What are the big 5 equations for physics

A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect

Stels [109]

Answer:

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

Explanation:

Given that

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field at point P

$E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta$

$sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}$

$r^2=d^2+x^2$

So

$E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}$

Now by integrating above equation

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

4 0

3 years ago

A student connects an object with mass m to a rope with a length r and then rotates the rope around her head parallel to the gro

Alexandra [31]

The object takes 0.5 seconds to complete one rotation, so its rotational speed is 1/0.5 rot/s = 2 rot/s.

Convert this to linear speed; for each rotation, the object travels a distance equal to the circumference of its path, or 2π (1.2 m) = 2.4π m ≈ 7.5 m, so that

2 rot/s = (2 rot/s) • (2.4π m/rot) = 4.8π m/s ≈ 15 m/s

thus giving it a centripetal acceleration of

a = (4.8π m/s)² / (1.2 m) ≈ 190 m/s².

Then the tension in the rope is

T = (50 kg) a ≈ 9500 N.

7 0

3 years ago

An object is 15 cm in front of a diverging lens with a

Rainbow [258]

A) See ray diagram in attachment (-6.0 cm)

By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where

q is the distance of the image from the lens

f = -10 cm is the focal length (negative for a diverging lens)

p = 15 cm is the distance of the object from the lens

Solving for q,

$\frac{1}{q}=\frac{1}{-10 cm}-\frac{1}{15 cm}=-0.167 cm^{-1}$

$q=\frac{1}{-0.167 cm^{-1}}=-6.0 cm$

B) The image is upright

As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:

$h_i = - h_o \frac{q}{p}$

where $h_i, h_o$ are the size of the image and of the object, respectively.

Since q < 0 and p > o, we have that $h_i >0$ , which means that the image is upright.

C) The image is virtual

As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.

This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual

4 0

3 years ago

Sodium has an atomic mass of 23 and an atomic number of 11, so the

horrorfan [7]

Answer:

12 Neutrons

Explanation:

So the mass of sodium is 22.990. You round it up to get 23(as stated in the problem). So, what exactly is atomic mass?

Atomic Mass is the total amount of neutrons and protons added up to form a total mass. So when you subtract 23-11 you get 12 Neutrons.

Tip: Don't know if you need this but-

The neutrons and protons are typically close in number (unless it's an isotope). So say that you subtract and the numbers of protons and neutrons aren't close at all. Well if that's the case, it's probably wrong.

hope this helps!!

5 0

3 years ago

How can you justifies the age of earth being 4.5 billion years old

IceJOKER [234]

Answer:

ummmmmmmmmm through science??

8 0

3 years ago