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2 years ago

11

Brainliest whoever answers first!!! and right!!! Identify the situations that have an unbalanced force. Check all that apply. a

weight speeds up as it falls through the air a garbage can is at rest on the ground a hockey puck moves in a straight line at a constant speed a bumper car hit by another car moves off at an angle a balloon flies across the room when the air is released

2 answers:

Marta_Voda [28]2 years ago

8 0

Answer:

A baseball speeds up as it falls through the air.

Yes. Forces on the balloon are unbalanced.

The balloon is speeding up, so we know that the downward force

of gravity is stronger than the upward force of air resistance.

A soccer ball is at rest on the ground.

No. The ball is not accelerating, so we know that the forces on it

are balanced.

The downward force of gravity on the ball and the upward force

of the ground are equal.

An ice skater glides in a straight line at a constant speed.

No. The skater's speed and direction are not changing, so he is not

accelerating. That tells us that the forces on him are balanced.

A bumper car hit by another car moves off at an angle.

Yes. The direction in which the car was moving changed.

That's acceleration, so we know that the forces on it are unbalanced,

at least at the moment of impact.

A balloon flies across the room when the air is released.

Yes. The balloon was not moving. But when the little nozzle was

opened, it started to zip around the room. So its speed changed.

And, as it goes bloozing around the room, its direction keeps changing too.

There's a whole lot of acceleration going on, so we know the forces on it

are unbalanced.

Explanation:

adelina 88 [10]2 years ago

5 0

Answer:

its A,D,E

Explanation:

I got 2 right one of them wrong

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The diagram shows waves of sound travelling through the air. By which factor is the sound intensity decreased at 3 meters?

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The diagram is missing; however, we know that the intensity of a sound wave is inversely proportional to the square of the distance from the source:
$I(r)= \frac{1}{r^2}$
where I is the intensity and r is the distance from the source.

We can assume for instance that the initial distance from the source is r=1 m, so that we put
$I= \frac{1}{r^2}= \frac{1}{(1)^2}=1$
The intensity at r=3 m will be
$I= \frac{1}{r^2}= \frac{1}{(3)^2}= \frac{1}{9}$
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The equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.

The horizontal momentum is given by:

$p_{i} = p_{f}$

$m_{1}v_{1}_{i} + m_{2}v_{2}_{i} = m_{1}v_{1}_{f} + m_{2}v_{2}_{f}$

Where:

m₁: is the mass of the lab cart = 15 kg

m₂: is the <em>mass </em>of the object dropped = 2 kg

$v_{1}_{i}$ : is the initial velocity of the<em> lab cart </em>

$v_{2}_{i}$ : is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)

$v_{1}_{f}$ : is the final velocity of the<em> lab cart </em>

$v_{2}_{f}$ : is the <em>final velocity</em> of the <em>object </em>

Then, the horizontal momentum is:

$15v_{1}_{i} + 2*0 = 15v_{1}_{f} + 2v_{2}_{f}$

When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:

$15v_{1}_{i} + 2*0 = v_{f}(15 + 2)$

Therefore, the equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ represents the horizontal momentum (option 3).

Learn more about linear momentum here:

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