Answer:
The minimum speed required is 2.62m/s
Explanation:
The value of gravitational acceleration = g = 9.81 m/s^2
Radius of the vertical circle = R = 0.7 m
Given the mass of the pail of water = m
The speed at the highest point of the circle = V
The centripetal force will be needed must be more than the weight of the pail of water in order to not spill water.
Below is the calculation:
M1V1 + M2V2 = M1V1' + M2V2'
where:
M1 is the mass of the large marble = 0.05 kg
V1 is the initial velocity of the large marble = 0.6 m/sec
M2 is the mass of the small marble = 0.03 kg
V2 is the initial velocity of the small marble = 0 m/sec (marble is at rest)
V1' is the final velocity of the large marble = -0.2 m/sec
V2' is the final velocity of the small marble that we want to calculate
Substitute with the givens in the above equation to get V2' as follows:
M1V1 + M2V2 = M1V1' + M2V2'
(0.05)(0.6) + (0.03)(0) = (0.05)(-0.2) + 0.03V2'
0.03 = -0.01 + 0.03V2'
0.03V2' = 0.03+0.01 = 0.04
V2' = 0.04/0.03
V2' = 1.334 m/sec
Answer:
Ts=51.83C
Explanation:
First we calculate the surface area of the cylinder, neglecting the top and bottom covers as indicated by the question
Cilinder Area= A=πDL
L=200mm=0.2m
D=20mm=0.02m
A=π(0.02m)(0.2m)=0.012566m^2
we use the equation for heat transfer by convection
q=ha(Ts-T)
q= heat=2Kw=2000W
A=Area=0.012566m^2
Ts=surface temperature
T=water temperature=20C
Solving for ts
Ts=q/(ha)+T
Ts=2000/(5000*0.012566m^2)+20=51.83C
Answer:
The skater has mechanical/gravitational potential energy at the two meter mark. The skater gets to two meters high on the other end of the ramp. In terms of the conservation of energy, the skater will never go higher than two meter on the other end of the the ramp because energy can be neither created nor destroyed.
Explanation:
I hoping it is right!!!∪∧∪ ∪ω∪
