Answer:
ωB = 300 rad/s
ωC = 600 rad/s
Explanation:
The linear velocity of the belt is the same at pulley A as it is at pulley D.
vA = vD
ωA rA = ωD rD
ωD = (rA / rD) ωA
Pulley B has the same angular velocity as pulley D.
ωB = ωD
The linear velocity of the belt is the same at pulley B as it is at pulley C.
vB = vC
ωB rB = ωC rC
ωC = (rB / rC) ωB
Given:
ω₀A = 40 rad/s
αA = 20 rad/s²
t = 3 s
Find: ωA
ω = αt + ω₀
ωA = (20 rad/s²) (3 s) + 40 rad/s
ωA = 100 rad/s
ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s
ωB = ωD = 300 rad/s
ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s
Answer:
A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released? (Disregard air resistance.)
Show step by step please.
Note: The answer is given it's should be 1000 m ??
This what i can up with so see what it is kid
Explanation:
Gases at pressure are released by rockets as they travel towards space. According to Newton's third law, the combustion chamber's exhaust gases push the rocket with an accelerating force known as the thrust.
<h3>Explain exactly Newton's Third Law:</h3>
According to Newton's third law, if an object A pulls on an object B, then object B must exert an equal-sized and opposite-direction force on the first thing directed in the opposite direction. This law illustrates a symmetry in nature whereby forces always occur in pairs and whereby no body can exert a force without also being subjected to one.
<h3>What are Newton's 3rd law examples?</h3>
Action and response are always equal but always move in the opposite direction, according to Newton's third law of motion. A human walking on the ground, a hammer driving a nail, a magnet attracting a paper clip, and a horse pulling a cart are all examples of Newton's third rule of motion.
To know more about Newton Third Law visit:
brainly.com/question/23772134
#SPJ4
Answer:
a) 51.8 m, b) 27.4 m/s, c) 142 m
Explanation:
Given:
v₀ = 42.0 m/s
θ = 60.0°
t = 5.50 s
Find:
h, v, and H
a) y = y₀ + v₀ᵧ t + ½ gt²
0 = h + (42.0 sin 60.0) (5.50) + ½ (-9.8) (5.50)²
h = 51.8 m
b) vᵧ = gt + v₀ᵧ
vᵧ = (-9.8)(5.5) + (42.0 sin 60.0)
vᵧ = -17.5 m/s
vₓ = 42.0 cos 60.0
vₓ = 21.0 m/s
v² = vₓ² + vᵧ²
v = 27.4 m/s
c) vᵧ² = v₀ᵧ² + 2g(y - y₀)
0² = (42.0)² + 2(-9.8)(H - 51.8)
H = 142 m
