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amid [387]
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
Physics
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

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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
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