Answer:
The answer is C)The force of gravity from Earth acting on the spacecraft decreased because the distance from Earth increased.
Explanation:
Gravity, a force, is dependent on the mass of the object exerting the gravity and the distance of an outside object from that object. The larger the object, the more gravity it will exert on an outside object. This force decreases as you move away from the object, but it will always still exist and never be equal to 0.
Answer:
6.23x10^6Pa
Explanation:
Data obtained from the question include:
F (force) = 490N
r (radius) = 0.005m
A (area of the circlular heel) =?
P (pressure) =.?
First, we'll begin by calculating the area of the circlular heel. This is illustrated below:
Area of circle = πr^2
Area = 22/7 x (0.00)^2
Area = 7.86x10^-5m^2
Pressure is simply force per unit area. It represented mathematically as
Pressure = Force /Area
Pressure = 490/7.86x10^-5
Pressure = 6.23x10^6N/m2
Recall: 1N/m2 = 1Pa
Therefore, 6.23x10^6N/m2 = 6.23x10^6Pa
Therefore, the woman exert a pressure of 6.23x10^6Pa on the floor
Answer:
So, at the depth of 24 cm below the surface of the glycerine the pressure is 2970 Pa. Hence, this is the required solution.
Explanation:
Given that,
Pressure exerted by the surface of glycerine, P = 2970 Pa and it is greater than atmospheric pressure.
The density of glycerine,
We need to find the depth h below the surface of the glycerine. The pressure due to some depth is given by :
h = 0.24 meters
or
h = 24 cm
The answer & explanation for this question is given in the attachment below.
Answer:
517.5Ns
Explanation:
F=(MV - MU)/t
where MV - MU is the change in momentum,
therefore, MV - MU = Ft
= 345 X 1.
= 517.5Ns
