A 40 N crate rests on a rough horizontal floor. A 12 N horizontal force is then applied to it. If the coecients of friction are
1 answer:
Answer:
Frictional force, f = 20 N
Explanation:
It is given that,
Weight of the crate, W = 40 N
Horizontal force acting on the crate, F = 12 N
The coefficient of static friction,
The coefficient of kinetic friction,
Let f is the frictional force acting on the crate. Friction is an opposing force. It is equal to the product of coefficient friction and the normal force. It is given by :
f = 20 N
So, the frictional force on the crate is 20 N. Hence, this is the required solution.
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Answer:
The rate of transfer of heat is 0.119 W
Solution:
As per the question:
Diameter of the fin, D = 0.5 cm = 0.005 m
Length of the fin, l =30 cm = 0.3 m
Base temperature,
Air temperature,
k = 388 W/mK
h =
Now,
Perimeter of the fin, p =
Cross-sectional area of the fin, A =
A =
To calculate the heat transfer rate:
where
Now,
Answer:
B. About 12 degrees
Explanation:
The orbital period is calculated using the following expression:
T = 2π*()
Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.
Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.
Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:
=
Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be years, or 31.62 years.
We find the amount of degrees it moves in 1 year:
or about 12 degrees.