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Alla [95]
3 years ago
14

ANSWER PLEASEEE AS FAST AS POSSIBLE !!

Physics
2 answers:
KonstantinChe [14]3 years ago
6 0

Answer:

4 A

Explanation:

V=IR where V is voltage, R resistance and I as current hence per the question

V=12V, R = 3ohms

V=IR

12 = I x 3

12/3 = I

I =4 Amps

shusha [124]3 years ago
4 0

Answer:

V=IR

Explanation:

12=(I)*3

I=4A

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Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.
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Explanation:

We have,

Semimajor axis is 4\times 10^{12}\ m

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

T^2=\dfrac{4\pi ^2}{GM}a^3

G is universal gravitational constant

M is solar mass

Plugging all the values,

T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s

Since,

1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}

So, the orbital period of a dwarf planet is 138.52 years.

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How can citizen -scientists help with the prediction of future earthquakes
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Answer:

Avoid downed power lines and stay away from buildings and bridges from which heavy objects might fall during an aftershock. Stay away until local officials tell you it is safe. A tsunami is a series of waves that may continue for hours. Do not assume that after one wave the danger is over.

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A glass lens that has an index of refraction equal to 1.57 is coated with a thin layer of transparent material that has an index
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Why are atoms of lithium, sodium, and potassium almost never found alone together in nature?
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3 years ago
A solid disk of mass 2 kg and radius 2 m is given a horizontal push of 20N at a point .3 m above its center. a. What is the mini
Margaret [11]

Answer:

\mu_s=1.0205

Explanation:

Given:

  • mass of solid disk, m=2\ kg
  • radius of disk, r=2\ m
  • force of push applied to disk, F=20\ N
  • distance of application of force from the center, s=0.3\ m

<em>For the condition of no slip the force of  static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>

\therefore F

where:

f_s = static frictional force

\Rightarrow 20

\Rightarrow 20

\Rightarrow 20

\mu_s>1.0204

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