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:
The spring constant is 3750 N/m
Explanation:
Use the following two relationships:
(Work) = (Force) x (Displacement)
(Force) = (Spring constant) x (Displacement)
=>
(Spring constant) = (Force) / (Displacement) = (Work) / (Displacement)^2
(Spring constant) = 6.0 kg.(m^2/s^2) / 0.0016 m^2 = 3750 N/m
The spring constant is 3750 N/m
Answer:
B
Explanation:
OOf we are doing this stuff atm
So if its faster at the front and slow at the back you can tell that its not slowing down because less of a force is there however at the front there is more of a force. Friction is low which means that its not makimg much contact so no sudden change of forces thats also why its B
Becoming Cold Blooded im pretty sure
Answer:
Explanation:
85 = ½(0.43)t²
t = √(2(85)/0.43)
t = 19.883380...
t = 20 s
v→ 8.55 m/s initial, 0 m/s final
a← 0.43 m/s²
