Answer:
S = V0 t + 1/2 a t^2
S = 5 m/s * 300 s + 1/2 * 1.2 m/s * (300 s^2)
S = 1500 m + .6 * 90000 m = 55,500 m
Check: V0 = 5 m/s
V2 = V0 + a t = 5 + 1.2 * 300 = 365 m/s
Vav = (V1 + V2) / 2 = (5 + 365) / 2 = 185 m/s (note uniform motion)
S = 185 * 300 = 55,500 m
We calculated V2 above at 365 m/s the speed after 300 sec
Answer: the external agent must do work equal to -1.3 × 10⁻⁸ J
Explanation:
Given that;
Mass M1 = 7.0 kg
r = 3.0/2 m = 1.5 m
Mass M2 = 21 kg
we know that G = 6.67 × 10⁻¹¹ N.m²/kg²
work done by an external agent W = -2GM2M1 / r
so we substitute
W = (-2 × 6.67 × 10⁻¹¹ × 21 × 7) / 1.5
W = -1.96098 × 10⁻⁸ / 1.5
W = -1.3 × 10⁻⁸ J
Therefore the external agent must do work equal to -1.3 × 10⁻⁸ J
. In single particle problem whole mass is concentrated at a single point so it has a single displacement, single velocity and single acceleration. while, in rigid body mass is distributed
"<span>The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are" is the type of image among the choices given in the question that would be projected. The correct option among all the options that are given in the question is the first option. I hope it helps you.</span>
