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4 years ago

7

Would the energy of the wave increase or decrease if the speed of the wave increases? Why?

1 answer:

choli [55]4 years ago

4 0

Answer:

i hope this helps some

Explanation:

The time-averaged power of a sinusoidal wave is proportional to the square of the amplitude of the wave and the square of the angular frequency of the wave. This is true for most mechanical waves. If either the angular frequency or the amplitude of the wave were doubled, the power would increase by a factor of four.

The speed of a wave is dependant on four factors: wavelength, frequency, medium, and temperature. Wave speed is calculated by multiplying the wavelength times the frequency (speed = l * f).

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How to row a boat please help I’m stuck in the ocean

kvv77 [185]

Answer:

um well

Explanation:

get ur paddles and start rowing by putting ur hands on the paddles, and then rowing them by putting the paddle up, and then back into the water.

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4 years ago

Read 2 more answers

An object of mass(0.8kg) is attached to massless string of length (2.0),and swung with a tangential velocity of (3.0m/s) what is

Novay_Z [31]

Answer:

3.6 N

Explanation:

The magnitude of centripetal force in this case is equal to the magnitude of tension in the spring.

The formula is :

T= mv²/r ------where T is tension

m= mass of object =0.8 kg

v= velocity of object {tangential velocity} = 3.0 m/s

r= length of string = 2m

Applying the formula with real values;

T= mv²/r

T= {0.8 * 3²} / 2

T= { 7.2}/2 = 3.6 N

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3 years ago

Three-point charges are arranged on a line. Charge q3 = +5.00nC and is at the origin. Charge q2 = -3.00 nC and is at x = 5.00 cm

Anon25 [30]

Answer:

Explanation:

Given

Charge $q_3=+5\ nC$ is placed at origin

Charge $q_2=-3\ nC$ is placed at $x=5\ cm$

Charge $q_1$ is Placed at $x=2.5\ cm$

charge $q_1$ must be positive so as to balance the force on charge $q_3$

Force on $q_3$ due to $q_1$ is

$F_{31}=\frac{kq_1q_3}{r^2}$

here $r=2.5\ cm$

$F_{31}=\frac{kq_1q_3}{(2.5)^2}$

Force on $q_3$ due to $q_2$ is

$F_{32}=\frac{kq_3q_2}{r^2}$

here $r=5\ cm$

$F_{32}=\frac{kq_3q_2}{(5)^2}$

$F_{31}=F_{32}$

$\frac{kq_1q_3}{(2.5)^2}=\frac{kq_3q_2}{(5)^2}$

$q_1=q_2(\frac{2.5}{5})^2$

$q_1=3\times (\frac{2.5}{5})^2$

$q_1=3\times \frac{1}{4}$

$q_1=0.75\ nC$

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4 years ago

The majority of water on Earth exists as salt water, and only a very small percentage exists as fresh water. More than half of t

serg [7]

Answer:

B Crystallization

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8 0

2 years ago

In a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kineti

Sergio [31]

Answer:

Part a)

$v_f = 25.2 m/s$

$t = 5.48 s$

Part b)

$v_f = 25.32 m/s$

$t = 4.96 s$

Explanation:

Part a)

When ski start from rest

$v_f^2 - v_i^2 = 2 a d$

on this inclined plane we know that the acceleration is given as

$a = g sin\theta$

$a = 9.81 sin28$

$a = 4.6 m/s^2$

now for final speed

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(4.6)(69)$

$v_f = 25.2 m/s$

now time taken by the ski to reach the bottom is given as

$v_f = v_i + at$

$25.2 = 0 + 4.6 t$

$t = 5.48 s$

Part b)

Now when ski start with initial speed of 2.5 m/s

then we will have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 2.5^2 = 2(4.6)(69)$

$v_f = 25.32 m/s$

now time taken by the ski to reach the bottom is given as

$v_f = v_i + at$

$25.32 = 2.5 + 4.6 t$

$t = 4.96 s$

3 0

3 years ago