A vector has an x-component of 19.5m and a y-component of 28.4m. Find the magnitude and direction of the vector
1 answer:
Answer:
34.5 m at 55.5 degrees above the positive x-axis
Explanation:
Resolving a vector means finding its components along two perpendicular axis: most commonly, they are chosen as the x and y axis.
In this problem, we have a vector whose components are:
x -component: 19.5 m
y- component: 28.4 m
The two components are perpendicular to each other: this means that we can find the magnitude of the vector by using the Pythagorean theorem
The direction, instead, can be found by using the following formula:
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The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:
We can apply the first Newton's law in x and y-direction.
If we do a free body diagram of the system we will have:
x-direction
All the forces acting in this direction are:
(1)
Where:
- T(1) is the tension due to the rope 1
- T(2) is the tension due to the rope 2
Here we just conclude that T(1) = T(2)
y-direction
The forces in this direction are:
(2)
Here W is the weight of the steel beam.
We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.
Knowing that T(1) = T(2) and W = mg, we have:
T(1) must be equal to 5479 N, so we have:
Therefore, the maximum angle allowed is θ = 37.01°.
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I hope it helps you!
The initial potential energy of the wagon containing gold boxes will enable
it roll down the hill when cut loose.
The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.
Reasons:
Mass wagon and gold = 166 kg
Location of the wagon = 77 meters up the hill
Slope of the hill = 8°
Location of the rangers = 41 meters from the canyon
Mass of Lone Ranger, m₁ = 65 kg
Mass of Tonto m₂ = 66 kg
Solution;
Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m
Potential energy = m·g·h
Where;
g = Acceleration due to gravity ≈ 9.81 m/s²
Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912
Potential energy of wagon, P.E. ≈ 17457.0912 J
By energy conservation, P.E. = K.E.
Where;
v = The velocity of the wagon a the bottom of the cliff
Therefore;
Velocity of the wagon, v ≈ 14.5 m/s
Momentum = Mass, m × Velocity, v
Initial momentum of wagon = m·v
Final momentum of wagon and ranger = (m + m₁ + m₂)·v'
By conservation of momentum, we have;
m·v = (m + m₁ + m₂)·v'
Which gives;
The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s
The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the
gold and jump out of the wagon before the wagon heads over the cliff.
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