-- First question . . . first answer choice
-- Second question . . . second answer choice
-- Third question . . . third answer choice
Answer:
can you clear your question I can't understand
Answer:
Given that,
- Power = 2000 W
- time = 60 seconds
- distance= 10m
Power = work done ÷ time
Here, since the movement is vertical, w = mgh
So,
Power = mgh÷t
2000 = (m × 9.8 ×10) ÷ 60
m = (2000 ×60) ÷98
m = 1224.5kg
Answer:
The work done by the gravel to stop the truck is 520.44 kJ
Explanation:
<u>Step 1</u>: Data given
Mass of the truck = 3047.8 kg
The ramp has an angle of 9.5 °
Velocity of the truck = 20.68 m/s
distance = 26.6 meters
<u>Step 2:</u> Calculate initial kinetic energy
sin 9.5° = 0.165
h = ℓ*sin 9.5° = 26.6*0.165= 4.39 m
Ek = 1/2m*Vo² = 1/2*3047.8*20.68² = 651714.7 Joule = 651.7 kJ = initial kinetic energy
<u>Step 3: </u>Calculate potential energy
Epot = U = m*g*h = 3047.8*9.81*4.39 = 131256.25 Joule = 131.26 kJ
<u>Step 4:</u> What work is done by the truck on the gravel?
Frictional energy Ef = 651.7 kJ - 131.26 kJ = 520.44 kJ
Answer:
lateral splaying of echoes in the far field can be improved by Increasing the maximum number of transmit focal zones and optimize their location.