Answer:
42KVA
Explanation:
Given data
High Voltage (HV)= 480V
Low Voltage (LV)= 277V
Fo find
Size of transformer=?
Solution
To find the size of transformer here we use the co-ratio.The Co-ratio is given as:
Co-Ratio= (HV - LV)/HV
where
HV is High Voltage
LV is Low Voltage
Now put the values we get
Co- Ratio=(480-277)/480=.42
So the size of transformer is 42KVA
Answer:
1. Torque → F. Study of forces
2. C.O.G → D. Point of action of weight.
3. Plumb line → A. Line of C.O.G
What term do you mean? like what he did to the dog is he stopped the dog
(a) The stress in the post is 1,568,000 N/m²
(b) The strain in the post is 7.61 x 10⁻⁶
(c) The change in the post’s length when the load is applied is 1.9 x 10⁻⁵ m.
<h3>Area of the steel post</h3>
A = πd²/4
where;
d is the diameter
A = π(0.25²)/4 = 0.05 m²
<h3>Stress on the steel post</h3>
σ = F/A
σ = mg/A
where;
- m is mass supported by the steel
- g is acceleration due to gravity
- A is the area of the steel post
σ = (8000 x 9.8)/(0.05)
σ = 1,568,000 N/m²
<h3>Strain of the post</h3>
E = stress / strain
where;
- E is Young's modulus of steel = 206 Gpa
strain = stress/E
strain = (1,568,000) / (206 x 10⁹)
strain = 7.61 x 10⁻⁶
<h3>Change in length of the steel post</h3>
strain = ΔL/L
where;
- ΔL is change in length
- L is original length
ΔL = 7.61 x 10⁻⁶ x 2.5
ΔL = 1.9 x 10⁻⁵ m
Learn more about Young's modulus of steel here: brainly.com/question/14772333
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Answer:
Explanation:
We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.
1 m =100 cm
Surface area =S=
We have to find the charge Q on the positive plates of the capacitor.
V=Initial voltage between plates
d=Initial distance between plates
Initial Capacitance of capacitor
Capacitance of capacitor after moving plates
Potential difference between plates after moving
Hence, the charge on positive plate of capacitor=
