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3 years ago

An object on the moon will weigh 1/6th of what it does on earth but have the same mass. Why is that?

Physics

1 answer:

elena55 [62]3 years ago

8 0

Answer:

Because the weight depends of the gravity

Explanation:

This is because weight and mass are different, in order to better understand this problem we will apply an example with real values, which will help us to determine a person's weight.

A man has a mass of 80 [kg] on Earth when measuring his weight he realizes that it is 784.9 [N] and on the moon it is 130.8 [N]

On Earth

$g_{e} = 9.81[m/s^2]\\g_{m} = 1.635[m/s^2]$

Where:

g = gravity

Weight on the moon

Wm = 80 * 1.635

Wm = 130.8[N]

Weight on the earth

We = 80 * 9.81

We = 784.8[N]

In this way we can see that the weight depends on the gravity of where the person is located.

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A 360-nm thick oil film floats on the surface of a pool of water. the indices of refraction of the oil and the water are 1.5 and

jenyasd209 [6]

The correct answer is 432, and 720.
The thickness of a film is t= 360nm
the refractive index of oil n₀t = (m +1/2) λ
For m =0
λ = 4n₀t
= 4(1.50)(360)
= 2160nm
for m = 1
λ = 4n₀t
= 4(1.50)(360)/3
= 720nm
m = 2
λ = 4n₀t/5 = 4(1.50)(360)/5
= 432nm
The wavelength which are most strongly reflected are
432nm, 720nm.

7 0

3 years ago

Which of the following would be used in luminosity calculations?

Daniel [21]

Answer:

1) joule

2) $kgm^{2}/s^{2}$

3) $10\%$

Explanation:

1) Luminosity is the amount of light emitted (measured in Joule) by an object in a unit of time (measured in seconds). Hence in SI units luminosity is expressed as joules per second ( $\frac{J}{s}$ ), which is equal to Watts ( $W$ ).

This amount of light emitted is also called radiated electromagnetic power, and when this is measured in relation with time, the result is also called radiant power emitted by a light-emitting object.

Therefore, if we want to calculate luminosity the Joule as a unit will be used.

2) Work $W$ is expressed as force $F$ multiplied by the distane $d$ :

$W=F.d$

Where force has units of $kgm/s^{2}$ and distance units of $m$ .

If we input the units we will have:

$W=(kgm/s^{2})(m)$

$W=kgm^{2}/s^{2}$ This is 1Joule ( $1 J$ ) in the SI system, which is also equal to $1 Nm$

3) The formula to calculate the percent error is:

$\% error=\frac{|V_{exp}-V_{acc}|}{V_{acc}} 100\%$

Where:

$V_{exp}=7.34 (10)^{-11} Nm^{2}/kg^{2}$ is the experimental value

$V_{acc}=6.67 (10)^{-11} Nm^{2}/kg^{2}$ is the accepted value

$\% error=\frac{|7.34 (10)^{-11} Nm^{2}/kg^{2}-6.67 (10)^{-11} Nm^{2}/kg^{2}|}{6.67 (10)^{-11} Nm^{2}/kg^{2}} 100\%$

$\% error=10.04\% \approx 10\%$ This is the percent error

8 0

3 years ago

The question is in the picture. :)

NISA [10]

Player A needs the least amount of energy. The ball is light weight and she is closest to the goal so the momentum need to kick the ball will be the least and the distance is has to travel is the shortest. But player C needs the most amount of energy. The ball is heavy so it will take the most momentum to move the ball and over such a long distance. Hope this help idrk.

5 0

3 years ago

Give examples of plants that can reproduce by vegetative reproduction.

marishachu [46]

Answer:

Some examples of vegetative propagation are farmers creating repeated crops of apples, corn, mangoes or avocados through asexual plant reproduction rather than planting seeds. Vegetative propagation can be accomplished from side-shoots, slips, stems and sections of tubers, bulbs or rhizomes.

Explanation:

5 0

3 years ago

A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the

barxatty [35]

Answer:

2.2 meters

Explanation:

Potential energy, PE created by a charge, q at a radius r from the charge source, Q, is expressed as:

$KE=\frac{kQq}{r}\ \ \ \ \ \ \ ...i$