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Lubov Fominskaja [6]

3 years ago

5

PLEASE HELP!!!!!!!!Move on to electric force. Blow up the two balloons and knot them. Then tie a thread onto each balloon. Suspe

nd the two balloons using tape so that they’re about six inches apart, and check that they don’t move or interact. Rub both balloons with wool or fur. If wool or fur is not available, rub the balloons on your hair. Do they attract or repel (push away) each other?

2 answers:

Triss [41]3 years ago

8 0

Answer:

they attract

Explanation:

timama [110]3 years ago

6 0

They attract. This is because of the Stagnant you have created.

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Identify the equation used to calculate the perpendicular force (F⊥) acting on a block on an inclined plane.

pychu [463]

Using geometrical arguments, we can see that the angle of the inclined plane $\theta$ is equal to the angle between Fg and the perpendicular force.

But the perpendicular force is the projection of Fg along the perpendicular axis, and Fg=mg, so the correct answer is
<span>C) F=mg cosΘ </span>

6 0

3 years ago

Read 2 more answers

When a surface is illuminated with electromagnetic radiation of wavelength 480 nm, the maximum kinetic energy of the emitted ele

algol [13]

Answer:

Max kinetic energy for 340 nm wavelength will be $2.238\times 10^{-19}j$

Explanation:

In first case wavelength of electromagnetic radiation $\lambda =480nm=480\times 10^{-9}m$

Plank's constant $h=6.6\times 10^{-34}J-s$

Maximum kinetic energy = 0.54 eV

Energy is given by $E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{480\times 10^{-9}}=4.125\times 10^{-19}J$

We know that energy is given

$E=K_{MAX}+\Phi$ , here $\Phi$ is work function

So $4.125\times 10^{-19}=0.54\times 10^{-19}+\Phi$

$\Phi =3.585\times 10^{-19}J$

Now wavelength of second radiation = 340 nm

So energy $E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{340\times 10^{-9}}=5.823\times 10^{-19}J$

So $K_{MAX}=5.823\times 10^{-19}-3.585\times 10^{-19}=2.238\times 10^{-19}j$

6 0

4 years ago

Which factor affects elastic potential energy but not gravitational potential energy?

julsineya [31]

Explanation:

The energy stored in an elastic objects as a result of deformation is called elastic potential energy. The energy stored in a spring is given by :

$E=\dfrac{1}{2}kx^2$

Where

k = spring constant

x = compression or stretching in an spring

While gravitational potential energy is given by :

PE = mgh

where

m = mass

g = acceleration due to gravity

h = height from ground

So, the factor affecting elastic potential energy but not gravitational potential energy is " spring constant ".

6 0

4 years ago

Read 2 more answers

An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch

irinina [24]

Answer:

a) $\sigma_{\rm in} = -2.18~{\rm \mu C/m^2}$

b) $\sigma_{\rm out}= 1.12~{\rm \mu C/m^2}$

c) $E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}$

Explanation:

Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:

$Q_{\rm in} = \sigma A_{\rm in} = \sigma(2\pi r_1 h) = (-0.35)(2\pi (0.08) h) = -0.175h~{\rm \mu C}$

$Q_{\rm out} = \sigma A_{\rm out} = \sigma(2\pi r_2 h) = (-0.35)(2\pi (0.1) h) = -0.22h~{\rm \mu C}$

Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.

When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.

$Q_{\rm wire} = \lambda h = 1.1h~{\rm \mu C}$

a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:

$\sigma_2 = \frac{Q_{\rm in}}{2\pi r_1h} = \frac{-1.1h}{2\pi r_1 h} = \frac{-1.1}{2\pi(0.08)} = -2.18~{\rm \mu C/m^2}$

b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.

The new surface charge density can be calculated as follows:

$\sigma_{\rm out}= \frac{Q_{\rm out}}{2\pi r_2h} = \frac{0.705h}{2\pi r_2 h} = \frac{0.705}{2\pi(0.1)} = 1.12~{\rm \mu C/m^2}$

c) The electric field outside the cylinder can be found by Gauss' Law:

$\int{\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.

$E(2\pi r h) = \frac{Q_{\rm enc}}{\epsilon_0}\\E2\pi rh = \frac{\sigma 2\pi (r_1 + r_2)h}{\epsilon_0}\\E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}$

4 0

3 years ago

Electrons are ejected from sodium metal by any light that has a wavelength shorter than 544 nm. What is the kinetic energy of an

wel

Answer:

KE=2.3 x 10⁻¹⁹ J

Explanation:

Given that

λ = 544 nm

λ' = 485 nm

The kinetic energy KE given as

KE= E - Ф

Where

$E=\dfrac{hC}{\lambda'}$

$\phi=\dfrac{hC}{\lambda}$

h= 6.626 x 10⁻³⁴

C=3 x 10⁸ m/s

Now by putting the values

$KE=\dfrac{hC}{\lambda'}-\dfrac{hC}{\lambda}$

$KE=\dfrac{34.34\times 10^{-34}\times 3\times 10^8}{485\times 10^{-9}}- \dfrac{34.34\times 10^{-34}\times 3\times 10^8}{544\times 10^{-9}}$

KE=2.3 x 10⁻¹⁹ J

This is kinetic energy.

6 0

3 years ago