A package is dropped from a helicopter moving upward at 15m/s. If it takes 10 s before the package strikes the ground, how high
2 answers:
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Answer:
a) θ = 58.3º
b) vfh = 13.7 m/s
c) g = -9.8 m/s2
d) h = 22.2 m
e) vfb = 15.5 m/s
Explanation:
a)
- Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
- Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
- vₓ₀ = v * cos θ (1)
- where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
- Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:
- Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:
- ⇒θ = cos⁻¹ (0.526) = 58.3º (4)
b)
- At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
- Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
- vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)
c)
- At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)
d)
- Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:
- Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
- v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
- Replacing (7) in (6), we get:
e)
- When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
- The horizontal component, since it keeps constant, is just v₀x:
- v₀ₓ = 13.7 m/s
- The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:
- Replacing by the time t (a given), g, and v₀y from (7), we can solve (9) as follows:
- Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:
The thermal expansion of the materials allows to find the deflection of the bimetallist strip is Δy = 3.48 cm
given paramers
* Bimetallic brass / steel tape
* Initial temperature, room temperature T = 20ºC
* Final temperature, boiling water = 100ºC
* initial length L₀ = 222mm (1cm / 10mm) = 22.2cm
* thickness of bimetallic tape e = 0.036 inch (2.54 cm/1 inch) = 0.0914 cm
to find
* perpendicular deviation or deflection (Δy)
Thermal expansion is the phenomenon of change in the length of a body due to the change in temperature, due to the increase in the length of the atomic and molecular bonds, macroscopically it is described by
ΔL = α L₀ ΔT
ΔL and ΔT are the variation of the length and temperature respectively, L₀ is the initial length and α the coefficient of expansion ends.
In this case we have a strip formed by two materials with different coefficient of thermal expansion,
Brass α_{brass} = 19 10⁻⁶ ºC⁻¹
Steel α_{steel} = 11 10⁻⁶ ºC⁻¹
In the attached we can see a diagram of the process, as the temperature increases, the material with greater thermal expansion lengthens more, so the system must curve towards the center of the material with less
thermal expansion. Let's find the length of the strip for each material
brass L_{f brass} - L₀ = α_{brass} L₀ ΔT
Steel L_{f steel} - L₀ = \alpha_{steel} L₀ ΔT
Note that the initial length is the same for the two materials and that the strip is in thermal equilibrium at room temperature.
If we assume that we have an arc of circumference, we can write the length of the arc
θ = L / r
where θ is the angle in radines, L the length of the arc and r the radius of curvature, let's write this equation for each material
brass L_{f \ brass} =θ r₁
steel L_{f \ steel} = θ r₂
we substitute in our equations
θ r₁ - L₀ = α_{brass} L₀ ΔT
θ r₂ - L₀ = α_{steel} L₀ ΔT
let's subtract the two equations
θ (r₁- r₂) = L₀ ΔT (α_{brass} - α_{steel})
the thickness of the strip is
e = r₁ -r₂
θ =
we calculate
θ =
θ = 0.155 rad
Let's use trigonometry to find the perpendicular deflection
tan θ = Δy / L₀
Δy = L₀ tan θ
Δy = 22.2 tan 0.155
Δy = 3.48 cm
Using the thematic expansion of the two materials we find the deflection of the bimetallist strip is 3.38 cm
Learn more about thermal expansion here: brainly.com/question/18717902
For a cylinder that has both ends open resonant frequency is given by the following formula:
Where n is the resonance node, v is the speed of sound in air and L is the length of a cylinder.
The fundamental frequency is simply the lowest resonant frequency.
We find it by plugging in n=1:
To find what harmonic has to be excited so that it resonates at f>20Hz we simply plug in f=20 Hz and find our n:
We can see that any resonant frequency is simply a multiple of a base frequency.
Let us find which harmonic resonates with the frequency 20 Hz:
Since n has to be an integer, final answer would be 323.
Where n is the resonance node, v is the speed of sound in air and L is the length of a cylinder.
The fundamental frequency is simply the lowest resonant frequency.
We find it by plugging in n=1:
To find what harmonic has to be excited so that it resonates at f>20Hz we simply plug in f=20 Hz and find our n:
We can see that any resonant frequency is simply a multiple of a base frequency.
Let us find which harmonic resonates with the frequency 20 Hz:
Since n has to be an integer, final answer would be 323.