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3 years ago

What is an example of kinetic energy

Physics

1 answer:

Olenka [21]3 years ago

6 0

An airplane has a large amount of kinetic energy in flight due to its large mass and fast velocity.

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A 8.01-nC charge is located 1.87 m from a 4.50-nC point charge. (a) Find the magnitude of the electrostatic force that one charg

valkas [14]

Answer:

Explanation:

Given

$q_1=8.01\ nC$

$q_2=4.50\ nC$

distance between the charges is given by $d=1.87\ m$

Electrostatic force between the charges is given by

$F_{12}=F_{21}=\frac{kq_1q_2}{d^2}$

where k=constant $=9\times 10^{9}$

$F_{12}=F_{21}=\frac{9\times 10^9\times 8.01\times 4.50\times 10^{-18}}{(1.87)^2}$

$F_{12}=F_{21}=92.769\times 10^{-9}\ N$

as both the charges are of similar nature therefore they repel each other

8 0

3 years ago

An arrow strikes a target moving at 75 m/s and embeds itself 15 cm into the target. If the arrow stopped with constant accelerat

zhenek [66]

Answer:

4 ms

Explanation:

initial velocity, u = 75 m/s

final velocity, v = 0

distance, s = 15 cm = 0.15 m

Let the acceleration is a and the time taken is t.

Use third equation of motion

v² = u² + 2 a s

0 = 75 x 75 - 2 a x 0.15

a = - 18750 m/s^2

Use first equation of motion

v = u + at

0 = 75 - 18750 x t

t = 4 x 10^-3 s

t = 4 ms

thus, the time taken is 4 ms.

Explanation:

5 0

3 years ago

View this and help out??

Zina [86]

I would say b as well. I’m sorry if it’s wrong

5 0

3 years ago

Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 590 Hz. The peg of one string slips

Slav-nsk [51]

To solve this problem it is necessary to apply the concepts related to frequency and vibration of strings. Mathematically the frequency can be expressed as

$f = \frac{v}{\lambda}$

Then the relation between two different frequencies with same wavelength would be

$\frac{f'}{f} = \frac{v'/\wavelength}{v/\wavelength}$

$\frac{f'}{f} = \frac{v'}{v}$

The beat frequency heard when the two strings are sounded simultaneously is

$f_{beat} = f-f'$

$f_{beat} = f(1-\frac{f'}{f})$

$f_{beat} = f(1-\frac{v'}{v})$

We have the velocity of the transverse waves in stretched string as

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{200N}{\mu}}$

And,

$v' = \sqrt{\frac{196N}{\mu}}$

Therefore the relation between the two is,

$\frac{v'}{v} = \sqrt{\frac{192}{200}}$

$\frac{v'}{v} = \sqrt{0.96}$

Finally substituting this value at the frequency beat equation we have

$f_{beat} = 590(1-\sqrt{0-96})$

$f_{beat} = 11.92Hz$

Therefore the beats per second are 11.92Hz

4 0

2 years ago

You cross two pure tall (TT) plants with two pure short (tt) plants and produce four offspring. Which of the following statement

grandymaker [24]

All will have a dominant trait I can't see the following statments

4 0

2 years ago

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