Ask question

Ask question

All categories

3 years ago

15

true or false? J. J Thomson presented a plum pudding model of the atom. His model showed electrons randomly embedded in a cloud

of positive charge

2 answers:

Vikentia [17]3 years ago

4 0

The answer is true at least i believe it is

Pachacha [2.7K]3 years ago

3 0

Your answer for this question would be true. this is because his model did show electrons randomly embedded in a cloud of positive charge

You might be interested in

Somone help me please!!

Ierofanga [76]

The 2nd answer is the correct answer

6 0

3 years ago

you slide abox of books at constant speed up a30 degree ramp, applying a force of 200 newtons directed up the slope. The coeffic

Tomtit [17]

The work done is 400 J

Explanation:

The work done by you in pushing the box along the slope is given by

$W=Fd$

where

F is the magnitude of the force applied

d is the distance covered by the box along the slope

Here we have the following:

F = 200 N is the magnitude of the force applied

$d=\frac{h}{sin \theta}$ is the distance covered, where

h = 1 m is the vertical rise

$\theta=30^{\circ}$ is the slope of the plane

Substituting and solving, we find

$W=F\frac{h}{sin \theta}=\frac{(200)(1)}{sin 30^{\circ}}=400 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

5 0

2 years ago

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st

Black_prince [1.1K]

Answer:

Part A:

$E_{midpoint}=0$

Part B:

$E_{center}=2711.7558 N/C$

Explanation:

Part A:

Formula of Electric Field Strength:

$E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}$

Where:

x is the distance from the ring

R is the radius of the ring

$\epsilon$ is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C$

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C$

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

$E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0$

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C$

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C$

Electric Field Strength at center of left ring is same as that of right ring.

$E_{center}=2711.7558 N/C$

5 0

3 years ago

Typical Pressurized Water Reactors can produce 1100 to 1500

lozanna [386]

Answer: about 1,100,000,000 to 1,500,000,000 Joules/second

Explanation:

1 MW (megawatt) = 1,000,000.00 J/s (joules per second)

1100(1,000,000) = 1,100,000,000

1500(1,000,000) = 1,500,000,000

3 0

2 years ago

How do you make a iPad I can text and stuff and don't understand how

tigry1 [53]

Circuit board stuff

4 0

2 years ago

Read 2 more answers