In your own words describe how matter is defined

Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

Elena L [17]

Answer:

$2.62898\times 10^{-6}\ C/m^3$

$1979.99974\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$

The volume charge density for the sphere is $2.62898\times 10^{-6}\ C/m^3$

$E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C$

The magnitude of the electric field is $1979.99974\ N/C$

8 0

2 years ago

A wire with a resistance R is lengthened to 3.07 times its original length by pulling it through a small hole. Find the new resi

charle [14.2K]

Explanation:

Below is an attachment containing the solution.

4 0

2 years ago

Describe the point at which the kinetic energy of the pendulum is the highest

Arlecino [84]

The highest point of the pendulums swing is when the potential energy is at its highest and the kinetic energy is at its lowest.

5 0

3 years ago

In his motorboat, Bill Ruhberg travels upstream at top speed to his favorite fishing spot, a distance of 120120 mi, in 33 hr.

photoshop1234 [79]

Answer:

The rate of the boat in still water is 44 mph and the rate of the current is 4 mph

Explanation:

x = the rate of the boat in still water

y = the rate of the current.

Distance travelled = 120 mi

Time taken upstream = 3 hr

Time taken downstream = 2.5 hr

Speed = Distance / Time

Speed upstream

$\frac{120}{3}=x-y\\\Rightarrow 40=x-y$

Speed downstream

$\frac{120}{2.5}=x+y\\\Rightarrow 48=x+y$

Adding both the equations

$48+40=x-y+x+y\\\Rightarrow 88=2x\\\Rightarrow 44=x$

$40=44-y\\\Rightarrow 40-44=-y\\\Rightarrow y=4$

The rate of the boat in still water is 44 mph and the rate of the current is 4 mph

8 0

2 years ago

A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation betw

TEA [102]

Answer:

The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)

Explanation:

The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. $Q$ , the amount of charge stored in this capacitor, will stay the same.

The formula $\displaystyle Q = C\, V$ relates the electric potential across a capacitor to:

$Q$ , the charge stored in the capacitor, and
$C$ , the capacitance of this capacitor.

While $Q$ stays the same, moving the two plates apart could affect the potential $V$ by changing the capacitance $C$ of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:

$\displaystyle C = \frac{\epsilon\, A}{d}$ ,

where

$\epsilon$ is the permittivity of the material between the two plates.
$A$ is the area of each of the two plates.
$d$ is the distance between the two plates.

Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of $\epsilon$ . Neither will that change the area of the two plates.

However, as $d$ (the distance between the two plates) increases, the value of $\displaystyle C = \frac{\epsilon\, A}{d}$ will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.

On the other hand, the formula $\displaystyle Q = C\, V$ can be rewritten as:

$V = \displaystyle \frac{Q}{C}$ .

The value of $Q$ (charge stored in this capacitor) stays the same. As the value of $C$ becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.

3 0

2 years ago