Answer:
23376 days
Explanation:
The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.
where k is a constant.
From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.
Let the orbital period of the earth be 
and its mean distance of from the sun be 
.
Also let the orbital period of the planet be 
and its mean distance from the sun be 
.
Equation (2) therefore implies the following;
We make the period of the planet the subject of formula as follows;
But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore
Substituting equation (5) into (4), we obtain the following;
cancels out and we are left with the following;
Recall that the orbital period of the earth is about 365.25 days, hence;
Answer:
7.98 m
Explanation:
In the given question,
distance above surface= 2 m
Distance penny from person = 8 m
Since the swimming pool is filled with water and atmosphere has air therefore the refractive index phenomenon will occur.
The refractive index of water: air is 4/3 (1.33).
Using the formula, 4/3 = real depth, apparent depth
real depth= 4/3 x apparent depth
Now, calculating apparent depth = 8 - 2
= 6 m
therefore, real depth = 4/3 x apparent depth
= 1.33 x 6
= 7.98
thus, 7.98 m is the real depth of water.
There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03
Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
where
is the coefficient of friction, while N is the normal force. So we have:
since we know that F=300 N and
, we can find N, the magnitude of the normal force:
b) The problem is identical to that of the first part; however, this time the coefficienct of friction is
due to the presence of the oil. Therefore, we have:
Answer:
0 J
Explanation:
given,
mass of the ball = 5 kg
radius of the horizontal circle = 0.5 m
tension in the string = 10 N
Work done = ?
Work done by the tension in the circular path will be equal to zero.
This is because body moves in circular path, the centripetal force act along the radius of the circle and motion is right angle to the tension on the string.
so, work done = F s cos θ
θ = 90°,
work done = F s cos 90° ∵ cos 90° = 0
Work done = 0 J
The correct answer is D, Diamond
