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Darya [45]
3 years ago
10

A plane flies 1800 miles in 9 ​hours, with a tailwind all the way. the return trip on the same​ route, now with a​ headwind, tak

es 12 hours. assuming both remain​ constant, find the speed of the plane and the speed of the wind.​ [hint: if x is the​ plane's speed and y the wind speed​ (in mph), then the plane travels to its destination at xplusy mph because the plane and the wind go in the same​ direction; on the return​ trip, the plane travels at xminusy ​mph.]
Physics
1 answer:
Fittoniya [83]3 years ago
5 0

Initially its moving with tail wind so here the speed of wind will support the motion of the plane

so we can say

V_{plane} + v_{wind} = \frac{distance}{time}

V_{plane} + v_{wind} = \frac{1800}{9}

V_{plane} + v_{wind} = 200 mph

now when its moving with head wind we can say that wind is opposite to the motion of the plane

V_{plane} - v_{wind} = \frac{distance}{time}

V_{plane} - v_{wind} = \frac{1800}{12}

V_{plane} - v_{wind} = 150mph

now by using above two equations we can find speed of palne as well as speed of wind

V_{plane} = 175 mph

v_{wind} = 25 mph

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Dmitriy789 [7]

Answer:

0.17724 m/s²

Explanation:

D = Diameter of roll = Length of wing = 11 m

T = Time it takes to complete the circle = 35 s

Velocity

v=\frac{2\pi R}{T}\\\Rightarrow v=\frac{\pi D}{T}\\\Rightarrow v=\frac{\pi\times 11}{35}\\\Rightarrow v=0.98735\ m/s

Acceleration

a=\frac{v^2}{R}\\\Rightarrow a=\frac{0.98735^2}{\frac{11}{2}}\\\Rightarrow a=0.17724\ m/s^2

Acceleration of the tip of the plane is 0.17724 m/s²

3 0
3 years ago
what are the speeds of (a) a proton that is accelerated from rest through a potential difference of −1000 v−1000 v and (b) an el
Evgen [1.6K]

Answer:

This is the answer: The speed of a proton is about 5.0 × 10⁵ m/s

Explanation:

Because of the speeds of protons! :D

5 0
1 year ago
Consult Interactive Solution 10.37 to explore a model for solving this problem. A spring is compressed by 0.0647 m and is used t
padilas [110]

Answer:

\omega=32.14\ rad/s

Explanation:

Given that,

The compression in the spring, x = 0.0647 m

Speed of the object, v = 2.08 m/s

To find,

Angular frequency of the object.

Solution,

We know that the elation between the amplitude and the angular frequency in SHM is given by :

v=\omega\times A

A is the amplitude

In case of spring the compression in the spring is equal to its amplitude

\omega=\dfrac{v}{A}

\omega=\dfrac{2.08\ m/s}{0.0647\ m}

\omega=32.14\ rad/s

So, the angular frequency of the spring is 32.14 rad/s.

4 0
3 years ago
A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of ?5.20 rad/s2. during a 3.80-s time inte
ddd [48]

<span>We can answer this using the rotational version of the kinematic equations:</span><span>
θ = θ₀ + ω₀<span>t + ½αt²     -----> 1</span></span>

ω² = ω₀² + 2αθ            -----> 2

Where:

θ = final angular displacement = 70.4 rad

θ₀ = initial angular displacement = 0

ω₀ = initial angular speed

ω = final angular speed

t = time = 3.80 s

α = angular acceleration = -5.20 rad/s^2

Substituting the values into equation 1:<span>
70.4 = 0 + ω₀(3.80) + ½(-5.20)(3.80)² </span><span>

ω₀ = (70.4 + 37.544) / 3.80 </span><span>

ω₀ = 28.406 rad/s </span><span>


Using equation 2:
ω² = (28.406)² + 2(-5.2)70.4 


ω = 8.65 rad/s 


</span>

5 0
2 years ago
A cyclist maintains a constant velocity of 6.1 m/s headed away from point A. At some initial time, the cyclist is 242 m from poi
KengaRu [80]

Answer:

553.1m

Explanation:

When an object moves at constant velocity we can express this movement like V=x/t, where V is the velocity, x is the displacement and t is the time spent on it.

In that way, the expression x=V.t give us the displacement from t=0s until t=51s, but we have to sum the initial distance from the point A.

x=242m +V.t = 242m + (6.1m/s x 51s) = 553.1m

7 0
3 years ago
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