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kap26 [50]
3 years ago
8

Connecting many devices in a single socket does not affect the flow of current in a

Physics
1 answer:
eimsori [14]3 years ago
7 0

<u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>:</u><u>-</u><em> </em><em>F</em><em>a</em><em>l</em><em>s</em><em>e</em>

<u>E</u><u>x</u><u>p</u><u>l</u><u>a</u><u>n</u><u>a</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u>

<em>False because it can leads to overloading and further to short circut.</em>

<h2><em><u>H</u></em><em><u>o</u></em><em><u>p</u></em><em><u>e</u></em><em><u> </u></em><em><u>I</u></em><em><u>t</u></em><em><u> </u></em><em><u>W</u></em><em><u>i</u></em><em><u>l</u></em><em><u>l</u></em><em><u> </u></em><em><u>H</u></em><em><u>e</u></em><em><u>l</u></em><em><u>p</u></em><em><u> </u></em><em><u>Y</u></em><em><u>o</u></em><em><u>u</u></em><em><u>!</u></em></h2>
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what are the various ways in which lithospheric plates interact with each other as they move around on a dynamic earth??
Ivan
<span>The plates make up Earth's outer shell, called the lithosphere. (This includes the crust and uppermost part of the mantle.) Churning currents in the molten rocks below propel them along like a jumble of conveyor belts in disrepair. Most geologic activity stems from the interplay where the plates meet or divide.</span>
3 0
3 years ago
Is the time in Tennessee the same as florida
geniusboy [140]
I would say the same thing as the first answer
5 0
3 years ago
Read 2 more answers
An oxygen molecule consists of two oxygen nuclei, each of massm= 2.7×10−26kg,separated by a distance 1.2×10−10m, and surrounded
fenix001 [56]

Answer:

a)  I = 1,944 10⁻⁴⁶ Kg m²

, b)   I = 7,915 10⁻⁵¹ Kg m²

Explanation:

a) The moment of inertia of point masses is

                 I = m r²

   

The nuclei have a very small size (10-15 m) so we can consider them punctual.

The distance to the center of mass passing through the middle of the two nuclei of equal mass

       r = 0.6 10⁻¹⁰ m

The moment of total inertia

       I = I₁ + I₂ = 2 I₀

       I = 2 2.7 10⁻²⁶ (0.6 10⁻¹⁰)²

       I = 1,944 10⁻⁴⁶ Kg m²

The kinetic energy of the rotation is

      w = h / 2π

      K = ½ I w²

      K = ½ 1,944 10⁻⁴⁶ (h / 2π)² = ½ 1.944 10⁻⁴⁶ (6.63 10⁻³⁴ / 2π)²

      K = 2.16 10⁻¹¹⁴ J (1eV / 1.6 10⁻¹⁹ J)

      K = 2.16 10⁻⁹⁵ eV

B) the moment of inertia of the electron in the orbit, we can calculate it with the parallel axis theorem

       I =I_{cm} + m R²

       I = m_{e} r² +  m_{e} R²

Where R we calculate it by Pythagoras

      R² = (0.6 10⁻¹⁰)² + (0.5 10⁻¹⁰) 2

      R = √ (0.61 10⁻²⁰)

      R = 0.78 10⁻¹⁰ m

      I = 9.1 10⁻³¹ (0.5 10⁻¹⁰)² + 9.1 10⁻³¹ (0.78 10⁻¹⁰)²

      I = (2,275 +5.54) 10⁻⁵¹

      I = 7,915 10⁻⁵¹ Kg m²

The rotation energy of the electron

      K = ½ I w²

If the angular velocity is the electrons outside the core, its kinetic energy is much lower by an order 10⁵, but the angular velocity of the electrons is much higher.

7 0
3 years ago
Characteristics help scientists ________ objects.
Naddika [18.5K]

Answer:Identify

Explanation:

6 0
3 years ago
Read 2 more answers
A spring is used to stop a 50-kg package which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held b
Elina [12.6K]

Answer:

0.3 m

Explanation:

Initially, the package has both gravitational potential energy and kinetic energy.  The spring has elastic energy.  After the package is brought to rest, all the energy is stored in the spring.

Initial energy = final energy

mgh + ½ mv² + ½ kx₁² = ½ kx₂²

Given:

m = 50 kg

g = 9.8 m/s²

h = 8 sin 20º m

v = 2 m/s

k = 30000 N/m

x₁ = 0.05 m

(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²

x₂ ≈ 0.314 m

So the spring is compressed 0.314 m from it's natural length.  However, we're asked to find the additional deformation from the original 50mm.

x₂ − x₁

0.314 m − 0.05 m

0.264 m

Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.

8 0
3 years ago
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