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dsp73
3 years ago
15

Can something have energy without having momentum? explain. can something have momentum without having energy? defend your answe

r
Physics
1 answer:
marta [7]3 years ago
8 0
Momentum  is a product mass and velocity. If a certain object posses a kinetic energy, then it should have a momentum since it is moving which has a velocity. However, if the object is at rest and only has potential energy, then it would not have momentum. So, for the first question the answer would be yes, an object can have energy without having any momentum. For the second question, every object whether it is moving or at rest, possess some energy, potential for an object at rest and kinetic for an object that is moving. Thus, the answer would be no, an object having momentum would always have energy.
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Explanation:

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An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0
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Complete question:

An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.

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The value of its capacitance is 1.027 x 10⁻¹² F

Explanation:

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area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²

separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m

voltage of the battery, V = 18 V

The value of its capacitance is calculated as;

C = \frac{k\epsilon_0A}{d} \\\\C = \frac{(1)(8.85\times 10^{-12})(2.9 \times 10^{-4})}{2.5 \times 10^{-3}} \\\\C = 1.027 \times 10^{-12} \ F

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3 years ago
A microwave oven operating at 1.22 × 108 nm is used to heat 165 mL of water (roughly the volume of a teacup) from 23.0°C to 100.
ANTONII [103]

<u>Answer:</u> The number of photons are 3.7\times 10^8

<u>Explanation:</u>

We are given:

Wavelength of microwave = 1.22\times 10^8nm=0.122m    (Conversion factor:  1m=10^9nm  )

  • To calculate the energy of one photon, we use Planck's equation, which is:

E=\frac{hc}{\lambda}

where,

h = Planck's constant = 6.625\times 10^{-34}J.s

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

\lambda = wavelength = 0.122 m

Putting values in above equation, we get:

E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{0.122m}\\\\E=1.63\times 10^{-24}J

Now, calculating the energy of the photon with 88.3 % efficiency, we get:

E=1.63\times 10^{-24}\times \frac{88.3}{100}=1.44\times 10^{-24}J

  • To calculate the mass of water, we use the equation:

Density=\frac{Mass}{Volume}

Density of water = 1 g/mL

Volume of water = 165 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{165mL}\\\\\text{Mass of water}=165g

  • To calculate the amount of energy of photons to raise the temperature from 23°C to 100°C, we use the equation:

q=mc\Delta T

where,

m = mass of water = 165 g

c = specific heat capacity of water = 4.184 J/g.°C

\Delta T = change in temperature = T_2-T_1=100^oC-23^oC=77^oC

Putting values in above equation, we get:

q=165g\times 4.184J/g.^oC\times 77^oC\\\\q=53157.72J

This energy is the amount of energy for 'n' number of photons.

  • To calculate the number of photons, we divide the total energy by energy of one photon, we get:

n=\frac{q}{E}

q = 53127.72 J

E = 1.44\times 10^{-24}J

Putting values in above equation, we get:

n=\frac{53157.72J}{1.44\times 10^{-24}J}=3.7\times 10^{28}

Hence, the number of photons are 3.7\times 10^8

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