Find the pictures in the attachment
Vector A⃗ has a magnitude of 3.00 and is directed parallel to the negative y-axis and vector B⃗ has a magnitude of 3.00 and is d
irected parallel to the positive y-axis. Determine the magnitude and direction angle (as measured counterclockwise from the positive x-axis) of vector C⃗ , if C⃗ =A⃗ −B⃗ .
C = 6.00; θ = 270˚
C = 3.00; θ = 270˚
C = 6.00; θ = 90˚
C = 3.00; θ = 90˚
C = 6.00; θ = 270˚
C = 3.00; θ = 270˚
C = 6.00; θ = 90˚
C = 3.00; θ = 90˚
2 answers:
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Answer:
(a) The magnitude of the lift force is 52144.71 N, approximately.
(b) The magnitude of the air resistance force opposing the movement is 17834.54 N, approximately.
Explanation:
Since the helicopter is moving horizontally at a constant velocity, we can assume that the net force acting on it is zero, then
(a) in the vertical direction we have
.
(b) Now horizontally,
Newton's third law says:
"<span>For every action, there is an equal and opposite reaction. ".
So, the force that Tom does on the sister is equal to force the sister applies on Tom:
</span>
<span>where the label "t" means "on Tom", while the label "s" means "on the sister".
From Newton's second law, we also know
</span>
where m is the mass and a the acceleration. <span>so we can rewrite the first equation as
</span>
<span>And find Tom's acceleration:
</span>
<span>
</span>
"<span>For every action, there is an equal and opposite reaction. ".
So, the force that Tom does on the sister is equal to force the sister applies on Tom:
</span>
<span>where the label "t" means "on Tom", while the label "s" means "on the sister".
From Newton's second law, we also know
</span>
where m is the mass and a the acceleration. <span>so we can rewrite the first equation as
</span>
<span>And find Tom's acceleration:
</span>
</span>
Answer:
a = 0.5 m/s²
Explanation:
Applying the definition of angular acceleration, as the rate of change of the angular acceleration, and as the seats begin from rest, we can get the value of the angular acceleration, as follows:
ωf = ω₀ + α*t
⇒ ωf = α*t ⇒ α = =
The angular velocity, and the linear speed, are related by the following expression:
v = ω*r
Applying the definition of linear acceleration (tangential acceleration in this case) and angular acceleration, we can find a similar relationship between the tangential and angular acceleration, as follows:
a = α*r⇒ a = 0.067 rad/sec²*7.5 m = 0.5 m/s²