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saul85 [17]
3 years ago
7

A solid insulating sphere of radius R = 1.0 m that carries a positive charge Q1 = 1.0 mC uniformly distributed over it is concen

tric with a very thin insulating shell of radius 2R that carries the charge Q2 = -1.0 mC. The charge Q2 is distributed uniformly over the shell.
Use Gauss's law to obtain the magnitude of the electric field:

a. E1 inside the sphere, r < R

b. E2 between the sphere and the shell, R < r < 2R

c. E3 outside the shell, r > 2R

d. Plot the electric field in the radial direction on a graph. Be sure to label the actual magnitude of the electric field at r = 0, r= R, and r = 2R.

Physics
1 answer:
RoseWind [281]3 years ago
3 0

Answer: see attach file

Explanation: In order to solve this problem we have used the Gaussian law in the different regions. Details are included in the attach.

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For the love of God please help me before my mom gets angry about my horrible grades
Maru [420]

Answer:

Heat Lost = (120 - 12.6) deg * 170 g * S = `18260 gm-deg * S

Heat Gained = (12.6 - 10) deg * 200 gm* 1 cal/ gm-deg - 520 cal

S = 520 / 18260 cal / gm-deg = .0285 cal / gm-deg

Since Heat Lost = Heat Gained

6 0
3 years ago
two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i
NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 - 14.7t - 4.9t^2

4.9t^2 + 14.7t - 19.6 = 0

t^2 + 3t - 4 = 0

(t + 4)(t - 1) = 0

(t - 1) = 0

\boxed {t = 1 ~ second}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 + 14.7t - 4.9t^2

4.9t^2 - 14.7t - 19.6 = 0

t^2 - 3t - 4 = 0

(t - 4)(t + 1) = 0

(t - 4) = 0

\boxed {t = 4 ~ seconds}

The difference in the two ball's time in the air is:

\Delta t = 4 ~ seconds - 1 ~ second

\large {\boxed {\Delta t = 3 ~ seconds} }

<h2>Question B:</h2><h3>First Ball</h3>

v^2 = u^2 - 2gH

v^2 = (-14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

<h3>Second Ball</h3>

v^2 = u^2 - 2gH

v^2 = (14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 11.025 ~ m}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 25.725 ~ m}

The difference in the two ball's height after 0.500 s is:

\Delta h = 25.725 ~ m - 11.025 ~ m

\large {\boxed {\Delta h = 14.7 ~ m} }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0
3 years ago
The back wall of a home aquarium is a mirror that is a distance of 46.0 cm away from the front wall. The walls of the tank are n
monitta

Answer:

i know the questin but i got to try and find it

Explanation:

5 0
3 years ago
Which requires more work, lifting a 10.0kg load a vertical distance of 2m or lifting a 5.0kg load a distance of 4m?
Minchanka [31]

For each load,  Work = (mass) x (gravity) x (distance .

Bigger load:      Work = (10 kg) x (9.8 m/s²) x (2 m) = 196 joules .

Smaller load:    Work = (5 kg)  x  (9.8 m/s²)  x  (4 m) = 196 joules.

The work required is equal in both cases.

The mass ratio of  2:1  is exactly balanced by
the height ratio of  1:2 .

6 0
3 years ago
In a research facility, a person lies on a horizontal platform which floats on a film of air. When the person's heart beats, it
anastassius [24]

Answer: 3.48g

Explanation:

here, we will be using conservation of momentum to solve the problem. i.e the total momentum remains unchanged, unless an external force acts on the system. We'll in thus question, there is no external force acting in the system.

Remember, momentum = mass * velocity, then

mass of blood * velocity of blood = combined mass of subject and pallet * velocity of subject and pallet

Velocity of blood = 56.5cm = 0.565m

mass of blood * 0.565 = 54kg * (0.000063/0.160)

mass of blood * 0.565 = 54 * 0.00039375

mass of blood * 0.565 = 0.001969

mass of blood = 0.00348kg

Thus, the mass of blood that leaves the heart is 3.48g

7 0
3 years ago
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