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:
hmm
Explanation:
By increasing the number of turns in the coil, strength of magnetic field, speed of rotation of the coil in the magnetic field and by decreasing the distance between the coil and the magnet the magnitude of the induced e.m.f. can be increased in generator/dynamo.
6 . . . . . a crest
7 . . . . . the amplitude
8 . . . . . the wavelength
9 . . . . . a trough