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3 years ago

Tsunami waves have ?

Physics

1 answer:

mr Goodwill [35]3 years ago

4 0

I think it #3(not Sure) I think Its because of their long wavelengths that tsunamis behave as shallow-water waves.

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An object attached to one end of a spring makes 20 vibrations in 10 seconds. Its angular frequency is: 0. 79 rad/s 1. 57 rad/s 2

morpeh [17]

Angular frequency in radian per second for 20 vibrations in 10 seconds is 12.6 rad/s

<h3>What is Angular frequency?</h3>

Angular frequency is the number of vibrations in radian per second.

The total number of vibrations n is 20 and the time taken for these vibrations is 10 s

The frequency of the vibrations will be

f = 20 / 10 = 2 Hz

Angular frequency is related to the frequency as

ω = 2πf

ω=2π × 2

ω = 12.6 rad/s

Thus, the angular frequency is 12.6 rad/s.

Learn more about Angular frequency.

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5 0

2 years ago

Measure the amount of space an object takes up is what

kondaur [170]

That's what scientists and other technical people call the object's "<em>volume</em>".

5 0

3 years ago

An athlete practicing for a track meet ran 400 meter in 50 seconds. What was his average speed?

sattari [20]

Speed= Distance ÷ time

Speed= 400 ÷ 50 =8 m/s

7 0

3 years ago

What is the average speed of this car over the entire journey ?

lesya [120]

Answer: 0.83333333333 km/h

Explanation: give brainliest

4 0

2 years ago

An electron is released from rest at a distance of 0.470 m from a large insulating sheet of charge that has uniform surface char

ArbitrLikvidat [17]

Answer:

Part a)

$W = 1.58 \times 10^{-20} J$

Part b)

$v = 1.86 \times 10^5 m/s$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = 4.00 \times 10^{-12} C/m^2$

now the electric field is given as

$E = \frac{4.00 \times 10^{-12}}{2(8.85 \times 10^{-12})}$

$E = 0.225 N/C$

Now acceleration of an electron due to this electric field is given as

$a = \frac{eE}{m}$

$a = \frac{(1.6 \times 10^{-19})(0.225)}{9.1 \times 10^{-31}}$

$a = 3.97 \times 10^{10}$

Now work done on the electron due to this electric field

$W = F.d$

$d = 0.470 - 0.03$

$d = 0.44 m$

So work done is given as

$W = (ma)(0.44)$

$W = (9.11 \times 10^{-31})(3.97 \times 10^{10})(0.44)$

$W = 1.58 \times 10^{-20} J$

Part b)

Now we know that work done by all forces = change in kinetic energy of the electron

so we will have

$W = \frac{1}{2}mv^2 - 0$

$1.58 \times 10^{-20} = \frac{1}{2}(9.1\times 10^{-31})v^2$

$v = 1.86 \times 10^5 m/s$

7 0

3 years ago