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shepuryov [24]
3 years ago
14

A spherical shell contains three charged objects. The first and second objects have a charge of − 14.0 nC and 33.0 nC , respecti

vely. The total electric flux through the shell is − 953 N ⋅ m 2 / C . What is the charge on the third object?
Physics
1 answer:
matrenka [14]3 years ago
7 0

Explanation:

Formula depicting relation between total flux and total charge Q is as follows.

              \phi  = \frac{Q}{\epsilon_{o}}    (Gauss's Law)

Putting the given values into the above formula as follows.

            Q = \phi \times \epsilon_{o}

                = -953 Nm^{2}/C \times 8.854 \times 10^{-12}

                = -8.4 \times 10^{-9} C

                = -8.4 nC

Therefore, when the unknown charge is q  then,

         -14.0 nC + 33.0 nC + q = -8.4 nC

               q = -27.4 nC

Thus, we can conclude that charge on the third object is -27.4 nC.

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forsale [732]

Answer:

The extension of the wire is 0.362 mm.

Explanation:

Given;

mass of the object, m = 4.0 kg

length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

A = πr²

A = π(0.001)² = 3.142 x 10⁻⁶ m²

The downward force of the object on the wire is given by;

F = mg

F = 4 x 9.8 = 39.2 N

The Young's modulus of aluminum is given by;

Y = \frac{stress}{strain}\\\\Y = \frac{F/A}{e/L}\\\\Y = \frac{FL}{Ae} \\\\e = \frac{FL}{AY}

Where;

Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

e = \frac{FL}{AY} \\\\e = \frac{(39.2)(2)}{(3.142*10^{-6})(69*10^9)} \\\\e = 0.000362 \ m\\\\e = 0.362 \ mm

Therefore, the extension of the wire is 0.362 mm.

8 0
3 years ago
An archer releases an arrow toward a target. The arrow travels 170 meters in 2 seconds. What is the speed of the arrow?
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A 0.500-kg stone is moving in a vertical circular path attached to a string that is 75.0 cm long. The stone is moving around the
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Answer:

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Explanation:

The velocity of the stone when it leaves the circular path is its tangential velocity, v, which is given by

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\omega is given by

\omega = 2\pi f

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Thus

v=2\pi fr

Using values from the question,

v=2\pi\times 1.50\times0.75

<em>Note the conversion of 75 cm to 0.75 m</em>

v=2\times3.14\times 1.50\times0.75 = 9.42\times0.75 = 7.065=7.07

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The matter from the explosion can reach him, hitting him. He should be able to feel that.
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If the light wave has a wavelength of 10m what would be its velocity
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with f the frequency and lambda the wavelength. We are give a wavelength of 10m. The frequencies of the visible light can range between 400 to about 790 Terahertz, so let us pick a middle point of 600 THz ("green-ish") as a "representative."

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The speed of such a wave would have to be 6e+15 m/s (which would be 7 orders of magnitude higher than the universal speed of light constant)

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3 years ago
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