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sasho [114]
3 years ago
9

A sinusoidal, transverse wave that propagates in the positive x-direction can be described with the wave function y of xt is equ

al to A times cosine of start parenthesis k times x minus omega times t end parenthesis, where k is the wave number, x is the position of a point on the wave, omega is the angular frequency, and t is time. What is another way to express the angular frequency, omega
Physics
1 answer:
myrzilka [38]3 years ago
4 0

Answer:

\omega =\frac{2 \pi }{T} rad/s

Explanation:

The wave function is:

y(xt) = Acos(kx- \omega t)

where :

k = wave number

x = position of a point on the wave

\omega = angular frequency

t = time

What is another way to express the angular frequency (omega)

Angular frequency (omega) can be express as :

\omega =\frac{2 \pi }{T} rad/s ( i.e one repetition that it takes to repeats itself)

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one mole of water is equivalent to 18 grams of water. a glass of water has a mass of 200 g. how many moles of water is in this?
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1 mole = 18 g
200 g = glass of water
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11.1 moles of water in 200 g (glass of water)
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How much watt is called 1 H. P​
agasfer [191]

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746 watts

Explanation:

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Which of the following statements is false? Group of answer choices Light that is high in energy has a high frequency. The atomi
MArishka [77]

Answer:

The last statement is false.

Explanation:

Photons (Electromagnetic radiation) are released when electrons drop from a higher energy lever to a lower energy level. Therefore the opposite insinuated by the last statement is wrong.

4 0
3 years ago
The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons tr
Degger [83]

Answer:

e. 1.2 x 10²³

Explanation:

According to the problem, The current equation is given by:

I(t)=0.88e^{-t/6\times3600s}

Here time is in seconds.

Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.

The relation between current and number of charge carriers is:

q=\int\limits {I} \, dt

Here the limits of integration is from 0 to infinite. So,

q=\int\limits {0.88e^{-t/6\times3600s}}\, dt

q=0.88\times(-6\times3600)(0-1)

q = 1.90 x 10⁴ C

Consider N be the total number of charge carriers. So,

q = N e

Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.

N = q/e

Substitute the suitable values in the above equation.

N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}

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8 0
3 years ago
When an average force F is exerted over a certain distance on a shopping cart of mass m, its kinetic energy increases by 12mv2.
VMariaS [17]

Answer:

A)d=\dfrac{1}{2F}mv^2

B)\Delta KE'=2\times \dfrac{1}{2}mv^2

Explanation:

Given that

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\Delta KE=\dfrac{1}{2}mv^2

we know that

Work done by all the forces =change in the kinetic energy

a)

Lets distance = d

We know work done by force F

W= F .d

F.d=ΔKE

F.d=\dfrac{1}{2}mv^2

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b)

If the force become twice

F' = 2 F

F'.d=ΔKE'

2 F .d = ΔKE'                          ( F.d =Δ KE)

2ΔKE = ΔKE'

\Delta KE'=2\times \dfrac{1}{2}mv^2

Therefore the final kinetic energy will become the twice if the force become twice.

8 0
3 years ago
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