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:
patient receiving drug 25 MCG/minute
Explanation:
given data
infusing = 15 ml/hr
drug = 50 mg
D5W = 500 ml
to find out
How many MCG/minute
solution
we know infusing rate is 15 ml/hr = 0.25 ml/min
so 0.25 ml drug content = 50 /500 × 0.25
0.25 ml drug content = 0.025 mg
so here
rate of drug will be 0.025 mg
rate of drug = 0.025 mg = 25 × gm/min
rate of drug = 25 MCG/minute
so patient receiving drug 25 MCG/minute
To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.
From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by
Where,
Final velocity
Initial Velocity
a = Acceleration
x = Displacement
Acceleration can be expressed in terms of the drag coefficient by means of
Frictional Force
Force by Newton's second Law
Where,
m = mass
a= acceleration
Kinetic frictional coefficient
g = Gravity
Equating both equation we have that
Therefore,
Re-arrange to find x,
The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.