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Answer: vl = 2.75 m/s vt = 1.5 m/s
Explanation:
If we assume that no external forces act during the collision, total momentum must be conserved.
If both cars are identical and also the drivers have the same mass, we can write the following:
m (vi1 + vi2) = m (vf1 + vf2) (1)
The sum of the initial speeds must be equal to the sum of the final ones.
If we are told that kinetic energy must be conserved also, simplifying, we can write:
vi1² + vi2² = vf1² + vf2² (2)
The only condition that satisfies (1) and (2) simultaneously is the one in which both masses exchange speeds, so we can write:
vf1 = vi2 and vf2 = vi1
If we call v1 to the speed of the leading car, and v2 to the trailing one, we can finally put the following:
vf1 = 2.75 m/s vf2 = 1.5 m/s
On Earth, a cannonball with a mass of 20 kg would weigh 196 Newtons.
With the formula F=mg, where F is the weight in Newtons, m is the mass, and g is the acceleration due to gravity on the Earth which is 9.8m/s^2.
F=20kg x 9.8m/s^2= 196 Newtons
BUT on the moon, acceleration due to gravity is 1.6 m/s^2,
so F=mg=20kgx1.6m/s^2= 32 N
When an object's atoms move faster, its thermal energy increases and the object becomes warmer.
Answer:
The minimum force to start the block moving up the wall = 49 N
Explanation:
Friction: This is the force that tend to oppose the motion of two bodies in contact. The S.I unit of frictional force is Newton (N)
The minimum force required to start the block moving up the wall = Frictional Force.
I.e F = Frictional force.
And, F = μR..........................Equation 1
Where μ = coefficient of static friction, R = Normal reaction.
But R = mg ( on a level surface).................. Equation 2
Where m = mass, g = acceleration due to gravity.
Given: m = 10 kg,
Constant: g = 9.8 m/s²
substituting these values into Equation 2
R = 10 × 9.8
R = 98 N.
Also given: μ = 0.50
Substituting these values into equation 1
F = 98 × 0.5
F = 49 N.
Therefore The minimum force to start the block moving up the wall = 49 N
