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stiv31 [10]
3 years ago
6

Kyle lays a mirror flat on the floor and aims a laser at the mirror. The laser beam reflects off the mirror and strikes an adjac

ent wall. The plane of the incident and reflected beams is perpendicular to the wall. The beam from the laser strikes the mirror at a distance a=33.7 cm from the wall. The reflected beam strikes the wall at a height b=36.7 cm above the surface of the mirror. Find the angle of incidence θi at which the laser beam strikes the mirror.

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

Answer:

The angle of incidence is  \theta_i  =42.6^o

Explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

   The distance between the mirror and the wall is  a = 33.7 cm

    The height of the above the mirror is  b = 36.7 cm

Generally the angle which the reflected ray make with the mirror is mathematically evaluated as

             \alpha  =tan ^{-1} (\frac{b}{a})

substituting values

             \alpha = tan ^{-1}( \frac{36.7}{33.7})

            \alpha =47.4^o

From the diagram we can deduce that  the angle of incidence is

             \theta_i  = 90 - \alpha

So          \theta_i = 90 - 47.4

              \theta_i  =42.6^o

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Alex

Answer:

I = 0.2 A

Explanation:

Lamp is rated at 300 mA

I_lamp = 0.3 A

Voltage is; V = 3V

Thus; Resistance is given by;

R = V/I

R = 3/0.3

R = 10 ohms

Now, since the ammeter of 5 ohms is connected in series with the lamp. Thus equivalent resistance;

R_eq = 10 + 5

R_eq = 15 ohms

Ammeter current will be;

I = V/R_eq

I = 3/15

I = 0.2 A

3 0
3 years ago
ASAP please 25 points if right
garik1379 [7]

Answer:

1. The resistance of any physical object to any velocity

2. It continues in it's existing state of rest or uniform motion

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3 years ago
A 4.5 g coin sliding to the right at 23.8 cm/s makes an elastic head-on collision with a 13.5 g coin that is initially at rest.
Airida [17]

Answer:

a) v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} ), b) \Delta K = 9.559\times 10^{-5}\,J

Explanation:

a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:

(4.5\times 10^{-3}\,kg)\cdot (23.8\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot (0\,\frac{m}{s} ) = (4.5\times 10^{-3}\,kg)\cdot (-11.9\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot v

The final velocity is:

v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )

b) The change in the kinetic energy of the 13.5 g coin is:

\Delta K = \frac{1}{2}\cdot (13.5\times 10^{-3}\,kg)\cdot \left[(11.9\times 10^{-2}\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}\right]

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4 0
3 years ago
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6 0
3 years ago
Problem 8: Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric fi
yanalaym [24]

Answer:

10581.59 V

Explanation:

We are given that

Magnetic field=B=0.65 T

Speed of electron=v=6.1\times 10^7m/s

Charge on electron, q=e=1.6\times 10^{-19} C

Mass of electron,m_e=9.1\times 10^{-31} kg

We have to find the potential difference in volts required in the first part of the experiment to accelerate electrons.

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Where V=Potential difference

m_e=Mass of electron

v=Velocity of electron

Using the formula

V=\frac{(6.1\times 10^7)^2\times 9.1\times 10^{-31}}{2\times 1.6\times 10^{-19}}

V=10581.59 V

Hence, the potential difference=10581.59 V

8 0
3 years ago
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