Answer:
B. You would weigh the same on both planets because their masses and the distance to their centers of gravity are the same.
Explanation:
Given that Planets A and B have the same size, mass.
Let the masses of the planets A and B are
and
respectively.
As masses are equal, so
.
Similarly, let the radii of the planets A and B are
and
respectively.
As radii are equal, so
.
Let my mass is m.
As the weight of any object on the planet is equal to the gravitational force exerted by the planet on the object.
So, my weight on planet A, 
my weight of planet B, 
By using equations (i) and (ii),
.
So, the weight on both planets is the same because their masses and the distance to their centers of gravity are the same.
Hence, option (B) is correct.
Explanation:
You would lose extra body heat causing a little bit of cooling
Answer:
Net force, F = 44.66 N
Explanation:
It is given by,
Initial velocity of the person, u = 0
Final velocity of the person, v = 0.68 m/s
Distance, s = 0.428 m
Combined mass of the person and the kayak, m = 82.7 kg
We need to find the net force acting on the kayak i.e.
F = ma...........(1)
Firstly, we will calculate the value of "a" from third equation of motion as :




Put the value of a in equation (1) as :

F = 44.66 N
So, the net force acting on the kayak is 44.66 N. Hence, this is the required solution.
Answer:
2324 J
Explanation:
The formula for work is:

where
is the force applied, and
is the distance moved, in this case 
and we need to find 
Since the crate is not moving up or down, we conclude that the <u>normal force must be equal to the weight </u>of the object:

where
is the normal force and
is the weight, which is:
, where g is the gravitational acceleration
and
is the mass
.
---------
Thus the normal force is:

Now, the force due to the friction is defined as:

where
is the coefficient of friction, 
So, for the crate to move, the force applied must be equal to the frictional force:

And now that we know the force we can calculate the work:

substituting known values:

Answer:
The law of conservation of energy
Explanation: