Larger molecules will move slower and smaller molecules will move faster. Did this answer your question?
Answer:
<em> -18896.49 V/m</em>
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Explanation:
Distance between the two plates = 10 cm = 10 x 
m = 0.1 m
Also, one of the plates is taken as<em> zero volt.</em>
a. The potential strength between the zero volt plate, and 7.05 cm (0.0705 m) away is 393 V
b. The potential strength between the other plate, and 2.95 cm (0.0295 m) away is 393 V
<em>Potential field strength = -dV/dx</em>
where dV is voltage difference between these points,
dx is the difference in distance between these points
For the first case above,
potential field strength = -393/0.0705 = -5574.46 V/m
For the second case ,
potential field strength = -393/0.0295 = -13322.03 V/m
Magnitude of the field strength across the plates will be
-5574.46 + (-13322.03) = -5574.46 + 13322.03 =<em> -18896.49 V/m</em>
Answer: find the answer in the explanation as kinetic energy converts to potential energy.
Explanation:
Before the truck driver sees a dog running into the road, The mechanical energy state of the truck will be kinetic energy at maximum.
Immediately he applied the brakes, the mechanical energy of the truck will be combination of kinetic energy and potential energy.
The kinetic energy will gradually decrease as potential energy continue to increase till it reaches maximum potential energy.
The truck will come to a stop at maximum potential energy
Answer:
work done on the object by gravitational force = 0 joules
Explanation:
work = force × displacement
force = mass × acceleration
so,
work = mass × acceleration × displacement
we know that mass= 10 kg , gravitational acceleration= 9.8 
and displacement= 0 m since the object is not moving vertically.
so,
work = 10 × 9.8 × 0 = 0 joules
Answer:
when the object goes from the focal length to twice the focal length the image goes from infinity to twice the focal length, this image is real and inverted
Explanation:
Let's use the constructor equation to describe the image of a concave mirror
1 / f = 1 / p + 1q
where f is the focal length, p and q the distance to the object and the image, respectively
1 /q = 1/f - 1/p
tell us that the image is between the focal and twice the focal, let's calculate the position of the image
for both ends
case 1, distance to the object p = f
1 / q = 1 / f -1 / f
1 / q = 0
q = ∞
the image is in infinity
case2, distance to object p = 2f
1 / q = 1 / f - 1 / 2f
1 / q = 1 / 2f
q = 2f
the image is twice the focal length, the object and the image are at the same point
therefore the image when the object goes from the focal length to twice the focal length the image goes from infinity to twice the focal length, this image is real and inverted
