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:(10.69, 11.436)
Explanation:
Given
initial height of ball is 2 m
height of basket is 3.05 m
Launching angle
y=1.05
equation of trajectory of ball is given by
for x=12.27
u=10.69
for x=11.73
u=11.436 m/s
Thus range of speed is (10.69, 11.436)
Average speed = total distance travelled ÷ total time taken
AS = (75km + 68km) ÷ (1hr + 2hr)
As = 143km ÷ 3hr
AS = 47.66667 km/hr
AS = 47.7 km/hr (3sf)
