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serg [7]
3 years ago
5

A system releases 20 J of heat into its surroundings while the surroundings

Physics
1 answer:
Allushta [10]3 years ago
7 0

Answer:

The change of energy is 60J

Explanation:

We have a closed system, which receives 80 J of energy. It means this energy is entering the system. And also we have 20 J that is dissipating in the shape of heat.

Therefore the amount remaining in the system will be 80 - 20 = 60 J

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I’m not sure how to do 30, could someone pls help?
choli [55]

I think its C or D.

Einstein's Nobel winning photoelectric equation.

The higher the work function, the more energy needed to photoeject an electron from the surfface

Try D

8 0
2 years ago
1. The farthest star listed in the chart is? Arcturus, Capella, Aldebaran, Pollux, Regulus, Castor
Pachacha [2.7K]

Answer:

Regulus

Explanation:

Which is 77.63 light years away

8 0
2 years ago
A circular flat coil that has N turns, encloses an area A, and carries a current i, has its central axis parallel to a uniform m
gogolik [260]

Answer:

A. Zero

Explanation:

The force on a coil of N turns, enclosing an area, A and carrying a current I in the presence of a magnetic field B, is :

F = N * I * A * B * sinθ

Where θ is the angle between the normal of the enclosed area and the magnetic field.

Since the normal of the area is parallel to the magnetic field, θ = 0

Hence:

F = NIABsin0

F = 0 or Zero

3 0
2 years ago
The Earth revolves around the Sun once a year at an average distance of 1.50×1011m. Find the orbital radius that corresponds to
DedPeter [7]

Answer:

9.4\cdot 10^{10} m

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}

where

r_o is the distance of the new object from the sun (orbital radius)

T_o=180 d is the orbital period of the object

r_e = 1.50\cdot 10^{11} m is the orbital radius of the Earth

T_e=365 d is the orbital period the Earth

Solving the equation for r_o, we find

r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m

3 0
3 years ago
A coal - fired power station produces 100MJ of electrical energy when it is supplied with 400MJ of energy from its fuel. The eff
Alexus [3.1K]

Answer:

100 divide by 400 times 100 percent

8 0
2 years ago
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