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3 years ago

When the rate of motion is the same direction the speed will decrease. True False

Physics

1 answer:

slega [8]3 years ago

6 0

A)true is the answer

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A plane drops a hamper of medical supplies from a height of 5000 m during a practice run over the ocean. The plane’s horizontal

Marizza181 [45]

Answer:

346.70015 m/s

Explanation:

In the x axis speed is

$v_x=149\ m/s$

In the y axis

$v_y=\sqrt{2gh}\\\Rightarrow v_y=\sqrt{2\times 9.8\times 5000}$

The resultant velocity is given by

$v=\sqrt{v_x^2+v_y^2}\\\Rightarrow v=\sqrt{149^2+2\times 9.8\times 5000}\\\Rightarrow v=346.70015\ m/s$

The magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean is 346.70015 m/s

4 0

3 years ago

Qué ocurre con el espectro cuando se interpone el celofán de colores?

Travka [436]

Answer: You will only see the color that cellophane lets through

Explanation:

Let's begin by the fact the whole electromagnetic spectrum is known as "white light", which is composed by a range of colors (wavelengths).

Now, if we have a source with white light (the Sun, for example) and we interpose a cellophane of any color (let's choose red), this cellophane will act as a filter and will only let pass the color of the cellophane.

This is because the filter will absorb the other colors of the spectrum.

3 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]

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:

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):

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