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Degger [83]
3 years ago
5

What is the energy equivalent of an object with a mass of 2.5 kg?

Physics
2 answers:
Bad White [126]3 years ago
8 0
This is the answer on Edgenuity 2.25*10^17
N76 [4]3 years ago
5 0

Answer : Energy equivalent of an object with a mass of 2.5 kg is E=2.25\times 10^{17}\ J.

Explanation :

It is given that mass of an object is, m = 2.5 kg

It is required to find energy equivalence.

According to mass energy equivalence relation :

E=mc^2

Where,

m is the mas of an object.

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

E=2.5\ kg\times (3\times 10^8)^2\ m/s

E=22.5\times 10^{16}\ J

E=2.25\times 10^{17}\ J

Hence, this is the required solution.

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An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to v
maxonik [38]

Answer:

a) The final pressure is 1.68 atm.

b) The work done by the gas is 305.3 J.

Explanation:

a) The final pressure of an isothermal expansion is given by:

T = \frac{PV}{nR}

T_{i} = T_{f}

\frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR}

Where:

P_{i}: is the initial pressure = 5.79 atm

P_{f}: is the final pressure =?

V_{i}: is the initial volume = 420 cm³

V_{f}: is the final volume = 1450 cm³

n: is the number of moles of the gas

R: is the gas constant

P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm

Hence, the final pressure is 1.68 atm.

b) The work done by the isothermal expansion is:

W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J

Therefore, the work done by the gas is 305.3 J.

I hope it helps you!        

3 0
3 years ago
A negatively charged particle is moving to the right, directly above a wire have a current flowing to the right. In which direct
Varvara68 [4.7K]

Answer:

C) upward

Explanation:

The problem can be solved by using the right-hand rule.

First of all, we notice at the location of the negatively charged particle (above the wire), the magnetic field produced by the wire points out of the page (because the current is to the right, so by using the right hand, putting the thumb to the right (as the current) and wrapping the other fingers around it, we see that the direction of the field above the wire is out of the page).

Now we can apply the right hand rule to the charged particle:

- index finger: velocity of the particle, to the right

- middle finger: direction of the magnetic field, out of the page

- thumb: direction of the force, downward --> however, the charge is negative, so we must reverse the direction --> upward

Therefore, the direction of  the magnetic force is upward.

3 0
3 years ago
An air bubble has a volume of 2.0 cm3 when it is released by a submarine 100 m below the surface of a freshwater lake. What is t
UkoKoshka [18]

Answer:

21.35 cm^3

Explanation:

let the volume at the surface of fresh water is V.

The volume at a depth of 100 m is V' = 2 cm^3

temperature remains constant.

density of water, d = 1000 kg/m^3

Pressure at the surface of fresh water is atmospheric pressure,

P = Po = 1.013 x 10^5 N/m^2

The pressure at depth 100 m is P' = Po + hdg

P' = 1.013 \times 10^{5}+ 100 \times 1000 \times 9.8

P' = 10.813 x 10^5 N/m^2

Use the Boyle's law

P V = P' V'

1.013 \times 10^{5}\times V = 10.813 \times 10^{5}\times 2

V = 21.35 cm^3

Thus, the volume of air bubble at the surface of fresh water is 21.35 cm^3.

5 0
3 years ago
The section of the electromagnetic spectrum that humans can generally see is called _____ light.
Lisa [10]

Answer:

(A) Visible

Explanation:

  • The section of the electromagnetic spectrum that humans can generally see is called visible light.

White light is visible light and the range of visible wavelengths ranges from 400 - 700 nanometers.

8 0
3 years ago
Read 2 more answers
A 17-mm-wide diffraction grating has rulings of 530 lines per millimeter. White light is incident normally on the grating. What
Sedbober [7]

Answer:

377 nm

Explanation:

Number of lines per meter is, N &=530 \times 1000 \\ &=530000 \text { lines } / \mathrm{m} \end{aligned}

Grating element is, d=\frac{1}{N}

=1.8868 \times 10^{-6} \mathrm{~m}[tex]
Order is, n=5
Condition for maximum intensity is, [tex]d \sin \theta=n \lambda

 \lambda &=\frac{1.8868 \times 10^{-6}}{5(\sin 90)} \\
&=0.377 \times 10^{-6} \mathrm{~m} \\
&=377 \mathrm{~nm}

7 0
3 years ago
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