(5 bulbs) x (25 watt/bulb) x (6 hour/day) x (30 day/month) =
(5 x 25 x 6 x 30) watt-hour/month =
22,500 watt-hour/month .
The most common unit of electrical energy used for billing purposes
is the 'kilowatt-hour' = 1,000 watt-hours .
22,500 watt-hour/month = <em>22.5 kWh/month</em>.
(22.5 kWh/month) x (1.50 Rs/kWh) = <em>33.75 Rs / month
</em>
Answer:
B) Depends on launch speed.
Explanation:
This is true when considering the basketball player in a given basketball game. The speed at which the player jumps up is a strong factor which determines the acceleration of his acceleration. The direct co-relation show that, speed and acceleration of the basketball player are interrelated.
The nall rolled 20 cm in 5 seconds
Answer:
h = 618.64 m
Explanation:
First we need to calculate the height gained by rocket while the fuel is burning. We use 2nd equation of motion for that purpose:
h₁ = Vit + (1/2)at²
where,
h₁ = height gained during the burning of fuel
Vi = Initial Velocity = 0 m/s
t = time = 7 s
a = acceleration = 8 m/s²
Therefore,
h₁ = (0 m/s)(7 s) + (1/2)(8 m/s²)(7 s)²
h₁ = 196 m
Now we use 1st equation of motion to find final speed Vf:
Vf = Vi + at
Vf = 0 m/s + (8 m/s²)(7 s)
Vf = 56 m/s
Now, we calculate height covered in free fall motion. Using 3rd equation of motion:
2ah₂ = Vf² - Vi²
where,
a = - 3.71 m/s²
h₂ = height gained during free fall motion = ?
Vf = Final Velocity = 0 m/s (since, rocket will stop at highest point)
Vi = 56 m/s
Therefore,
(2)(-3.71 m/s²)h₂ = (0 m/s)² - (56 m/s)²
h₂ = 422.64 m
So the total height gained will be:
h = h₁ + h₂
h = 196 m + 422.64 m
<u>h = 618.64 m</u>
