Velocity of the mass after 11 seconds = ( value of the gravitational acceleration) * ( time )
velocity = ( 9.81 m / s^2 ) ( 11)
velocity = 107.91 meters per second
W = mg = 350 newton
m = W/g = 350/9.8 = 35.71 kg
on mars
W = mg = 134 newton
g = W/m = 134/35.71 = 3.75 meters/second2
Question:
A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?
Answer:
Time for the race will be t = 9.26 s
Explanation:
Given data:
As the sprinter starts the race so initial velocity = v₁ = 0
Distance = s₁ = 20 m
Acceleration = a = 4.20 ms⁻²
Distance = s₂ = 100 m
We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.
Using 3rd equation of motion
(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)
v₂ = 12.96 ms⁻¹
Time for 20 m distance = t₁ = (v₂ - v ₁)/a
t₁ = 12.96/4.2 = 3.09 s
He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be
Time for 100 m distance = t₂ = s₂/v₂
t₂ = 80/12.96 = 6.17 s
Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s
T = 9.26 s
A transfer of charge is actually a gross movement of electrons. Charged objects have a normal or "balanced" state. This state is balanced in a sense of positive charges (protons) and negative charges (electrons). When an object has an excess of deficiency of electrons, it will try to regain its balance by releasing or accepting electrons.
Answer:
the final energy of the system is 35.5 kJ.
Explanation:
Given;
initial energy of the system, E₁ = 10 kJ
heat transferred to the system, q₁ 30 kJ
Heat lost to the surrounding, q₂ = 5kJ
heat gained by the system, Q = q₁ - q₂ = 30 kJ - 5kJ = 25 kJ
work done on the system, W = 500 J = 0.5 kJ
Apply first law of thermodynamic,
ΔU = Q + W
where;
ΔU is change in internal energy
Q is the heat gained by the system
W is work done on the system
ΔU = 25kJ + 0.5 kJ
ΔU = 25.5 kJ
The final energy of the system is calculated as;
E₂ = E₁ + ΔU
E₂ = 10 kJ + 25.5 kJ
E₂ = 35.5 kJ.
Therefore, the final energy of the system is 35.5 kJ.
