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2 days ago

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The magnitude of the electrical force acting between a +2.4 × 10–8 C charge and a +1.8 × 10–6 C charge that are separated by 0.0

08 m is N, rounded to the tenths place.

2 answers:

Keith_Richards [1K]2 days ago

8 0

On Ed, it's 6.1 N. I just completed the test and got it right!

kicyunya [1K]2 days ago

8 0

To solve this problem, Coulomb's law will be applied as follows:
F = k*q1*q2 / r^2 where:
F indicates the force magnitude between the charges
k is a constant = 9.00 * 10^9 N.m^2/C^2
q1 = <span>+2.4 × 10–8 C
q2 = </span><span>+1.8 × 10–6 C
r represents the distance separating the charges = </span><span>0.008 m

By substituting these values, we derive:
F = (9*10^9)(2.4*10^-8)(1.8*10^-6) / (0.008)^2 = 6.075, which rounds to 6.1 Newtons

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Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet

ValentinkaMS [1149]

Answer:

The rate at which the root beer level is decreasing is 0.08603 cm/s.

Explanation:

The formula for the volume of the cone is:

$V=\frac {1}{3}\times \pi\times r^2\times h$

Where V denotes the cone's volume

r indicates the radius

h signifies the height

The ratio of radius to height remains consistent throughout the cone.

Thus, we have r = d / 2 = 10 / 2 cm = 5 cm

h is 13 cm

Consequently, r / h = 5 / 13

r = {5 / 13} h

$V=\frac {1}{3}\times \frac {22}{7}\times ({{{\frac {5}{13}\times h}}})^2\times h$

$V=\frac {550}{3549}\times h^3$

Additionally, we differentiate the volume expression in relation to time:

$\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}$

Given that $\frac {dV}{dt}$ = -4 cm³/sec (the negative sign indicates outflow)

h equals 10 cm

Hence,

$-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}$

$\frac{55000}{1183}\times \frac {dh}{dt}=-4$

$\frac {dh}{dt}=-0.08603\ cm/s$

The rate at which the root beer level is decreasing is 0.08603 cm/s.

3 0

11 days ago

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

Sav [1105]

Respuesta:

11.4 m/s

Explicación:

La fórmula para la aceleración centrípeta es:

$a=\frac{v^2}{R}$

donde, a es la aceleración, v la velocidad alrededor de la circunferencia y R el radio del círculo.

En este problema,

a = g = aceleración debida a la gravedad en la cima = $9.81\ m/s^2$

v = ?

R = 13.2 m

Por lo tanto,

$9.81=\frac{v^2}{13.2}$

$v^2=9.81\times {13.2}$

v = 11.4 m/s

8 0

13 days ago

An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

Maru [1056]

1) For x = 6.6 cm, $E_x=3.47\cdot 10^6 N/C$

2) For x = 6.6 cm, $E_y=0$

3) For x = 1.45 cm, $E_x=-3.76\cdot 10^6N/C$

4) For x = 1.45 cm, $E_y=0$

5) Surface charge density at b = 4 cm: $+62.75 \mu C/m^2$

6) At x = 3.34 cm, the x-component of the electric field equals zero

7) Surface charge density at a = 2.9 cm: $+65.25 \mu C/m^2$

8) None of these regions

Explanation:

1)

The electric field from an infinite charge sheet is perpendicular to it:

$E=\frac{\sigma}{2\epsilon_0}$

where

$\sigma$ is the surface charge density

$\epsilon_0=8.85\cdot 10^{-12}F/m$ represents vacuum permittivity

Outside the slab, the electric field behaves like that of an infinite sheet.

Consequently, the electric field at x = 6.6 cm (situated to the right of both the slab and sheet) results from the combination of the fields from both:

$E=E_1+E_2=\frac{\sigma_1}{2\epsilon_0}+\frac{\sigma_2}{2\epsilon_0}$

where

$\sigma_1=-2.5\mu C/m^2 = -2.5\cdot 10^{-6}C/m^2\\\sigma_2=64 \muC/m^2 = 64\cdot 10^{-6}C/m^2$

The field from the sheet points left (negative, inward), and the slab’s field points right (positive, outward).

Thus,

$E=\frac{1}{2\epsilon_0}(\sigma_1+\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}+64\cdot 10^{-6})=3.47\cdot 10^6 N/C$

and the negative sign indicates a rightward direction.

