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s344n2d4d5 [400]

3 years ago

6

If a neutral material like fur is rubbed on another neutral object like a rubber ball what is the resulting charges of the objec

ts?
A) The balloon becomes negatively charged and the fur becomes positively charged.

B) They both remain neutral.

C) They both become negatively charged.

D) The balloon becomes negatively charged and the fur stays neutral.

E) They both become positively charged.

1 answer:

mrs_skeptik [129]3 years ago

7 0

B is the right answer I hope I helped

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Hire rubs an inflated balloon on his hair and his hair sticks to the balloon as he moves the balloon away from his head how does

sergey [27]

The options are missing and they are;

A) the electric force increases because the balloon loses its charge.

B) the electric force increases because the distance increases.

C) the electric force decreases because the distance increases.

D) the electric force decreases because his hair loses its charge.

Answer:

Correct answer is option C - the electric force decreases because the distance increases.

Explanation:

The formula for electric force is;

F = k•q1•q2/r²

Where;

K is coulombs constant

q1 and q2 are particle charges

r is distance

So,looking at the formula given earlier, if we increase the distance, the denominator will increase and thus the Force will decrease.

So the correct option is option C

8 0

3 years ago

Read 2 more answers

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle

LenKa [72]

Answer : The de-Broglie wavelength of this electron, $0.101\AA$

Explanation :

The formula used for kinetic energy is,

$K.E=\frac{1}{2}mv^2$ ..........(1)

According to de-Broglie, the expression for wavelength is,

$\lambda=\frac{h}{mv}$

or,

$v=\frac{h}{m\lambda}$ ...........(2)

Now put the equation (2) in equation (1), we get:

$\lambda=\frac{h}{\sqrt{2\times m\times K.E}}$ ...........(3)

where,

$\lambda$ = wavelength = ?

h = Planck's constant = $6.626\times 10^{-34}Js$

m = mass of electron = $9.11\times 10^{-31}Kg$

K.E = kinetic energy = $4.71\times 10^{-15}J$

Now put all the given values in the above formula (3), we get:

$\lambda=\frac{6.626\times 10^{-34}Js}{\sqrt{2\times 9.11\times 10^{-31}Kg\times 4.71\times 10^{-15}J}}$

$\lambda=1.0115\times 10^{-11}m=0.101\AA$

conversion used : $(1\AA=10^{-10}m)$

Therefore, the de-Broglie wavelength of this electron, $0.101\AA$

7 0

3 years ago

Large electric fields in cell membranes cause ions to move through the cell wall. The field strength in a typical membrane is 1.

serg [7]

Explanation:

The given data is as follows.

Electric field (E) = $1 \times 10^{7} N/C$

Charge (e) = $1.6 \times 10^{-19}$ C

Formula to calculate the magnitude of force is as follows.

F = qE

= $1.6 \times 10^{-19} \times 1 \times 10^{7} N/C$

= $1.6 \times 10^{-12} N$

Therefore, we can conclude that magnitude of the force on a calcium ion with charge +e is $1.6 \times 10^{-12} N$ .

6 0

3 years ago

The rate at which light energy is radiated from a source is measured in which of the following units?

Over [174]

The correct option is D.
Lumen is used to quantify the amount of total light energy that a source is putting out in all direction, thus, it refers to luminous output of a light source. Initial lumen refers to the luminosity of a light when it was first turned on; the luminosity is highest at this point.

3 0

3 years ago

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Light of wavelength 648.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 57.5 cm from the slit.

boyakko [2]

Answer:

$71.0 \mu m$

Explanation:

The formula for the single-slit diffraction is

$y=\frac{n\lambda D}{d}$

where

y is the distance of the n-minimum from the centre of the diffraction pattern

D is the distance of the screen from the slit

d is the width of the slit

$\lambda$ is the wavelength of the light

In this problem,

$\lambda=648.0 nm=6.48\cdot 10^{-7}m$

$D=57.5 cm=0.575 m$

$y=1.05 cm=0.0105 m$ , with n=2 (this is the distance of the 2nd-order minimum from the central maximum)

Solving the formula for d, we find:

$d=\frac{n\lambda D}{y}=\frac{2(6.48\cdot 10^{-7} m)(0.575 m)}{0.0105 m}=7.10\cdot 10^{-5} m= 71.0 \mu m$

6 0

3 years ago