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3 years ago

The location where an earthquake begins

Physics

1 answer:

Ivahew [28]3 years ago

3 0

Answer:

Explanation:

the location where an earthquake begin is called

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A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

PLEASE HELLLLLLLLLLLLPPPPPPPPPP!!!!!! I will give brainliest

alexandr1967 [171]

Answer:

its C. The north pole of one magnet attracts the south pole of another

Explanation:

I JUST TOOK THE TEST

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3 years ago

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An ambulance driver traveling at 31.0 m/s (69.3 mph) honks his horn as he sees a motorist ahead on the highway traveling in the

IrinaVladis [17]

Answer:

fo = 378.52Hz

Explanation:

Using Doppler effect formula:

$f'=\frac{C-Vb}{C-Va}*fo$

where

f' = 392 Hz

C = 340m/s

Vb = 20m/s

Va = 31m/s

Replacing these values and solving for fo:

fo = 378.52Hz

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3 years ago

What is the total resistance in this circuit?

saul85 [17]

Answer:B

Explanation:

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3 years ago

Suppose a free-fall ride at an amusement park starts at rest and is in free fall. What is the velocity of the ride after 2.3 s?

horsena [70]

Answer:

V = a * t = 9.8 m/s^2 * 2.3 s = 22.5 m/s velocity after 2.3 s

S = 1/2 g t^2 since initial speed is zero

S = 1/2 * 9.8 m/s^2 * 5.29 s^2 = 25.9 m

8 0

3 years ago