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mariarad [96]
3 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]3 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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4 0
3 years ago
Read 2 more answers
How does an increase in cold working effect Modulus of Elasticity and why?
VladimirAG [237]

Answer:

There is a decrease in modulus of elasticity

Explanation:

Young's Modulus of elasticity also known as elastic modulus is the deformation of a body along a particular axis under the action of opposing forces along that axis. at atomic levels, it depends on bond energy or strength.

In cold working processes, plastic deformation a metal occurs below its re-crystallization temperature due to which crystal structure of metal gets distorted and as a result of dislocations fractures also occur resulting in hardening of metal but bonds at atomic levels defining elasticity are temporarily affected.

Thus an increase in cold working results in a decrease in modulus of elasticity.

7 0
3 years ago
A ship anchored at sea is rocked by waves that have crests Lim apart the waves travel at 70m/S, at what frequency do the waves r
inessss [21]

Question: A ship anchored at sea is rocked by waves that have crests 100 m apart the waves travel at 70m/S, at what frequency do the waves reach the ship?

Answer:

0.7 Hz

Explanation:

Applying,

v = λf............... Equation 1

Where v = velocity of the wave, f = frequency fo the wave, λ = wavelength of the wave

make f the subject of the equation

f = v/λ................. Equation 2

From the question,

Given: v = 70 m/s, λ = 100 m ( distance between successive crest)

Substitute these values into equation 2

f = 70/100

f = 0.7 Hz

Hence the frequency at which the wave reach the ship is 0.7 Hz

3 0
2 years ago
A student wishes to record a 7.5-kilogram watermelon colliding with the ground. Calculate how far the watermelon must fall freel
algol [13]

Answer:

42.86m

Explanation:

The first thing we should keep in mind is that the watermelon moves with uniform acceleracion equal to gravity (9.81m / s ^ 2)

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

\frac {Vf^{2}-Vo^2}{2.g} =Y

where

Vf=29m/s= final speed

Vo= initial speed=0m/S

g=gravity=9.81m/s^2

Y= distance traveled(m)

solving

\frac {Vf^{2}-Vo^2}{2.g} =Y\\\frac {(29m/s)^{2}-(0m/s)^2}{2(9.8m/s^2)} =Y\\Y=42.86m\\

the distance traveled by watermelon is 42.86m

       

5 0
3 years ago
A car drives around a horizontal, circular track at constant speed. Consider the following three forces that act on the car: (1)
musickatia [10]

Answer: 1, 2 and 3.

Explanation:

By definition, work, is the process through which a force, applied on an object, produces a displacement of this object, and can be expressed as the dot product of the force vector and the displacement vector.

It can be also understood as the projection of the force in the direction of the displacement, so, if both vectors are perpendicular, as the projection of one vector along the other is nul, if a force and the displacement are perpendicular, no work is done.

In our case, the 3 forces mentioned, are perpendicular respect the displacement.

For normal force and gravity, as they act vertically, and the car moves along an horizontal trajectory, both are perpendicular, so no work is done.

For the friction force, as it is the only horizontal force present, it is just the centripetal force that keeps the car moving in a circular path.

It always aims to the center of the circle, along the radius at any point, so it is always perpendicular to the displacement also.

6 0
3 years ago
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