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If the gravitational pull is zero and I multiply by mass I get a zero
Answer: minimum speed of launch must be 7.45m/s
Explanation:
Given the following:
Height or distance (s) = 2.83m
The final velocity(Vf) at maximum height = 0
Upward motion, acceleration due to gravity(g) us negative = -9.8m/s^2
From the 3rd equation of motion:
V^2 = u^2 - 2gs
Where V = final velocity
u = initial velocity
Therefore, u = Vi
u = √Vf^2 - 2gs
u = √0^2 - 2(-9.8)(2.83)
u = √0 + 55.468
u = √55.468
u = 7.4476 m/s
u = 7.45m/s
Answer:
The angular speed after 6s is 
.
Explanation:
The equation
relates the moment of inertia 
of a rigid body, and its angular acceleration 
, with the force applied 
at a distance 
from the axis of rotation.
In our case, the force applied is 
, at a distance 
, to a ring with the moment of inertia of 
; therefore, the angular acceleration is
Therefore, the angular speed 
which is
after 6 seconds is
According to the statement " Collision <span>between two bodies in which the total kinetic energy of the two bodies after the collision is equal to their total kinetic energy before the collision."
The best answer is :
Option A " </span><span>BODY A COMES TO REST BODY B STARTS MOVING WITH INITIAL VELOCITY OF BODY A "</span>
