The car at 60 kph has 9 times more kinetic energy than the car traveling at 20 kph. This assumes that both cars have the same mass. Kinetic energy depends on the square of thee speed so if one car is going 3 times faster, its kinetic energy will be 3^2 ( = 9 ) greater. The car going at 60 kph will have 4 times the KE of the car going at 30 kph ( again assuming that the cars have the same mass.)
A) See ray diagram in attachment (-6.0 cm)
By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:
where
q is the distance of the image from the lens
f = -10 cm is the focal length (negative for a diverging lens)
p = 15 cm is the distance of the object from the lens
Solving for q,
B) The image is upright
As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:
where 
are the size of the image and of the object, respectively.
Since q < 0 and p > o, we have that 
, which means that the image is upright.
C) The image is virtual
As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.
This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual
A) No, the equations presented above are the product of the derivation of position and velocity when the acceleration is constant.
The equations change to polynomial function of the second degree for the description of the acceleration when described as a function of time.
B) Yes, when the acceleration is zero it is concluded that the velocity is constant, therefore they could be used to describe the position as a function of the change in velocity.
Answer:
Given: V = 220V, Pmin = 360W, Pmax = 840W
For minimum heating case:
We know that
Pmin = VI
360 = 220 X I
I = 1.63 amp
R = V/I
R = 220/1.63
R = 134.96ohms
For maximum heating case:
We know that
Pmax = VI
840 = 220 X I
I = 3.81 amp
R = V/I
R = 220/3.81
R = 57.74 ohms
Answer:
31302 Volts and 55/111 Amps (≈0.5)
Explanation:
Secondary voltage / 141 = 1110 / 5
Secondary voltage = (1110*141) / 5
Secondary voltage = 31302
Amperage = 110/ (31302/141) = 55/111
