3750 seconds to travel that far
Answer: minimum speed of launch must be 7.45m/s
Explanation:
Given the following:
Height or distance (s) = 2.83m
The final velocity(Vf) at maximum height = 0
Upward motion, acceleration due to gravity(g) us negative = -9.8m/s^2
From the 3rd equation of motion:
V^2 = u^2 - 2gs
Where V = final velocity
u = initial velocity
Therefore, u = Vi
u = √Vf^2 - 2gs
u = √0^2 - 2(-9.8)(2.83)
u = √0 + 55.468
u = √55.468
u = 7.4476 m/s
u = 7.45m/s
Answer: energy
Explanation: So the input and output of the power grid system is energy.
Answer:
(a) 0.345 T
(b) 0.389 T
Solution:
As per the question:
Hall emf, 
Magnetic Field, B = 0.10 T
Hall emf, 
Now,
Drift velocity, 
Now, the expression for the electric field is given by:
(1)
And
Thus eqn (1) becomes
where
d = distance
(2)
(a) When 
(b) When 
Answer:
reduced
Explanation:
The use of bearing surfaces that are themselves sacrificial, such as low shear materials, of which lead/copper journal bearings are an example
