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2 years ago

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At maximum speed an airplane travels 1720 miles against the wind in 5 hours. Flying with the wind, the plane can travel the same

distance in 4 hours. Let X be the maximum speed of the plane and Y be the speed of the wind. What is the speed of the plane with no wind?

2 answers:

Basile [38]2 years ago

7 0

An airplane at a maximum speed of 1720 miles against the wind for 5 hours so wha could be the speed of the plane with no wind if the wind could fly the plane with the same distance for 4 hrs and i think the best answer to this problem is 387 and i hope your satisfied with my answer.

Ede4ka [16]2 years ago

4 0

Answer:

387 miles / hour

Explanation:

X be the maximum speed of plane and Y be the speed of wind.

Speed is the ratio of distance traveled to the time taken by the body.

When the plane travels against the wind, the speed of the plane is X - Y and when the plane is moving with the wind the speed is X + Y.

According to the formula

$speed = \frac{distance}{time}$

$X - Y = \frac{1720}{5}$

X - Y = 344 ....(1)

And

$X + Y = \frac{1720}{4}$

X - Y = 430 ......(2)

Adding both the equations,

2X = 774

X = 387 miles/hour

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Olenka [21]

Missing graph. I attach it in the answer.

In a uniformly accelerated motion, the velocity at time t is given by:
$v(t)=at$
where a is the acceleration and t is the time.

Given the previous equation, if we plot v(t) versus t, we find a straight line; moreover, a (the acceleration) represents the slope of the curve.

Looking at the graph, we see that when the time goes from 10 s to 20 s, the velocity increases from 4 m/s to 6 m/s. Therefore the slope of the curve is
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3 years ago

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distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm.

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Answer:

The speed of the galaxy relative to the Earth is $3.09\times 10^6\ m/s$ .

Explanation:

We have,

(a) Wavelength emitted by light at distant galaxy is 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm. It can be seen that the wavelength of light reduces as it reaches Earth. It is called Red shift. As per Doppler's effect, we can say that the galaxy is receding from the Earth.

(b) Let v is the speed of the galaxy relative to the Earth. It can be given by :

$v=c(\dfrac{\lambda'}{\lambda}-1)\\\\v=3\times 10^8\times (\dfrac{438.6 }{434.1 }-1)\\\\v=3\times 10^8\times (\dfrac{438.6}{434.1}-1)\\\\v=0.0103\cdot3\cdot10^{8}\\\\v=3.09\times 10^6\ m/s$

So, the speed of the galaxy relative to the Earth is $3.09\times 10^6\ m/s$ .

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2 years ago

A wheel is rotating at 30.0 rpm. The wheel then accelerates uniformly to 50.0 rpm in 10.0 seconds. Determine the – a) rate of an

Mademuasel [1]

Answer:

The angular acceleration is $0.209\ rad/s^2$

Explanation:

Given that,

Angular velocity, $\omega_{i} = 30.0\ rpm$

Angular velocity, $\omega_{f} = 50.0\ rpm$

Time t = 10.0 sec

We need to calculate the angular acceleration

Using formula of angular acceleration

$\alpha=\dfrac{\Delta \omega}{\Delta t}$

$\alpha=\dfrac{\omega_{f}-\omega_{i}}{\Delta t}$

$\alpha=\dfrac{50.0-30.0}{10.0}$

Now, we change the angular velocity in rad/s.

$\omega=20\times\dfrac{2\pi}{60}$

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2 years ago

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The only force acting on a 3.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a

bekas [8.4K]

Answer:

The work done on the canister by the 5.0 N force during this time is

54.06 Joules.

Explanation:

Let the initial kinetic energy of the canister be

KE₁ = $\frac{1}{2} mv_1^{2}$ = $\frac{1}{2} *3*3.6^{2}$ = 19.44 J in the x direction

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KE₂ = $\frac{1}{2} mv_2^{2}$ = $\frac{1}{2} *3*7.0^{2}$ = 73.5 J in the y direction

Therefore from the Newton's first law of motion, the effect of the force is the change of momentum and the difference in energy between the initial and the final

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3 years ago

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Answer:

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Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

<em>"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle </em><em>a)</em><em> el período, </em><em>b) </em><em>la velocidad angular, </em><em>c) </em><em>la fuerza con la que gira la prenda y </em><em>d) </em><em>la velocidad lineal de la lavadora."</em>

El tambor gira a velocidad angular constante ( $\omega$ ), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga ( $a$ ), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor ( $T$ ), en segundos:

$T = \frac{1}{f}$ (1)

Donde $f$ es la frecuencia, en hertz.

( $f = 4\,hz$ )

$T = \frac{1}{4\,hz}$

$T = 0.25\,s$

Ahora determinamos la fuerza aplicada sobre la prenda ( $F$ ), en newtons:

$F = m\cdot a$ (2)

$F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}}$ (2b)

Donde:

$m$ - Masa de la prenda, en kilogramos.

$r$ - Radio interior del tambor, en metros.

( $m = 0.32\,kg$ , $r = 0.4\,m$ , $T = 0.25\,s$ )

$F = \frac{4\pi^{2}\cdot (0.32\,kg)\cdot (0.4\,m)}{(0.25\,s)^{2}}$

$F \approx 80.852\,N$

La velocidad lineal de la lavadora es:

$v = \frac{2\pi\cdot r}{T}$ (3)

( $r = 0.4\,m$ , $T = 0.25\,s$ )

$v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}$

$v \approx 10.053\,\frac{m}{s}$

Y la velocidad angular del tambor de la lavadora:

$\omega = \frac{2\pi}{T}$

( $T = 0.25\,s$ )

$\omega = \frac{2\pi}{0.25\,s}$

$\omega \approx 25.133\,\frac{rad}{s}$

7 0

2 years ago