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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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According to the Biot-Savart law, the magnetic field produced by a moving charge will be zero at a point _______ the moving char
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<span>According to the Biot-Savart law, the magnetic field produced by a moving charge will be zero at a point in back of. The correct option among all the options that are given in the question is the second option or option "b". I hope that this answer has actually come to your help.</span>
7 0
3 years ago
a projectile with mass m is fired with initial horizontal velocity vx from height h above level ground. which change would have
Elza [17]

Explanation:

We know that in a projectile motion time of flight

t= \sqrt{\frac{h}{4.9} }

So, we can notice that time of flight is dependent only on height from which the projectile is thrown. The more the height the more will be time of flight. So, if we increase the height the time of flight of the projectile will certainly increase.

8 0
4 years ago
Read 2 more answers
An electron is released in a uniform electric field, and it experiences an electric force of 2.36 x10--13 N directed toward the
Andrei [34K]

Answer:

The magnitude of the electric field is 1475000 N/C and the  direction of the electric field is toward north.

Explanation:

Given that,

Electric force F=2.36\times10^{-13}\ N

The direction of force is toward the north.

We need to calculate the magnitude of the electric field

Using formula of electric field

E=\dfrac{F}{q}

Where, f = electric force

q = charge

Put the value into the formula

E=\dfrac{2.36\times10^{-13}}{1.6\times10^{-19}}

E=1475000\ N/C

The direction of the electric field is toward the direction of the force so,

The direction of the electric field is toward north.

Hence, The magnitude of the electric field is 1475000 N/C and the  direction of the electric field is toward north.

7 0
3 years ago
2 dm expressed in milimeters
asambeis [7]
The answer is 200 mm
8 0
3 years ago
A 3.1 kg dog stands on an 18 kg flatboat and is 6.1 m from the shore. He walks 2.5 m on the boat toward shore and then stops. As
Pachacha [2.7K]

Answer:

The distance of dog from the shore is 3.97 m

Explanation:

given,

mass of the dog = 3.1 Kg

mass of the flatboat = 18 Kg

Distance from the shore = 6.1 m

dog moves on boat = 2.5 m

dog moves leftward and boat moves rightward.

If it's a frictionless system with no initial velocity, the center of mass doesn't move. Conservation of momentum, which = 0 in this case.

When dog move toward the shore a reactive force will act on the boat.

m_b x_b + m_d x_d = 0

m_b i and m_d is mass of boat and mass of dog.

x_b and x_d is distance moved by the boat and the dog.

x_b = \dfrac{m_dx_d}{m_b}

neglecting negative sign

x_b + x_d = 2.5

\dfrac{m_dx_d}{m_b} + x_d = 2.5

x_d = \dfrac{2.5}{1+\dfrac{m_d}{m_b}}

x_d = \dfrac{2.5}{1+\dfrac{3.1}{18}}

x_d = \dfrac{2.5}{1.172}

x_d = 2.133 m

The distance of dog from the shore is 6.1 – 2.133 = 3.97 m

The distance of dog from the shore is 3.97 m

4 0
3 years ago
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