A. <span>I .................
</span>
Answer:
-5 m/s
Explanation:
The linear velocity of B is equal and opposite the linear velocity of E.
vB = -vE
vB = -ωE rE
10 m/s = -ωE (12 m)
ωE = -0.833 rad/s
The angular velocity of E is the same as the angular velocity of D.
ωE = ωD
ωD = -0.833 rad/s
The linear velocity of Q is the same as the linear velocity of D.
vQ = vD
vQ = ωD rD
vQ = (-0.833 rad/s) (6 m)
vQ = -5 m/s
(a) The acceleration of the system is determined as 1.58 m/s².
(b) The relative weight of P is pounds is determined as 0.14 lb.
<h3>
Acceleration of the system</h3>
The acceleration of the system is calculated as follows;
W - T = m₂a --- (1)
T = m₁a ----(2)
μmgsinθ - m₁a = m₂a
(0.3 x 3 x 9.8 x sin40) - (0.4 + 0.2)a = 3a
5.67 - 0.6a = 3a
5.67 = 3.6a
a = 5.67/3.6
a = 1.58 m/s²
<h3>
Relative Weight of P</h3>
W = ma
W = 0.4 x 1.58
W = 0.632 N = 0.14 lb
Learn more about weight here: brainly.com/question/2337612
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Answer:
Scientific theories and laws develop from the acquisition of scientific knowledge.
Explanation:
t
Answer:
The force applied on one wheel during braking = 6.8 lb
Explanation:
Area of the piston (A) = 0.4 
Force applied on the piston(F) = 6.4 lb
Pressure on the piston (P) = 
⇒ P = 
⇒ P = 16 
This is the pressure inside the cylinder.
Let force applied on the brake pad = 
Area of the brake pad (
)= 1.7
Thus the pressure on the brake pad (
) = 
When brake is applied on the vehicle the pressure on the piston is equal to pressure on the brake pad.
⇒ P =
⇒ 16 =
⇒ = 16 ×
Put the value of we get
⇒ = 16 × 1.7
⇒ = 27.2 lb
This the total force applied during braking.
The force applied on one wheel = 
= 
= 6.8 lb
⇒ The force applied on one wheel during braking.
