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Lubov Fominskaja [6]
3 years ago
12

A bomb of mass 6kg initially at rest explodes into two fragments of masses 4kg and2kg respectively. If the greater mass moves wi

th a velocity of 5m_s, find the velocity of 2kg mass​
Physics
1 answer:
Katen [24]3 years ago
7 0

Answer:

v = 10 [m/s]

Explanation:

The largest mass is that of 4 [kg], in this way the momentum can be calculated by means of the product of the mass by velocity.

P=m*v\\

where:

P = momentum [kg*m/s]

m = mass = 4 [kg]

v = velocity = 5 [m/s]

Now the momentum:

P=4*5\\P=20[kg*m/s]

This same momentum is equal for the other mass, in this way we can find the velocity.

P=m*v\\20=2*v\\v=10[m/s]

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Yes! Thinking about it graphically a position vs time graph models meters per second in most cases, making every point on the line have the units m/s. If we want the find the slope we are finding the change between each point and those units would change to m/s/s or m/s^2 giving us the same units for acceleration. Simply put, slope of a velocity graph gives us acceleration.

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Since the coefficient of friction is less than 1, what does that mean about the normal force and the force of friction?
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there is friction between the two things

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3 years ago
A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
never [62]

The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

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2 years ago
If a rock sample has a mass of 1.17 g and a volume of 0.33 cm3, what type of rock is it?
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The correct answer is diamond

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