Force = mass × acceleration
To find acceleration, we can divide the speed by the time it took:
acceleration = 2.40×10^7 / 1.8×10^-9
acceleration = 1.33×10^16
the mass is equal to the mass of an electron
force = (9.11×10^-31)(1.33×10^16)
force = 1.21×10^-14 N
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.
Answer: A.AB
Explanation:
This Velocity vs Time graph shows the acceleration of a body or object, since acceleration is the variation of velocity in time.
As we can see in the attached image, the graph can be divided in four segments:
OA: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a positive slope, hence we are dealing with a positive uniform acceleration.
AB: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be positive. This means the <u>acceleration is nonuniform and positive.</u>
BC: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be negative. This means the acceleration is nonuniform and negative.
CD: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a negative slope, hence we are dealing with a negative uniform acceleration.
From all these segments, the only one that fulfils the nonuniform positive acceleration condition is option A:
Segment AB
Answer:
The heat transferred through the wall that day is 13728 BTUs
Explanation:
Here, we have the area of the wall given as
Area of wall = 2 × Length × Height + 2 × Width × Height
Length = 15 feet
Width = 11 Feet and
Height = 9 feet
Therefore, the area = 2×15×9 + 2×11×9 = 468 ft²
Temperature difference is given by
Average outside temperature - Wall temperature = 40 - 18 = 22 °F
Therefore the heat transferred through the wall that day (24 hours) at 18 sq.ft. hr/BTU is given by;
468 × 22 × 24/18 = 13728 = 13728 BTUs.
