Your being on the moon has no effect on the moon's
gravitational field strength, or on the Earth's for that
matter.
However, YOU notice a change on YOU when YOU move
from one to the other, because of the effect of the gravitational
field strength on you and your internal organs.
If you could stand on the moon, you would experience an incredible
sense of lightness, since the forces of attraction between the moon
and anything else are only 16% as great as the same forces are on
Earth.
The frictional force is 39.4 N
Explanation:
We can solve this problem by applying Newton's 2nd law of motion: in fact, the net force acting on the block is equal to the product between its mass and its acceleration. So we can write

where
is the net force
m is the mass
a is the acceleration
Here we know that the box is moving with constant velocity, so its acceleration is zero:

This means that the net force is also zero:

The net force on the block is given by the applied force, forward, and the frictional force, backward:

where
is the applied force
is the frictional force
Therefore, solving for
,

Learn more about friction:
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Physics is an important element in the education of chemists, engineers and computer scientists as well as practitioners of the other physical and biomedical sciences .
Answer:
e% = 3.4%
Explanation:
This is a calorimetry problem where the heat released equals the heat absorbed
m
(T₀ - T_f) = M c_{e2} (T₁ - T_f)
Index 1 refers to water and index 2 to metal, in this case it asks for the specific heat of the metal (c_{e2})
c_{e2} = m / M c_{e1} (T_f -T₀) / (T₁ - T_f)
Let's calculate
c_{e} = 60/100 4.19 (24-20) / (100-24)
c_{e2} = 0.1323 j / gC
This metal is possibly lead, which is its specific heat is 0.128 J / gC
The percentage error is
e% = (c_{e2} - 0.128) /0.128 100
e% = 3.4%
Answer:
Photons strike the CD 
Explanation:
First of all, it is necessary to calculate the energy produced by a semiconductor laser for 69 minutes:


Next, it is possible to calculate the number of photons striking the CD surface during this time:

long in CD
light velocity
Solve to n'

