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<span>torque = rF
= 0.1(10)
=1 Nm</span>
Answer:
The reading of the scale during the acceleration is 446.94 N
Explanation:
Given;
the reading of the scale when the elevator is at rest = your weight, w = 600 N
downward acceleration the elevator, a = 2.5 m/s²
The reading of the scale can be found by applying Newton's second law of motion;
the reading of the scale = net force acting on your body
R = mg + m(-a)
The negative sign indicates downward acceleration
R = m(g - a)
where;
R is the reading of the scale which is your apparent weight
m is the mass of your body
g is acceleration due to gravity, = 9.8 m/s²
m = w/g
m = 600 / 9.8
m = 61.225 kg
The reading of the scale is now calculated as;
R = m(g-a)
R = 61.225(9.8 - 2.5)
R = 446.94 N
Therefore, the reading of the scale during the acceleration is 446.94 N
Answer:
a) w = 4.24 rad / s
, b) α = 8.99 rad / s²
Explanation:
a) For this exercise we use the conservation of kinetic energy,
Initial. Vertical bar
Emo = U = m g h
Final. Just before touching the floor
Emf = K = ½ I w2
As there is no friction the mechanical energy is conserved
Emo = emf
mgh = ½ m w²
The moment of inertial of a point mass is
I = m L²
m g h = ½ (m L²) w²
w = √ 2gh / L²
The initial height h when the bar is vertical is equal to the length of the bar
h = L
w = √ 2g / L
Let's calculate
w = RA (2 9.8 / 1.09)
w = 4.24 rad / s
b) Let's use Newton's equation for rotational motion
τ = I α
F L = (m L²) α
The force applied is the weight of the object, which is at a distance L from the point of gro
mg L = m L² α
α = g / L
α = 9.8 / 1.09
α = 8.99 rad / s²
Answer:
the index of refraction of the second medium is lower
Explanation:
take an exaple of a light ray from air to water that is optically denser the ray is refracted to the normal thus lowering its index of refraction
