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dsp73
2 years ago
11

What ate the effects of gravitional force​

Physics
1 answer:
faust18 [17]2 years ago
4 0

Answer:

Gravity is what holds the planets in orbit around the sun and what keeps the moon in orbit around Earth. The gravitational pull of the moon pulls the seas towards it, causing the ocean tides. Gravity creates stars and planets by pulling together the material from which they are made.

Explanation:

I was learning about this in class

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A current of 0.4 A flows through a wire. How many electrons flow through a cross section of
Free_Kalibri [48]

9 × 10²¹ electrons flow through a cross section of the wire in one hour.

<h3>What is the relation between current and charge?</h3>
  • Mathematically, current = charge / time
  • In S.I. unit, Charge is written in Coulomb and time in second.

<h3>What is the amount of charge flown through a wire for one hour if it carries 0.4 A current?</h3>
  • Charge= current × time
  • Current= 0.4 A, time = 1 hour= 3600 s
  • Charge= 0.4× 3600

= 1440 C

<h3>How many numbers of electrons present in 1440C of charge?</h3>
  • One electron= 1.6 × 10^(-19) C
  • So, 1440 C = 1440/1.6 × 10^(-19)

= 9 × 10²¹ electrons

Thus, we can conclude that the 9 × 10²¹ electrons flow through a cross section of the wire in one hour.

Learn more about current here:

brainly.com/question/25922783

#SPJ1

4 0
2 years ago
According to ohm's law if you don't change the value of the resistor &amp; you double the voltage in a circuit the amount of cur
Nana76 [90]

They double!! Hope im not too late!!

3 0
3 years ago
Convert planks constant in cgs system
dezoksy [38]

in cgs system, plank's constant= h=6.626 x10⁻²⁶ erg s

Value of Plank's constant in SI system= 6.626 x10⁻³⁴ Js

now 1 Joule= 10⁷ ergs

so h= 6.626 x10⁻³⁴ Js (10⁷ ergs/1J)

h=6.626 x10⁻²⁷ erg s

7 0
3 years ago
A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien
n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

Put the value into the formula

\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2

\phi=7.85\times10^{-7}\ Tm^2

We need to calculate the induced emf

Using formula of induced emf

\epsilon=\dfrac{d\phi}{dt}

Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

\epsilon =IR

IR=\dfrac{7.85\times10^{-7}}{dt}

Idt=\dfrac{7.85\times10^{-7}}{R}

Idt=\dfrac{7.85\times10^{-7}}{0.50}

Idt=0.00000157=1.57\times10^{-6}\ C

We know that,

Idt=dq

dq=1.57\times10^{-6}\ C

We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

dV=1.57\ V

Hence, The voltage across the capacitor is 1.57 V.

5 0
3 years ago
PLEASE HELP ASAP!!!! A huge thanks to anyone who can help me with 14 problems. I'll do anything to return the favor. All true an
snow_lady [41]
Hello, I see you are in a jam. Lemme help.

1.) True
2.) True
3.) True
4.) True
5.) True

LOL these are all true ;)
4 0
3 years ago
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