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3 years ago

14

How does the balloon react to the cloth item?

1 answer:

almond37 [142]3 years ago

8 0

Answer: When rubbing a balloon with a wool cloth, it puts negative charges on the balloon. Negative charges attract to positive charges. If a balloon is not rubbed with the wool cloth, it has an equal amount of negative to positive charges, so it will attract to a rubbed balloon.

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I NEED THIS ANSWER TODAY PLEASE HELP ME

Anastaziya [24]

The answer is "B" - If there are no windows then there will be no light coming in, and therefore you don't have to worry about what time of day you do the experiment at.

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3 years ago

Read 2 more answers

What does it mean that are bodies are in osmotic equilibrium but chemical and electrical disequilibrium? Why? What molecules are

Whitepunk [10]

Explanation:

The point is that water is moving smoothly but that the solutes are not.Even though the containers are chemically different (chemical disequilibrium), once all the solutes in one container are contrasted to all the solutes in another container, both have the same total solutes concentrations (this means that they are in osmotic balance).

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3 years ago

Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

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3 years ago

A student places his hand in front of a plane mirror as shown in the diagram. Which terms correctly describe the image that the

ANTONII [103]

Either A or D but maybe D

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2 years ago

Positive Charge is distributed along the entire x axis with a uniform density 12 nC/m. A proton is placed at a position of 1.00

lions [1.4K]

Answer:

b. $\Delta KE = 390 eV$

Explanation:

As we know that the electric field due to infinite line charge is given as

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

here we can find potential difference between two points using the relation

$\Delta V = \int E.dr$

now we have

$\Delta V = \int(\frac{\lambda}{2\pi \epsilon_0 r}).dr$

now we have

$\Delta V = \frac{\lambda}{2\pi \epsilon_0}ln(\frac{r_2}{r_1})$

now plug in all values in it

$\Delta V = \frac{12\times 10^{-9}}{2\pi \epsilon_0}ln(\frac{1+5}{1})$

$\Delta V = 216ln6 = 387 V$

now we know by energy conservation

$\Delta KE = q\Delta V$

$\Delta KE = (e)(387V) = 387 eV$

3 0

3 years ago