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Yuki888 [10]
3 years ago
15

Prove that the weight of an object on moon is 1/6th of that on earth​

Physics
1 answer:
Elena L [17]3 years ago
4 0

Answer:

The mass of moon is 1/100 times and its radius 1/4 times that of earth. As a result, the gravitational attraction on the moon is about one sixth when compared to earth. Hence, the weight of an object on the moon is 1/6th its weight on the earth.

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What happens when a ray of light is directed at a mirror, a glass block and a prism?
zimovet [89]
When it is directed at a mirror it is reflected
When it is directed at a glass block it refracts
When it is directed through a prism the light splits to form a spectrum ( red orange yellow green blue indigo and violet )

Hope this is ok
5 0
3 years ago
Using an example, explain how small atoms are.
Aliun [14]
Unlike a ball, an atom doesn't have a fixed radius. The radius of an atom can only be found by measuring the distance between the nuclei of two touching atoms, and then halving that distance... Does that answer your question?
4 0
3 years ago
The research of social psychologist Vincent Hsu determined that Americans tend to fulfill their ideal of self-reliance through
dsp73
I think the correct answer from the choices listed above is option C. The research of social psychologist vincent hsu determined that americans tend to fulfill their ideal of self-reliance through <span> developing personal ideals. Hope this answers the question.</span>
3 0
3 years ago
A football punter accelerates a .55 kg football
Ronch [10]

Answer:

17.6 N

Explanation:

The force exerted by the punter on the football is equal to the rate of change of momentum of the football:

F=\frac{\Delta p}{\Delta t}

where

\Delta p is the change in momentum of the football

\Delta t=0.25 s is the time elapsed

The change in momentum can be written as

\Delta p = m(v-u)

where

m = 0.55 kg is the mass of the football

u = 0 is the initial  velocity (the ball starts from rest)

v = 8.0 m/s is the final velocity

Combining the two equations and substituting the values, we  find the force exerted on the ball:

F=\frac{m(v-u)}{\Delta t}=\frac{(0.55)(8.0-0)}{0.25}=17.6 N

5 0
3 years ago
Two friends are playing a version of proton golf where the hole is marked by a single proton. The first friend reads his meter,
yarga [219]

Answer:

the electric field strength on the second one is 2.67 N/C.

Explanation:

the electric fiel on the first one is:

E1 = k×q/(r^2)

r^2 = k×q/(E1)

     = (9×10^9)×(q)/(24.0)

     = 375000000q

then the electric field on the second one is:

E2 = k×q/(R^2)

we know that R = 3r

                       R^2 = 9×r^2

E2 = k×q/(9×r^2)

     = k×q/(9×375000000q)

     = k/(9×375000000)

     = (9×10^9)/(9×375000000)

     = 2.67 N/C

Therefore, the electric field strength on the second one is 2.67 N/C.

5 0
3 years ago
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