Answer:
The average power the student expended to overcome gravity is 560W, Watts is a units of work it is Joules /time or kg*
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Explanation:
Weight = 700N
The power is the work in (Joules) or (N*m) in a determinate time (s) to get Watts (W) units for work
Answer:
(a) 
Explanation:
We have given speed of light 
Given time = 1 year =365 days
We know that in 1 minute = 60 sec
1 hour = 60×60 = 3600 sec
In one day = 24 hour = 24×60×60=86400 sec
So in 365 days = 365×86400
We know that distance = speed ×time
We have given that 1 mi = 1609 m
So 
So option (a) is correct
Answer:
2.52 m/s
Explanation:
When the man takes a step, his foot is stationary while his body revolves around it. At the point when his body is directly above his foot, there will be no normal force at his maximum speed.
Sum of the forces in the radial direction:
∑F = ma
mg = m v² / r
g = v² / r
v = √(gr)
Given that r = 0.650 m:
v = √(9.8 m/s² × 0.650 m)
v = 2.52 m/s
Answer:
Explanation:
Let the velocity of projectile be v and angle of throw be θ.
The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m
considering its vertical displacement
h = - ut +1/2 g t²
100 = - vsinθ x 5 + .5 x 9.8 x 5²
5vsinθ = 222.5
vsinθ = 44.5
It covers 160 horizontally in 5 s
vcosθ x 5 = 160
v cosθ = 32
squaring and adding
v²sin²θ +v² cos²θ = 44.4² + 32²
v² = 1971.36 + 1024
v = 54.73 m /s
Answer:
0.94 m³/s
Explanation:
From the question given above, the following data were obtained:
Air flow (in ft³/min) = 2×10³ ft³/min
Air flow (in m³/s) =.?
Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:
35.315 ft³/min = 1 m³/min
Therefore,
2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min
2×10³ ft³/min = 56.63 m³/min
Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:
1 m³/min = 1/60 m³/s
Therefore,
56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min
56.63 m³/min = 0.94 m³/s
Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.
