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photoshop1234 [79]
3 years ago
13

At what speed, as a fraction of c, does a moving clock tick at four fifth the rate of an identical clock at rest?

Physics
1 answer:
ololo11 [35]3 years ago
6 0

Answer:

The seed as a fraction of the speed of light is \frac{3}{5}c

Solution:

As per the question:

Suppose, t_{i} be the rate of an identical clock between two time intervals.

For a moving clock, moving with velocity 'v', at the clock tick of four-fifth:

t = \frac{5}{4}t_{i}

Now,

Using the relation of time dilation, from Einstein's relation:

t = \frac{t_{i}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

\frac{5}{4}t_{i} = \frac{t_{i}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

Squaring both sides:

(\frac{5}{4})^{2} = (\frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}})^{2}

\frac{25}{16} = \frac{1}{{1 - \frac{v^{2}}{c^{2}}}}

1 - \frac{16}{25} = \frac{v^{2}}{c^{2}}

\frac{v}{c} = \sqrt{\frac{9}{25}}

\frac{v}{c} = \frac{3}{5}

v = \frac{3}{5}c

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Lynna [10]
In BPC

tan\theta =a/b = 3/4

\theta = tan^-1(0.75)

\theta = 36.87 deg

BP = sqrt(a^2 + b^2) = sqrt((3)^2 + (4)^2) = 5 m

Eb = k Q/BP^2 = (9 x 10^9) (16 x 10^-9)/5^2 = 5.76 N/C

Ea = k Q/AP^2 = (9 x 10^9) (16 x 10^-9)/4^2 = 9 N/C

Ec = k Q/CP^2 = (9 x 10^9) (16 x 10^-9)/3^2 = 16 N/C

Net electric field along X-direction is given as

Ex = Ea + Eb Cos36.87 = (9) + (5.76) Cos36.87 = 13.6 N/C

Net electric field along X-direction is given as

Ey = Ec + Eb Sin36.87 = (16) + (5.76) Sin36.87 = 19.5 N/C

Net electric field is given as

E = sqrt(Ex^2 + Ey^2) = sqrt((13.6)^2 + (19.5)^2) = 23.8 N/C
8 0
3 years ago
What is the outcome of the Training and Exercise Planning Workshop (TEPW)?
balandron [24]

Answer / Explanation:

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An essential factor for the exercise management process is to create a collaborative environment where a whole community stakeholders can engage in a forum to discuss and coordinate training and exercise activities across local organizations to maximize the use of available resources and prevent duplication of effort.

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3 years ago
What is 25.61 cm to Gm?
Soloha48 [4]

Explanation:

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Question 4 (18 marks) (a) During a Physics Lab experiment, 1 st year SFY students analyzed the behavior of capacitors by connect
Nataly_w [17]

Answer:

1.) 274.5v

2.) 206.8v

Explanation:

1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.

The potential difference and charge across EACH capacitor will be

V = Voe

Where Vo = initial voltage

e = natural logarithm = 2.718

For the first capacitor 2.50 µF,

V = Vo × 2.718

746 = Vo × 2.718

Vo = 746/2.718

Vo = 274.5v

To calculate the charge, use the below formula.

Q = CV

Q = 2.5 × 10^-6 × 274.5

Q = 6.86 × 10^-4 C

For the second capacitor 6.80 µF 

V = Voe

562 = Vo × 2.718

Vo = 562/2.718

Vo = 206.77v

The charge on it will be

Q = CV

Q = 6.8 × 10^-6 × 206.77

Q = 1.41 × 10^-3 C

B.) Using the formula V = Voe again

165 = Vo × 2.718

Vo = 165 /2.718

Vo = 60.71v

Q = C × 60.71

Q = C

4 0
3 years ago
Which of the following calculations is equal to 1?
lutik1710 [3]

Answer:

we need to see the answers but probably 1 or -1

5 0
3 years ago
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