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2 years ago

How do you work out Potential Difference??? Please can you make it simple :) Thanks.

Physics

1 answer:

patriot [66]2 years ago

7 0

Answer:

Potential difference is the work done in moving a positive test charge from infinity to the point in question.

Voltage is an expression of PD. (Joules / Coulomb)

Say that a capacitor has a PD of 5 Volts. The work in moving a positive test charge from the positive plate to the negative plate is -5 Joules/Coulomb or -5 volt. (At the positive plate the positive test charge (1 Coulomb) already has a PD of + 5 Volts.)

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The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the Sun, namely 1.496x10^(11) m. The p

Rudiy27

Answer:

a) How many parsecs are there in one astronomical unit?

$4.85x10^{-6}pc$

(b) How many meters are in a parsec?

$3.081x10^{16}m$

$9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$63325AU$

(e) How many light-years in a parsec?

3.26ly

Explanation:

The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun ( $1.496x10^{11} m$ ) is defined as 1 astronomical unit:

$\tan{p} = \frac{1AU}{d}$

Where d is the distance to the star.

Since p is small it can be represent as:

$p(rad) = \frac{1AU}{d}$ (1)

Where p(rad) is the value of in radians

However, it is better to express small angles in arcseconds

$p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}$

$p('') = 2.06x10^5 p(rad)$

$p(rad) = \frac{p('')}{2.06x10^5}$ (2)

Then, equation 2 can be replace in equation 1:

$\frac{p('')}{2.06x10^5} = \frac{1AU}{d}$

$\frac{d}{1AU} = \frac{2.06x10^5}{p('')}$ (3)

From equation 3 it can be see that $1pc = 2.06x10^5 AU$

a) How many parsecs are there in one astronomical unit?

$1AU . \frac{1pc}{2.06x10^5AU}$ ⇒ $4.85x10^{-6}pc$

(b) How many meters are in a parsec?

$2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU}$ ⇒ $3.081x10^{16}m$

(c) How many meters in a light-year?

To determine the number of meters in a light-year it is necessary to use the next equation:

$x = c.t$

Where c is the speed of light ( $c = 3x10^{8}m/s$ ) and x is the distance that light travels in 1 year.

In 1 year they are 31536000 seconds

$x = (3x10^{8}m/s)(31536000s)$

$x = 9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m}$ ⇒ $63325AU$

(e) How many light-years in a parsec?

$2.06x10^{5}AU . \frac{1ly}{63235AU}$ ⇒ $3.26ly$

5 0

3 years ago

Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes th

Airida [17]

F = G m1*m2 / r^2 => [G] = [F]*[r]^2 /([m1]*[m2]) = N * m^2 / kg^2

That is one answer.

Also, you can use the fact that N = kg*m/s^2

[G] = kg * m / s^2 * m^2 / kg^2 = m^3 /(s^2 * kg)

5 0

2 years ago

A length of wire is cut into five equal pieces. if each piece has a resistance of 0.45 ω, what was the resistance of the origina

azamat

Calculation, add them. 0.85 x 5 = 4.25 Ω

5 0

2 years ago

It is known that birds can detect the earth's magnetic field, but the mechanism of how they do this is not known. It has been su

Lubov Fominskaja [6]

Answer:

A) 0.50 mV

Explanation:

In this problem, we can think the wings of the bird as a metal rod moving across a magnetic field. So, and emf will be induced into the wings of the bird, according to the formula:

$\epsilon = BvL sin \theta$

where

$B=5\cdot 10^{-5} T$ is the strength of the magnetic field

v = 13 m/s is the speed of the bird

L = 1.2 m is the wingspan of the bird

$\theta=40^{\circ}$ is the angle between the direction of motion and the direction of the magnetic field

Substituting numbers into the formula, we find

$\epsilon = (5.0\cdot 10^{-5} T)(13 m/s)(1.2 m) sin 40^{\circ}=0.00050 V = 0.50 mV$

8 0

2 years ago

The landing of a spacecraft is cushioned with the help of airbags. During its landing on Mars, the velocity of downward fall is

statuscvo [17]

You can use the impulse momentum theorem and just subtract the two momenta.
P1 - P2 = (16-1.2)(11.5e4)=1702000Ns
If you first worked out the force and integrated it over time the result is the same

5 0

2 years ago

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