Answer:
a. 120 W
b. 28.8 N
Explanation:
To a good approximate, the only external force that does work on a cyclist moving on level ground is the force of air resistance. Suppose a cyclist is traveling at 15 km/h on level ground. Assume he is using 480 W of metabolic power.
a. Estimate the amount of power he uses for forward motion.
b. How much force must he exert to overcome the force of air resistance?
(a)
He is 25% efficient, therefore the cyclist will be expending 25% of his power to drive the bicycle forward
Power = efficiency X metabolic power
= 0.25 X 480
= 120 W
(b)
power if force times the velocity
P = Fv
convert 15 km/h to m/s
v = 15 kmph = 4.166 m/s
F = P/v
= 120/4.166
= 28.8 N
definition of terms
power is the rate at which work is done
force is that which changes a body's state of rest or uniform motion in a straight line
velocity is the change in displacement per unit time.
The force exerted by student A with his scooter is 306 N and that of student B is 204 N.
<h3>
Force applied by each student</h3>
The force exerted by each student is calculated from Newton's second law of motion.
F = ma
where;
- m is mass
- a is acceleration
F(A) = 127.5 x 2.4
F(A) = 306 N
F(B) = 120 x 1.7
F(B) = 204 N
Thus, the force exerted by student A with his scooter is 306 N and that of student B is 204 N.
Learn more about force here: brainly.com/question/12970081
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Answer:
At a distance of 1376.49 candle emits 0.2 watt power
Explanation:
Distance between Sun and earth 
Sun emits a power of 
Power emitted by candle = 0.20 watt
We know that brightness is given by
So 
So at a distance of 1376.49 candle emits 0.2 watt power
Because when your running any experiment there will always be an experimental error so always be ready for it so that you can correct that error in your experiment
Answer:
Δx = 39.1 m
Explanation:
- Assuming that deceleration keeps constant during the braking process, we can use one of the kinematics equations, as follows:
where vf is the final velocity (0 in our case), v₀ is the initial velocity
(25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance
traveled since the brakes are applied.
- Solving (1) for Δx, we have:
