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gregori [183]
3 years ago
11

Seven seconds after a brilliant flash of lightning, thundershakes the house. How far was the lightning strike from thehouse?

Physics
1 answer:
OLEGan [10]3 years ago
6 0

Seven seconds after a brilliant flash of lightning, thunder shakes the house.

The lightning strike was <em>about 2 kilometers away</em>. <em>(c)</em>

The speed of sound is roughly 340 m/s. So in 7 seconds, the sound of the thunder would have traveled roughly 2,380 meters from the location of the strike.  But in the tangled mess of warm air, cold air, and rain surrounding a thunderstorm, the sound doesn't travel a straight line, so the actual location of the lightning was a little closer to the house than the total distance that the sound traveled.  "About 2 kilometers" is a reasonable estimate.

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Rzqust [24]

Given

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m2 (mass of the second object): 55 Kg

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v2 (velocity of the second object): ?

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The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:

Pa= Pb

Where Pa is the momentum before collision and Pb is the momentum after collision.

Now applying this law for the above problem we get

Momentum before collision= momentum after collision.

Momentum before collision = (m1+m2) x v1 =(55+5)x 4.5 = 270 Kgm/s

Momentum after collision = (m1+m2+m3) x v2 =(55+5+2.5) x v2

Now we know that Momentum before collision= momentum after collision.

Hence we get

270 = 62.5 v2

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3 years ago
A bicycle travels 6.10 km due east in 0.210 h, then 11.30 km at 15.0° east of north in 0.560 h, and finally another 6.10 km due
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Answer:

Explanation:

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after that he finally travel 6.10 km due to east in 0.21 h

suppose its position vector is r_3

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so position of final position is given by

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\vec{v_{avg}}=15.42\hat{i}+11.13\hat{j}

|v_{avg}|=\sqrt{361.71}=19.01 km/hr

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tan\theta =\frac{11.13}{15.42}=0.721

\theta =35.791^{\circ} w.r.t to x axis

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