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solong [7]
2 years ago
13

QUICK ITS DUE IN 6 MINUTES. create a scenario that applies the 3 laws of motion. Explain in complete sentences how each law if d

emonstrated
Physics
1 answer:
yKpoI14uk [10]2 years ago
5 0

Answer:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

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1. A ball with a mass of 2.3 kg travels with a velocity of 10 m/s. What is it's kinetic energy?
monitta

Answer:

KE=½mv²

KE=½2.3kg×(10m/s)²

KE=½2.3kg×100m²/s²

KE=2.3kg×50m²/s²

KE=115joules

7 0
2 years ago
How does a rubber rod become negatively charged through friction?
Artist 52 [7]
Thank you for posting your question here and Brainly!~

Even though you have not provided answer choices, I believe the answer is whatever letter corresponds with the answer: "It is rubbed with another object, and electrons move onto the rod."

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8 0
3 years ago
Read 2 more answers
Calculate the aceleration of a vehicle wich start with a zero meter per second, and acelerates to 34 m/s in 21 s.
Leviafan [203]

Answer:

For the aceleration we have:

Vf = Vo + a * t

Clearing "a":

a = (Vf - Vo) / t

Replacing and resolving:

a = (34 m/s - 0 m/s) / 21 s

a = 34 m/s / 21 s

a = 1,61 m/s^2

The aceleration of the vehicle is<u> 1,61 meters per second squared</u>

8 0
3 years ago
a pistol fires a bullet towards a target located 175m away. The bullet is traveling at 320 m/s. How long does it take for the bu
inn [45]

Time = distance/speed

Time = (175 m) / (320 m/s)

<em>Time = 0.547 second</em>

7 0
3 years ago
A 1.89 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4.77 N
Maksim231197 [3]

Answer:

Magnitude F(t)=26.6 N

Direction: -x

Explanation:

Given data

Spring constant K=4.77 N/m

Mass m=1.89 kg

Displace A=5.56m

Time t=3.96s

To find

Magnitude of force F

Solution

The angular frequency is given as

w=\sqrt{\frac{K}{m} } \\w=\sqrt{\frac{4.77N/m}{1.89kg} }\\w=1.59rad/s

Force on object is

F(t)=-mAw^{2}Cos(wt)

Substitute given values

So

F(t)=-(1.89kg)(5.56m)(1.59rad/s)^{2}Cos(1.59*3.96)\\F(t)=-26.6N

So

Magnitude F(t)=26.6 N

Direction: -x

4 0
3 years ago
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