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Alona [7]
3 years ago
11

If the light rays in the incident ray originate at 20º from the surface...

Physics
1 answer:
Ainat [17]3 years ago
5 0

Given :

The light rays in the incident ray originate at 20º from the surface.

To Find :

The angle for the incident ray.

Solution :

We know angle of incidence is measured from perpendicular to the surface .

So,

\theta_i = 90^o - 20^o\\\\\theta_i = 70^o

Now, we know by law of reflection :

Angle of reflection = Angle of incident

\theta_r = 70^o

Therefore, the angle of incident and reflection is 70° each.

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What has an ionised atom lost or gained?<br> Protons<br> Gamma Rays<br> Electrons<br> Neutrons
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Read 2 more answers
A horizontal circular platform (m = 119.1 kg, r = 3.23m) rotates about a frictionless vertical axle. A student (m = 54.3kg) walk
Murrr4er [49]

Answer:

\omega_2=5.1rad/s

Explanation:

Since there is no friction angular momentum is conserved. The formula for angular momentum thet will be useful in this case is L=I\omega. If we call 1 the situation when the student is at the rim and 2 the situation when the student is at r_2=1.39m from the center, then we have:

L_1=L_2

Or:

I_1\omega_1=I_2\omega_2

And we want to calculate:

\omega_2=\frac{I_1\omega_1}{I_2}

The total moment of inertia will be the sum of the moment of intertia of the disk of mass m_D=119.1 kg and radius r_D=3.23m, which is I_D=\frac{m_Dr_D^2}{2}, and the moment of intertia of the student of mass m_S=54.3kg at position r (which will be r_1=r=3.23m or r_2=1.39m) will be I_{S}=m_Sr_S^2, so we will have:

\omega_2=\frac{(I_D+I_{S1})\omega_1}{(I_D+I_{S2})}

or:

\omega_2=\frac{(\frac{m_Dr_D^2}{2}+m_Sr_{S1}^2)\omega_1}{(\frac{m_Dr_D^2}{2}+m_Sr_{S2}^2)}

which for our values is:

\omega_2=\frac{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(3.23m)^2)(3.1rad/s)}{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(1.39m)^2)}=5.1rad/s

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3 years ago
A car starts from rest and accelerates for 2.3 s
gtnhenbr [62]

Answer:

1.2 miles

Explanation:

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