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3 years ago

For a given wave IF frequency doubles the wavelength

Physics

1 answer:

motikmotik3 years ago

7 0

Answer:

C is halved

Explanation:

The frequency and the wavelength of a wave are related by the equation:

$v=f\lambda$

where

v is the speed of the wave

f is the frequency

$\lambda$ is the wavelength

From the equation above, we see that for a given wave, if the wave is travelling in the same medium (and so, its speed is not changing), then the frequency and the wavelength are inversely proportional to each other.

Therefore, if the frequency doubles, the wavelength will halve in order to keep the speed constant:

$\lambda = \frac{v}{f}\\\lambda' = \frac{v}{f'}=\frac{v}{2f}=\frac{1}{2}(\frac{v}{f})=\frac{\lambda}{2}$

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A soft-drink bottler purchases glass bottles from a vendor. The bottles are required to have internal pressure strength of at le

maria [59]

Answer: K =24 psi

Explanation:

Given: Standard deviation =3psi

Internal pressure strength =157psi

Number of random bottle =n=64

K= 3 × square root of 64

K= 3×8=24 psi

If mean internal pressure K fall below K,

157-1.3=155.7psi

At 2%:

0.16×64 = 10.24

5 0

3 years ago

1. How much energy is needed just to melt 0.56kg of ice at 0◦ C?

ivann1987 [24]

Answer:

This will require 266.9 of heat energy.

Explanation:

To calculate the energy required to raise the temperature of any given substance, here's what you require:The mass of the material, m The temperature change that occurs, ΔT The specific heat capacity of the material,

(which you can look up). This is the amount of heat required to raise 1 gram of that substance by 1°C.

Here is a source of values of

c for different substances:

Once you have all that, this is the equation:

Q=m×c×ΔT(Q is usually used to symbolize that heat required in a case like this.)For water, the value of c is 4.186g°C So, Q=750×4.186×85=266=858=266.858

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3 years ago

A water wheel rotates with a period of 2.26 s. If the water wheel has a radius of l.94 m,

Dafna11 [192]

Answer:

The correct answer is B)

Explanation:

When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel. So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.

The formula for calculating the velocity of a point on the edge of the wheel is given as

$V_{r}$ = 2π r / T