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2 years ago

5

Lightning flashes and you hear a thunderclap 4 seconds later. The velocity of sound is 340 m/s. How far away did the lightning s

trike?

1 answer:

Kipish [7]2 years ago

3 0

Answer:

1,360 m

Explanation:

340 meters per second, and it was 4 seconds later.

340 * 4 is how to find teh distance where the lightning struck.

340 * 4 = 1,360 meters

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How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.

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Answer:

4.2s

Explanation:

Given parameters:

Power = 2190W

Mass of box = 1.47 x 10⁴g

distance = 6.34 x 10⁴mm

Unknown:

Time = ?

Solution:

Power is the rate at which work is done;

Mathematically;

Power = $\frac{work done}{time}$

Time = $\frac{work done}{power}$

Work done = weight x height

convert mass to kg;

100g = 1kg;

1.47 x 10⁴g = 14.7kg

convert the height to m;

1000mm = 1m

6.34 x 10⁴mm gives 63.4m

Work done = 14.7 x 9.8 x 63.4 = 9133.4J

Time taken = $\frac{9133.4}{2190}$ = 4.2s

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What is a disadvantage of using nuclear power to produce electricity?

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A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t

enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

$v_{avg} = \frac{6000}{20*60}$

$v_{avg} = 5m/s$

now since effect of wind is in opposite direction then we can say

$V_{net} = v_{bird} - v_{wind}$

$5 = 9 - v_{wind}$

$v_{wind}= 4 m/s$

Part b)

now if bird travels in the same direction of wind then we will have

$v_{net}= v_{bird} + v_{wind}$

$v_{net} = 9 + 4 = 13 m/s$

now we can find the time to go back

$time = \frac{distance}{speed}$

$time = \frac{6000}{13}$

$time = 7.7 minutes$

Part c)

Total time of round trip when wind is present

$T = t_1 + t_2$

$T = 20 + 7.7 = 27.7 min$

now when there is no wind total time is given by

$T = \frac{6000}{9} + \frac{6000}{9}$

$T = 22.22 min$

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How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to an altitude of 5r miles above t

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Let the data is as following

mass of payload = "m"

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now we know that we place the payload from the position on the surface of moon to the position of 5r from the surface

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so it is given by

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we know that

$U_f = -\frac{GMm}{6r}$

$U_i = -\frac{GMm}{r}$

now from above formula

$W = -\frac{GMm}{6r} + \frac{GMm}{r}$

$W = \frac{5GMm}{6r}$

so above is the work done to move the mass from surface to given altitude

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