Answer:
Explanation:
height of the building h = 15 m
initial vertical velocity u = 0
v² = u² + 2 g h
u = 0 , final velocity after falling height h
h = 15
v² = 2 x 9.8 x 15
v = 17.14 m /s
. Time of fall be t
h = 1/2 g t²
15 = 1/2 x 9.8 x t²
t = 1.749 s
This time will be required for truck to come under the building .
speed = distance / time
distance = speed x time
= 22 x 1.749
38.5 m
Truck should be at a distance of 38.5 m . when the stuntman starts falling .
If you shine a light straight down onto that vector, then the length of its shadow on the x-axis is
x-component = 88 · cosine(13°)
x-component = 88 · (0.9744)
<em>x-component = 85.74</em>
net force = applied force + gravity = 18N+(1.5kg)(-9.8m/s^2) = 3.3N
Force=mass times acceleration
acceleration=Force/mass
acceleration=3.3/1.5 m/s^2
acceleration=2.2m/s^s
Answer:
The difference between the cost of operating LED and incandescent bulb is $5.1
Explanation:
We are given the cost of electricity that is 12.75 cents per kWh. We want to find out the difference in the operating cost of an incandescent and LED bulb for a time period of 2,000 hours.
Since we are not given the rating of the incandescent bulb and LED bulb, we will assume their ratings.
For a light intensity of 250 Lumens;
The average rating of an LED bulb is approximately 5 Watts.
The average rating of an incandescent bulb is approximately 25 Watts.
Now lets find out the kWh of each bulb.
Energy = Power×Time
For LED bulb:
E = 5×2,000 = 10,000 Wh
Divide by 1000 to convert into kWh
E = 10,000/1000 = 10 kWh
Cost = 12.75×10 = 127.5 cents
Cost = $1.27
For Incandescent bulb:
E = 25×2,000 = 50,000 Wh
Divide by 1000 to convert into kWh
E = 50,000/1000 = 50 kWh
Cost = 12.75×50 = 637.5 cents
Cost = $6.37
Difference in Cost:
Difference = $6.37 - $1.27 = $5.1
Therefore, the difference between the cost of operating LED and incandescent bulb is $5.1.
Answer:
1.25Hz
Explanation:
For waves on a string, the second harmonic is obtained from;
2fo = 1/l √T/M
Where;
l = length of the string
M= mass in kilograms
T = tension in the string
2fo = 1/2√5/0.8
2f0 = 1.25Hz
