He should a step-up transformer with k=220/120=1.83 so output coil must have 240*1.83=440 turns
Answer:
3.49 seconds
3.75 seconds
-43200 ft/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration

Time the parachutist falls without friction is 3.19 seconds

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds


Now the initial velocity of the last half height will be the final velocity of the first half height.

Since the height are equal


Time taken to fall the first half is 2.65 seconds
Total time taken to fall is 2.65+1.1 = 3.75 seconds.
When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

Magnitude of acceleration is -43200 ft/s²
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Answer:
A, total.
<em>The </em><em>total</em><em> energy in a mechanical system is determined by adding the potential and kinetic enters together.</em>
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<u><em>i hope this helped at all.</em></u>
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Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa
<u>Explanation:</u>
Given -
Stress Direction, A = [1 0 0 ]
Slip plane = [ 1 1 1]
Normal to slip plane, B = [ 1 1 1 ]
Critical stress, Sc = 2.92 MPa
Let the direction of slip on = [ 1 1 0 ]
Let Ф be the angle between A and B
cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3
cos Ф = 1/√3
σ = Sc / cosФ cosλ
For slip along [ 1 1 0 ]
cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1
cos λ = 1/√2
Therefore,
σ = 2.92 / 1/√3 1/√2
σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa
Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa