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Luden [163]
3 years ago
6

How does the speed of a wave relate to its wavelength and frequency?*

Physics
1 answer:
grigory [225]3 years ago
7 0

Answer: Wavespeed (V) = Frequency F x wavelength λ (V = F λ)

Explanation:

The wavespeed is the distance covered by a wave in one second. It is measured in metre per second, and represented by the symbol V. It is directly proportional to the wavelength and frequency

i.e Velocity (V) = Frequency F x wavelength λ

V = F λ

For instance:

Assume wavelength (λ)= 20 m

Frequency = 10 Hz.

To get the wavespeed, use the formula

V = F λ

V = 20 metres x 10 hertz

V = 200 metres per second

Thus, the wave travels at a speed of 200 metres per second

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What is a cataclysmic comet?
iren [92.7K]
The comet which is violent and large scale, causing sudden and violent upheaval is known as "Cataclysmic Comet"

Hope this helps!
7 0
3 years ago
Read 2 more answers
an 1150kg elevator moving down speeds up at a rate of 3.5m/s. what is the tension in the supporting cables?
gtnhenbr [62]

Answer:

The tension force in the supporting cables is 7245N

Explanation:

There are two forces acting on the elevator: the force of gravity pointing down (+) with magnitude (elevator mass) x (gravitational acceleration), and the tension force of the cable pointing up (-) with an unknown magnitude F. The net force is the sum of these forces:

F_{net} = F_g - F = m\cdot g - F\\

We are given the resulting acceleration along with the mass, i.e., we know the net force, allowing us to solve for F:

1150kg\cdot 3.5\frac{m}{s^2}= 1150kg \cdot 9.8\frac{m}{s^2}-F\\\implies F = 1150kg\cdot(9.8-3.5)\frac{m}{s^2}= 7245N

The tension force F in the supporting cables is 7245N


3 0
3 years ago
Practice questions, will mark brainliest!
andrew-mc [135]

Answer:

266.67Watts

Explanation:

Time = 2.5hr to seconds

3600s = 1hr

2.5hrs = 3600×2.5= 9000s

Force = 32N

Distance = 75km  to m

1000m = 1km

75km = 1000×75 = 75000m

Power = workdone / time

Work = force × distance

Therefore work = 32N × 75000m

Work = 2400000Nm

Power = work ➗ time

Power = 2400000Nm ➗ 9000s

Power = 266.67Watts

Watts is the S. i unit of power

I hope this was helpful, please mark as brainliest

4 0
3 years ago
Two large conducting parallel plates A and B are separated by 2.4 m. A uniform field of 1500 V/m, in the positive x-direction, i
Lapatulllka [165]

Answer:

a. 1.027 x 10^7 m/s b. 3600 V c. 0 V and d. 1.08 MeV

Explanation:

a. KE =1/2 (MV^2) where the M is mass of electron

b. E = V/d

c. V= 0 V (momentarily the pd changes to zero)

d KE= 300*3600 v = 1.08 MeV

6 0
3 years ago
A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial sp
Dimas [21]

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m =  0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2mU^{2}

K.E = 1/2 x 0.17 x 6^{2}

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/s^{2}

To calculate the final velocity, let us use third equation of motion.

V^{2} = U^{2} + 2as

V^{2}  = 6^{2} + 2 x 0.3 x 61

V^{2} = 36 + 36

V^{2} = 72

V = \sqrt{72}

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

5 0
2 years ago
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