Answer:
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Explanation:
Let L represent Moe's height during the leap.
Moe's velocity v at any point in time during the leap is;
v = dL/dt = u - gt .......1
Where;
u = it's initial speed
g = acceleration due to gravity on Mars
t = time
The determine how far the cricket was off the ground when it became Moe’s lunch.
We need to integrate equation 1 with respect to t
L = ∫dL/dt = ∫( u - gt)
L = ut - 0.5gt^2 + L₀
Where;
L₀ = Moe's initial height = 0
u = 105m/s
t = 56 s
g = 3.75 m/s^2
Substituting the values, we have;
L = (105×56) -(0.5×3.75×56^2) + 0
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Answer:
19.74 N
Explanation:
mass of ball (m) = 0.25 kg
radius (r) = 0.5 m
time (t) = 2 revolutions per seconds = 1/2 = 0.5 second per revolution
find the tension in the string
tension (T) = 
- where velocity (v) =

tension now becomes (T) = 
tension (T) = 
- now substituting the values of mass (m), time (t) and radius (r) into the equation above we have
tension (T) = 
tension (T) =
=
= 19.74 N
If no frictional work is considered, then the energy of the system (the driver at all positions is conserved.
Let
position 1 = initial height of the diver (h₁), together with the initial velocity (v₁).
position 2 = final height of the diver (h₂) and the final velocity (v₂).
The initial PE = mgh₁ and the initial KE = (1/2)mv₁²
where g = acceleration due to gravity,
m = mass of the diver.
Similarly, the final PE and KE are respectively mgh₂ and (1/2)mv₂².
PE in position 1 is converted into KE due to the loss in height from position 1 to position 2.
Therefore
(KE + PE) ₁ = (KE + PE)₂
Evaluate the given answers.
A) The total mechanical energy of the system increases.
FALSE
B) Potential energy can be converted into kinetic energy but not vice versa.
TRUE
C) (KE + PE)beginning = (KE + PE) end.
TRUE
D) All of the above.
FALSE
For the most part aesthetics but you could argue they reduce air turbulence in some cases which would lead to better gas mileage. On sports cars however the goal is usually down force which is not wanted on economy cars as it can actually worsen gas mileage.
I hope that answered your question.
Answer:
70 W
Explanation:
Power = work / time
P = mgh / t
P = (80 kg) (9.8 m/s²) (3000 m) / (36000 s)
P = 65.3 W
Rounding to one significant figure, the average power output is 70 W.