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Alexxx [7]
3 years ago
10

what will be the work done if a Unit chargeis moved from infinity to a distance 'R' from infinitely large charged sheet? CUT TO

POINT ANSWER PLEASE
Physics
1 answer:
fredd [130]3 years ago
4 0
We know, W = k. q₁q₂/ [1/r₁² - r₂² ]

Here, r₁ = R
r₂ = ∞

Substitute their values, 
 W = k. q₁q₂/ [1/R - 1/∞]
 W = k. q₁q₂/ [1/R]
 W = k. q₁q₂/ R

Hope this helps!
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The centripetal force acting on the space shuttle as it orbits Earth is equal to the shuttles momentum
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What makes a clinical thermometer suitable for measuring small changes in body temperature? *
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Answer: Because of the fine bore of the tube.

Explanation:

Temperature is the degree of hotness and coldness. And thermometer is the instrument use to measure temperature.

The two most common types of themometric fluids for thermometer are alcohol and mercury.

What makes a clinical thermometer suitable for measuring small changes in body temperature is because of the fine bore of the tube which makes it possible for small temperature changes to cause large changes in the length of mercury columns, making the thermometer very sensitive to temperature changes.

The most prominent feature of the thermometer is the kink or constriction of bore near the bulb.

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A giant armadilo moving northward with a constant acceleration covers the distance between two points 60m apart in 6 seconds. It
Naddika [18.5K]
Acceleration is the rate of change of the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. To determine acceleration, we need to know the initial velocity and the final velocity and the time elapsed. From the given values, we need t o calculate for the initial velocity. We use some kinematic equations. We do as follows:

 x = v0t + at^2/2
60 = v0(6) + a(6)^2/2
60 = 6v0 + 18a          (EQUATION 1)

vf = v0 + at
15 = v0 + a(6)
15 = v0 + 6a             (EQUATION 2)

Solving for v0 and a,
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Which of the following is not a part of the appendicular skeleton
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Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle
Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

v = \frac{2 \pi r}{T}  

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

T^{2} = r^{3}

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

T^{2} = r^{3}

T = \sqrt{(133370 Km)^{3}}

T = \sqrt{(2.372x10^{15} Km)}

T = 4.870x10^{7} Km

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( 1.50x10^{8} Km )

1 AU is defined as the distance between the earth and the sun.

\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU

T = 0.324 AU

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

T = \frac{0.324 AU}{1 AU} . 1 year

T = 0.324 year

That can be expressed in units of days

T = \frac{0.324 year}{1 year} . 365.25 days  

T = 118.60 days

<em>Circular velocity for the particle in the </em><em>Encke Division</em><em>:</em>

v = \frac{2 \pi r}{T}

v = \frac{2 \pi (133370 Km)}{(118.60 days)}

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

118.60 days .\frac{86400 s}{1 day} ⇒ 10247040 s

133370 Km .\frac{1000 m}{1 Km} ⇒ 133370000 m

v = \frac{2 \pi (133370000 m)}{(10247040 s)}

v = 81.778 m/s

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

T^{2} = r^{3}

T = \sqrt{(69000 Km)^{3}}

T = \sqrt{(3.285x10^{14} Km)}

T = 1.812x10^{7} Km

\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU

T = 0.120 AU

T = \frac{0.120 AU}{1 AU} . 1 year

T = 0.120 year

T = \frac{0.120 year}{1 year} . 365.25 days  

T = 43.83 days

<em>Circular velocity for the particle in </em><em>D Ring</em><em>:</em>

v = \frac{2 \pi r}{T}

v = \frac{2 \pi (69000 Km)}{(43.83 days)}

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

43.83 days . \frac{86400 s}{1 day} ⇒ 3786912 s

69000 Km . \frac{1000 m}{ 1 Km} ⇒ 69000000 m

v = \frac{2 \pi (69000000 m)}{(3786912 s)}

v = 114.483 m/s

 

\frac{114.483 m/s}{81.778 m/s} = 1.399            

The particle in the D ring is 1399 times faster than the particle in the Encke Division.  

7 0
2 years ago
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