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Julli [10]
3 years ago
15

Two balls of mass m1 and m2, with velocities v1 and v2 collide head on. Is there any way for both balls to have zero velocity af

ter the collision? If so, find the conditions under which this can occur.
Physics
1 answer:
nordsb [41]3 years ago
8 0

Answer:

Explanation:

As the final Kinetic energy is zero or less than initial kinetic energy, the collision must be inelastic.  

In Inelastic collision both the bodies must stick together as final velocity is zero for both the bodies.

To conserve the momentum, momentum associated before the collision of first must be equal and opposite to the momentum associated with the second ball.

i.e.

m_1v_1=m_2v_2

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An engine absorbs 1749 J from a hot reservoir and expels 539 J to a cold reservoir in each cycle. a. What is the engine’s effici
Masja [62]

Answer:

a)η = 69.18 %

b)W= 1210 J

c)P=3967.21 W

Explanation:

Given that

Q₁ = 1749 J

Q₂ = 539  J

From first law of thermodynamics

Q₁   = Q₂ +W

W=Work out put

Q₂=Heat rejected to the cold reservoir

Q₁ =heat absorb by hot  reservoir

W= Q₁- Q₂

W= 1210 J

The efficiency given as

\eta=\dfrac{W}{Q_1}

\eta=\dfrac{1210}{1749}

\eta=0.6918

η = 69.18 %

We know that rate of work done is known as power

P=\dfrac{W}{t}

P=\dfrac{1210}{0.305}\ W

P=3967.21 W

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3 years ago
Keisha is making a diagram of a simple machine in the body.
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Answer:

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3 years ago
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Which two changes would decrease the electric force between two charged
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Answer:

it is b and e

Explanation:

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2 years ago
A hollow cylinder with an inner radius of and an outer radius of conducts a 3.0-A current flowing parallel to the axis of the cy
Artemon [7]

Complete Question:

A hollow cylinder with an inner radius of 4.0mm and an outer radius of 30mm conducts a 3.0-A current flowing parallel to the axis of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 12mm from its center?

Answer:

The magnitude of the magnetic field = 7.24 μT

Explanation:

Inner radius, a = 4.0 mm = 0.004 m

Outer radius, b = 30 mm = 0.03 m

Radius, r = 12 mm = 0.012 m

let h² = b² - a²

h² = 0.03² - 0.004²

h² = 0.000884

Let d² = r² - a²

d² = 0.012² - 0.004²

d² = 0.000128

Current I = 3A

μ = 4π * 10⁻⁷

The magnitude of the magnetic field is given by:

B = \frac{\mu I d^{2} }{2\pi r h^{2} } \\B =  \frac{4\pi * 10^{-7}   * 3* 0.000128^{2} }{2\pi *0.012* 0.000884^{2} }

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B = 7.24 μT

7 0
3 years ago
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