Work = (force) x (distance.
The force required to lift the load is its weight.
Weight = (mass) x (gravity)
so Work = (mass) x (gravity) x (distance)
Now Power = (work) / (time)
so Power = (mass) x (gravity) x (distance) / (time)
= (700kg) x (9.8 m/s²) x (2 m) / (0.4 sec)
= ( 700 x 9.8 x 2) / (0.4) (kg-m²/sec²) / (sec)
= ( 34,300 ) (joule) / (sec)
= 34,300 watts .
This is one of those exercises where the math and the physics
are air-tight and bullet-proof but the answer is absurd.
34,300 watts is about 46 horsepower. I don't care how many
Wheaties Power Lifter Paul had for breakfast today, he is NOT
snatching a barbell that weighs 1,543 pounds (0.77 ton !)
to the height of the top of his head in less than 1/2 second !
Answer:
Velocity, V = 3t²- 28t+6
Displacement, s = t³ - 14t²+6t -8
At t = 5.8 s
s = -249.05 m
v = -55.48 m/s
At t = 12.7 s
s = -141.48 m
v = 134.27 m/s
Explanation:
We have acceleration of a particle is given by a = 6t - 28
Velocity
At t = 0 we have v₀ = 6 m/s
v₀ = 6 = 3 x 0 ²-28 x 0+C
C = 6
So velocity, V = 3t² - 28t+6
Displacement
At t = 0 we have s₀ = -8 m
s₀ = -8 = 0³ + 14 x 0² + 6 x 0 + C
C = -8
So displacement, s = t³ - 14t²+6t -8
At t = 5.8 s
s = 5.8³ - 14 x 5.8²+6 x 5.8 - 8 = -249.05 m
v = 3 x 5.8² - 28 x 5.8 + 6 = -55.48 m/s
At t = 12.7 s
s = 12.7³ - 14 x 12.7²+6 x 12.7 - 8 = -141.48 m
v = 3 x 12.7² - 28 x 12.7 + 6 = 134.27 m/s
Average speed = change in position / total time = 115 - 10 / 1.75 = 60 miles/hr. D is the correct option.
Answer:
9.8 m / s^2
Explanation:
Assuming free fall====> there is no initial downward/upward velocity
Assuming metric units 78.4<u> m/s </u>
vf = a t
78.4 = a (8) shows a = 9.8 m/s^2
Answer:
82.7 m
Explanation:
u= 22m/s
a= 2.4 m/s^2.
t= 3.2 secs
Therefore the distance travelled can be calculated as follows
S= ut + 1/2at^2
= 22 × 3.2 + 1/2 × 2.4 × 3.2^2
= 70.4 + 1/2×24.58
= 70.4 + 12.29
= 82.7 m
Hence the distance travelled by the truck is 82.7 m