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BlackZzzverrR [31]
3 years ago
10

If you used 650 W of power lifting a 300 N weight in 2 seconds how high did you lift?

Physics
1 answer:
dimulka [17.4K]3 years ago
7 0

Answer:

4.33m

Explanation:

Power = work done/ time

work done = power × time =650 × 2 = 1300J

work done = force × distance

distance = work done/force

distance = 1300/300 = 4.33m

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What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?
adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, 3.03\times 10^{-19}J

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant  = 10973731.6 m⁻¹

n_f = Higher energy level = 3

n_i= Lower energy level = 2

Putting the values, in above equation, we get:

\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )

\lambda=6.56\times 10^{-7}m

Now we have to calculate the energy.

E=\frac{hc}{\lambda}

where,

h = Planck's constant = 6.626\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 6.56\times 10^{-7}m

Putting the values, in this formula, we get:

E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}

E=3.03\times 10^{-19}J

Therefore, the energy of one photon of hydrogen atom is, 3.03\times 10^{-19}J

3 0
3 years ago
What is the volume of a cube with length = 3 centimeters, width = 3
suter [353]

Answer:

D. 27 cubic centimeters

Explanation:

l*w*h

3*3*3= 27

4 0
3 years ago
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