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4 years ago

WHAT are the highest frequency and the lowest frequency pets of the EM spectrum?

Physics

1 answer:

Rus_ich [418]4 years ago

8 0

Highest frequency EM waves: cosmic rays and gama rays

Lowest frequency EM waves:
Radio and Tv waves

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Gerry is looking at salt under a powerful microscope and notices a crystalline structure. What can be known about the salt sampl

malfutka [58]

i know it is c

here is the answer

7 0

2 years ago

A potassium atom (atomic number 19) and a bromine atom (atomic number 35) can form a chemical bond through a transfer of one ele

Nookie1986 [14]

Answer:

it will give one electron to the bromine atom, so the bromine atom will have 36 electrons. this is an ionic bonding

Explanation:

3 0

3 years ago

A small segment of wire contains 10 nC of charge. The segment is shrunk to one-third of its original length. A proton is very fa

Alik [6]

To solve this problem we will apply the concepts related to the electric field, linear charge density and electrostatic force.

The electric field is

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Here,

$\lambda$ = Linear charge density

$\epsilon_0$ = Permittivity of free space

r = Distance

The linear charge density can be written as,

Linear charge density is given as

$\lambda = \frac{q}{L}$

Replacing,

$E = \frac{\frac{q}{L}}{2\pi \epsilon_0 r}$

$E = \frac{q}{2\pi \epsilon_0 rL}$

The initial and final electric Force can be written as function of the charge and the electric field as

$F_i = E_i q$

$F_f = E_f q$

If we replace the value for the electric field we have,

$F_i = (\frac{q}{2\pi \epsilon_0 rL})q = (\frac{q^2}{2\pi \epsilon_0 rL})$

Length is one third at the end, then

$F_f = (\frac{q}{2\pi \epsilon_0 r(L/3)})q = (\frac{3q^2}{2\pi \epsilon_0 rL})$

The ratio of the force is

$\frac{F_f}{F_i} = \frac{(\frac{3q^2}{2\pi \epsilon_0 rL})}{(\frac{q^2}{2\pi \epsilon_0 rL})}$

$\frac{F_f}{F_i} = 3$

Therefore the required ratio is 3

7 0

3 years ago

Consider two vectors A~ and B~ and their resultant A~ +B~ . The magnitudes of the vectors A~ and B~ are, respectively, 18.2 and

saw5 [17]

Answer:

15.76°

Explanation:

Hi!

Let C~ = A~ + B~

Then

C~ dot A~ = |C| |A| cos Ф

C~ dot A~ = (A~ + B~) dot A~ = |A|^2 + |B| |A| cos(120)

Therefore

cos Ф = (|A| + |B| cos(120)) / |C| = 15.65/|C|

and

|C|^2 = |A|^2 + |B|^2 + 2 |B| |A| cos(120)

|C| = 16.261

cos Ф = 15.65/16.261

Ф = 15.76°

7 0

4 years ago

g A bowling ball with a mass of 3.86 kg and a radius of 0.161 m starts from rest at a height of 2.5 m and rolls down a 48.4 o sl

Fynjy0 [20]

Answer:

$v=1.5m/s$

Explanation:

The gravitational potential energy gets transformed into translational and rotational kinetic energy, so we can write $mgh=\frac{mv^2}{2}+\frac{I\omega^2}{2}$ . Since $v=r\omega$ (the ball rolls without slipping) and for a solid sphere $I=\frac{2mr^2}{5}$ , we have:

$mgh=\frac{mv^2}{2}+\frac{2mr^2\omega^2}{2*5}=\frac{mv^2}{2}+\frac{mv^2}{5}=\frac{7mv^2}{10}$

So our translational speed will be:

$v=\sqrt{\frac{10gh}{7}}=\sqrt{\frac{10(9.8m/s^2)(0.161m)}{7}}=1.5m/s$

6 0

3 years ago