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AURORKA [14]
2 years ago
15

A cart with a mass of 120 kg and a velocity

Physics
1 answer:
klasskru [66]2 years ago
6 0

Answer:

dumpster mass

Explanation:

you will need the mass of the dumpster to calculate using conservation of momentum

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How are balanced and unbalanced forces related to motion?
Aliun [14]

Answer:

Balanced forces do not result in any change in motion. forces: forces applied to an object in opposite directions that are not equal in size. Unbalanced forces result in a change in motion.

Explanation:

4 0
3 years ago
Read 2 more answers
Help please (20 pts) I think it's C
gregori [183]

Answer:

I'm not sure it is c I'm sure it is d

6 0
3 years ago
Find the acceleration a body whose velocity increases from 11m/s to 33m/s in 10 seconds
solong [7]

Answer:

1.1 m/(s)^2

Explanation:

u=11 m/s

v=33 m/s

t=10s

v=u+at

=> 33=22+(a)(10)

=> 33-22=10a

=> 10a=11

=> a=11/10=1.1 m/(s)^2

7 0
2 years ago
Construct a graph of position versus time for the motion of a dog, using the data in the table below. Explain how the graph indi
Lynna [10]

Answer:

The dog is moving at a constant speed

Explanation:

Given that,

Position : 5, 10, 15, 20, 25

Time = 5. 10, 15, 20, 25

We need to draw a position time graph

Using given data

A graph of position and time shows the speed.

According to graph,

The graph indicates that the dog is moving at a constant speed because the graph is straight line.

Hence, The dog is moving at a constant speed

6 0
3 years ago
Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

3 0
4 years ago
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