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Damm [24]
3 years ago
14

Find the weight of a 80 kg man on the surface of the moon? What should be his mass on the moon and the earth? (ge =9.8 m/s2, gm

=1.63 m/s2 )
Physics
1 answer:
Tcecarenko [31]3 years ago
6 0

If a man's mass is 80 kg, then ...

-- He weighs 258 newtons on Mars.

-- He weighs 785 newtons (about 176.4 pounds) on Earth.

-- He weighs 130 newtons on the Moon.

-- He weighs different amounts in other different places.

His mass doesn't change.  It's 80 kg no matter where he goes.

You might be interested in
Two charges, each 9 µC, are on the x axis, one at the origin and the other at x = 8 m. Find the electric field on the x axis at
bearhunter [10]

a) Electric field at x = -2 m: 21,060 N/C to the left

b) Electric field at x = 2 m: 18,000 N/C to the right

c) Electric field at x = 6 m: 18,000 N/C to the left

d) Electric field at x = 10 m: 21,060 N/C to the right

e) Electric field is zero at x = 4 m

Explanation:

a)

The electric field produced by a single-point charge is given by

E=k\frac{q}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

Here we have two charges of

q=9\mu C = 9\cdot 10^{-6} C

each. Therefore, the net electric field at any point in the space will be given by the vector sum of the two electric fields. The two charges are both positive, so the electric field points outward of the charge.

We call the charge at x = 0 as q_0 , and the charge at x = 8 m as q_8.

For a point located at x = -2 m, both the fields E_0 and E_8 produced by the two charges point to the left, so the net field is the sum of the two fields in the negative direction:

E=-\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=-kq(\frac{1}{(-2)^2}+\frac{1}{(8-(-2))^2})=-21060 N/C

b)

In this case, we are analyzing a point located at

x = 2 m

The field produced by the charge at x = 0 here points to the right, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, so:

E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(2)^2}-\frac{1}{(8-2)^2})=18000 N/C

And since the sign is positive, the direction is to the right.

c)

In this case, we are considering a point located at

x = 6 m

The field produced by the charge at x = 0 here points to the right again, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, as before; so:

E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(6)^2}-\frac{1}{(8-6)^2})=-18000 N/C

And the negative sign indicates that the electric field in this case is towards the left.

d)

In this case, we are considering a point located at

x = 10 m

This point is located to the right of both charges: therefore, the field produced by the charge at x = 0 here points to the right, and the field produced by the charge at x = 8 m here points to the right as well. Therefore, the net field is given by the sum of the two fields:

E=\frac{kq_0}{(0-x)^2}+\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(10)^2}+\frac{1}{(8-(10))^2})=21060 N/C

And the positive sign means the field is to the right.

e)

We want to find the point with coordinate x such that the electric field at that location is zero. This point must be in between x = 0 and x = 8, because that is the only region where the two fields have opposite directions. Therefore, te net field must be

E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(-x)^2}-\frac{1}{(8-x)^2})=0

This means that we have to solve the equation

\frac{1}{x^2}-\frac{1}{(8-x)^2}=0

Re-arranging it,

\frac{1}{x^2}-\frac{1}{(8-x)^2}=0\\\frac{(8-x)^2-x^2}{x^2(8-x)^2}=0

So

(8-x)^2-x^2=0\\64+x^2-16x-x^2=0\\64-16x=0\\64=16x\\x=4 m

So, the electric field is zero at x = 4 m, exactly halfway between the two charges (which is reasonable, because the two charges have same magnitude)

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0
3 years ago
Read 2 more answers
A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotat
vladimir2022 [97]

Answer:

f = 0.4 Hz

Explanation:

The frequency of rotation of an object in order to achieve required centripetal or radial acceleration, can be found out by using the following formula:

f = (1/2π)√(ac/r)

where,

f = frequency of rotation = ?

ac = radial acceleration = 34.1 m/s²

r = radius = length of beam = 5.55 m

Therefore,

f = (1/2π)√[(34.1 m/s²)/(5.55 m)]

<u>f = 0.4 Hz</u>

6 0
3 years ago
Before you conduct your experiment, you need to form a hypothesis. A hypothesis is a prediction of what you think will happen in
Stolb23 [73]

Example: A apple rotting.

If I put my apple in a fridge, then it would not rot as fast because it is in a cooled area. (example)

Hope it helps! Brainiest Answer would be amazing!

5 0
4 years ago
Accurate _______ is needed in a valid experiment.<br><br> Hurry, I need help.
Lorico [155]

Answer: repeatable

Explanation: Accurate repeatable is needed in a valid experiment. I think

4 0
3 years ago
How often do spring and neap tides occur
iragen [17]

Answer:

Without thought to the season, Spring tides happen [two times] every [lunar] month the whole year. When the Sun and the Moon are quadratic, Neap tides also happen two times during a month.

Here is a picture to help show you:

(With the Spring tide, the moon is toward the "left" or "right" of the Earth. With the Neap tide, the moon is "above" or "below/under" the Earth. It's kind of hard to explain).

6 0
3 years ago
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