You add 50 mL of water at 20°C to 200 mL of water at 70°C. What is the most

An excited hydrogen atom releases an electromagnetic wave to return to its normal state. You use your futuristic dual electric/m

Norma-Jean [14]

Answer:

Explanation:

Direction of velocity of electromagnetic wave is given by the formula

E = E j ( vector form )

B = - Bi ( vector form )

Direction of velocity = direction of vector E X B

= E X B

= E j x -Bi

= - EB -k

v = EBk

So the direction of velocity will be along z direction.

6 0

3 years ago

How does the heat from sun reach earth

kakasveta [241]

Answer:

The sun heats the earth through radiation. Since there is no medium (like the gas in our atmosphere) in space, radiation is the primary way that heat travels in space. When the heat reaches the earth it warms the molecules of the atmosphere, and they warm other molecules and so on.

3 0

3 years ago

A bungee jumper has a mass of 60kg and uses a 25m long bungee cord (unstretched length) with an elastic coefficient of 800N/m. a

KonstantinChe [14]

Answer:

a ) 2.68 m / s

b ) 1.47 m

Explanation:

The jumper will go down with acceleration as long as net force on it becomes zero . Net force of (mg - kx ) will act on it where kx is the restoring force acting in upward direction.

At the time of equilibrium

mg - kx = 0

x = mg / k

= (60 x 9.8 ) / 800

= 0.735 m

At this moment , let its velocity be equal to V

Applying conservation of energy

kinetic energy of jumper + elastic energy of cord = loss of potential energy of the jumper

1/2 m V² + 1/2 k x² = mg x

.5 x 60 x V² + .5 x 800 x .735 x .735 = 60 x 9.8 x .735

30 V² + 216.09 = 432.18

V = 2.68 m / s

b ) At lowest point , kinetic energy is zero and loss of potential energy will be equal to stored elastic energy.

1/2 k x² = mgx

x = 2 m g / k

= (2 x 60 x 9.8) / 800

= 1.47 m

3 0

3 years ago

Read 2 more answers

Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance

siniylev [52]

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

E_ {total} = E₁ + E₂

E_{ total} = $k \frac{Q}{x_1^2} + k \frac{Q}{x_2^2}$

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

x₂ = x₁ -d

E total = $k \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}$

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

E_ {total} = $k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )$

E_ {total} = $k \frac{Q}{x_1^2}$ $( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )$

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

E_ {total} = 0

x₁² = (x₁ + d)²

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

E_ {total} = $-k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}$

this expression holds for the points located at

-r₁ <x₁ <r₁

3) Point between the two spheres

E_ {total} = $k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}$

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

E_ {total} = $+ k \frac{Q}{(d-x_1)^2}$ + k Q / (d-x1) 2

point range

-r₂ <x₂ <r₂

5) Right side point

E_ {total} = $k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}$

E_ {total} = $- k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )$ - k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

8 0

2 years ago

Does Anybody Know The Answers?

ch4aika [34]

Answer:

I was going to give you the paper where I saw it but since you are not giving enough points I can not give you so I am only going to give you some of these that are here sorry

Explanation:

1.

$9^{2} + 12^2 = x^2\\81 + 144= x^2\\\sqrt{225} = \sqrt{x} \\ 15=x\\\\ 2.\\x^2+12^2+=13^2\\x^2+144 =169\\x^2 = 25\\\sqrt{x^2 =\sqrt{25\\\\$