Answer: The wheel's average rotational acceleration is -0.4 radians per second squared (rad/s^2)
Explanation: Please see the attachments below
Answer:
<h2>a) 496N</h2><h2>b) 50.56kg</h2><h2>c) 1.90m/s²</h2>
Explanation:
According to newton's secomd law, ∑F = ma
∑F is the summation of the force acting on the body
m is the mass of the body
a is the acceleration
Given the normal force when the elevator starts N1 = 592N
Normal force after the elevator stopped N2 = 400N
When the elevator starts, its moves upward, the sum of force ∑F = Normal (N)force on the elevator - weight of the person( Fg)
When moving up;
N1 - Fg = ma
N1 = ma + Fg ...(1)
Stopping motion of the elevator occurs after the elevator has accelerates down. The sum of forces in this case will give;
N2 - Fg = -ma
N2 = -ma+Fg ...(2)
Adding equation 1 and 2 we will have;
N1+N2 = 2Fg
592N + 400N = 2Fg
992N 2Fg
Fg = 992/2
Fg = 496N
The weight of the person is 496N
<em>\b) To get the person mass, we will use the relationship Fg = mg</em>
g = 9.81m/s
496 = 9.81m
mass m = 496/9.81
mass = 50.56kg
c) To get the magnitude of acceleration of the elevator, we will subtract equation 1 from 2 to have;
N1-N2 = 2ma
592-400 = 2(50.56)a
192 = 101.12a
a = 192/101.12
a = 1.90m/s²
I think the answer is discovery.
Answer:
D. 49N
Explanation:
Because there is no acceleration the sum of the forces in the x-direction is zero. Your weight force is mass x gravity which is 98N. The x-component of which is sin(30). 98sin30 = 49N in the negative x-direction. -49N + The force applied by the worker must equal zero. So, the force applied by the worker must be 49N.
