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mariarad [96]
4 years ago
9

A wave has frequency of 50 Hz and a wavelength of 10 m. What is the speed of the wave?

Physics
1 answer:
son4ous [18]4 years ago
5 0

Answer:

500 m/s

Explanation:

Speed of wave = Frequency × Wavelength

Speed of wave = 50 Hz × 10 m

Therefore the speed is 500 m/s

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What do you understand by 'weightlessness due to freefall'? ​
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Four identical particles of mass 0.514 kg each are placed at the vertices of a 4.70 m x 4.70 m square and held there by four mas
Alona [7]

In order to solve the problem it is necessary to take into account the concepts related to the moment of inertia, and the center of mass of the object.

Our values are given by:

m = 0.514kg

I = 4.7m (each side)

A) For the case when the axis passes through the midpoints of opposite sidesand lies in the plane of the square. The formula is given by,

I = 4m (\frac{l}{2})^2\\I = 4*(0.514)(\frac{4.7}{2})^2\\I =  11.35kgm^2\\

B) For the case when the axis passes through the midpoint of one of the sides and is perpendicular to the plane of the square. The formula is given by,

I = 2m(\frac{l}{2})^2+2m(\frac{l}{2}^2+l^2)\\I = 2*(0.514)*(\frac{4.7}{2})^2+2*(0.514)*((\frac{4.7}{2})^2+4.7^2)\\I = 34.06Kgm^2

C) For the case when the axis lies in the plane of square f passes through two diagonally opposite particles,

I = \frac{1}{2}*l^2

I = \frac{1}{2}*(4.7)^2

I = 11.045kgm^2

7 0
3 years ago
HELP PLEASE GIVING 30 POINTS AND BRAINLIEST DUE VERY SOON!!
olga55 [171]

Answer:

3 is your anwser

Explanation:

5 0
3 years ago
19) A child on a sled starts from rest at the top of a 15.0° slope. If the trip to the bottom takes 15.2 s,
sveticcg [70]

Answer: 288.8 m

Explanation:

We have the following data:

t=15.2 s is the time it takes to the child to reach the bottom of the slope

V_{o}=0 is the initial velocity (the child started from rest)

\theta=15\° is the angle of the slope

d is the length of the slope

Now, the Force exerted on the sled along the ramp is:

F=ma (1)

Where m is the mass of the sled and a its acceleration

In addition, if we draw a free body diagram of this sled, the force along the ramp will be:

F=mg sin \theta (2)

Where g=9.8 m/s^{2} is the acceleration due gravity

Then:

ma=mg sin \theta (3)

Finding a:

a=g sin \theta (4)

a=9.8 m/s^{2} sin(15\°) (5)

a=2.5 m/s^{2} (6)

Now, we will use the following kinematic equations to find d:

V=V_{o}+at (7)

V^{2}=V_{o}^{2}+2ad (8)

Where V is the final velocity

Finding V from (7):

V=at=(2.5 m/s^{2})(15.2 s) (9)

V=38 m/s (10)

Substituting (10) in (8):

(38 m/s)^{2}=2(2.5 m/s^{2})d (11)

Finding d:

d=288.8 m

6 0
3 years ago
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