a force vector f has a magnitude of 12.0 n. it is oriented 60° to the left of the y ax what are its x and y components?
1 answer:
The force vector that has a magnitude of 12.0 N. and is oriented 60° to the left of the (y) has the followings components:
- v(x) =6 N
- v(y) = 10.39 N
To solve this exercise the formulas and procedures we will use are:
- v(x) = v * cosine (angle)
- v(y) = v * sine (angle).
Where:
- v= magnitude of the vector
- v(x) = component of the vector on the (x) axis
- v(y) = component of the vector on the (y) axis
- angle = angle
Information about the problem:
- angle = 60º
- v = 12.0 N
- v(x)= ?
- v(y)= ?
Applying the formula of the component of the vector in the (x) axis we have:
v(x) = v * cosine (angle).
v(x) = 12.0 N * cosine (60º)
v(x) =6 N
Applying the formula of the component of the vector in the (y) axis we have:
v(y) = v * sine (angle)
v(y) = 12.0 N * sine (60º)
v(y) = 10.39 N
<h3>What is a vector?</h3>It can be said to be a straight line described by a point (a) and (b) that has direction and sense.
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Answer:
The velocity of the plane at take off is 160 m/s.
The distance travel by the plane in that time is 3200 meter.
Explanation:
Given:
Acceleration, a = 4 m/s²
Time, t = 40 s
u = 0 i .e initial velocity
To Find:
velocity , v = ?
distance , s =?
Solution:
we have first Kinematic equation
v = u + at
∴ v = 0 + 4×40
∴ v = 160 m/s
Now by Third Kinematic equation
∴ s = 0 + 0.5 × 4× 40²
∴ s = 3200 meter
Answer:
F=5449 N
Explanation:
Work done is a product of force and displacement ie
Work done, W, = Force*Displacement
Power, P, is Work done/Time
where P is power, W is work done, F is force, S is displacement and t is time
In this case, F is the frictional force. Converting the power from hp to W, we multiply by 746 hence P=746*168=125328 W
Since displacement/time is velocity, then
P=FV where V is velocity in m/s
Making F the subject
F=5449 N