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alex41 [277]
3 years ago
15

EM Waves have crests, troughs, and wavelengths True or false?

Physics
2 answers:
Nezavi [6.7K]3 years ago
4 0
True.........................................
3241004551 [841]3 years ago
4 0
The answer is True :)
You might be interested in
A weather balloon has a volume of 90.0 l when it is released from sea level. What
frozen [14]

The final atmospheric pressure is 5.19\cdot 10^4 Pa

Explanation:

Assuming that the temperature of the air does not change, we can use Boyle's law, which states that for a gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume. In formula,

pV=const.

where

p is the gas pressure

V is the volume

The equation can also be rewritten as

p_1 V_1 = p_2 V_2

where in our problem we have:

p_1= 1.03\cdot 10^5 Pa is the initial pressure (the atmospheric pressure at sea level)

V_1 = 90.0 L is the initial volume

p_2 is the final pressure

V_2 = 175.0 L is the final volume

Solving the equation for p2, we find the final pressure:

p_2 = \frac{p_1 V_1}{V_2}=\frac{(1.01\cdot 10^5)(90.0)}{175.0}=5.19\cdot 10^4 Pa

Learn more about ideal gases:

brainly.com/question/9321544

brainly.com/question/7316997

brainly.com/question/3658563

#LearnwithBrainly

3 0
3 years ago
a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone​
Dafna1 [17]

Answer:

The centripetal acceleration of the stone is 5 m/s²

Explanation:

The length of the string to which the stone is attached, r = 1 m

The speed with which the string is rotated, v = 5 m/s

The centripetal acceleration, a_c, is given as follows;

a_c = \dfrac{v^2}{r}

Therefore, the centripetal acceleration of the stone found as follows;

a_c = \dfrac{(5 \ m/s)^2}{1 \ m} = 5 \ m/s^2

The centripetal acceleration of the stone, a_c = 5 m/s².

5 0
3 years ago
Consider a 100 g object dropped from a height of 1 m. Assuming no air friction (drag), when will the object hit the ground and a
Katyanochek1 [597]

Answer:

speed and time are Vf = 4.43 m/s and  t = 0.45 s

Explanation:

This is a problem of free fall, we have the equations of kinematics

      Vf² = Vo² + 2g x

As the object is released the initial velocity is zero, let's look at the final velocity with the equation

      Vf = √( 2 g X)

      Vf = √(2 9.8  1)

      Vf = 4.43 m/s

This is the speed with which it reaches the ground

 

Having the final speed we can find the time

      Vf = Vo + g t

       t = Vf / g

       t = 4.43 / 9.8

       t = 0.45 s

This is the time of fall of the body to touch the ground

3 0
4 years ago
uniform ladder of length 6.0 m and weight 300 N leans against a frictionless vertical wall. The foot of the ladder isplaced 3.0
olganol [36]

Answer:

Fx1 (6 m) sin 60 = 300 (3 m) cos 60  balancing torques about floor

Fx1 = 900 * 1/2 / 5.20 = 86.6 N  this is the horizontal force that must be supplied by the wall to balance torques about the floor

This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.

Fx1 = Fx2 = 86.6 N

3 0
3 years ago
(m/s)<br> 20<br> 10.<br> 100<br> -10<br> total distance
Sergio039 [100]

Answer:

sum of them 4

Explanation:

8 0
3 years ago
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