2. A car going 35 km/hr takes 23.0 s to come to a stop at the red light. What is it's

Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R.From the express

AlladinOne [14]

Answer:

Vb = k Q / r r <R

Vb = k q / R³ (R² - r²) r >R

Explanation:

The electic potential is defined by

ΔV = - ∫ E .ds

We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product

VB - VA = - ∫ E dr

Let's substitute every equation they give us and we find out

r> R

Va = - ∫ (k Q / r²) dr

-Va = - k Q (- 1 / r)

We evaluate with it Va = 0 for r = infinity

Vb = k Q / r r <R

We perform the calculation of the power with the expression of the electric field that they give us

Vb = - int (kQ / R3 r) dr

We integrate and evaluate from the starting point r = R to the final point r <R

Vb = ∫kq / R³ r dr

Vb = k q / R³ (R² - r²)

This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity

8 0

3 years ago

HIII DROPPING COINS YALL

suter [353]

Answer:

tysm

Explanation:

8 0

2 years ago

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An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-ΩΩ resistor, and a switch S, all in ser

s2008m [1.1K]

Answer:

1.6 s

Explanation:

To find the time in which the potential difference of the inductor reaches 24V you use the following formula:

$V_L=V_oe^{-\frac{Rt}{L}}$

V_o: initial voltage = 60V

R: resistance = 24-Ω

L: inductance = 42H

V_L: final voltage = 24 V

You first use properties of the logarithms to get time t, next, replace the values of the parameter:

$\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s$

hence, after 1.6s the inductor will have a potential difference of 24V

3 0

3 years ago

A force of 44 N will stretch a rubber band 88 cm (0.080.08 m). Assuming that Hooke's law applies, how far will aa 11-N forc

Setler79 [48]

Answer:

The rubber band will be stretched 0.02 m.

The work done in stretching is 0.11 J.

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = 0.02 m

The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.

potential energy stored = $\frac{1}{2}ke^{2}$

==> $\frac{1}{2}* 550* 0.02^{2}$ = 0.11 J

3 0

3 years ago

Which constants and directions always apply when using gravity?

Lostsunrise [7]

Answer:

.

Explanation:

6 0

3 years ago

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