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Tomtit [17]
3 years ago
12

Seven seconds after a brilliant flash of lightning, thunder shakes the house. How far was the lightning strike from the house? S

even seconds after a brilliant flash of lightning, thunder shakes the house. How far was the lightning strike from the house? Much farther away than two kilometers Much closer than one kilometer About two kilometers away About one kilometer away It is impossible to predict.
Physics
1 answer:
zalisa [80]3 years ago
6 0

Answer:

About two kilometers away

\rm distance=2.401\ km

Explanation:

Given:

The time gap between the light and sound to travel to the house, t=7\ s

<em>Since the clouds are formed in the troposphere region of the atmosphere which extends from 8 kilometers to 12 kilometers above the earth-surface and the velocity of light is 300000 kilometers per second so it is visible almost instantly, hence we neglect the time taken by the light to travel to the house from the clouds.</em>

<u>∴Distance between the lightning-strike and the house:</u>

\rm distance=v\times t

we have the speed of sound as: v=343\ m.s^{-1}

So,

\rm distance=343\times 7

\rm distance=2401\ m

\rm distance=2.401\ km

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0.18216 T

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Maximum emf is given by

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3 years ago
Bolt starts the race not moving, but then increased his speed until he reaches a top speed. Once Bolt reaches his top speed he m
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Answer:

a

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Hence, for Bolt to be able to maintain the top speed for a few seconds, he needs to move at a constant velocity.

The correct option is a.

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3 years ago
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SIZIF [17.4K]
C is the first & the second question is A
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A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a
Nimfa-mama [501]

Answer:

F_0 = 393 N

Explanation:

As we know that amplitude of forced oscillation is given as

A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}

here we know that natural frequency of the oscillation is given as

\omega_0 = \sqrt{\frac{k}{m}}

here mass of the object is given as

m = \frac{W}{g}

\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}

\omega_0 = 8.48 rad/s

angular frequency of applied force is given as

\omega = 2\pi f

\omega = 2\pi(10.5) = 65.97 rad/s

now we have

0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}

F_0 = 393 N

6 0
3 years ago
Help me to solve it . It’s urgent
Artist 52 [7]

Answer: 0°

Explanation:

Step 1: Squaring the given equation and simplifying it

Let θ be the angle between a and b.

Given: a+b=c

Squaring on both sides:

... (a+b) . (a+b) = c.c

> |a|² + |b|² + 2(a.b) = |c|²

> |a|² + |b|² + 2|a| |b| cos 0 = |c|²

a.b = |a| |b| cos 0]

We are also given;

|a+|b| = |c|

Squaring above equation

> |a|² + |b|² + 2|a| |b| = |c|²

Step 2: Comparing the equations:

Comparing eq( insert: small n)(1) and (2)

We get, cos 0 = 1

> 0 = 0°

Final answer: 0°

[Reminders: every letters in here has an arrow above on it]

7 0
3 years ago
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