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finlep [7]
3 years ago
9

A plane flying against the wind covers the 900-kilometer distance between two aerodromes in 2 hours. The same plane flying with

the wind covers the same distance in 1 hour and 48 minutes. If the speed of the wind is constant, what is the speed of the wind?
Physics
1 answer:
Mashcka [7]3 years ago
4 0

Answer:

The speed of the wind is 25 km/hr.

Explanation:

Let us call v_p the speed of the plane and v_w the speed of the wind. When the plane is flying against the wind, it covers the distance of 900-km in 2 hours (120 minutes); therefore;

(1). v_p - v_w = \dfrac{900km}{120min}

And when the plane is flying with the wind, it covers the same distance in 1 hour 48 minutes (108 minutes)

(2). v_p+v_w= \dfrac{900km}{108min}

From equation (1) we solve for v_p and get:

v_p = \dfrac{900km}{120min}+v_w,

and by putting this into equation (2) we get:

\dfrac{900km}{120min}+v_w+v_w= \dfrac{900km}{108min}

2v_w= \dfrac{900km}{108min}-\dfrac{900km}{120min}

2v_w = 8.3km/min - 7.5km/min

2v_w = 0.83km/min

v_w = 0.4165km/min

or in km/hr this is

\boxed{v_w= 25km/hr }

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F' = F

Explanation:

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m₂ = mass of second object

r = distance between objects

Now, if the masses and the distance between them is doubled:

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<u>F' = F</u>

7 0
2 years ago
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d ≈ 7,6 g/cm³  

Explanation:

d = m/V = 40g/5,27cm³ ≈ 7,6 g/cm³

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3 0
3 years ago
The core of a star must be at temperature of _____ degrees Celsius for hydrogen fusion to take place. 10,000 100,000 1,000,000 1
disa [49]
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6 0
3 years ago
Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103
son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

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As M=m, then

PE = -\frac{Gm^2}{R}

Where,

m = Mass

G =Gravitational Universal Constant

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PART A) As half its initial value is u'=2u, then

U = -\frac{2Gm^2}{R}

dU = -\frac{2Gm^2}{R}

dKE = -dU

Therefore replacing we have that,

\frac{1}{2}mv^2 =\frac{Gm^2}{2R}

Re-arrange to find v,

v= \sqrt{\frac{Gm}{R}}

v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}

v = 816.7m/s

Therefore the  velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

2r = 2*10^3m

Therefore

dU = Gm^2(\frac{1}{R}-\frac{1}{2r})

v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}

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7 0
3 years ago
The tip of the second hand of a clock moves in a circle of 20 cm circumference. In one minute the hand makes a complete revoluti
Cerrena [4.2K]

Answer:

v_{avg} = 0

Explanation:

As we know that average velocity is defined as the ratio of total displacement of the object and its time interval.

so here we can say

v_{avg} = \frac{displacement}{time}

now we know that in one complete revolution the total displacement of the tip of the seconds hand is zero

because it will have same position after one complete revolution from where it starts

so here we can say that the average velocity will be zero

v_{avg} = 0

7 0
3 years ago
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