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3 years ago

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A positive charge distribution exists within a nonconducting spherical region of radius a. The volume charge density ρ is not un

iform but varies with the distance r from the center of the spherical charge distribution, according to the relationship ρ=βr for 0<=0<=a, where β is a positive constant, and ρ=0, and r>a.A. Show that the total charge Q in the spherical region of radius a is βπa^4B. In terms of β,r,a and fundamential constants, determine the magnitude of the electric field at a point a from distance r from the center of the spherical charge distribution for each of the following cases.i. r>aii. r=aiii 0

1 answer:

baherus [9]3 years ago

5 0

A negative charge and a positive charge

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enot [183]

Answer:

B

Explanation:

I just took the test

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3 years ago

1) You slam on the brakes of your car in a panic, and skid a certain distance on a straight level road. If you had been travelin

aleksandr82 [10.1K]

Answer:

d = 4 d₀o

Explanation:

We can solve this exercise using the relationship between work and the variation of kinetic energy

W = ΔK

In that case as the car stops v_f = 0

the work is

W = -fr d

we substitute

- fr d₀ = 0 - ½ m v₀²

d₀ = ½ m v₀² / fr

now they indicate that the vehicle is coming at twice the speed

v = 2 v₀

using the same expressions we find

d = ½ m (2v₀)² / fr

d = 4 (½ m v₀² / fr)

d = 4 d₀o

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2 years ago

Please answer :><br> 40 POINTS

LenaWriter [7]

Answer:

rotates

Explanation:

I'm so bored

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3 years ago

Read 2 more answers

A baseball pitcher throws the ball towards the batter at 90 mph. His bat connects with the ball for a line drive, after which th

forsale [732]

Answer:

$F=-18412.9N$ , where the minus indicates the direction is opposite to that of the throw.

Explanation:

a)

Since MKS stands for meter-kilogram-second and we know that:

$1\ hour = 3600\ seconds$

$1\ mile = 1600\ meters$

$1000g = 1kg$

We can write that:

$\frac{1\ hour}{3600\ seconds}=1$

$\frac{1600\ meters}{1\ mile}=1$

$\frac{1kg}{1000g}=1$

These are conversion factors, equal to 1, so multiplying our results by them won't change their value, only their units.

So we have that:

$90 mph=90 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=40m/s$

$110 mph=110 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=48.89m/s$

$145 g=145 g(\frac{1kg}{1000g})=0.145kg$

b)

Newton's 2nd Law tells us that F=ma, and the definition of acceleration is $a=\frac{\Delta v}{\Delta t}$ , so we have:

$F=m\frac{\Delta v}{\Delta t}=m\frac{v_f-v_i}{t}$

Taking the throw direction as the positive one, for our values we have:

$F=m\frac{v_f-v_i}{t}=(0.145kg)\frac{(-48.89m/s)-(+40m/s)}{0.0007s}=-18412.9N$

4 0

3 years ago

The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

Simora [160]

Answer:

The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

$F = k \frac{q_1 q_2}{d^2}$

where k is Coulomb's constant, $q_1$ and $q_2$ are the charges and d is the distance between the charges.

Working a little the equation, we can take:

$d^2 = k \frac{q_1 q_2}{F}$

$d = \sqrt{ k \frac{q_1 q_2}{F}}$

And this equation will give us the distance between the charges. Taking the values of the problem

$k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00$

(the force has a minus sign, as its attractive)

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 28,462,500 \ m^2}}$

$d = 5,335.026 m$

And this is the distance between the charges.

3 0

3 years ago