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alexdok [17]
3 years ago
6

A negative charge of 4.0 x 10 C and a positive charge of 7.0 x 10 C are separated by 0.15 m. What is

Physics
1 answer:
Burka [1]3 years ago
5 0

Answer:

112×10^12N

Explanation:

We use the following equation to find the force between two charges that are away from each other at a particular distance:

<h2> F=k×q1×q2/r^2</h2><h2> F=9×10^9×4×10×7×10/(0.15)^2 F=252×10^11/0.225 </h2><h2> F=1120×10^11=112×10^12</h2>
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If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.
grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution.  By definition is defined as:

\theta = 1.22\frac{\lambda}{d}

Where,

\lambda= Wavelength

d = Width of the slit

\theta= Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

r = l\theta

Relacing \theta

r = l(\frac{1.22\lambda}{d})

r = 1.22\frac{l\lambda}{d}

Replacing with our values we have that,

r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})

r = 3680.91m

Therefore the diameter of the beam on the moon is

d = 2r

d = 2 * (3690.91)

d = 7361.8285m

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0
3 years ago
The square loop shown in the figure moves into a 0.80T magnetic field at a constant speed of 10m/s. The loop has a resistance of
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7 0
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Determine the minimum size of glass tubing that can be used to measure water level. If the capillary rise in the tube does not e
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Answer:

Explanation:

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h = 2•T•Cos θ / r•ρ•g

Where

h = height of liquid level

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ρ = density of liquid

θ = angle of contact = 0°

g =acceleration due to gravity=9.81m/s²

The liquid is water then,

ρ = 1000 kg / m³

Given that,

T = 0.0735 N/m

h = 0.25mm = 0.25 × 10^-3m

Then,

r = 2•T•Cos θ / h•ρ•g

r = 2 × 0.0735 × Cos0 / 2.5 × 10^-3 × 1000 × 9.81

r = 5.99 × 10^-3m

Then, r ≈ 6mm

The radius of the capillary tube is 6mm

So, the minimum size is

Volume = πr²h

Volume = π × 6² × 0.25

V = 2.83 mm³

The minimum size of the capillary tube is 2.83mm³

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