Answer:
The Gravitational Force is reduced 4 times
Explanation:
The equation of Gravitational force follows:
F = (G*m1*m2)/r^2
Assume that G*m1*m2 = 1 and r = 1:
F = 1/1^2 = 1 N
Multiply the radius by 2
F = 1/2^2 = 1/4 N
So doubling the distance reduces the force 4 times.
Answer:
a The kinetic energy is 
b The height of the center of mass above that position is
Explanation:
From the question we are told that
The length of the rod is 
The mass of the rod
The angular speed at the lowest point is 
Generally moment of inertia of the rod about an axis that passes through its one end is
Substituting values


Generally the kinetic energy rod is mathematically represented as



From the law of conservation of energy
The kinetic energy of the rod during motion = The potential energy of the rod at the highest point
Therefore



Sunshine will help prode vitamin C
Answer:
a. P = nRTV
Explanation:
The question is incomplete. Here is the complete question.
"All of the following equations are statements of the ideal gas law except a. P = nRTV b. PV/T = nR c. P/n = RT/v d. R = PV/nT"
Ideal gas equation is an equation that describes the nature of an ideal gas. The molecule of an ideal gas moves at a particular velocity depending on the temperature. This gases collides with one another elastically. The collision that an ideal gas experience is a perfectly elastic collision.
The ideal gas equation is expressed as shown:
PV = nRT where:
P is the pressure of the gas
V is the volume
n is the number of moles
R is the ideal gas constant
T is the temperature.
Based on the formula given for an ideal gas, it can be inferred that the equation. P = nRTV is not a statement of an ideal gas equation.
The remaining option will results to an ideal gas equation if they are cross multipled.
Answer:
51.96 m/s^-1
Explanation:
a) see the attachment
b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be
v_ox=v_o*cosФ
=60*cos (30)
= 51.96 m/s^-1