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atroni [7]
3 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]3 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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The correct question is:

"Find the energy each gains"

The energy gained by a charged particle accelerated through a potential difference is given by

\Delta U = q\Delta V

where

q is the charge of the particle

\Delta V is the potential difference

For a proton,

q=+e=1.6\cdot 10^{-19}C

And since \Delta V=100 V

The energy gained by the proton is

\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J

For an alpha particle,

q=+2e=3.2\cdot 10^{-19}C

Therefore, the energy gained is

\Delta U=(3.2\cdot 10^{-19})(100)=3.2\cdot 10^{-17}J

Finally, for a singly ionized helium nucleus (a helium nucleus that has lost one electron)

q=+e=1.6\cdot 10^{-19}C

So the energy gained is the same as the proton:

\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J

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A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent
azamat

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

  • <em>Length of the string, L = 100 cm</em>

<em />

The wavelengths of the constituent travelling waves is calculated as follows;

L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}

for first mode: n = 1

\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm

for second mode: n = 2

\lambda = \frac{2L}{2} = L = 100 \ cm

For the third mode: n = 3

\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm

For fourth mode: n = 4

\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50  \ cm

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

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3 years ago
If a small rock is dropped from a height of 3.1 m how fast will it be moving when it reaches the ground 0.80 seconds later
garik1379 [7]
X=1/2 at^2
3.1=1/2 a *0.64
a=9.68
v=at
v=0.8*9.6875=7.75

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3 years ago
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