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morpeh [17]
2 years ago
12

You are moving at a speed 2/3 c toward randy when randy shines a light toward you. at what speed do you see the light approachin

g you
Physics
1 answer:
yarga [219]2 years ago
3 0
I see the light moving exactly at speed equal to c.

In fact, the second postulate of special relativity states that:
"The speed of light in free space has the same value c<span> in all inertial frames of reference."
</span>
The problem says that I am moving at speed 2/3 c, so my motion is a uniform motion (constant speed). This means I am in an inertial frame of reference, so the speed of light in this frame must be equal to c.
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A source of emf is connected by wires to a resistor and electrons flow in the circuit the wire diameter is teh same throughout t
vladimir2022 [97]
I think is E
I hope that helps
3 0
2 years ago
A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init
Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

h=\frac{gt^2}{2}

For second stone

h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}---2

Equating 1 &2 we get

\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}

\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

26.46t-35.721-52.92t+142.884=0

t=4.05 s

4 0
2 years ago
Your low-flow showerhead is delivering water at 1.2×10−4m3/s, about 2.0 gallons per minute.
Oksanka [162]

To solve this problem it is necessary to apply the fluid mechanics equations related to continuity, for which the proportion of the input flow is equal to the output flow, in other words:

Q_1 = Q_2

We know that the flow rate is equivalent to the velocity of the fluid in its area, that is,

Q = VA

Where

V = Velocity

A = Cross-sectional Area

Our values are given as

Q_2 = 1.2*10^{-4}m^3/s

d = 0.021m

r = \frac{0.021}{2} = 0.0105m

Since there is continuity we have now that,

V_1A_1 = Q_2

V_1A_1 = 1.2*10^{-4}

V_1 = \frac{1.2*10^{-4}}{A_2}

V_1 = \frac{1.2*10^{-4}}{\pi r^2}

V_1 = \frac{1.2*10^{-4}}{\pi (0.0105)^2}

V_1 =0.347m/s

Therefore the speed of the water's house supply line is 0.347m/s

7 0
2 years ago
Imagine that you replace the block in the video with a happy or sad ball identical to the one used as a pendulum, so that the sa
gizmo_the_mogwai [7]

Answer: a) The Answer to the question is option a) None of it.

Explanation:

The reason is because according to the law of conservation of energy Energy can neither be created nor destroyed but can be transformed from one form to the other. Therefore none of the kinetic energy was dissipated, rather it was transformed to another form of energy.

6 0
2 years ago
You have a 2m long wire which you will make into a thin coil with N loops to generate a magnetic field of 3mT when the current i
Anni [7]

Answer:

<em>radius of the loop =  7.9 mm</em>

<em>number of turns N ≅ 399 turns</em>

Explanation:

length of wire L= 2 m

field strength B = 3 mT = 0.003 T

current I = 12 A

recall that field strength B = μnI

where n is the turn per unit length

vacuum permeability μ  = 4\pi *10^{-7}  T-m/A = 1.256 x 10^-6 T-m/A

imputing values, we have

0.003 = 1.256 x 10^−6 x n x 12

0.003 = 1.507 x 10^-5 x n

n = 199.07 turns per unit length

for a length of 2 m,

number of loop N = 2 x 199.07 = 398.14 ≅ <em>399 turns</em>

since  there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.

this length is also equal to the circumference of each loop

the circumference of each loop = 2\pi r

0.005 = 2 x 3.142 x r

r = 0.005/6.284 = 7.9*10^{-4} m = 0.0079 m =<em> 7.9 mm</em>

8 0
2 years ago
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