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3 years ago

How much work is done when you push a crate horizontally with 130

1 answer:

3 0

Work= Force in the direction of displacement*displacement.

You know the force in the direction of displacement (horizontally) and the displacement. So,

W=130*11=1430

Therefore, the work done is 1,430 Joules

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What type of electromagnetic radiation is being used in the picture?

pashok25 [27]

Answer: b i think

Explanation:

6 0

3 years ago

Hydrogen line spectrum lies entirely within visible range

mart [117]

No, that's silly.

You've got your Pfund series where electrons fall down to the 5th level,
your Brackett series where they fall to the 4th level, and your Paschen
series where they fall to the 3rd level. All of those transitions ploop out
photons at Infrared wavelengths.

THEN next you get your Balmer series, where the electrons fall in
to the 2nd level. Most of those are at visible wavelengths, but even
a few of the Balmer transitions are in the Ultraviolet.

And then there's the Lyman series, where electrons fall all the way
down to the #1 level. Those are ALL in the ultraviolet.

6 0

3 years ago

What is the power of an electric with a current of 0.5 A and a voltage of 120 V ? The formula for power is P =IV

zaharov [31]

Answer:

60Watts

Explanation:

Given parameters:

Current = 0.5A

Voltage = 120V

Unknown

Power = ?

Solution:

The power in the electric circuit is the product of current and voltage;

P = IV

Insert the given parameters and solve;

P = 0.5 x 120

P = 60Watts

7 0

2 years ago

Read 2 more answers

Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W

jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

$T^{2}=\frac{4\pi^{2}}{GM}r^{3}$ (1)

Where;

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24}kg$ is the mass of the Earth

$r$ is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period $T_{X}$ is:

$T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}$ (2)

Where $r_{X}=1.2(10)^{6}m$

$T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}$ (3)

$T_{X}=413.712 s$ (4)

For satellite Y, the orbital period $T_{Y}$ is:

$T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}$ (5)

Where $r_{Y}=1.9(10)^{5}m$

$T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}$ (6)

$T_{Y}=26.064 s$ (7)

This means $T_{X}>T_{Y}$

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

$V_{X}=\sqrt{\frac{GM}{r_{X}}}$ (8)

$V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}$

$V_{X}=18224.783 m/s$ (9)

<u>For Satellite Y:</u>

$V_{Y}=\sqrt{\frac{GM}{r_{Y}}}$ (10)

$V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}$

$V_{Y}= 45801.13 m/s$ (11)

This means $V_{Y}>V_{X}$

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

4 0

3 years ago

Read 2 more answers

Why is the answer C?

4vir4ik [10]

Explanation:

We want to find the statement that is proven by the fact that the balls reach the same height.

A isn't supported by the evidence. Balls can reach the same height without having the same initial speed.

B isn't supported by the evidence. Balls can reach the same height without having the same launch angle.

C is supported. Projectiles spend the same amount of time going up as they do coming down, so if two projectiles reach the same height, then they must spend the same amount of time in the air.

D isn't supported by the evidence. Balls thrown at the same speed and complementary angles have the same range but different heights.

E isn't supported by the evidence. The mass of the ball doesn't affect the height.

7 0

3 years ago