Answer:
60 cm
Explanation:
We are given;
- Focal length of a concave mirror as 30.0 cm
- Object distance is 15.0 cm
We are required to determine the radius of curvature.
We need to know that the radius of a curvature is the radius of a circle from which the curved mirror is part.
We also need to know that the radius of curvature is twice the focal length of a curved mirror.
Therefore;
Radius of curvature = 2 × Focal length
Therefore;
Radius of curvature = 2 × 30 cm
= 60 cm
Answer:
(a). The reactive power is 799.99 KVAR.
(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.
Explanation:
Given that,
Power factor = 0.6
Power = 600 kVA
(a). We need to calculate the reactive power
Using formula of reactive power
...(I)
We need to calculate the 
Using formula of 

Put the value into the formula


Put the value of Φ in equation (I)


(b). We draw the power triangle
(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95
Using formula of reactive power


We need to calculate the difference between Q and Q'

Put the value into the formula


Hence, (a). The reactive power is 799.99 KVAR.
(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.
V = [4/3]π r^3 => [dV / dr ] = 4π r^2
[dV/dt] = [dV/dr] * [dr/dt]
[dV/dt] = [4π r^2] * [ dr/ dt]
r = 60 mm, [dr / dt] = 4 mm/s
[dV / dt ] = [4π(60mm)^2] * 4mm/s = 180,955.7 mm/s
Answer:
A: In all cases, the acceleration was the same.
Explanation:
I know this because its a clear obvious answer not only that it was one of my USA TESTPREP questions and it was right.
All you mainly have to do is the math - F=ma , In each case , the acceleration is 5 m/s squared
Given:
Height of tank = 8 ft
and we need to pump fuel weighing 52 lb/
to a height of 13 ft above the tank top
Solution:
Total height = 8+13 =21 ft
pumping dist = 21 - y
Area of cross-section =
=
=16

Now,
Work done required = 
= 
= 832
)
= 113152
= 355477 ft-lb
Therefore work required to pump the fuel is 355477 ft-lb