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777dan777 [17]
3 years ago
8

What is the name often given to the force exerted by the first object on a second object​

Physics
1 answer:
sukhopar [10]3 years ago
3 0
I think it’s action force please tell me if I’m wrong I hope you have a wonderful day
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Two blocks of clay, one of mass 1.00 kg and one of mass 7.00 kg, undergo a completely inelastic collision. Before the collision
RideAnS [48]

Answer:8 m/s

Explanation:

Given

m_1=1 kg

m_2=7 kg

kinetic Energy of m_1=32 J

initially m_2 is at rest and let say m_1 is moving with velocity u

kinetic Energy of m_1 is =\frac{m_1u^2}{2}=32

u^2=64

u=8 m/s

In Completely inelastic collision both mass stick together and move with common velocity

Suppose v is the common velocity

m_1u+0=(m_1+m_2)v

v=\frac{1\times 8}{1+7}=1 m/s

therefore Final velocity with which both blocks moves is 1 m/s

7 0
3 years ago
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Ede4ka [16]

Answer:

wassup

Explanation:

3 0
3 years ago
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What is the total wavelength if one-half of the wave is 3?
seraphim [82]

Answer:

6

Explanation:

Half the wave = 3

Wavelength = 3 x 2 = 6

3 0
3 years ago
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him
juin [17]

Answer:

The velocity of the student has after throwing the book is 0.0345 m/s.

Explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s

Angle = 31°

We need to calculate the magnitude of the velocity of the student has after throwing the book

Using conservation of momentum along horizontal  direction

m_{b}v_{b}\cos\theta= m_{c}v_{c}

v_{s}=\dfrac{m_{b}v_{b}\cos\theta}{m_{c}}

Put the value into the formula

v_{c}=\dfrac{1.25\times3.61\times\cos31}{112}

v_{c}=0.0345\ m/s

Hence, The velocity of the student has after throwing the book is 0.0345 m/s.

3 0
3 years ago
T-Joe (65 kg) is running at 3 m/s. T-Brud (50 kg) is running at 4 m/s. What would be T-Joe's momentum?
Debora [2.8K]

Answer:

P_J=195N

Explanation:

From the question we are told that

Mass\ of T-joe\ M_J=65\\Velocity\ of T-joe\ V_J=3m/s\\Mass of\ T-Brud\ M_B=50kg\\Velocity\ of T-Brud\ V_B=3m/s\\

Generally the equation for momentum is mathematically given by

P=mv

Therefore

T-Joe momentum P_J

P_J=65*3

P_J=195N

5 0
3 years ago
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