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Gre4nikov [31]
3 years ago
7

Two massive, positively charged particles are initially held a fixed distance apart. When they are moved farther apart, the magn

itude of their mutual gravitational force changes by a factor of n. Which of the following indicates the factor by which the magnitude of their mutual electrostatic force changes?
a. 1/n2
b. 1/n
c. n
d. n2
Physics
1 answer:
Aneli [31]3 years ago
6 0

c. n

The magnitude of the gravitational force is given by Newton's law of universal gravitation:

F_g=-G\frac{m_1m_2}{d^2}

Here G is the gravitational constant, m_1, m_2 are the masses of the particles and d is the distance between them.

The magnitude of the electrostatic force is given by Coulomb's law:

F_e=k\frac{q_1q_2}{d^2}

Here k is the Coulomb constant, q_1, q_2 are the chargues of the particles and d is the distance between them.

As can be seen, both forces are inversely proportional to the square of the distance. Thus, the factor is the same in both cases.

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What type of modulation is typically used by broadcasting stations to transmit pictures on television screens?
erik [133]

Answer:

The correct option is;

Amplitude

Explanation:

When transmitting picture signals over the air by broadcasting stations, the signals are shifted into high frequency channels of Very High Frequency (VHF) or Ultra High Frequency (UHF) carrier currents and imposing the the television signal by changing the amplitude of the high frequency carrier current to match the transmitted television signal waveform shape

4 0
2 years ago
Read 2 more answers
49. \ A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner
faust18 [17]

Answer:

    \frac{L_1}{L_2} = \sqrt{(n^2 - 1)}

Explanation:

For this interesting problem, we use the definition of centripetal acceleration  

      a = v² / r  

angular and linear velocity are related  

     v = w r  

we substitute  

    a = w² r

the rectangular body rotates at an angular velocity w  

We locate the points, unfortunately the diagram is not shown. In this case we have the axis of rotation in a corner, called O, in one of the adjacent corners we call it A and the opposite corner A  

the distance OB = L₂  

the distance AB = L₁

the sides of the rectangle  

It is indicated that the acceleration in in A and B are related  

      a_A = n \ a_B  

we substitute the value of the acceleration  

    w² r_A = n r_B  

the distance from the each corner is  

    r_B = L₂  

    r_A = \sqrt{L_1^2 + L_2^2}  

we substitute  

   \sqrt{L_1^2 + L_2^2} = n L₂  

    L₁² + L₂² = n² L₂²  

    L₁² = (n²-1) L₂²  

4 0
3 years ago
A ball rolls off the edge of a table with a fairly large horizontal velocity. Which of the following statements are true? (Selec
Degger [83]

Answer:

A.The vertical velocity is constantly increasing as the ball falls.

B.The horizontal velocity does not noticeably change as the ball falls.

G.The horizontal velocity does not affect how long it will take the ball to fall to the floor.

H.The velocity vector of the ball changes as it travels through the air.

Explanation:

As the ball is projected horizontally so here the vertical component of the velocity is zero

So the time to reach the ground is given as

H = \frac{1}{2} gt^2

so we will have

t = \sqrt{\frac{2H}{g}}

so this is the same time as the ball is dropped from H height

Since there is no force in horizontal direction so its horizontal velocity will always remain constant while vertical velocity will change at constant rate which is equal to acceleration due to gravity.

So overall the velocity vector will change due to net acceleration g

4 0
3 years ago
If the radius of an atom is 60 pm and the radius of the Earth is 6000 km, by how many orders of
kherson [118]

Answer:

1 x 10¹⁷

               

Explanation:

Given data:

        Radius of the earth  = 6000km

        Radius of an atom  = 60pm

Now, how many orders is the radius of the earth larger than an atom

Solution:

To solve this problem, let us express both quantity as the same unit;

         1000m  = 1km

    6000km  = 6000 x 10³m   =  6 x 10⁶m

    60pm;

            1 x 10⁻¹²m  = 1pm

    60pm  = 60 x  1 x 10⁻¹²m  = 6 x 10⁻¹¹m

Now;

The order:   \frac{6 x 10^{6} }{6 x 10^{-11} }   = 1 x 10¹⁷

               

6 0
2 years ago
The alpha line in the balmer series of the hydrogen spectrum consists of light having a wavelength of 6.56. calculate the freque
guajiro [1.7K]
The alpha line in the Balmer series is the transition from n=3 to n=2 and with the wavelength of λ=656 nm = 6.56*10^-7 m. To get the frequency we need the formula: v=λ*f where v is the speed of light, λ is the wavelength and f is the frequency, or c=λ*f. c=3*10^8 m/s. To get the frequency: f=c/λ. Now we input the numbers: f=(3*10^8)/(6.56*10^-7)=4.57*10^14 Hz. So the frequency of the light from alpha line is f= 4.57*10^14 Hz. 
5 0
3 years ago
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