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Greeley [361]
3 years ago
14

Explain why is not advisable to use small values of I in performing an experiment on refraction through a glass prism?

Physics
2 answers:
Svetradugi [14.3K]3 years ago
6 0
Explain<span> why it is </span>not advisable to use small values<span> of incident ray in </span>performing experiment<span> on the</span>refraction through a glass prism<span>.</span>
Colt1911 [192]3 years ago
3 0

Answer:

Explanation:

Prism is a glass object that is used to study the pattern of refraction.

During this experiment it is not advised to used small values of incident angle , as when a small amount of incident ray is passed the refracted pattern observed at the end will not provide satisfactory results.

The amount of light that comes out of prism which is the refracted light will also be small hence study of complete refracted pattern will not be possible.

The small incident angle will give rise to more errors in measurement.

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3 years ago
bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. what is the veloc
USPshnik [31]

Answer:

The final velocity of the car A is -1.053 m/s.

Explanation:

For an elastic collision both the kinetic energy and the momentum of the system are conserved.

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m_A = mass of car A;

v_{A1} = the initial velocity of car A;

v_{A2} = the final velocity of car A;

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m_B = mass of car B;

v_{B1} = the initial velocity of car B;

v_{B2} = the final velocity of car B.

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m_Av_{A1}+m_Bv_{B1} =m_Av_{A2}+m_Bv_{B2}

And the conservation of kinetic energy says that

\dfrac{1}{2} m_Av_{A1}^2+\dfrac{1}{2}m_Bv_{B1}^2=\dfrac{1}{2}m_Av_{A2}^2+\dfrac{1}{2}m_Bv_{B2}^2

These two equations are solved for final velocities  v_{A2} and v_{B2} to give

$v_{A2} =\frac{m_A-m_B}{m_A+m_B} v_{A1}+\frac{2m_B}{m_A+m_B} v_{B1}$

$v_{B2} =\frac{2m_A}{m_A+m_B} v_{A1}+\frac{m_B-m_A}{m_A+m_B} v_{B1}$

by putting in the numerical values of the variables we get

$v_{A2} =\frac{281-209}{281+209} (2.82)+\frac{2*209}{281+209} (-1.72)$

\boxed{v_{A2} = -1.05m/s}

and

$v_{B2} =\frac{2*281}{281+209} (2.82)+\frac{209-281}{281+209} (-1.72)$

\boxed{v_{B2} = 3.49m/s}

Thus, the final velocity of the car A is -1.053 m/s and of car B is 3.49 m/s.

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