What is the amplitude of a wave related to

Rashid [163]

One crest of a wave to another.

5 0

3 years ago

Q.1 Lorna built the circuit diagram below. All the bulbs are identical. (a) Complete the table below by writing on or off for ea

irga5000 [103]

Answer:

on and off

Explanation:

if there are switches, it can change if the electricity can get to the bulb or not. if it appears that there is no pathway for the electricity to get to the light bulb, it is of, if there is a pathway, its on

8 0

2 years ago

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

An object in projectile motion will follow which path? a]curved up from the ground b]curved down toward the ground c]straight do

Lesechka [4]

It's B because when you throw something it doesn't go up it slowly descends downward

7 0

3 years ago

Read 2 more answers

You place a 0.17 kg can of soup and a 0.31 kg jar of pickles on the kitchen counter, separated by a distance of 0.42 m. What is

tangare [24]

$1.984 \times 10^{-11} \mathrm{N} \text { is the force of gravity exerted on the jar of pickles. }$

<u>Explanation</u>:

According to Newton's third law that each force has an equal and opposite reaction force in this case both of the jars will exert the same force an each other

. The force is given by

$\mathrm{F}=\frac{G \times M_{1} \times M_{2}}{r^{2}}$

Where, F = force, $G=\text { gravitational constant }=\left(6.67 \times 10^{-11}\right)$ , $mass \left(\mathrm{M}_{1}\right)=0.17 \mathrm{kg}$ , $mass \left M_{2}= 0.31 \mathrm{kg}$ and Distance(r) = 0.42 m.

Substitute the values in the formula.

$\mathrm{F}=\frac{6.67 \times 10^{-11} \times 0.17 \times 0.31}{0.42^{2}}$

$\mathrm{F}=\frac{3.51 \times 10^{-12}}{0.176}$

$\mathrm{F}=1.984 \times 10^{-11} \mathrm{N}$

$\text { The force of gravity exerted on the jar of pickles is } 1.984 \times 10^{-11} \mathrm{N} \text { . }$

3 0

3 years ago