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solmaris [256]
3 years ago
12

A beam oflight is reflected off a mirror. If the angle of incidence is 30 degrees, according to the Law of Reflection the angle

of the reflected light beam will bea.15 degreesb.30 degreesc.60 degreesd.90 degrees
Physics
1 answer:
zaharov [31]3 years ago
7 0

Answer:

30°

Explanation:

According to the second law of reflection, it States that the angle of incidence i is equal to the angle of reflection r.

The angle of incidence is known to be the angle between the incident ray and the normal.

The Angle of reflection is the angle between the reflected ray and the normal.

This normal ray is a ray that is perpendicular to the surface.

According to the question, if the beam of light is reflected off the surface and its angle of incidence is 30°, its angle of reflection will also be 30° i.e i=r = 30°

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3 years ago
A floating ice block is pushed through a displacement d = (23 m) i - (9 m) j along a straight embankment by rushing water, which
Sav [38]

Answer:

Work = 5941 J

Explanation:

As we know that work done is given by the equation

W = F. d

here we know that

F  = (200 N)\hat i - (149 N) \hat j

also we have

d = (23m) \hat i - (9 m)\hat j

now from above formula we have

W = (200 N\hat i - 149 N \hat j).(23 m\hat i - 9m \hat j)

W = 5941 J

3 0
3 years ago
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What could be used as another word for electrical potential
tekilochka [14]

Answer:

hey mate

answer is probably voltage as per me

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Explanation:

Voltage, electric potential difference, electric pressure or electric tension is the difference in electric potential between two points, which is defined as the work needed per unit of charge to move a test charge between the two points

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2 years ago
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I have a voltage source of 12V but a light that only burns at 5V. The lamp works on 18 mA. Calculate the resistance that you EXT
ratelena [41]

Answer:

The resistance that will provide this potential drop is 388.89 ohms.

Explanation:

Given;

Voltage source, E = 12 V

Voltage rating of the lamp, V = 5 V

Current through the lamp, I = 18 mA

Extra voltage or potential drop, IR =  E- V  

                                                    IR = 12 V - 5 V = 7 V

The resistance that will provide this potential drop (7 V) is calculated as follows:

IR = V

R = \frac{V}{I} = \frac{7 \ V}{18 \times 10^{-3} A} \ = 388.89 \ ohms

Therefore, the resistance that will provide this potential drop is 388.89 ohms.

7 0
3 years ago
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates
34kurt

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance C_{0} = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \times 10^{4} V/m?

Explanation:

It is known that relation between electric field and the voltage is as follows.

             V = Ed

Now,  

              Q = CV

or,           Q = C \times Ed

Therefore, substitute the values into the above formula as follows.

              Q = C \times Ed

                  = 8.50 pF \times (\frac{10^{-12} F}{1 pF})(3 \times 10^{4} m/s)(1 mm)(\frac{10^{-3} m}{1 mm})

                  = 2.55 \times 10^{-10} C

Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is 2.55 \times 10^{-10} C.

3 0
3 years ago
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