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3 years ago

PLS URGENT!!

Physics

1 answer:

Mama L [17]3 years ago

3 0

Wavelength = velocity / frequency = 3 x 10^8 / 100.5 x 10^6 = 2.99 m

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How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume

Svetach [21]

Answer:

$9.56\cdot 10^{-7} C$

Explanation:

A parallel-plate capacitors consist of two parallel plates charged with opposite charge.

Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.

The electric field between two infinite sheets with opposite charge is:

$E=\frac{\sigma}{\epsilon_0}$

where

$\sigma=\frac{Q}{A}$ is the surface charge density, where

Q is the charge on the plate

A is the area of the plate

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

In this problem:

- The side of one plate is

L = 19 cm = 0.19 m

So the area is

$A=L^2=(0.19)^2=0.036m^2$

Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:

$E=3\cdot 10^6 N/C$

Substituting this value into the previous formula and re-arranging it for Q, we find the charge:

$E=\frac{Q}{A\epsilon_0}\\Q=EA\epsilon_0 = (3\cdot 10^6)(0.036)(8.85\cdot 10^{-12})=9.56\cdot 10^{-7} C$

7 0

3 years ago

Please help! I'm not sure what equation or the process to do this question.

lions [1.4K]

Answer:

The momentum is 1.94 kg m/s.

Explanation:

To solve this problem we equate the potential energy of the spring with the kinetic energy of the ball.

The potential energy $U$ of the compressed spring is given by

$U = \dfrac{1}{2} kx^2$ ,

where $x$ is the length of compression and $k$ is the spring constant.

And the kinetic energy of the ball is

$K.E = \dfrac{1}{2}mv^2.$

When the spring is released all of the potential energy of the spring goes into the kinetic energy of the ball; therefore,

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2,$

solving for $v$ we get:

$v = x \sqrt{\dfrac{k}{m} }.$

And since momentum of the ball is $p=mv$ ,

$p =mx \sqrt{\dfrac{k}{m} }.$

Putting in numbers we get:

$p =(0.5kg)(0.25m) \sqrt{\dfrac{(120N/m)}{0.5kg} }.$

$\boxed{p=1.94kg\: m/s}$

5 0

2 years ago

Please help on this one?

qwelly [4]

Answer:

D. An image that is smaller than the object and is behind the mirror

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3 years ago

A periodic wave has a wavelength of 0.60 m and a speed of 40 m/s. What is the<br> wave frequency?

levacccp [35]

The correct answer is .05

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3 years ago

Which one is it cause I have no idea

Stella [2.4K]

it depends on the game but in most cases, it will be the opposing team.

ans is a.

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2 years ago

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