Moon rocks resemble rocks from which of the following layers of the earth?

A boat is travelling due west at a speed of 47 miles per hour. If the boat started off 170 miles directly north of the city of S

ankoles [38]

Answer is miles sir your welcome it was simple

6 0

3 years ago

The phase change in which a substance changes from a gas directly to a solid is

vlabodo [156]

<span>Condensation is the change of the substance from liquid to solid phase. Example of this is the formation of ice. Vaporization is the change of substance from liquid to gas phase. Example of this is the boiling of water. Deposition is the change of a substance from gas to solid phase. Example of this is the formation of ice on a winter day. Sublimation is the change of a substance from solid to gas phase. Example of this is dry ice. The answer is letter C.</span>

7 0

3 years ago

Read 2 more answers

In any problems involving circular motion, which way does the tangential speed vector point?

Anton [14]

In what may be one of the most remarkable coincidences in
all of physical science, the tangential component of circular
motion points along the tangent to the circle at every point.

The object on a circular path is moving in that exact direction
at the instant when it is located at that point in the circle. The
centripetal force ... pointing toward the center of the circle ...
is the force that bends the path of the object away from a straight
line, toward the next point on the circle. If the centripetal force
were to suddenly disappear, the object would continue moving
from that point in a straight line, along the tangent and away from
the circle.

4 0

3 years ago

A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.

bija089 [108]

a)

i) Potential for r < a: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

ii) Potential for a < r < b: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

iii) Potential for r > b: $V(r)=0$

b) Potential difference between the two cylinders: $V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

c) Electric field between the two cylinders: $E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

Explanation:

a)

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

$E=\frac{\lambda}{2\pi \epsilon_0 r}$

where

$\lambda$ is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

$V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)$

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

$E=0$

So the potential where the electric field is zero is constant:

$V=const.$

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density $+\lambda$ and an equal negative charge density $-\lambda$ . Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

$\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)$

However, we know that the potential at b is zero, so

$V(r)=V(b)=0$

ii) The electric field in the region a < r < b instead it is given only by the positive charge $+\lambda$ distributed over the surface of the inner cylinder of radius a, therefore it is

$E=\frac{\lambda}{2\pi r \epsilon_0}$

And so the potential in this region is given by:

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$ (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

And so, for r<a,

$V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

b)

Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.

We have:

- Potential at the surface of the inner cylinder:

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

- Potential at the surface of the outer cylinder:

$V(b)=0$

Therefore, the potential difference is simply equal to

$V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

c)

Here we want to find the magnitude of the electric field between the two cylinders.

The expression for the electric potential between the cylinders is

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

The electric field is just the derivative of the electric potential:

$E=-\frac{dV}{dr}$

so we can find it by integrating the expression for the electric potential. We find:

$E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

So, this is the expression of the electric field between the two cylinders.

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0

3 years ago

A ball is thrown vertically upward from the ground. Its distance in feet from the ground in t seconds is s equals negative 16 t

marishachu [46]

Answer:

t₁ = 3 s

Explanation:

In this exercise, the vertical displacement equation is not given

y = 240 t + 16 t²

Where y is the displacement, 240 is the initial velocity and 16 is half the value of the acceleration

Let's replace

864 = 240 t + 16 t²

Let's solve the second degree equation

16 t² + 240 t - 864 = 0

Let's divide by 16

t² + 15 t - 54 = 0

The solution of this equation is

t = [-15 ± √(15 2 - 4 1 (-54)) ] / 2 1

t = [-15 ±√(225 +216)] / 2

t = [-15 + - 21] / 2

We have two solutions.

t₁ = [-15 +21] / 2

t₁ = 3 s

t₂ = -18 s

Since time cannot have negative values, the correct t₁ = 3s

4 0

3 years ago