The train is moving at 50 m/s and Emma is walking down the aisle with 1 m/s speed in the same direction of train. The relative velocity of Emma with respect to other passengers pf the train would be 1 m/s. This is because, the train is not moving relative to them and only emma is moving at 1 m/s. If a person observes from outside, Emma would have (50 +1) m/s = 51 m/s velocity.
relative velocity when two objects are moving in same direction as oberved from outside observer:
relative velocity when two objects are moving in the opposite directionas oberved from outside observer:
A 5.00 A current runs through a 12 gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5*10^28 free electrons per cubic metre.
a) How many electrons pass through the light bulb each second?
b) What is the current density in the wire? (answer in A/m^2)
<span>c) At what speed does a typical electron pass by any given point in the wire? (answer in m/s)
</span>a) 5.0 A = 5.0 C/s
. Number of electrons in 5.0C = 5.0 / 1.60^-19 = 3.125^19
. 5.0 A ►= 3.125^19 electrons/s
b) A/m² = 5.0 / π(1.025^-3 m)² .. .. ►= 1.52^6 A/m²
c) Charge density (q/m³) = 8.50^28 e/m³ x 1.60^-19 = 1.36^10 C/m³
(q/m³)(m²)(m/s) = q/s (current i in C/s [A])
(m²) = Area
(m/s) = mean drift speed
(q/m³)(A)(v) = i
v = i.[(q/m³)A]ˉ¹
<span>v = 5.0 [1.36^10 * π(1.025^-3 m)²]ˉ¹.. .. ►v = 1.10^-4 m/s</span>
Answer:
<h3>13,976.23Joules</h3>
Explanation:
Workdone by the rope is expressed using the formula;
W = Fd sin(theta)
F is the tension in the rope = 180
d is the displacement = 300m
theta is the angle of inclination = 15°
Substitute the given parameters into the formula;
W = 180(300)sin15
W = 54000sin 15
W = 13,976.23
Hence the workdone by the rope is 13,976.23Joules
(a) -48.0 cm, diverging
We can use the lens equation:
where
f is the focal length
p = 16.0 cm is the object distance
q = -12.0 cm is the image distance (with a negative sign because the image is on the same side as the object, so it is virtual)
Solving for f, we find the focal length of the lens:
The lens is diverging, since the focal length is negative.
(b) 6.38 mm, erect
We can use the magnification equation:
where
y' is the size of the image
y = 8.50 mm is the size of the object
Substituting p and q that we used in the previous part of the problem, we find y':
and the image is erect, since the sign is positive.
(c)
See attached picture.
