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3 years ago

If Charge A is positive, then Charge B must be:

Physics

1 answer:

Eduardwww [97]3 years ago

4 0

It must be positive also
Explanation The energy is coming out of A back into b and then going into A again and out

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vladimir2022 [97]

Answer:

6118N

Explanation:

7 0

4 years ago

Calculate the pressure in bar of 2520 moles of hydrogen gas stored at 27 °c in the 180-l storage tank of a modern hydrogen-power

marin [14]

Answer: 347 bar

Explanation:

1) Data:

n = 2520 moles

T = 27 + 273.15 = 298.15 K

V = 180 liter

p = ?

2) Formula

Ideal gas equation: pV = nRT

3) Solution:

p = nRT/V = 2520 mol × 0.0821 atm×liter/K×mol ×298.15K / 180 liter = 342.7 atm

Conversion: 342.7 atm × 1,01325 bar / atm = 347 bar

7 0

3 years ago

4000 ml to kl plis help mee

Damm [24]

It is .004000 using this chart

7 0

4 years ago

Calculate the speed of a satellite moving in a stable circular orbit about the Earth at a height of 4000 km from the surface of

evablogger [386]

v^2 =GM/r^2

the speed of satellite is 3.12*10^6

8 0

4 years ago

A planet of mass m 6.75 x 1024 kg is orbiting in a circular path a star of mass M 2.75 x 1029 kg. The radius of the orbit is R 8

Umnica [9.8K]

Answer:

The orbital period of the planet is 387.62 days.

Explanation:

Given that,

Mass of planet $m =6.75\times10^{24}\ kg$

Mass of star $m'=2.75\times10^{29}\ kg$

Radius of the orbit $r =8.05\times10^{7}\ km$

Using centripetal and gravitational force

The centripetal force is given by

$F = \dfrac{mv^2}{r}$

$F=m\omega^2r$

We know that,

$\omega=\dfrac{2\pi}{T}$

$F=m(\dfrac{2\pi}{T})^2r$ ....(I)

The gravitational force is given by

$F = \dfrac{mm'G}{r^2}$ ....(II)

From equation (I) and (II)

$m(\dfrac{2\pi}{T})^2r=\dfrac{mm'G}{r^2}$

Where, m = mass of planet

m' = mass of star

G = gravitational constant

r = radius of the orbit

T = time period

Put the value into the formula

$T^2=\dfrac{4\pi^2R^3}{m'G}$

$T^2=\dfrac{4\times(3.14)^2\times(8.05\times10^{7})^3}{2.75\times10^{29}\times6.67\times10^{-11}}$

$T=2\times3.14\times\sqrt{\dfrac{(8.05\times10^{10})^3}{2.75\times10^{29}\times6.67\times10^{-11}}}$

$T =3.34\times10^{7}\ s$

$T= 387.62\days$

Hence, The orbital period of the planet is 387.62 days.

4 0

3 years ago

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