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julia-pushkina [17]
3 years ago
15

An airplane capable of an airspeed of 100 km/hr is 60 km off the coast above the sea. If the wind is blowing from the coast out

to sea at 40 km/hr, what is the least amount of time it will take for the plane to get to shore ?
Physics
1 answer:
Nady [450]3 years ago
6 0

To solve this problem we will apply the concepts related to relative speed. We will obtain it from the deduction made on the aircraft as a speed of the two components that act on it. Through the kinematic equations of motion, we can then calculate the time required.

The airspeed of airplane is 100km/h  while the wind is blowing from the coast out to sea at 40km/h. Wind is blowing from the coast out to sea means that it opposes the airspeed. Therefore, resultant relative speed of airplane is

v_r = 100-40=60km/h

Total distance is 60km then with this net velocity we have that the required time is

v = \frac{x}{t} \rightarrow t = \frac{x}{v}

Where,

x = Displacement

t = Time

v = Velocity

Replacing,

t = \frac{60km}{60km/h} = 1hour

t = 60 minutes

Therefore the time taken by the plane to reach the shore is 60 minutes

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8 0
3 years ago
Transverse, surface, and longitudinal waves are all __________ waves because they __________.
algol [13]

Answer:

a. mechanical; require a medium to travel through

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And an example of a transverse wave is the waves that form in the water when a rock is thrown (ripples), these waves need a medium (the water) to propagate.

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5 0
3 years ago
A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.
Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

\omega_f = Final angular velocity

\omega_i = Initial angular velocity = 1.5 ra/s

\alpha = Angular acceleration

\theta = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2

Acceleration while slowing down is -0.0047 rad/s²

t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s

Time taken to slow down is 335.103 seconds

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0

Solving the equation

t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0
3 years ago
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