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:
Explanation:
Given that,
Magnetic field of 0.24T
B = 0.24T
Field perpendicular to plane i.e 90°
Rate of decrease of length of side of square is 5.4cm/s
dL/dt = 5.4cm/s = 0.054m/s
Since it is decreasing
Then, dL/dt = -0.054m/s
When L is 14cm, what is the EMF induced?
L = 14cm = 0.14m
EMF is give as
ε = - dΦ/dt
Where flux is given as
Φ = BA
Where A is the area of the square
A = L²
Then, Φ = BL²
Substituting this into the EMF
ε = - dΦ/dt
ε = - d(BL²)/dt
B is constant
ε = - Bd(L²)/dt
ε = -2BL dL/dr
ε = -2 × 0.24 × 0.14 × -0.054
ε = 3.63 × 10^-3 V
ε = 3.63mV
Answer:
the force of attraction between the two charges is 3.55 N.
Explanation:
Given;
first charge carried by the object, q₁ = 15.5 µC
second charge carried by the q₂ = -7.25 µC
distance between the two charges, r = 0.525 m
The force of attraction between the two charges is calculated as;

Therefore, the force of attraction between the two charges is 3.55 N.
Answer: 4,438.96m
Explanation:
(kindly find attachment below)
From the attachment below, it can be seen that the resultant displacement and the other 2 displacements form a right angle triangle, with A+B as the hypotenus, 3.2km as the opposite and the displacement B as the adjacent.
By using phythagoras theorem
H² = O² + A²
(5.38)² = (3.20)² + B²
28.944 = 10.24 + B²
B² = 28.944 - 10.24
B² = 18.7044
B = √18.7044
B = 4.439km to meter is 4.439 * 1000 = 4,438. 96m
Fe=K Q1/Q2/d2
Q1 is the first charge
Q2 is the second charge
d is the distance
K= 9x10^9 NM^2/C2
Now let’s plug the numbers
Fe=9x10^9NM^2/C2 (2x10^-4C)(8x10^-4C) / (0.3m^2) you notice we took away the negative charges when we plugged the charges
Ok now we notice that we have C2 which is C to the power 2 we can write it as C^2 and we have two CSU’s beside each one of the charges we can get rid of them all by curtailment
And we can curtailment the M^2and the other M^2
Now we left with only 9x10^9N (2x10^-4)(8x10^-4)/ 0.3
Let’s multiply the (9)(2)(8)=144
And add the exponents (9)+(-4)+(-4)=1
So now we got 144x10N divide by the distance which is 0.3
144x10N / 0.3 = 4800N
Hope it helps u understand :)