/s$

We need to calculate the angular momentum

Using formula of angular momentum

$l_{after}=I_{t}\omega$

$l_{after}=(\dfrac{ml^2}{12}+m_{1}r^2+m_{2}r^2)\omega$

Put the value into the formula

$l_{after}=(\dfrac{5\times4.00^2}{12}+6.00\times2.0^2+6.00\times2.0^2)\omega$

$183.96=54.66\omega$

$\omega=\dfrac{183.96}{54.66}$

$\omega=3.36\ rad/sec$

After collision the ball leaves with velocity

We need to calculate the velocity after collision

Using formula of the velocity

$v= r\omega$

$v=2.0\times3.36$

$v=6.72\ m/s$

We need to calculate the height

Using formula of height

$h=\dfrac{v^2}{2g}$

Put the value into the formula

$h=\dfrac{(6.72)^2}{2\times9.8}$

$h=2.304\ m$

Hence, The height of the other ball go after the collision is 2.304 m.

5 0

3 years ago

What is the motion of the particles in this kind of wave?

densk [106]

The particles will move up and down over small areas.

Option B.

<u>Explanation:</u>

The type of the wave that has been discussed in the question is the transverse kind of wave. The transverse kind of wave are the waves where the particle motion is the perpendicular to the motion of the wave.

In the transverse kind of wave the which is a moving kind of a wave where the oscillations are perpendicular to the direction of the movement of the wave.

8 0

2 years ago