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SOVA2 [1]
2 years ago
13

Please select the word from the list that best fits the definition

Physics
2 answers:
Kobotan [32]2 years ago
6 0

Answer:

false

Explanation:

took test

alexdok [17]2 years ago
4 0

Answer:

yuuuhhhhhhh

Explanation:

it maxed Imao

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Vlad [161]
You need to have the Mass and velocity
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3 years ago
A student drops a rubber ball onto a surface. Assume that this is a closed system. The ball bounces, but each successive bounce
spayn [35]
I would say D.) The ball bounces many times suggesting the energy is used up efficiently
5 0
2 years ago
Read 2 more answers
At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f
jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp =  change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s,  0.65 kg m/s)

The magnitude of this vector is calculated as follows:

|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

4 0
3 years ago
Natalie accelerate her skateboard along a straight path from 0 m/s to 4.0 m/s in 2.5 s. find her average acceleration
Vikentia [17]
Acceleration=(speed end - speed start)/ time
Data:
speed end=4 m/s
speed start=0 m/s
time=2.5 s

acceleration=(4 m/s - 0 m/s)/2.5 s=1.6 m/s²

Answer: the acceleration would be 1.6 m/s²
8 0
3 years ago
Can someone explain which of Newton’s Law is demonstrated in part 1 and which is demonstrated in part 2? (Picture)
rewona [7]

Answer:

Every action has an equal and opposite reaction. If the student doesn't push, nothing moves, is one student pushes, both move which is an example of newtons third law.

Explanation:

3 0
3 years ago
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