1. The velocity decreases, and the kinetic energy decreases.
2. An increase in temperature difference between the inside and outside of the building.
3. The total kinetic energy remains the same.
4. 76,761 J
5. The energy loss must increase.
Answer:
The final velocity of the vehicle is 10.39 m/s.
Explanation:
Given;
acceleration of the vehicle, a = 2.7 m/s²
distance moved by the vehicle, d = 20 m
The final velocity of the vehicle is calculated using the following kinematic equation;
v² = u² + 2ah
v² = 0 + 2 x 2.7 x 20
v² = 108
v = √108
v = 10.39 m/s
Therefore, the final velocity of the vehicle is 10.39 m/s.
210 Pb ---> -ie + 210 B:
84 8.3
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