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erica [24]
3 years ago
15

What is the potential energy of an object that has mass of 410 kg at a height of 12m

Physics
1 answer:
serious [3.7K]3 years ago
8 0

PE = (weight)              x (height)

     = (mass) x (gravity) x (height)

     = (410 kg) x (9.8 m/s²) x (12 m)

     =  (410 x 9.8 x 12)  kg-m²/s²

     =    48,210 joules  (rounded)
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If a photon of Light leaves the sun than reachers the earth 494 second later. What is the distance from the earth to sun
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Conclusion of experiment for determine the magnitude and direction of magnetic field of earth What we will write in our experime
Tema [17]

Answer:

The magnitude and direction of the earth's magnetic field is between the range of 0.25 to 0.65 Gauss considering the angle of the earth's north pole is eleven degrees.

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3 years ago
Explain nuclear reactions, nuclear fission and nuclear fusion.
SVETLANKA909090 [29]

Answer:

Nuclear reaction is a reaction in which nucleus of the atom undergoes a change.

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8 0
3 years ago
If the frequency of the 13C signal of TMS is 201.16 MHz, the two 13C signals of acetic acid at 179.0 and 20.0 ppm are separated
Lelu [443]

The difference in frequency of the two signals is 1.33 \times 10^{10} \ kHz.

The given parameters;

  • <em>frequency of the 13 C signal = 201.16 MHz</em>

The energy of the 13 C signal located at 20 ppm is calculated as follows;

E = hf\\\\E_1 = h \frac{c}{\lambda} \\\\E_1 =  \frac{(6.626 \times 10^{-34})\times 3\times 10^8}{20 \times 10^{-6}} \\\\E_1 = 9.94 \times 10^{-21} \ J

The energy of the 13 C signal located at 179 ppm is calculated as follows;

E_2 = \frac{hc}{\lambda} \\\\E_2 = \frac{(6.626\times 10^{-34})\times (3\times 10^{8})}{179 \times 10^{-6} } \\\\E_2 = 1.11 \times 10^{-21} \ J

The difference in frequency of the two signals is calculated as follows;

E_1- E_2 = hf_1 - hf_2\\\\E_1 - E_2 = h(f_1 - f_2)\\\\f_1 - f_2 = \frac{E_1 - E_2 }{h} \\\\f_1 - f_2 = \frac{(9.94\times 10^{-21}) - (1.11 \times 10^{-21})}{6.626\times 10^{-34}} \\\\f_1 - f_2 = 1.33 \times 10^{13} \ Hz\\\\f_1 - f_2 = 1.33\times 10^{10} \ kHz

Thus, the difference in frequency of the two signals is 1.33 \times 10^{10} \ kHz.

Learn more here:brainly.com/question/14016376

4 0
3 years ago
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