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The capacitance of the variable capacitor of a radio can be changed from 100 to 350 pF by turning the dial from 0° to 180°. With

tangare [24]

Answer:

0.7 mJ

Explanation:

Identify the unknown:

The work required to turn the dial from 180° to 0°

List the Knowns:

Capacitance when the dial is set at 180°: C = 350 pF = 350 x 10^-12 F Capacitance when the dial is set at 0°: C = 100 pF = 100 x 10^-12 F

Voltage of the battery: V = 130 V

Set Up the Problem:

Energy stored in a capacitor:

U_c=1/2*V^2*C

=1/2*Q^2/C

When the dial is set at 180°:

U_c=1/2*(130)^2*350*10^-12=10^-4

Q=√2*U_c*C=4*10^-7

When the dial is set at 0°:

U_c=1/2*(4*10^-7)^2/100*10^-12

=8*10^-4 J

Solve the Problem:

ΔU_c=7*10^-4 J

=0.7 mJ

note:

there maybe error in calculation but method is correct

4 0

3 years ago

Read 2 more answers

The work function of a metal surface is 4.80 × 10-19 J. The maximum speed of the electrons emitted from the surface is vA = 7.7

poizon [28]

Answer:

$\lambda_A=2.65177\times 10^{-7}\ m$

$\lambda_B=3.32344\times 10^{-7}\ m$

Explanation:

h = Planck's constant = $6.63\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

m = Mass of electron = $9.11\times 10^{-31}\ kg$

$W_0$ = Work function = $4.8\times 10^{-19}\ J$

$v_A$ = Velocity of A particle = $7.7\times 10^5\ m/s$

$v_B$ = Velocity of B particle = $5.1\times 10^5\ m/s$

The wavelength is given by

$\lambda=\frac{hc}{\frac{1}{2}mv^2+W_0}$

$\lambda_A=\frac{6.63\times 10^{-34}\times 3\times 10^8}{\frac{1}{2}9.11\times 10^{-31}(7.7\times 10^5)^2+4.8\times 10^{-19}}\\\Rightarrow \lambda_A=2.65177\times 10^{-7}\ m$

The wavelength $\lambda_A=2.65177\times 10^{-7}\ m$

$\lambda_B=\frac{6.63\times 10^{-34}\times 3\times 10^8}{\frac{1}{2}9.11\times 10^{-31}(5.1\times 10^5)^2+4.8\times 10^{-19}}\\\Rightarrow \lambda_B=3.32344\times 10^{-7}\ m$

The wavelength $\lambda_B=3.32344\times 10^{-7}\ m$

5 0

4 years ago

What is meant by input and output work

N76 [4]

Answer:

the x and y values

Explanation:

because on the table the x is the input and y is the output

5 0

3 years ago

When percussing the liver, the sound should be: hyperresonant resonant dull flat submit answer?

OleMash [197]

Percuss means gently tap. This is a part of examining the condition of organs in thorax or abdomen. The purpose of percussion in the liver is to measure its size. A normal liver makes a dull(consolidated) sound. A dull sound indicates the presence of solid mass(dense areas) under the surface, which is normal. Dullness indicates borderline of the liver and hence find its length. A hollow and air filled region beneath sounds resonant. A hyper-resonant is heard on region where it is hyper-inflated with air Flat or extremely dull sounds are normally heard over solid areas such as bones.

4 0

4 years ago

Describe an experiment to show that liquids expand and contract at different rates

Arte-miy333 [17]

Answer:

Depend on the rate of temperature applied.

Explanation:

Liquid expands on heating.

When liquid is heated then there is an increase in the average speed of it's molecules as a result molecules starts moving faster and hence kinetic energy of the liquid also get increased. This kinetic energy is directly proportional to the temperature applied. So, increase in temperature result in increase in kinetic energy which reduces the inter-molecular force of fluid and as a result liquid expands.

Liquid contracts on cooling.

Fall in temperature results in reduced kinetic energy which in turn also decreases the kinetic energy of the molecules of liquid thereby increasing the inter-molecular force of attraction of fluid that's why liquid contracts on cooling.

8 0

3 years ago