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Grace [21]
3 years ago
6

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Physics
1 answer:
Scorpion4ik [409]3 years ago
7 0

Answer:

protect equipment by stopping the flow of electric current; protect your house from fire.

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The distance between two stations is 1995 Km. How much time will it take to cover the distance at an average speed of 19KM/hour
Flura [38]
105 hours or 4.375 days
5 0
4 years ago
At the end of the adiabatic expansion, the gas fills a new volume V₁, where V₁ > V₀. Find W, the work done by the gas on the
tino4ka555 [31]

Answer:

W=\frac{p_0V_0-p_1V_1}{\gamma-1}

Explanation:

An adiabatic process refers to one where there is no exchange of heat.

The equation of state of an adiabatic process is given by,

pV^{\gamma}=k

where,

p = pressure

V = volume

\gamma=\frac{C_p}{C_V}

k = constant

Therefore, work done by the gas during expansion is,

W=\int\limits^{V_1}_{V_0} {p} \, dV

=k\int\limits^{V_1}_{V_0} {V^{-\gamma}} \, dV

=\frac{k}{\gamma -1} (V_0^{1-\gamma}-V_1^{1-\gamma})\\

(using pV^{\gamma}=k )

=\frac{p_0V_0-p_1V_1}{\gamma-1}

4 0
3 years ago
a crane lifts a 75 kg mass at a height of 8m . calculate the gravitational potential energy gained by the mass (g= 10N)
Bogdan [553]
Advice: Search up your questions in the Brainly search bar before asking your questions.
5 0
3 years ago
Resolving power of a telescope whose objective lens has diameter 1.22m for a wavelength 4000 Å.​
Mamont248 [21]

\sf \huge \purple{ Answer : - }

Diameter of objective lens = 1.22m

Wavelength of light = 4000Å

We have to find resolving power of telescope ..

★ Resolving power of telescope is given by

RP = D/1.22λ

  • D denotes diameter of lens
  • λ denotes wavelength of light

RP = D/1.22λ

RP = (1.22×10⁷)/(1.22×4)

RP = 0.25 × 10⁷

RP = 2.5 × 10⁶

★ Resolving power of microscope is given by

RP = 2μsinθ/λ

5 0
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

4 0
4 years ago
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