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timofeeve [1]
2 years ago
6

1. Use Coulomb’s Law (equation below) to calculate the approximate force felt by an electron at point A in the schematic below.

For the sake of simplicity, assume that the charge reservoirs, Q- and Q+ , can be treated as point charges each 2.5 cm away from point A and 5.0 cm away from each other. Make sure to denote which direction the force will act. The red and blue regions indicated in the schematic are conductors. Here, Q- is equal to the charge of six electrons, and Q+ is equal to the charge of six protons.
Physics
1 answer:
Amiraneli [1.4K]2 years ago
8 0

Answer:

Explanation:

From the data it appears that A is the middle point between two charges.

First of all we shall calculate the field at point A .

Field due to charge -Q ( 6e⁻ ) at A

= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴

= 13.82 x 10⁻⁶ N/C

Its direction will be towards Q⁻

Same field will be produced by Q⁺ charge . The direction will be away

from Q⁺  towards Q⁻ .

We shall add the field  to get the resultant field  .

= 2 x 13.82 x 10⁻⁶

= 27.64 x 10⁻⁶ N/C

Force on electron put at A

= charge x field

= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶

= 44.22 x 10⁻²⁵ N

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You throw a baseball (mass 0.145 kg) vertically upward. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected.
Andru [333]

Answer:

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Explanation:

Hi there!

The equations of height and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

v = velocity of the ball at time t.

Placing the origin at the throwing point, y0 = 0.

Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.

v = v0 + g · t

6.00 m/s = 12.0 m/s -9.81 m/s² · t

(6.00 - 12.0)m/s / -9.81 m/s² = t

t = 0.612 s

Now, let´s calculate the height of the baseball at that time:

y = y0 + v0 · t + 1/2 · g · t²     (y0 = 0)

y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²

y = 5.51 m

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Have a nice day!

4 0
2 years ago
A simple watermelon launcher is designed as a spring with a light platform for the watermelon. When an 8.00 kg watermelon is put
olga_2 [115]

To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.

Since the forces are balanced the Spring force is equal to the force of the weight that is

F_s = F_g

kx = mg

Where,

k = Spring constant

x = Displacement

m = Mass

g = Gravitational Acceleration

Re-arrange to find the spring constant

k = \frac{mg}{x}

k = \frac{8*9.8}{0.1}

k = 784N/m

Just before launch the compression is 40cm, then from Potential Elastic Energy definition

PE = \frac{1}{2} kx^2

PE =\frac{1}{2} 784*0.4^2

PE = 63.72J

Therefore the energy stored in the spring is 63.72J

6 0
2 years ago
A wooden piece is made in different shapes take length (l) = radius (r) = 2m Calculate its volume as a:
In-s [12.5K]

This question deals with the volume of different shapes.

a) volume of the sphere is "33.51 m³".

b) volume of the cylinder is "25.13 m³".

a)

The volume of a sphere is given by the following formula:

Volume = \frac{4}{3}\pi r^3\\\\Volume = \frac{4}{3}\pi (2\ m)^3

<u>Volume = 33.51 m³</u>

<u />

b)

The volume of a cylinder is given by the following formula:

Volume = \pi r^2l\\\\Volume =\pi (2\ m)^2(2\ m)

<u>Volume = 25.13 m³</u>

<u />

Learn more about <em>volume </em>here:

brainly.com/question/16686115?referrer=searchResults

The attached picture shows the formulae of the <em>volume</em> of different shapes.

7 0
2 years ago
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined
MrRa [10]

Incomplete question as the car's  speed is missing.I have assumed car's  speed as 6.0m/s.The complete question is here

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s

Answer:

F_{B}=-5755N

Explanation:

Set up force equation

∑F=ma

∑F=W+FB

\frac{mv^{2} }{R}=W+F_{B}\\  F_{B}=\frac{mv^{2} }{R}-W\\F_{B}=\frac{(W/g)v^{2} }{R}-W\\F_{B}=\frac{(6000N/9.8m/s^{2} )(6m/s)^{2} }{(15m)}-6000N\\F_{B}=-5755N

The minus sign for downward direction

6 0
2 years ago
Sophia is planning on going down an 8-m water slide. Her weight is 50 N. She knows that she has gravitational potential energy (
RideAnS [48]

Answer:

Explanation:

graph would be a straight line from (0, 0) to (400, 8)

Plot points are

PE = mgh

50(0) = 0 J

50(2) = 100 J

50(4) = 200 J

50(6) = 300 J

50(8) = 400 J

4 0
2 years ago
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