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ollegr [7]
3 years ago
15

Point R is 0.4 m from a 2C charge. Find the electric field at point R.

Physics
1 answer:
Travka [436]3 years ago
8 0
We know, E = 1/4π∈ * q/r²
Here, 1/4π∈ = 9 × 10⁹
q = 2 C
r = 0.4 

Substitute their values, 
E = 9 × 10⁹ * 2/(0.4)²
E = 9 × 10⁹ * 2/0.16
E = 9 × 10⁹ * 12.5
E = 112.5 × 10⁹ N/C

So, Your Final Answer would be 1.125 × 10¹¹ N/C

Hope this helps!
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A hockey puck with mass 0.3 kg is shot across an ice-covered pond. Before the hockey puck was hit, the puck was at rest. After t
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Answer:

The net friction force is 8.01 N

Explanation:

Net friction force = mass of hockey puck × acceleration

From the equations of motion

v^2 = u^2 + 2as

v = 40 m/s

u = 0 m/s (puck was initially at rest)

s = 30 m

40^2 = 0^2 + 2×a×30

60a = 1600

a = 1600/60 = 26.7 m/s^2

The acceleration of the puck is 26.7 m/s^2

Net friction force = 0.3 × 26.7 = 8.01 N

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4 years ago
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Observers in a distant solar system are watching our Sun and using the radial velocity method to try to determine whether the Su
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Answer: almost 12 years

Explanation:

Using the radial velocity method, planet hunters can track a sun's spectrum, to determine whether the sun has planets. The spectrum appears first slightly blue-shifted, and then slightly red-shifted. If the shifts are regular, repeating themselves at fixed intervals of days, months, or even years, it  is almost certainly caused by a body orbiting the sun, tugging it back and forth over the course of its orbit. Because of the gravitational pull of Jupiter (about 3,000 times the mass of Earth), In which its average distance from the sun is 480 millions miles and takes nearly 12 years to make one revolution.

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An object is in circular motion. How will the object behave if the centripetal force is removed
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The object will sail away in a straight line ... continuing in the same direction it was going when the centripetal force stopped.

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What happens when magnesium reacts with fluorine
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It crates Magnesium Fluoride

Explanation:

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3 years ago
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A fillet weld has a cross-sectional area of 25.0 mm2and is 300 mm long. (a) What quantity of heat (in joules) is required to acc
HACTEHA [7]

Answer:

77362.56 J

163730.28571 J

Explanation:

A = Area = 25 mm²

l = Length = 300 mm

K = Constant = 3.33\times 10^{-6}

\eta = Heat transfer factor = 0.75

f_m = Melting factor = 0.63

T = Melting point of low carbon steel = 1760 K

Volume of the fillet would be

V=Al\\\Rightarrow V=25\times 300\\\Rightarrow V=7500\ mm^3=7500\times 10^{-9}\ m^3

The unit energy for melting is given by

U_m=KT^2\\\Rightarrow U_m=3.33\times 10^{-6}\times 1760^2\\\Rightarrow U_m=10.315008\ J/mm^3

Heat would be

Q=U_mV\\\Rightarrow Q=10.315008\times 7500\\\Rightarrow Q=77362.56\ J

Heat required to weld is 77362.56 J

Amount of heat generation is given by

Q_g=\dfrac{Q}{\eta f_m}\\\Rightarrow Q_g=\dfrac{77362.56}{0.75\times 0.63}\\\Rightarrow Q_g=163730.28571\ J

The heat generated at the welding source is 163730.28571 J

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