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Dafna11 [192]
2 years ago
14

A plane has an average air speed (this is the speed the plane moves through air) of 750 mph. The plane flies a route of 5000 mil

es from Los Angeles to London. The average speed of the wind along this route is 30 mph blowing east from Los Angeles towards London. How much longer does it take the plane to fly from London to Los Angeles than the reverse direction?
Physics
1 answer:
Digiron [165]2 years ago
6 0

Answer:

6 hours 15 minutes

Explanation:

On the trip from L.A. to London, the plane travels at 750 mph against a headwind of 50 mph, and that makes the net 700 mph (in aviation speak, 750 is the airspeed, while 700 is the groundspeed).  5000 miles divided by 700 mph results in about 7.14 hours, or about 7 hours and 9 minutes.  On the return trip, ASSUMING THE SAME WIND, the plane travels at 750 mph, but this time the wind of 50 mph is a tail wind.  So the net (groundspeed) is 800 mph.  Traveling 5000 miles at 800 mph only takes 6.25 hours, or 6 hours and 15 minutes.  

Outbound flight 7 hours 9 minutes

Return flight 6 hours 15 minutes

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6 0
2 years ago
A giant wall clock with diameter d rests vertically on the floor. The minute hand sticks out from the face of the clock, and its
Katyanochek1 [597]

Answer:

d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)

Explanation:

We can try writing the equation of the horizontal component of the length of the minute hand in terms of distance and the angle, that depends of time in this particular case.

The x-component of the length of the minute hand is:

d_{x}(t)=dcos(\theta (t)) (1)

  • d is the length of the minute hand (d=D/2)
  • D is the diameter of the clock
  • t is the time (min)

Now, using the angular kinematic equations we can express the angle in term of angular velocity and time. As we know, the minute hand moves with a constant angular velocity, so we can use this equation:

\theta (t)=\omega *t (2)

Also we know, that the minute hand moves 90 degrees or π/2 rad in 15 min, so using the definition of angular velocity, we have:

\omega=\frac{\Delta \theta}{\Delta t}=\frac{\theta_{f}-\theta_{i}}{t_{f}-t{i}}=\frac{\pi/2-0}{15-0}=\frac{\pi}{30}

Now, let's put this value on (2)

\theta (t)=\frac{\pi}{30}*t

Finally the length x(t) of the shadow of the minute hand as a function of time t, will be:

d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)

I hope it helps you!

6 0
2 years ago
Stacy measures two quantities: the mass of each washer and the force that the washers exert on the force meter. In general, how
geniusboy [140]

Answer:

When she adds more washers to the meter, the magnitude of force that is shown on the force meter increases.

Explanation:

The force that the washers exert on the force meter is actually the weight of the washers. Weight is actually a force with gravitation acceleration.

                       F = W = mg

Where g is gravitational acceleration and its value is 9.81 m/s² and m is the mass of any object. As she adds more washers to the meter so the total mass of the washers increases. As the mass of the washers increases, magnitude of the force (Weight) shown on the force meter increases.

6 0
2 years ago
A system contains a perfectly elastic spring, with an unstretched length of 20 cm and a spring constant of 4 N/cm.
mote1985 [20]

Answer:

a) When its length is 23 cm, the elastic potential energy of the spring is

0.18 J

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Explanation:

Hi there!

a) The elastic potential energy (EPE) is calculated using the following equation:

EPE = 1/2 · k · x²

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k = spring constant.

x = stretched lenght.

Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).

First, let´s convert the spring constant units into N/m:

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b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:

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EPE = 0.72 J

When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

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sasho [114]
Assumption both thunder and misty are pulling in same direction,
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7 0
2 years ago
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