Question 1 (1 point)
1 answer:
Answer:
The resultant velocity of the plane relative to the ground is;
150 kh/h north
Explanation:
The flight speed of the plane = 210 km/h
The direction of flight of the plane = North
The speed at which the wind is blowing = 60 km/h
The direction of the wind = South
Therefore, representing the speed of the plane and the wind in vector format, we have;
The velocity vector of the plane = 210.
The velocity vector of the wind = -60.
Where, North is taken as the positive y or direction
The resultant velocity vector is found by summation of the two vectors as follows;
Resultant velocity vector = The velocity vector of the plane + The velocity vector of the wind
Resultant velocity vector = 210. + (-60.
) = 210.
- 60.
= 150.
The resultant velocity vector = 150.
Therefore, the resultant velocity of the plane relative to the ground = 150 kh/h north.
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Answer:
v = 9.04 m / s
Explanation:
For this exercise we can use the relation that the work of the non-conservative force (friction) is equal to the variation of the mechanical energy of the system.
W = Em_f - Em₀ (1)
Starting point. Lower slope
Em₀ = K = ½ m v²
highest point. Where is the skier at a height h
Em_f = U = m g h
The work of rubbing
W = -fr x
the negative sign is because the friction force opposes the movement.
Let's set a reference system where the x axis is parallel to the slope and the y axis is perpendicular
let's use trigonometry to break down the weight
cos θ = W_y / W
sin θ = Wₓ / W
W_y = W cos θ
Wₓ = W sin θ
Y axis
N - Wₓ = 0
N = mg sin θ
X axis
fr = m a
the friction force has the expression
fr = μ N
fr = μ mg sin θ
we look for the job
W = - μ mg sin θ x
where x is the distance along the slope
we substitute in 1
-μ mg sin θ x = mg h - ½ m v²
let's use trigonometry to find the distance x
tan 30 = h / x
x = h / tan 30
we substitute
- = m gh - ½ m v²
we use
tan 30 = sin30 / cos30
v² = 2g h + 2 μ g h cos 30
v =
let's calculate
v =
v = 9.04 m / s
Magnetic moment of loop is given by the formula
now magnetic field is given here
Now the torque on the loop is given by
so above is the torque on the loop.