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3 years ago

Can someone help me with this whole page? i can't seem to figure it out. Is physics homework

Physics

1 answer:

Roman55 [17]3 years ago

7 0

Go on khan academy they help you with all of this stuff and you will know more

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A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

5 0

3 years ago

Ohms law is A.(R=E/W). B.(R=E/1). C.(E/Z). D.none of them

nikklg [1K]

Answer:

D. none of them.

Explanation:

This is because Ohm's law is:

Voltage = Current × Resistance

or,

V = IR

6 0

3 years ago

An object travels 20 m in 10s What is its speed?

ella [17]

speed=distance/time

distance=20m

time=10s

speed=?

speed=20×10

speed=200m/s

3 0

3 years ago

A constant force is applied to an object, causing the object to accelerate at 10 m/s^2. What will the acceleration be if: a) The

Liula [17]

What you need to know is that the force is

F=ma

The force is the product of mass and acceleration

this means that the acceleration is

a=F/m

a) The force is halved?
this means that f will be $\frac{F}{2}$ now:

a= $\frac{F}{2m}$

So the accelaration will also he halved (it's the original acceleratation divided by 2)

b) The object's mass is halved?
a= $\frac{F}{m/2}$ =a= $\frac{F2}{m}$

which is the original acceleration times two!! so it will double

c) The force and the object's mass are both halved?
now we have

a= $\frac{F/2}{m/2}$ =a= $\frac{2F}{2m}$ =a= $\frac{F}{m}$

so they will cancel each other out and the acceleration will stay the same!

5 0

3 years ago

Why might political parties prefer closed primaries to open primaries

sweet [91]

In a closed primary, only voters registered with a given party can vote in that party's primary.

5 0

3 years ago