Answer:
B
Explanation:
Work done can be said to be positive if the applied force has a component to be in the direction of the displacement and when the angle between the applied force and displacement is positive.
In 1 and 2 work done is positive
Answer:
0.0002°, 0.1691°, 0.338°
Explanation:
Difference between the two line = 5.97 * 10-⁸m
d = 1 / N
N = 5.0 * 10³
d = 2.0 * 10⁴m
nL = Nsin¤
For first order
588.995 * 10-⁹ = 2.0 * 10-⁴ sin ¤
Sin¤ = 2.944*10-³
¤ = sin-¹ 0.002944
¤ = 0.1687°
First order ¤ =
Sin-¹(589.592*-⁹ / 2.0 * 10-⁴)
Sin-¹ (0.002947) = 0.1689°
Angular separation = 0.1689 - 0.1687 = 0.0002°
Second order ¤ = sin-¹ [2 (589.59*10-⁹ / 2.0*10-⁴)] = sin-¹ (0.005895)
Second order ¤ = 0.3378°
Angular difference = 0.3378° - 0.1687° = 0.1691°
Third order ¤ = sin-¹ [3(589.59*10-⁹ /2.0*10-⁴] = 0.5067°
Angular difference = 0.5067° - 0.1687° = 0.338°
If you drop a <span>6.0x10^-2 kg ball from height of 1.0m above hard flat surface, and a</span>fter the ball had bounce off the flat surface, the kinetic energy of the ball would be mgh - 0.14 = 0.45.
A. 4 cm behind the mirror
<span> For any mirror, </span><span><span>so</span><span>si</span>=<span>f^2</span></span>. Therefore, by plugging in the values, you get <span>18<span>si</span>=144. 144/18 = 8, so </span><span><span>si</span>=8</span>
<span> The focal point is located 12 cm from the mirror, and </span><span>si</span><span> is the distance of the image from the focal point, so the image is 4 cm from the mirror. The mirror is convex, so then the focal point and the image are both behind the mirror.</span>