2)

Both the sheet’s and slab’s fields are perpendicular to their surfaces, directing along the x-axis, hence there's no y-component for the total field.

<pThus, the y-component totals zero.

This happens because both the sheet and slab stretch infinitely along the y-axis. Choosing any x-axis point reveals that the y-component of the field, generated by a surface element dS of either the sheet or slab, $dE_y$ , will be equal and opposite to the corresponding component from the opposite side, $-dE_y$ . Thus, the combined y-direction field is always zero.

3)

This scenario resembles part 1), but the point here is

x = 1.45 cm

which lies between the sheet and the slab. The fields from both contribute leftward as the slab has a negative charge (resulting in an outward field). Thus, the total field computes to

$E=E_1-E_2$

Replacing with expressions from part 1), we get

$E=\frac{1}{2\epsilon_0}(\sigma_1-\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}-64\cdot 10^{-6})=-3.76\cdot 10^6N/C$

where the negative illustrates a leftward direction.

4)

This portion parallels part 2). Since both fields remain perpendicular to the slab and sheet, no component exists along the y-axis, thus the electric field's y-component is zero.

5)

Notably, the slab behaves as a conductor, signifying charge mobility within it.

The net charge on the slab is positive, indicating a surplus of positive charge. With the negatively charged sheet on the left of the slab, positive charges shift towards the left slab edge (at a = 2.9 cm), while negative charges move to the right edge (at b = 4 cm).

The surface charge density per unit area of the slab is

$\sigma=+64\mu C/m^2$

This average denotes the surface charge density on both slab sides at points a and b:

$\sigma=\frac{\sigma_a+\sigma_b}{2}$ (1)

Additionally, the infinite sheet at x = 0 negatively charged $\sigma_1=-2.5\mu C/m^2$ , induces an opposite net charge on the slab's left surface, thus

$\sigma_a-\sigma_b = +2.5 \mu C/m^2$ (2)

Having equations (1) and (2) allows for solving the surface charge densities at a and b, yielding:

$\sigma_a = +65.25 \mu C/m^2\\\sigma_b = +62.75 \mu C/m^2$

6)

We aim to compute the x-component of the electric field at

x = 3.34 cm

This point lies inside the slab, bounded at

a = 2.9 cm

b = 4.0 cm

In a conducting slab, the electric field remains at zero owing to charge equilibrium; thus, the x-component thereof in the slab is zero

7)

From part 5), we determined the surface charge density at x = a = 2.9 cm is $\sigma_a = +65.25 \mu C/m^2$

8)

As mentioned in part 6), conductors have zero electric fields internally. Since the slab is conductive, the electric field inside remains zero; therefore, the regions where the electric field is null are

2.9 cm < x < 4 cm

Thus, the suitable answer is

"none of these regions"

Learn more about electric fields:

8 0

6 days ago

An object is attached to a hanging unstretched ideal and massless spring and slowly lowered to its equilibrium position, a dista

Sav [1105]

Answer:

h = 12.8 cm

Explanation:

The initial parameters are as follows:

distance = 6.4 cm

when the object descends, its weight matches the spring's force

weight = spring force

mg = ky... equation 1

potential energy stored in a stretched spring = work done by the spring

mgh = 0.5 x k x h^{2}....equation 2

Substituting from equation 1 into equation 2

kyh = 0.5 x k x h^{2}

y = 0.5 x h

2y = h

where y is 6.4, yielding the maximum elongation as

h = 2 x 6.4 = 12.8 cm

6 0

5 days ago

A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of

Sav [1105]

Answer:

A rock weighing 50kg should be positioned at a distance of 0.5m from the pivot of the seesaw.

Explanation:

τchild=τrock

We will utilize the formula for torque:

(F)child(d)child)=(F)rock(d)rock)

The gravitational force acts equally on both objects.

(m)childg(d)child)=(m)rockg(d)rock)

We can eliminate gravity from both sides of the equation for simplification.

(m)child(d)child)=(m)rock(d)rock)

Now employing the given masses for the rock and child. The seesaw's total length is 2 meters, with the child sitting at one end, placing them 1 meter from the center of the seesaw.

(25kg)(1m)=(50kg)drock

Solve for the distance where the rock should be positioned in relation to the seesaw's center.

drock=25kg⋅m50kg

drock=0.5m

6 0

5 days ago