Given:
distance from the projector lens to the image, di
projector lens focal length, f
distance from the transparency to the projector lens, do
thin lens equation: 1/f = 1/di + 1/do
do = 4 inches
di = 8 feet
convert feet to inches, for uniformity.
1 foot = 12 inches
8 feet * 12 inches/ft = 96 inches
1/f = 1/96 inches + 1/4 inches
Adding fractions, denominator must be the same.
1/f = (1/96 * 1/1) + (1/4 * 24/24)
1/f = 1/96 + 24/96
1/f = 25/96
to find the value of f, do cross multiplication
1*96 = f * 25
96 = 25f
96/25 = f
3.84 = f
The focal length of the project lens is 3.84 inches
The average velocity of the car for the whole journey is 69.57 km/h.
The given parameters:
- <em>Length of the road, L = 320 km</em>
- <em>Distance covered = 240 km at 75 km/h</em>
- <em>time spent refueling, t₂ = 0.6 hr</em>
- <em>Final velocity, = 100 km/hr</em>
The time spent by the before refueling is calculated as follows;
The time spent by the car for the remaining journey;
The total time of the journey is calculated as follows;
The average velocity of the car for the whole journey is calculated as follows;
Learn more about average velocity here: brainly.com/question/6504879
I am pretty sure the answer to your question is B
The car at 60 kph has 9 times more kinetic energy than the car traveling at 20 kph. This assumes that both cars have the same mass. Kinetic energy depends on the square of thee speed so if one car is going 3 times faster, its kinetic energy will be 3^2 ( = 9 ) greater. The car going at 60 kph will have 4 times the KE of the car going at 30 kph ( again assuming that the cars have the same mass.)
