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Monica [59]
3 years ago
6

A man flies a small airplane from Fargo to Bismarck, North Dakota --- a distance of 180 miles. Because he is flying into a head

wind, the trip takes him 2 hours. On the way back, the wind is still blowing at the same speed, so the return trip takes only 1 hours. What is the plane's speed in still air, and how fast is the wind blowing?
Physics
1 answer:
love history [14]3 years ago
4 0

Answer:

airplane speed 135mph windspeed 45 mph

Explanation:

This information helps us to write down a system of linear equations

When going head wind, the speed of the wind is substracted from that of the airplane and on the return trip it is added, then:

A:=Airlplane speed

W:= Wind speed

(A+W)*1h=180mi (1)

(A-W)*2h=180mi (2)

then from (1) A=180-W (3), replacing this in (2) we get (180-W-W)*2h =180mi, then

360-4W=180, or 180=4W, then W=45 mph. Replacing this in (3) we have that A=180-45=135 mph.

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7 0
3 years ago
An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in
Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

T {V}^{\gamma - 1} = \text{constant}

\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}

Given:-

{T}_{1} = 275 \; K  

{T}_{2} = T \left( \text{say} \right)

{V}_{1}  = V

{V}_{2} = 6V

\gamma = \cfrac{5}{3} \;    (the gas is monoatomic)

\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}

 

\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}  

T  =  275 \times 0.30

T  =  83 K.

3 0
3 years ago
skateboarder, starting from rest, rolls down a 13.5 m ramp. When she arrives at the bottom of the ramp her speed is 7.37 m/s. If
scZoUnD [109]

Answer:

1.7 m/s²

Explanation:

d = length of the ramp = 13.5 m

v₀ = initial speed of the skateboarder = 0 m/s

v = final speed of the skateboarder = 7.37 m/s

a = acceleration

Using the equation

v² = v₀² + 2 a d

7.37² = 0² + 2 a (13.5)

a = 2.01 m/s²

θ = angle of the incline relative to ground = 29.9

a' = Component of acceleration parallel to the ground

Component of acceleration parallel to the ground is given as

a' = a Cosθ

a' = 2.01 Cos29.9

a' = 1.7 m/s²

7 0
3 years ago
A certain superconducting magnet in the form of a solenoid of length 0.56 m can generate a magnetic field of 6.5 T in its core w
Gnom [1K]

Answer:

The value  is N =36203 \  turns

Explanation:

From the question we are told that

   The length of the solenoid is  l = 0.56 \  m

   The magnetic field is B  =  6.5 \ T

    The current is I = 80 \ A

     The desired temperature is  T = 4.2 \ K

Generally the magnetic field is mathematically represented as

       B  = \frac{\mu_o * N * I }{L }

=>     N = \frac{B  * L }{\mu_o * I }

Here  \mu_o is the permeability of free space with value  

      \mu_o =  4\pi * 10^{-7} N/A^2

So

     N = \frac{6.5  * 0.56 }{ 4\pi * 10^{-7} *  80  }

=>   N =36203 \  turns

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qwelly [4]

Answer:

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