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quester [9]
2 years ago
6

A particular electromagnetic wave travelling in vacuum is detected to have a frequency of 3 × 10 12 Hz. How much time will it ta

ke to complete one cycle of this wave? Answer with solution
Engineering
1 answer:
irina1246 [14]2 years ago
7 0

3×10^-12 seconds

Explanation:

T=1/f

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Answer:

All Brake lights are dimmer than normal because high resistance in the brake switch could be the cause according to Technician B.

Explanation:

According to Technician A

When the bulb is faulty then no current will flow through bulb and it will be open circuit.So no light will produce in bulb .

According to Technician B

When a high resistance inserted in series  circuit the voltage across each resistance is reduced and this cause the light glow dimly.

Formula of resistance in series circuit

Rt=r1+r2+r3......

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8. Describe and correct the error in stating the domain. Xf * (x) = 4x ^ (1/2) + 2 and g(x) = - 4x ^ (1/2) The domain of (f + g)
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Which of the following is an example of an observation?
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The monkey is brown!
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Car B is traveling a distance dd ahead of car A. Both cars are traveling at 60 ft/s when the driver of B suddenly applies the br
vagabundo [1.1K]

Answer:

Explanation:

Using the kinematics equation v = v_o + a_ct to determine the velocity of car B.

where;

v_o = initial velocity

a_c = constant deceleration

Assuming the constant deceleration is = -12 ft/s^2

Also, the kinematic equation that relates to the distance with the time is:

S = d + v_ot + \dfrac{1}{2}at^2

Then:

v_B = 60-12t

The distance traveled by car B in the given time (t) is expressed as:

S_B = d + 60 t - \dfrac{1}{2}(12t^2)

For car A, the needed time (t) to come to rest is:

v_A = 60 - 18(t-0.75)

Also, the distance traveled by car A in the given time (t) is expressed as:

S_A = 60  * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2

Relating both velocities:

v_B = v_A

60-12t = 60 - 18(t-0.75)

60-12t =73.5 - 18t

60- 73.5 = - 18t+ 12t

-13.5 =-6t

t = 2.25 s

At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars

i.e.

S_B = S_A

d + 60 t - \dfrac{1}{2}(12t^2) = 60  * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2

d + 60 (2.25) - \dfrac{1}{2}(12*(2.25)^2) = 60  * 0.75 +60((2.25)-0.75) -\dfrac{1}{2}*18*((2.25)-0.750)^2

d + 104.625 = 114.75

d = 114.75 - 104.625

d = 10.125 ft

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