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3 years ago

Penny has two dogs: Fluffy (left) and Butch (right).

Physics

2 answers:

Monica [59]3 years ago

3 0

I think it would be D

They would both move in the same way

Kipish [7]3 years ago

3 0

None knows how dogs looks like. One may be 4kg - another 30kg

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A uniform charge density of 509 nC/m3 is distributed throughout a spherical volume of radius 6.03 cm. Consider a cubical Gaussia

Artemon [7]

Explanation:

(a) It is known that relation between charge and volume is as follows.

$q_{enclosed} = \rho V_{cube}$

= $(509 \times 10^{-9} C/m^{3}) \times (0.04 m)^{3}$

= $509 \times 10^{-9} \times 6.4 \times 10^{-5}$

= $3.26 \times 10^{-11} C$

Now, according to Gauss's law

$\phi = \frac{q_{enclosed}}{\epsilon_{o}}$

= $\frac{3.26 \times 10^{-11} C}{8.85 \times 10^{-12}C^{2}N^{-1}m^{-2}}$

= 3.68 $N m^{2}/C$

Hence, the electric flux through this cubical surface if its edge length is 4.00 cm is 3.68 $N m^{2}/C$ .

(b) Similarly, we will calculate the electric flux when edge length is 16.8 cm as follows.

$q_{enclosed} = \rho V_{cube}$

= $(509 \times 10^{-9} C/m^{3}) \times (0.168 m)^{3}$

= $509 \times 10^{-9} \times 4.74 \times 10^{-3}$

= $2.41 \times 10^{-11} C$

Now, according to Gauss's law

$\phi = \frac{q_{enclosed}}{\epsilon_{o}}$

= $\frac{2.41 \times 10^{-11} C}{8.85 \times 10^{-12}C^{2}N^{-1}m^{-2}}$

= 2.72 $N m^{2}/C$

Therefore, the electric flux through this cubical surface if its edge length is 4.00 cm is 2.72 $N m^{2}/C$ .

4 0

4 years ago

How do I disable internal laptop display on Linux? So that I can project to external monitor! video=LVDS-1:d and video=eDP-1:d w

Salsk061 [2.6K]

You must shoot the laptop!!!

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4 years ago

Jared walks 120 m east, 150 m south, and then 40 m west. Find the total

shusha [124]

Answer:

310 m

Explanation:

120+150+40=310

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3 years ago

A 1.3 kg booster is attached to a 6.0 kg rocket is initially travelling at a velocity of 175 m/s. If the booster is ejected in b

nikklg [1K]

Answer:

It will travel to the infinity and beyond

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4 years ago

If a circuit contains a 12v battery and two 3 bulbs in series

Paha777 [63]

If the resistance of each bulb is 3 ohms, and they are connected across the battery, then current will flow through them.

In series, the two 3-ohm bulbs will look like a single 6-ohm resistance.

Current = (voltage) / (resistance) = (12v) / (6 ohms) = 2 Amperes.

The battery will need to deliver some power to the circuit.

Power = (voltage) x (current) = (12v) x (2 Amperes) = 24 watts .

Each bulb will dissipate 12 watts, in the form of light and heat.

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6 0

4 years ago