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atroni [7]
4 years ago
7

A migrating bird flew across a lake at an average speed of 18 meters per second. Was the distance that the bird flew across the

lake less than 21 kilometers? (1 kilometer = 1,000 meters) (1) It took the bird less than 20 minutes to fly across the lake. (2) It took the bird less than 0.4 hour to fly across the lake.
Physics
1 answer:
Natasha_Volkova [10]4 years ago
7 0

Answer:

It cannot be said with 100% guarantee.

Explanation:

Since bird is travelling with 18 meters per second, that means that it covers a distance of 18 meters in 1 second.

for part 1 we convert 20 minutes into seconds, thus, 20 * 60 = 1200 seconds.

Distance covered in 1200 seconds will be:

S = v * t (S is distance , v is speed, t is time)

18 * 1200 = 21600 meters = 21.6 Kilometers

If the bird took less than 20 minutes than distance covered would be less than 21.6 Kilometer which could be greater than or lesser than 21 Kilometers. On average we can say that the distance bird travelled will be less than 21 Kilometer. But it is not definite.

For part B convert 0.4 hours into seconds to find the distance it travelled.

0.4 * 60 * 60 = 1440 seconds

Distance covered will be

18 * 1440 = 25920 meters = 25.9 Kilo meters.

Again, if the bird takes less than 0.4 hours to cover the lake, the lake will be less than 25.9 kilometers long. This means that it could be greater than or lesser than 21 kilometers. It cannot be definitive

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Answer:

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H=\frac{\bigtriangleup Q}{\bigtriangleup t}=kA\frac{T_H-T_C}{L}\\\\L_f=334\times10^3J/kj\\\\H_{ice}=\frac{Q_{ice}}{t}=\frac{m_{ice}L_f}{t} #Heat flowing through melting ice.

H_{ice}=\frac{5.0\times10^-^3kg)334\times10^3J/kg)}{10\times60s}\\\\=2.7833J/s

To solve for k:

H=\frac{\bigtriangleup Q }{\bigtriangleup t}\\\\=kA\frac{T_H-T_C}{L}H=\frac{\bigtriangleup Q }{\bigtriangleup t}\\\\=kA\frac{T_H-T_C}{L}H=\frac{\bigtriangleup Q }{\bigtriangleup t}\\\\=kA\frac{T_H-T_C}{L}\\\\k=\frac{H}{A\frac{T_H-T_C}{L}}=\frac{HL}{A(T_H-T_C)}

=\frac{(2.7833J/s\times 0.55)}{1.00\times10^-^4m^2\times 100K}\\\\=153.0815W/m.K

The thermal conductivity k of the metal is 153.0815W/m.K

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Workers do 8000 J of work on a 2000-N crate to push it up a ramp. If the ramp is 2 m high, what is the efficiency of the ramp?
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