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Ipatiy [6.2K]
3 years ago
10

Consider a horizontal layer of the dam wall of thickness dx located a distance x above the reservoir floor. What is the magnitud

e dF of the force on this layer that results from adding the water to the reservoir? Express your answer in terms of x, dx, the magnitude of the acceleration due to gravity g, and any quantities from the problem introduction.

Physics
1 answer:
adoni [48]3 years ago
6 0

Answer:

Explanation:

Attached is the solution

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WGVU-AM is a radio station that serves the Grand Rapids, Michigan area. The main broadcast frequency is 1480kHz. At a certain di
Yuki888 [10]

Answer:

a

  \lambda  =  202.7 \  m

b

  w =  9.3 *10^{6} \  rad/s

c

  k = 0.031 m^{-1}

d

  E_{max} = 9.0 *10^{-3} \  V/m

Explanation:

From the question we are told that  

       The frequency of the radio station is  f=  1480 \  kHz  =  1480 *10^{3}\ Hz

         The magnitude of the magnetic field is  B  =  3.0* 10^{-11} \  T

Generally the wavelength is mathematically represented as

          \lambda  =  \frac{c}{f}

Here c is the speed of light with value  c =  3.0 *10^{8} \ m/s

So

        \lambda  =  \frac{3.0 *10^{8}}{ 1480 *10^{3}}

=>      \lambda  =  202.7 \  m

Generally the angular frequency is mathematically represented as

       w =  2 \pi * f

=>   w =  2 * 3.142 *  1480 *10^{3}

=>   w =  9.3 *10^{6} \  rad/s

Generally the wave number is mathematically represented as        

=>      k = \frac{2 \pi }{\lambda}

=>      k = \frac{2  *  3.142  }{ 202.7}

=>      k = 0.031 m^{-1}

Generally the amplitude of the electric field at this distance from the transmitter is mathematically represented as

         E_{max} = c *  B

=>      E_{max} = 3.0 *10^{8} *   3.0* 10^{-11}

=>      E_{max} = 9.0 *10^{-3} \  V/m

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2 years ago
A mysterious, bright, starlike object in the galaxy core?
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7 0
3 years ago
A small propeller airplane can comfortably achieve a high enough speed to take off on a runway that is 1/4 mile long. A large, f
QveST [7]

Answer:

1 mile

Explanation:

We can use the following equation of motion to solve for this problem:

v^2 - v_0^2 = 2a\Delta s

where v m/s is the final take-off velocity of the airplane, v_0 = 0 initial velocity of the can when it starts from rest, a is the acceleration of the airplanes, which are the same, and \Delta s is the distance traveled before takeoff, which is minimum runway length:

v^2 - 0^2 = 2a\Delta s

\Delta s = \frac{v^2}{2a}

From here we can calculate the distance ratio

\frac{\Delta s_1}{\Delta s_2} = \frac{v_1^2/2a_1}{v_2^2/2a_2}

\frac{\Delta s_1}{\Delta s_2} = \left(\frac{v_1}{v_2}\right)^2\frac{a_2}{a_1}

Since the 2nd airplane has the same acceleration but twice the velocity

\frac{\Delta s_1}{\Delta s_2} = 0.5^2* 1

\Delta s_2 = 4 \Delta s_1 = 4*(1/4) = 1 mile

So the minimum runway length is 1 mile

6 0
3 years ago
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