1.
m = mass of Mr. Ure = 65 kg
g = acceleration due to gravity = 9.8 m/s²
force of earth's gravity on Mr. Ure is given as
F = mg
F = 65 x 9.8
F = 637 N
2.
F = force of gravity on car = 3050 N
m = mass of the car = ?
g = acceleration due to gravity = 9.8 m/s²
force of gravity on car is given as
F = mg
3050 = m (9.8)
m = 3050/9.8
m = 311.22 kg
3.
m = mass of Mr. Rees = 90 kg
g = acceleration due to gravity = 9.8 m/s²
force of earth's gravity on Mr. Rees is given as
F = mg
F = 90 x 9.8
F = 882 N
Answer:
The answer to your question is: F = 0.4375 N. The force will be 16 times lower than with the first conditions.
Explanation:
Data
F = 7 N
F = ? if the masses is quartered
Formula
Process
Normal conditions F = Km₁m₂/r² = 7
When masses quartered F = K(m₁/4)(m₂/4)/r² = ?
F = K(m₁m₂/16)/r²
F = K(m₁m₂/16r² = 7/16 = 0.4375 N
Answer:
Tangential acceleration is in the direction of velocity - along the circumference of a circle if the object is undergoing circular motion
a = (V2 - V1) / T
Radial acceleration is perpendicular to the direction of motion if the object is not moving in a straight line (perhaps along the circumference of a circle)
a = m V^2 / R = m ω^2 R where R is the radius vector of the velocity - note that the Radius vector is directed from the center of motion to the object and for circular motion would be constant in magnitude but not in direction
Answer:
K = ρL²g
Explanation:
Consider L as the length of the raft inside the water when the raft is displaced through additional distance y;
Then:
F = upthrust ( restoring force) = weight of the liquid displaced.
where;
A = L²
F = ky.
Then,
Divide both sides by y
K = ρL²g
