Help me please forget about you and she was

On a cold day, snakes usually lie very still and eat little or nothing while birds usually move around and eat a lot of food. Wh

RoseWind [281]

Birds are warm-blooded, which means they need to move and eat more in the cold to stay warm; snakes are cold-blooded, so they don't have to.

5 0

3 years ago

A ball thrown horizontally from a point 30.0 m above the ground strikes the ground after travelling a horizontal distance of 18.

AURORKA [14]

The components of the ball's position $r$ at time $t$ are

$\begin{cases}r_x(t)=v_{0x}t\\v_y(t)=30.0-4.9t^2\end{cases}$

The ball stops 18.0 m from where it began, so that

$\begin{cases}18.0=v_{0x}t\\0=30.0-4.9t^2\end{cases}$

From the second equation, we can show that the ball travels for about $t=2.47$ seconds, which means it was initially thrown with a horizontal velocity of

$18.0\,\mathrm m=v_{0x}(2.47\,\mathrm s)\implies v_{0x}=7.29\,\dfrac{\mathrm m}{\mathrm s}$

7 0

4 years ago

Smaller bits of rock that pass through the atmosphere and slam into Earth's surface are called

bezimeni [28]

I think that they are called meteorites

6 0

3 years ago

Iwoweieeieeieeoeeoeeoeeoeeoeoeooeeowwowwoeeo

ELEN [110]

Answer:

What???????????????

8 0

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:

A)

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.

B)

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

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

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

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