Answer:
86584N
132.306 mm
Explanation:
Q = 274
Modulus of elasticity = 118 gpa
1.
Area = 316mm² without plastic deformation
F = QA
= 274x10⁶x316x10^-6
= 274000000 x 0.000316
= 86584 N
This is the maximum load.
2.
Max length =
L = 132(1 + 274x10⁶/118x10⁹)
L = 132(1+274000000/118000000000)
L = 132(1+0.002322)
L = 132(1.002322)
L = 132.306
This is the maximum length to which it may be stretched without causing plastic deformation.
Answer:
The speed is the same at 1.5 m/s while
The work done by the force F is 0.4335 J
Explanation:
Here we have angular acceleration α = v²/r
Force = ma = 2.8 × 1.5²/r₁
and ω₁ = v₁/r₁ = ω₂ = v₁/r₂
The distance moved by the force = 600 - 300 = 300 mm = 0.3 m
If the velocity is constant
The speed is 1.5 m/s while the work done is
2.8 × 1.5²1/(effective radius) ×0.3
r₁ = effective radius
2.8*9.81 = 2.8 × 1.5²/r₁
r₁ = 0.229
The work done by the force = 2.8 × 1.5²*1/r₁ *0.3 = 0.4335 J
Answer:
I hope that it will be help full
Explanation:
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<h2>
Answer:</h2>
7532V
<h2>
Explanation:</h2>
For a given transformer, the ratio of the number of turns in its primary coil () to the number of turns in its secondary coil () is equal to the ratio of the input voltage () to the output voltage () of the transformer. i.e
= ----------------(i)
<em>From the question;</em>
= number of turns in the primary coil = 8 turns
= number of turns in the secondary coil = 515 turns
= input voltage = 117V
<em>Substitute these values into equation (i) as follows;</em>
=
<em>Solve for </em><em>;</em>
= 117 x 515 / 8
= 7532V
Therefore, the output voltage (in V) of the transformer is 7532