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harkovskaia [24]

2 years ago

7

What do you mean by specific heat capacity?

1 answer:

Dennis_Churaev [7]2 years ago

6 0

Please mark me as Brainliest ......

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The longest wavelength within the visible spectrum is _______ light.

Goryan [66]

<span>The longest wavelength within the visible spectrum is the red light. The answer is letter C. It is called visible light because it is the only light that can be seen by the human eye. Red light is the longest wavelength around 620 to 750 nanometer. It is followed by orange which has a wavelength of 590 t 620 nanometer. And then blue which has a wavelength of 450 to 495 nanometer. And the shortest wavelength is violet which has a wavelength of 380 to 459 nanometer. </span>

5 0

2 years ago

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Minchanka [31]

Answer:

B)

Explanation:

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5 0

2 years ago

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Help me please I can't get the final step

inna [77]

Answer:

$\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}$

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

$\displaystyle A=\frac{3}{2}B^mC^n$

and the dimensions of each variable is:

$A=L^2T^2$

$B=LT^{-1}$

$C=LT^2$

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

$\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n$

Operating:

$L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)$

$L^2T^2=L^{m+m}T^{-m+2n}$

Equating the exponents:

$m+n=2$

$-m+2n=2$

Adding both equations:

$3n=4$

Solving:

$n=4/3$

$m=2-4/3=2/3$

Answer:

$\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}$

6 0

2 years ago

Calculate the orbital speed (in m/s) of a satellite that circles the Earth with a time period of 12.00 hours. The mass of the Ea

Hoochie [10]

Answer:

$v = 3869 m/s$

Explanation:

As we know that the orbital speed of the satellite is given as

$v = \sqrt{\frac{GM}{r}}$

also we know that

time period of the revolution is given as

$T = \frac{2\pi r}{v}$

now from above equation we know that

$T = \frac{2\pi (\frac{GM}{v^2})}{v}$

$T = \frac{2\pi GM}{v^3}$

so we will have

$v = (\frac{2\pi GM}{T})^{1/3}$

now plug in all data in this equation

$v = (\frac{2\pi (6.67 \times 10^{-11})(5.97 \times 10^{24})}{12 \times 3600})^{1/3}$

$v = 3869 m/s$

3 0

2 years ago

A. If an electron in a hydrogen atom has an energy of −6.06 × 10^−20 J, which Bohr orbit is it in?

Orlov [11]

Explanation:

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3 0

3 years ago