Work= (force)(distance)
178= m(9.81)x0.5
178=m(4.905)
178/4.905=m
His mass is 36.3 kg
Answer:
115, 80, 15m
Explanation
t1 = 14s
t2 = 18s
change in time = 4s (18-14)
r(final) = r(initial) + (average velocity) x (change in time)
multiply the average velocity with the change in time
= (4, 0, -3) x 4 = 16, 0, -12
now we'll add this value to the initial position of the car
(99, 80, 27)m + (16, 0, -12)m = (115, 80, 15)m
Answer:
A kid becoming an adult
A leg becoming bruised
A person's blood pressure raising because they are running
need a picture to answer specific questions.
Explanation:
All of these are physical changes. Hope that this helps you and have a great day :)
Answer:
Explanation:
Parameters given:
Mass of Puck 1, m = 1 kg
Mass of Puck 2, M = 1 kg
Initial velocity of Puck 1, u = 20 m/s
Initial velocity of Puck 2, U = 0 m/s
Final velocity of Puck 1, v = 5 m/s
Since we are told that momentum is conserved, we apply the principle of conservation of momentum:
Total initial momentum of the system = Total final momentum of the system
mu + MU = mv + MV
(1 * 20) + (1 * 0) = (1 * 5) + (1 * V)
20 = 5 + V
V = 20 - 5 = 15 m/s
Puck 2 moves with a velocity of 15 m/s
<span>Answer:
Therefore, x component: Tcos(24°) - f = 0 y component: N + Tsin(24°) - mg = 0 The two equations I get from this are: f = Tcos(24°) N = mg - Tsin(24°) In order for the crate to move, the friction force has to be greater than the normal force multiplied by the static coefficient, so... Tcos(24°) = 0.47 * (mg - Tsin(24°)) From all that I can get the equation I need for the tension, which, after some algebraic manipulation, yields: T = (mg * static coefficient) / (cos(24°) + sin(24°) * static coefficient) Then plugging in the values... T = 283.52.
Reference https://www.physicsforums.com/threads/difficulty-with-force-problems-involving-friction.111768/</span>
