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aleksklad [387]
3 years ago
15

During science class, Ava and Justin were discussing the rotation of the Sun and debating over which model would best illustrate

how the Suns rotates. Which of the models shown below would best represent the rotation of the Sun? с B D​
Physics
1 answer:
agasfer [191]3 years ago
3 0

Answer:

it would b B

Explanation:

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Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11
tiny-mole [99]

Answer:

The line charge density is 1.59\times10^{-4}\ C/m

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

V=EA

V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2

\lambda=\dfrac{V\times2\epsilon_{0}}{r}

Where, r = radius

V = potential difference

Put the value into the formula

\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}

\lambda=1.59\times10^{-4}\ C/m

Hence, The line charge density is 1.59\times10^{-4}\ C/m

4 0
3 years ago
What happens when an object with a lower density is placed in a container with an
maksim [4K]
The only thing that definitely happens in every such case is:
The container becomes heavier.
5 0
3 years ago
Why do we feel the force of gravity from the Earth but we don’t feel the force of gravity from the moon?
Zigmanuir [339]

Answer:

The gravitational force on the moon is less than on Earth because the strength of gravity is determined by an object's mass. The bigger the object, the bigger the gravitational force. Gravity is pretty much everywhere. We just feel it in different ways depending on our state of motion.

Explanation:

Hope this helped!!

8 0
2 years ago
Read 2 more answers
Please help me I am desperate this is due now
elena55 [62]

Answer:

Distance travelled is 7 meters and the displacement is 3 meters

7 0
3 years ago
Read 2 more answers
A lamp is 10% efficient.How much electrical energy must be supplied to the lamp each second if it produces 20 J of light energy
k0ka [10]

If it produces 20J of light energy in a second, then that 20J is the 10% of the supply that becomes useful output.

20 J/s = 10% of Supply

20 J/s = (0.1) x (Supply)

Divide each side by 0.1:

Supply = (20 J/s) / (0.1)

<em>Supply = 200 J/s  </em>(200 watts)

========================

Here's something to think about:  What could you do to make the lamp more efficient ?  Answer:  Use it for a heater !

If you use it for a heater, then the HEAT is the 'useful' part, and the light is the part that you really don't care about.  Suddenly ... bada-boom ... the lamp is 90% efficient !

6 0
3 years ago
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