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lapo4ka [179]
2 years ago
5

Two blocks are on a horizontal frictionless surface. Block a is moving with an initial velocity

Physics
1 answer:
AfilCa [17]2 years ago
5 0

The final velocity of the block A will be 2.5 m/sec. The principal of the momentum conversation is used in the given problem.

<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.

In a given concern, mass m₁ is M, mass m₂ is 3M. Initial speed for the mass m₁ and  m₂ will be u₁=5 and u₂=0 m/s respectively,

According to the law of conservation of momentum

Momentum before collision =Momentum after collision

m₁u₁+m₂u₂=(m₁+m₂)v

M×5+3M×0=[M+3M]v

The final velocity is found as;

V=51.25 m/s

The velocity of block A is found as;

\rm V_f=\frac{(m_1-m_2)u_1}{m_1+m_2} +\frac{2m_2u_2}{m_1+m_2} \\\\  V_f=\frac{(M-3M)5+2\times3M\times0}{m_1+m_2} \\\\ V_f=2.5 m/s

Hence, the final velocity of the block A will be 2.5 m/sec.

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

brainly.com/question/1113396

#SPJ4

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The radio waves have the highest photon energy and the lowest is microwaves.

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4 0
2 years ago
A magnetic field of magnitude 0.550 T is directed parallel to the plane of a circular loop of radius 43.0 cm. A current of 5.80
Assoli18 [71]

Answer:

\mu = 3.36\times 10^{-3}\ A-m^2

Explanation:

Given that,

The magnitude of magnetic field, B = 0.55 T

The radus of the loop, r = 43 cm = 0.43 m

The current in the loop, I = 5.8 mA = 0.0058 A

We need to find the magnetic moment of the loop. It is given by the relation as follows :

\mu = AI\\\\\mu=\pi r^2\times I

Put all the values,

\mu=\pi \times (0.43)^2\times 0.0058\\\\=3.36\times 10^{-3}\ A-m^2

So, the magnetic moment of the loop is equal to3.36\times 10^{-3}\ A-m^2.

3 0
3 years ago
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