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.
Answer:
<h3>a stationary electric charge, typically produced by friction, which causes sparks or crackling or the attraction of dust or hair.Static electricity has several uses, also called applications, in the real world. One main use is in printers and photocopiers where static electric charges attract the ink, or toner, to the paper. Other uses include paint sprayers, air filters, and dust removal. Static electricity can also cause damage.Static electricity is an imbalance of electric charges within or on the surface of a material. The charge remains until it is able to move away by means of an electric current or electrical discharge.</h3>
Answer:
a,)3.042s
b)4.173s
c)3.281s
Explanation:
For a some pendulum the period in seconds T can be calculated using below formula
T=2π√(L/G)
Where L = length of pendulum in meters
G = gravitational acceleration = 9.8 m/s²
Then we are told to calculate
(a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at 3.00 m/s2?
Since oscillations for this pendulum is located in the elevator that is accelerating upward at 3.00 then
use G = 9.8 + 3.0 = 12.8 m/s²
Period T=2π√(L/G)
T= 2π√(3/12.8)
T=3.042s
b) (b) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating downward at 3.00 m/s2?
G = 9.8 – 3.0 = 6.8 m/s²
T= 2π√(3/6.8)
T=4.173s
C)(c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 3.00 m/s2?
Net acceleration is
g'= √(g² + a²)
=√(9² + 3²)
Then period is
T=2π√(3/11)
T=3.281s
