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Bogdan [553]
3 years ago
6

Is energy conserved during a fission/fusion reaction?

Physics
1 answer:
avanturin [10]3 years ago
3 0

Answer:

No,but it can be changed into other forms.

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if a person can jump 2m in earth surface how high can he jump in the moon (g of moon = 1.66m/s, g of earth = 9.8 m/s) [hint: use
julia-pushkina [17]

Answer:

h_{moon} = 11.8\ m

Explanation:

Since the work done is same everywhere in the universe. Hence, the work done in jumping will be same for the person on moon and earth:

W_{moon} = W_{earth}\\\\P.E_{moon} = P.E_{earth}\\\\mg_{moon}h_{moon} = mg_{earth}h_{earth}\\\\g_{moon}h_{moon} = g_{earth}h_{earth}\\\\(1.66\ m/s^2)h_{moon} = (9.8\ m/s^2)(2\ m)\\\\h_{moon} = \frac{(9.8\ m/s^2)(2\ m)}{1.66\ m/s^2}\\\\h_{moon} =  11.8\ m

5 0
3 years ago
Question 7 of 10
Lyrx [107]

Answer:

A. They can transfer energy through a vacuum

C. They vibrate in two directions that are perpendicular to each other

D. They radiate outward in all directions

Explanation:

8 0
3 years ago
Read 2 more answers
Express in words AND mathematically the relationship between…<br> Period and frequency
castortr0y [4]
Frequency = how many waves you get per sec
Period = how long each wave takes

Period = 1/frequency
6 0
3 years ago
A cubic box of volume 5.1 x 10 -2m3 is
I am Lyosha [343]

Answer:

21544 N

Explanation:

Let the atmospheric pressure be 101325 Pa

The side length of the cubic box:

V = d^3 = 0.051m^3

d = \sqrt[3]{V} = \sqrt[3]{0.051} = 0.371 m

The area of the cubic box:

A = d^2 = 0.371^2 = 0.1375 m^2

20 C = 20 + 273 = 293 K

180 C = 180 + 273 = 453 K

As the volume of the air inside the closed cube is not changed, assume the ideal gas law we have

\frac{P_1}{T_1} = \frac{P_2}{T_2}

Where P1 = 101325 Pa and T1 = 293K are the original atmospheric pressure and temperature. P2 and T2 = 453 are the new pressure and temperature after the cube gets heat up

P_2 = P_1\frac{T_2}{T_1} = 101325\frac{453}{293} = 156656 Pa

The net force on each side of the box it its pressure times side area

F = P_2A = 156656 * 0.1375 = 21544 N

8 0
3 years ago
Light enters an equilateral prism with an incident angle of 35° to the normal of the surface. Calculate the angle at which the
julia-pushkina [17]

Answer:

65.9°

Explanation:

When light goes through air to glass

angle of incidence, i = 35°

refractive index, n = 1.5

Let r be the angle of refraction

Use Snell's law

n=\frac{Sini}{Sinr}

1.5=\frac{Sin35}{Sinr}

Sin r = 0.382

r = 22.5°

Now the ray is incident on the glass surface.

A = r + r'

Where, r' be the angle of incidence at other surface

r' = 60° - 22.5° = 37.5°

Now use Snell's law at other surface

\frac{1}{n}=\frac{Sinr'}{Sini'}

Where, i' be the angle at which the light exit from other surface.

\frac{1}{1.5}=\frac{Sin37.5'}{Sini'}

Sin i' = 0.913

i' = 65.9°

4 0
3 years ago
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