Answer:
the money that would be saved is $13.14.
Explanation:
Given;
power consumed by the light bulb, P = 100 W = 0.1 kW
time of running the bulb, t = 3 hours for 365 days = 1,095 hours
cost rate of power consumption, C = $0.12 per kWh
Energy consumed by the light bulb for the given days;
E = Pt
E = 0.1 kW x 1,095 hr
E = 109.5 kWh
Cost of energy consumed = 109.5 kWh x $0.12 / kWh
= $13.14
Therefore, the money that would be saved is $13.14.
Answer:
is the time taken by the car to accelerate the desired range of the speed from zero at full power.
Explanation:
Given:
Range of speed during which constant power is supplied to the wheels by the car is
.
- Initial velocity of the car,

- final velocity of the car during the test,

- Time taken to accelerate form zero to 32 mph at full power,

- initial velocity of the car,

- final desired velocity of the car,

Now the acceleration of the car:



Now using the equation of motion:


is the time taken by the car to accelerate the desired range of the speed from zero at full power.
Answer:
The internet is most useful to them because they use it to communicate.
Explanation:
If I were to send a message to my brother in Florida, through the internet, while I'm in Pennsylvania he would get it in minutes. On the other hand if I were going to meet him and then explain what I wanted to tell him in person it would take a much longer time.
===> Distance fallen from rest in free fall =
(1/2) (acceleration) (time²)
(122.5 m) = (1/2) (9.8 m/s²) (time²)
Divide each side by (4.9 m/s²): (122.5 m / 4.9 m/s²) = time²
(122.5/4.9) s² = time²
Take the square root of each side: 5.0 seconds
===> (Accelerating at 9.8 m/s², he will be dropping at
(9.8 m/s²) x (5.0 s) = 49 m/s
when he goes 'splat'. We'll need this number for the last part.)
===> With no air resistance, the horizontal component of velocity
doesn't change.
Horizontal distance = (10 m/s) x (5.0 s) = 50 meters .
===> Impact velocity = (10 m/s horizontally) + (49 m/s vertically)
= √(10² + 49²) = 50.01 m/s arctan(10/49)
= 50.01 m/s at 11.5° from straight down,
away from the base of the cliff.