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mario62 [17]
3 years ago
11

At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what

distance from the line will the field strength to be 2000 N/C
Physics
1 answer:
blsea [12.9K]3 years ago
8 0

Answer:

Distance from the line with the field strength of 2000N/C = 0.7071 D or D√(0.5) or D/(√2)

Explanation:

Electric field is given by,

E = kQ/r²

where k = constant, Q = total charges, R = distance from the line of charge = D

Provided, total charges are constant, then,

The electric field is inversely proportional to the square of D.

E ∝1/D²

E = c/D²

c represents kQ, the two constant parameters.

1000 = c/D²

c = 1000D²

For E = 2000 N/C

Let the distance from the line with that field strength be x

E = kQ/x²

E ∝1/x²

E = c/x²

2000 = c/x²

c = 1000D²

2000 = 1000D²/x²

x² = 0.5D²

x = √(0.5D²)

x = D√0.5

x = 0.7071 D or D√(0.5) or D/(√2)

Hope this Helps!!!

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On a hot day, the temperature of a 65,000-L swimming pool increases by 1.20°C. What is the net heat transfer during this heating
vichka [17]

Answer:

326149.2 KJ

Explanation:

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Q = m*cv*ΔT

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A 65000 L swimming pool will have a mass of:

65000L *\frac{1m^3}{1000L} * \frac{1000kg}{1m^3} = 65000 kg

The specific heat capacity at constant volume of water is equal to 4.1814 KJ/KgC.

We replace the data and get:

Q = m*cv*ΔT = 65000 kg * 4.1814 KJ/KgC * 1.2°C = 326149.2 KJ

3 0
3 years ago
Sound travels at a rate of 340 m/s in all directs through the air. Matt rings a very loud bell at one location, and Steve hears
DENIUS [597]

Answer:

It will take about 1.32 seconds to travel to his location.

Explanation:

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velocity=\frac{distance}{time} \\340\,\frac{m}{s} =\frac{450\,\,m}{t} \\t=\frac{450}{340} \,s\\t\approx 1.32\,\,s

4 0
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Would the energy of the wave increase or decrease if the speed of the wave increases? Why?
choli [55]

Answer:

i hope this helps some

Explanation:

The time-averaged power of a sinusoidal wave is proportional to the square of the amplitude of the wave and the square of the angular frequency of the wave. This is true for most mechanical waves. If either the angular frequency or the amplitude of the wave were doubled, the power would increase by a factor of four.

The speed of a wave is dependant on four factors: wavelength, frequency, medium, and temperature. Wave speed is calculated by multiplying the wavelength times the frequency (speed = l * f).

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zysi [14]

Given that,

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We need to calculate the mass of two washers in kg

Using conversion of unit

Mass of each washer m= 4.9\times10^{-3}\ Kg

So, Mass of two washers is

M=2\times m

Put the value of m

M=2\times4.9\times10^{-3}\ k

M=0.0098\ Kg

If 4 washer are attached to the spring

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Using formula of force

F=mg

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F=4\times4.9\times10^{-3}\times9.8

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6 0
4 years ago
Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.
Ksivusya [100]

|v| =\sqrt{ G \cdot M / r}, where

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Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

<em>Equation 1</em>  (see below) relates net force the object experiences, \Sigma F to its orbit velocity v and its mass m required for it to stay in orbit :

\Sigma F = m \cdot v^{2} / r <em>(equation 1)</em>

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force \Sigma F acting on the soccer ball shall equal to its weight, W = m \cdot g where g the gravitational acceleration constant. Thus

\Sigma F = W = m \cdot g <em>(equation 2)</em>

Substitute equation 2 to the left hand side of <em>equation 1</em> and solve for v; note how the mass of the soccer ball, m, cancels out:

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g = G \cdot M/ r^{2} <em>(equation 4)</em>

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Thus

\begin{array}{lll}|v| &=& \sqrt{g \cdot r}\\ & =&\sqrt{ G \cdot M / r}\end{array}

3 0
4 years ago
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