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3 years ago

This is an example of using a hammer as a(n)

Physics

2 answers:

barxatty [35]3 years ago

3 0

, to hit nails into a piece of wood or a wall, or to break things into pieces.

Burka [1]3 years ago

3 0

Answer:

It's A

Explanation:

the answer on e2020 is A, a lever

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Numbers in exponential form

kaheart [24]

Do the same thing you were doing

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3 years ago

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 18.0 cm , giving it a ch

vladimir1956 [14]

A) The electric field inside the paint layer is zero

B) The electric field just outside the paint layer is $3.2\cdot 10^7 N/C$ (radially inward)

C) The electric field at 6.00 cm from the surface is $1.2\cdot 10^7 N/C$ (radially inward)

Explanation:

We can solve the problem by applying Gauss Law, which states that the electric flux through a Gaussian surface must be equal to the charge contained in the surface divided by the vacuum permittivity:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the magnitude of the electric field

dS is the element of the surface

q is the charge contained within the surface

$\epsilon_0$ is the vacuum permittivity

By taking a sphere centered in the origin,

$\int E dS = E \cdot 4\pi r^2$

where $4\pi r^2$ is the surface of the Gaussian sphere of radius r.

In this problem, we want to find the electric field just inside the paint layer, so we take a value of r smaller than

$R=9.0 cm = 0.09 m$ (radius of the plastic sphere is half of the diameter)

Since the charge is all distributed over the plastic sphere, the charge contained within the Gaussian sphere is zero:

$q=0$

And therefore,

$E4\pi r^2 = 0\\\rightarrow E = 0$

So, the electric field inside the plastic sphere is zero.

Here we apply again Gauss Law:

$E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}$

In this case, we want to calculate the electric field just outside the paint layer: this means that we take r as the radius of the plastic sphere, so

$r=R=0.18 m$

The charge contained within the Gaussian sphere is therefore

$q=-29.0 \mu C = -29.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.09)^2}=-3.2\cdot 10^7 N/C$

And the negative sign indicates that the direction of the field is radially inward (because the charge that generates the field is negative). However, the text of the question says "Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward", so the answer to this part is

$E=3.2\cdot 10^7 N/C$

For this part again, we apply Gauss Law:

$E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}$

In this case, we want to calculate the field at a point 6.00 cm outside the surface of the paint layer; this means that the radius of the Gaussian sphere must be

r = 9 cm + 6 cm = 15 cm = 0.15 m

While the charge contained within the sphere is again

$q=-29.0 \mu C = -29.0\cdot 10^{-6}C$

Therefore, the electric field in this case is

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.15)^2}=-1.2\cdot 10^7 N/C$

And again, this is radially inward, so according to the sign convention asked in the problem,

$E=1.2\cdot 10^7 N/C$

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

3 0

3 years ago

On the scale of the cosmic calendar, in which the history of the universe is compressed to 1 year, how long has human civilizati

HACTEHA [7]

Answer:

D. A few seconds

Explanation:

Cosmic Calendar is a way of visualizing the history/timeline of the whole universe in just one year. It starts with Big Bang happening on January 1st midnight. The current moment is taken on the December 31st just before the midnight. Every second on this calendar accounts for around 435 years.

The earliest remains of life are thought to have surfaced on September 14. The primitive humans arrived at 22:24 on December 31st. The ancient Egyptian civilization arrived only at 23:59:48 on December 31st. So seen at a cosmic scale the human civilization is just 12 seconds old.

6 0

3 years ago

The absolute potential at a distance of 2.0 m from a negative point charge is-100. V. What is the absolute potential 4.0 m away

jolli1 [7]

Answer:

The absolute potential 4.0 m away from the same point charge is -50 V.

(A) is correct option.

Explanation:

Given that,

Distance = 2.0 m

Potential = -100 V

Absolute potential = 4.0 m

We need to calculate the charge

Using formula of potential

$V=\dfrac{kq}{r}$

Where, V = potential

q = charge

r = distance

Put the value into the formula

$-100=\dfrac{9\times10^{9}\times q}{2.0}$

$q=\dfrac{200}{9\times10^{9}}$

$q=-22.2\times10^{-9}\ C$

We need to calculate the potential

Using formula of potential

$V=\dfrac{9\times10^{9}\times(-22.2\times10^{-9})}{4.0}$

$V=-49.95\ V$

$V=-50\ V$

Hence, The absolute potential 4.0 m away from the same point charge is -50 V.

6 0

3 years ago

Anybody ? Know how to do this

Nesterboy [21]

Just look at the periodic table and google how to read an element from the periodic table. A picture of a single cell from the table will pop up with arrows of what's what and it will help you answer this in no time! (Brainliest please)

4 0

3 years ago