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Karo-lina-s [1.5K]
2 years ago
6

An airplane is flying on a flight path that will take it directly over a radar tracking station. If the distance between the tra

cking station and the airplane is decreasing at a rate of 400 miles per hour when the distance between them is 10 miles, what is the speed of the plane if the plane is always 3 miles above the ground?

Physics
1 answer:
Wewaii [24]2 years ago
6 0

Answer:

The speed of planes is 419.31 miles per hours

Explanation:

Given that

\dfrac{dS}{dt}=-400\ miles/hr

S= 10 miles

And we have to find \dfrac{dx}{dt}.

From triangle ABC

AC^2=AB^2+BC^2

S^2=x^2+3^2}

by differentiating with respect to time t

2S\dfrac{dS}{dt}=2x\dfrac{dx}{dt}

S\dfrac{dS}{dt}=x\dfrac{dx}{dt}

Now when S=10 miles then

10^2=x^2+3^2}

x= 9.53 miles

So

S\dfrac{dS}{dt}=x\dfrac{dx}{dt}

10\times (-400)=9.53\times \dfrac{dx}{dt}

\dfrac{dx}{dt}=-419.31\ miles/hr

So the speed of planes is 419.31 miles per hours

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RoseWind [281]

Answer:

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Explanation:

A) Let's use a ball for the nucleus, the electron is at a farther distance the sphere for the electron must be at a distance of

Let's use proportions rule

                x_ electron = 0.529 10⁻¹⁰ /1.2 10⁻¹⁵ 1.5

               x _electron = 0.66 10⁵ mm = 0.66 10² m

B) the radii of the Earth and the sun are

               R_{E} = 6.37 10⁶ m

                tex]R_{Sum}[/tex] = 6.96 10⁸ m

                Distance = 1.5 10¹¹ m

                x_Earth = 1.5 10¹¹ / 6.96 10⁸  1.5

                x _Eart = 1.13 10² m

C) The radius of a sphere that represents the earth, if the sphere that represents the sun is 1.5 mm, let's use another rule of proportions

            d_sphere = 1.5 / 6.96 10⁸  6.37 10⁶

            d_sphere = 1.37 10⁻² mm

5 0
2 years ago
A person is working on a steel structure while standing on the ground. An accident occurred where 5 A pass through the structure
matrenka [14]

Answer:

35mA

Explanation:

Hello!

To solve this problem we must use the following steps

1. Find the electrical resistance of the metal rod using the following equation

R=\alpha  \frac{l}{a}

WHERE

α=

metal rod resistivity=2x10^-4 Ωm

l=leght=2m

A=  Cross-sectional area

A=\frac{\pi }{4} d^2=\frac{\pi }{4} (0.06)^2=0.00283

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R=(2x10^-4)\frac{2}{0.00283} =0.14

2. Now we model the system as a circuit with parallel resistors, where we will call 1 the metal rod and 2 the man(see attached image)

3.we know that the sum of the currents in 1 and 2 must be equal to 5A, by the law of conservation of energy

I1+I2=5

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V1=V2

(0.14I1)=2000(i2)

solving for i1

I1=14285.7i2

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14285.7i2+i2=5

i2=\frac{5}{14285.7+1} =3.5x10^-4A=35mA

6 0
2 years ago
An electron moves with a constant horizontal velocity of 3.0 × 106 m/s and no initial vertical velocity as it enters a deflector
Ghella [55]

Answer:

a = 5.05 x 10¹⁴ m/s²

Explanation:

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v_{x} = velocity along the horizontal direction = 3.0 x 10⁶ m/s

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Time of travel can be given as

t = \frac{X}{v_{x}}

inserting the values

t = 0.11/(3.0 x 10⁶)

t = 3.67 x 10⁻⁸ sec

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Y = vertical distance traveled = 34 cm = 0.34 m

a = acceleration = ?

t = time of travel  = 3.67 x 10⁻⁸ sec

v_{y} = initial velocity along the vertical direction = 0 m/s

Using the kinematics equation

Y = v_{y} t + (0.5) a t²

0.34 = (0) (3.67 x 10⁻⁸) + (0.5) a (3.67 x 10⁻⁸)²

a = 5.05 x 10¹⁴ m/s²

7 0
2 years ago
In an experiment to measure the wavelength of light using a double slit, it is found that the fringes are too close together to
Mandarinka [93]

Answer:

halve the slit separation

Explanation:

As we know that

In YDS experiment, the equation of fringe width is as follows

\beta = \frac{\lambda D}{d}

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4 0
2 years ago
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Dima020 [189]
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where
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diameter = 2.05mm=.00205 m
current = 2.75 A
</span>get first the current density:
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find the cross section area
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cross section = 3.3 006x10-6 m^2
substitute the values 
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current density=35.55 x1 0^2 A/m^2
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Electric field stregnth= 46.415 Volts/m

The electric field strength of copper is 46.415 V/m.


6 0
2 years ago
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