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erastovalidia [21]
2 years ago
9

The two particles are both moving to the right. Particle 1 catches up with particle 2 and collides with it. The particles stick

together and continue on with velocity vf. Which of these statements is true?
A. vf=v1
B. vf=v2
C. vf is less than v2
D. vf is less than v2 but greater than v1
E. vf>v1

Physics
1 answer:
Irina18 [472]2 years ago
8 0

Answer:

See bolded below.

Explanation:

Consider the " Before " and " After. " " Before, " this particle 1 was trying to catch up with this particle 2, and " after " particle one had collided with particle two. Take a look at the attachment below for a more detailed examination.

Here is how this will play out. Particle 1, with great velocity, will hit particle 2, which would mean that Particle 2 has less velocity than Particle 1. Now after the collision, energy is transferred to Particle 2, and while Particle 1 has now stopped in it's tracks, Particle 2 - with more energy than before - will continue as long as it has to before friction eventually brings it to a stop.

_______________________________________________________

From this we can conclude that Vf, from the picture below, must have less energy than V1, but more energy than V2 - and vice versa.

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Butoxors [25]

Answer:5 m/s^{2}

Explanation:

The described situation is is related to vertical motion (and free fall). So, we can use the following equation that models what happens with this rock:

y=y_{o}+V_{o}sin\theta t-\frac{1}{2}gt^{2} (1)

Where:

y=0m is the rock's final height

y_{o}=100 m is the rock's initial height

V_{o}=15 m/s is the rock's initial velocity

\theta=90\° is the angle at which the rock was thrown (directly upwards)

t=10 s is the time

g is the acceleration due gravity in Planet X

Then, isolating g and taking into account sin(90\°)=1:

g=(-\frac{2}{t^{2}})(y-y_{o}-V_{o}t) (2)

g=(-\frac{2}{(10 s)^{2}})(0 m-100 m-(15 m/s)(10 s)) (3)

Finally:

g=5 m/s^{2} (4) This is the acceleration due gravity in Planet X

7 0
1 year ago
A 145-g baseball is thrown so that it acquires a speed of 25 m/s. What was the net work done on the ball to make it reach this s
inysia [295]

When the ball has left your hand and is flying on its own, its kinetic energy is

KE = (1/2) (mass) (speed²)

KE = (1/2) (0.145 kg) (25 m/s)²

KE = (0.0725 kg) (625 m²/s²)

<em>KE = 45.3 Joules</em>

If the baseball doesn't have rocket engines on it, or a hamster inside running on a treadmill that turns a propeller on the outside, then there's only one other place where that kinetic energy could come from:  It MUST have come from the hand that threw the ball.  The hand would have needed to do  <em>45.3 J</em>  of work on the ball before releasing it.

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2 years ago
The food calorie, equal to 4186J , is a measure of how much energy is released when food is metabolized by the body. A certain b
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<h2>The hiker will go up to 850 m on the hill</h2>

Explanation:

The total energy gained  by the hiker = 140 x 4186 J

This energy is consumed in the potential energy acquired , while climbing up the hill.

The potential energy P.E = mass of hiker x acceleration due to gravity x height

Thus

140 x 4186 = 69 x 10 x h

or h = \frac{4186x140}{69x10}  = 850 m

If the 20% of the total energy is used

the height h₀ = \frac{0.2x4186x140}{69x10} = 170 m

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2 years ago
A 36,287 kg truck has a momentum of 907,175 kg • m/s. What is the trucks velocity.
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In this case, since Mass is being multiplied by Velocity, to solve for be Velocity you would divide both sides by Velocity.  The velocity will equal the momentum divided by the mass.

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13. An aircraft heads North at 320 km/h rel:
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The velocity of the aircraft relative to the ground is 240 km/h North

Explanation:

We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.

Mathematically:

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Taking north as positive direction, we have:

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Therefore, we find

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Learn more about vector addition:

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#LearnwithBrainly

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