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.
Answer:
d = 0.38 m
Explanation:
As we know that the person due to the airbag action, comes to a complete stop, in 36 msec or less, and during this time, is decelerated at a constant rate of 60 g, we can find the initial velocity (when airbag starts to work), as follows:
vf = v₀ -a*t
If vf = 0, we can solve for v₀:
v₀ = a*t = 60*9.8 m/s²*36*10⁻³s = 21.2 m/s
With the values of v₀, a and t, we can find Δx, applying any kinematic equation that relates all of some of these parameters with the displacement.
Just for simplicity, we can use the following equation:
where vf=0, v₀ =21.2 m/s and a= -588 m/s².
Solving for d:
⇒ d = 0.38 m
Answer:
a). The velocity of the first log is -1.65 m/s.
(b). The velocity of the second log is 1.07 m/s.
Explanation:
Given that,
Mass of lumberjack M= 110 kg
Mass of log m= 206 kg
Final velocity = 3.09 m/s
(a). We need to calculate the velocity of the first log just before the lumberjack jumps off
Using conservation of momentum
Put the value into the formula
The velocity of the first log is -1.65 m/s.
(b). If the lumberjack comes to rest relative to the second log
We need to calculate the velocity of the second log
Put the value into the formula
The velocity of the second log is 1.07 m/s.
Hence, (a). The velocity of the first log is -1.65 m/s.
(b). The velocity of the second log is 1.07 m/s.