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Artemon [7]
3 years ago
11

3.00 kg block moving 2.09 m/s right hits a 2.22 kg block moving 3.92 m/s left. afterwards, the 3.00 kg block moves 1.71 m/s left

. find the velocity of the 2.22 kg block afterwards
Physics
1 answer:
Sloan [31]3 years ago
6 0

Answer:

1.215 m/s to the right

Explanation:

The overall momentum of the system is conserved before and after collision.

Before the collision:

Momentum = 3.00*2.09 - 2.22*3.92 = -2.4324 (keep in mind that the motions are opposite in directions so they are supposed to have opposite signs. I chose leftward motion as negative and rightward motion as positive)

After the collision:

<h3>Momentum = -3.00*1.71 + 2.22*x;</h3>

Since momentum is conserved:

<h3>-3.00*1.71 +2.22x = -2.4324</h3>

Solve for x to find the velocity of the 2nd block.

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Pushing a broke down car, even done by more than one person, is difficult especially if the distance to be covered is quite far. A car is heavy and it requires a lot of force to start the car moving. This is because the inertia of the car to remain at rest is great. Additionally, the force applied in pushing the car must be greater than the frictional force to cause it to accelerate. The frictional force is dependent on the mass of the object which means that the frictional force acting on the car is also great. Finally, with every push of the car, the frictional force will always be present and acting on the opposite direction. The push that will be supplied must be sustained all throughout.
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4 years ago
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Example of the center of the gravity<br>​
velikii [3]

Answer:

The example of the center of the gravity is the middle of a seesaw

Explanation:

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5 0
3 years ago
A box having a mass of 0.2 kg is dragged across a horizontal floor by
Lyrx [107]

you can check attachment for answer.

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7 0
3 years ago
An energy source forces a constant current of 2A to flow through a light bulbfilament for twenty seconds. If 4.6 kJ is given off
QveST [7]

Answer:

The voltage drop across the bulb is 115 V

Explanation:

The voltage drop equation is given by:

V=\frac{\Delta W}{\Delta q}

Where:

ΔW is the total work done (4.6kJ)

Δq is the total charge

We need to use the definition of electric current to find Δq

I=\frac{\Delta q}{\Delta t}

Where:

I is the current (2 A)

Δt is the time (20 s)

2=\frac{\Delta q}{20}

q=40 C

Then, we can put this value of charge in the voltage equation.

V=\frac{4600}{40}=115 V

Therefore, the voltage drop across the bulb is 115 V.

I hope it helps you!

6 0
3 years ago
A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th
aliina [53]

Answer:

58.5 m

Explanation:

First of all, we need to find the total time the ball takes to reach the water. This can be done by looking at the vertical motion only.

The initial vertical velocity of the ball is

u_y = u sin \theta

where

u = 21.5 m/s is the initial speed

\theta=33.5^{\circ} is the angle

Substituting,

u_y = (21.5) sin 33.5^{\circ} =11.9 m/s

The vertical position of the ball at time t is given by

y = h + u_y t + \frac{1}{2}gt^2

where

h = 13.5 m is the initial heigth

g = -9.8 m/s^2 is the acceleration of gravity (negative sign because it points downward)

The ball reaches the water when y = 0, so

0 = h + u_yt +\frac{1}{2}gt^2\\0 = 13.5 +11.9 t - 4.9t^2

Which gives two solutions: t = 3.27 s and t = -0.84 s. We discard the negative solution since it is meaningless.

The horizontal velocity of the ball is

u_y = u cos \theta = (21.5) cos 33.5^{\circ} =17.9 m/s

And since the motion along the horizontal direction is a uniform motion, we can find the horizontal distance travelled by the ball as follows:

d= u_x t = (17.9)(3.27)=58.5 m

3 0
3 years ago
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