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MrRa [10]
3 years ago
14

A rifle is fired and recoils when the bullet leaves the gun. This is an example of Newton's third law. The acceleration of the b

ullet is called _______
Physics
2 answers:
yuradex [85]3 years ago
6 0
He size of the forces on the first object equals the size of the force on the second object. 
sweet [91]3 years ago
5 0
<span>The acceleration of the bullet is called ACTION.

</span>Formally stated, Newton's third law<span> is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.
</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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A sound wave with a frequency of 400 Hz is moving through a solid object. If the wavelength of the sound wave is 8 m, what is th
Paraphin [41]

Answer:

v = 3200 m/s

Explanation:

As we know that the frequency of the sound wave is given as

f = 400 Hz

wavelength of the sound wave is given as

\lambda = 8 m

so now we have

speed = wavelength \times frequency

so we will have

v = (8m) \times (400 Hz)

v = 3200 m/s

4 0
3 years ago
The frequency of sound wave A is 250 Hz. Find the time period.
olga2289 [7]
Time period = 1 / frequency
Time period = 1 / 250 th of a second
7 0
4 years ago
PLEASE HELP ME WITH A PHYSICS QUESTION!!!!!!!!!!
stira [4]

Answer:

C. 3.00 s

Explanation:

Given:

Δy = 1.80 m − 46.0 m = -44.2 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

-44.2 m = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 3.00 s

6 0
3 years ago
Read 2 more answers
) you carry a 7.0 kg bag of groceries 1.2 m above the ground at constant velocity across a 2.7 m room. how much work do you do o
lapo4ka [179]
M = 7.0 kg, the mass of the groceries
h = 1.2 m, the elevation of the bag of groceries

The bag of groceries moves a constant velocity over the 2.7-m room.
At constant velocity, there is no applied force, and the kinetic energy remains constant.

At an elevation of 1.2 m, there is an increase in PE (potential energy) given by
V = m*g*h
    = (7.0 kg)*(9.8 m/s²)*(1.2 m)
    = 82.32 J

The change in PE is equal to the work done.

Answer: 82.3 J

3 0
3 years ago
Two carts are free to slide along the frictionless track shown in figure below. Cart A of mass m1 = 8 kg is released from 12m. A
Harrizon [31]

The height reached by the two carts after collision is determined as 5.34 m.

<h3>Initial velocity of Cart A</h3>

Apply the principle of conservation of mechanical energy.

K.E = P.E

v = √2gh

v = √(2 x 9.8 x 12)

v = 15.34 m/s

<h3>Final velocity of the two carts after the collision</h3>

Apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

8(15.34) + 4(0) = v(8 + 4)

122.72 = 12v

v = 10.23 m/s

<h3>Height reached by both carts</h3>

Apply the principle of conservation of mechanical energy.

P.E = K.E

mgh = ¹/₂mv²

h = v²/(2g)

h = (10.23²) / (2 x 9.8)

h = 5.34 m

Learn more about linear momentum here: brainly.com/question/7538238

#SPJ1

3 0
2 years ago
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