If riding a lawnmower engine exerts 19 hp in one minute to move the lawnmower how much work is done
1 answer:
Answer:
the work done by the lawnmower is 236.14 J.
Explanation:
Given;
power exerted by the lawnmower engine, P = 19 hp
time in which the power was exerted, t = 1 minute = 60 s.
1 hp = 745.7 watts
The work done by the lawnmower is calculated as follows;
Therefore, the work done by the lawnmower is 236.14 J.
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Answer:
Explanation:
Energy of signal being radiated per second on all sides = 71 x 10³ J .
At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.
So energy crossing per unit area
=
= 11.67 x 10⁻² Wm⁻²s⁻¹.
This is the intensity of the signal.
At 2200 m this intensity will further reduce by 100 times
So there it becomes equal to
11.67 x 10⁻⁴ Wm⁻² s⁻¹.
Answer:
32 seconds
Explanation:
m1 = 80 kg
m2 = 10 kg
v2 = 5m/s
According to the property of conservation of momentum, assuming that both you and the bag are stationary before the safety rope comes lose:
Since the space station is 20 meters away, the time taken to reach it is given by:
It takes you 32 seconds to reach the station.
Answer:
a) -1.25 rev/s² and 23.3 rev
b) 2.67s
Explanation:
a) ω = (500 rev/min)(1min/ 60s) => 8.333 rev/s
ω= (200 rev/min)(1min/ 60s) => 3.333rev/s
time 't'= 4 s
angular acceleration 'α'=?
constant angular acceleration equation is given by,
ω= ω
+ α
t
α= (ω
- ω
)/t => (3.333-8.333)/4
α= -1.25 rev/s²
θ-θ = ω
t + 1/2α
t²
=(8.333)(4) + 1/2 (-1.25)(4)²
=23.3 rev
b) ω=0 (comes to rest)
ω = 3.333 rev/s
α= -1.25 rev/s²
ω= ω
+ α
t
t= (ω - ω
)/α
=> (0- 3.333)/-1.25
t= 2.67s