PHYSICS COLLEGE

A flat disk of radius 0.50 m is oriented so that the plane of the disk makes an angle of 30 degrees with a uniform electric field. If the field strength is 713.0 N/C find the electric Tiux through the surface A) 560 Nm2/C B) 620 N·m2/C C) 160 n N.m2/C D) 280 N.m2/C

Answers

Answer 1

Answer:

Answer:

electric flux is 280 Nm²/C

so correct option is D 280 Nm²/C

Explanation:

radius r = 0.50 m

angle = 30 degree

field strength = 713 N/C

to find out

the electric flux through the surface

solution

we find here electric flux by given formula that is

electric flux = field strength × area× cos∅ .......1

here area = πr² = π(0.50)²

put here all value in equation 1

electric flux = field strength × area× cos∅

electric flux = 713 × π(0.50)² × cos60

we consider the cosine of the angle between the direction of the field and the normal to the surface of the disk

so we use cos60

electric flux = 280 Nm²/C

so correct option is D 280 Nm²/C

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wo parallel plates of area 100cm2are given charges of equal magnitudes 8.9 ×10−7C but opposite signs. The electric field within the dielectric material filling the space between the plates is 1.4 ×106V/m. (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.

Answers

(a) 7.18

The electric field within a parallel plate capacitor with dielectric is given by:

$E=(\sigma)/(k \epsilon_0)$ (1)

where

$\sigma$ is the surface charge density

k is the dielectric constant

$\epsilon_0$ is the vacuum permittivity

The area of the plates in this capacitor is

$A=100 cm^2 = 100\cdot 10^(-4) m^2$

while the charge is

$Q=8.9\cdot 10^(-7)C$

So the surface charge density is

$\sigma = (Q)/(A)=(8.9\cdot 10^(-7) C)/(100\cdot 10^(-4) m^2)=8.9\cdot 10^(-5) C/m^2$

The electric field is

$E=1.4\cdot 10^6 V/m$

So we can re-arrange eq.(1) to find k:

$k=(\sigma)/(E \epsilon_0)=(8.9\cdot 10^(-5) C/m^2)/((1.4\cdot 10^6 V/m)(8.85\cdot 10^(-12) F/m))=7.18$

(b) $7.66\cdot 10^(-7)C$

The surface charge density induced on each dielectric surface is given by

$\sigma' = \sigma (1-(1)/(k))$

where

$\sigma=8.9\cdot 10^(-5) C/m^2$ is the initial charge density

k = 7.18 is the dielectric constant

Substituting,

$\sigma' = (8.9\cdot 10^(-5) C/m^2) (1-(1)/(7.18))=7.66\cdot 10^(5) C/m^2$

And by multiplying by the area, we find the charge induced on each surface:

$Q' = \sigma' A = (7.66\cdot 10^(-5) C/m^2)(100 \cdot 10^(-4)m^2)=7.66\cdot 10^(-7)C$

If I am given a total capacitence of two capacitors, their capacitence togather is 22 F. What capacitence would the individual capacitors have if they are connected in parallel or connected in series.

Answers

Answer:

In parallel combination, the capacity of each capacitor is 11 F.

In series combination, the capacity of each capacitor is 44 F.

Explanation:

Let there are two capacitors each of capacitance C.

When they are connected in parallel:

In parallel combination, the effective capacitance is Cp.

Cp = C1 + C2 = C + C

22 = 2 C

C = 11 F

When they are connected in series:

In parallel combination, the effective capacitance is Cs.

1 / Cs = 1 / C1 + 1 / C2 = 1 / C + 1 / C = 2 / C

1 / 22 = 2 / C

C = 44 F

A softball player swings a bat, accelerating it from rest to 2.6 rev/srev/s in a time of 0.20 ss . Approximate the bat as a 0.90-kgkg uniform rod of length 0.95 mm, and compute the torque the player applies to one end of it.

Answers

Answer:

$\tau=22.13Nm$

Explanation:

information we have:

mass: $m=0.9kg$

lenght: $L=0.95m$

frequency: $f=2.6rev/s$

time: $t=0.2s$

and from the information we have we can calculate the angular velocity $\omega$ . which is defined as

$\omega=2\pi f$

$\omega=2\pi (2.6rev/s)\n\omega=16.336 rev/s$

----------------------------

Now, to calculate the torque

We use the formula

$\tau=I \alpha$

where $I$ is the moment of inertia and $\alpha$ is the angular acceleration

moment of inertia of a uniform rod about the end of it:

$I=(1)/(3)mL^2$

substituting known values:

$I=(1)/(3) (0.9kg)(0.95m)^2\nI=0.271kg/m^2$

for the torque we also need the acceleration $\alpha$ which is defined as:

$\alpha=(\omega)/(t)$

susbtituting known values:

$\alpha=(16.336rev/s)/(0.2s) \n\alpha=81.68rev/s^2$

and finally we substitute $I$ and $\alpha$ into the torque equation $\tau=I \alpha$ : $\tau=(0.271kg/m^2)(81.68rev(s^2)\n\tau=22.13Nm$

Final answer:

To calculate the torque, we need to use the formula: Torque = Moment of Inertia * Angular Acceleration. By approximating the bat as a uniform rod and using its length and mass, we can find the moment of inertia. Then, using the given angular velocity, we can calculate the angular acceleration. Finally, we can determine the torque by multiplying the moment of inertia by the angular acceleration.

Explanation:

To compute the torque the player applies to one end of the bat, we need to use the formula:

Torque = Moment of Inertia * Angular Acceleration

Given that the bat is approximated as a uniform rod and we know its length and mass, we can calculate the moment of inertia. Then, using the given angular velocity, we can compute the angular acceleration. Finally, we can find the torque by multiplying the moment of inertia by the angular acceleration.

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One of the primary visible emissions from a distant planet occurs at 425 nm. Calculate the energy of a mole of photons of this emission.]

Answers

Answer:

Explanation:

Given

Wavelength of incoming light $\lambda =425\ nm$

We know

$speed\ of\ wave=frequency* wavelength$

$frequency=(speed)/(wavelength)$

$\mu =(3* 10^8)/(425* 10^(-9))$

$\mu =7.058* 10^(14)\ Hz$

Energy associated with this frequency

$E=h\mu$

where h=Planck's constant

$E=6.626* 10^(-34)* 7.058* 10^(14)$

$E=46.76* 10^(-20)\ Hz$

Energy of one mole of Photon $=N_a* E$

$=6.022* 10^(23)* 46.76* 10^(-20)$

$=281.58* 10^(3)$

$=281.58\ kJ$

Final answer:

To calculate the energy of a mole of photons of the emission at 425 nm, use the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Convert the wavelength to meters, substitute the values into the equation, and calculate to find the energy of a single photon. Multiply this by Avogadro's number to find the energy of a mole of photons.

Explanation:

To calculate the energy of a mole of photons of the emission at 425 nm, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength (in meters).

Converting the wavelength to meters, we have 425 nm = 425 x 10^-9 m.

Substituting the values into the equation, we get E = (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m). Calculating this gives us the energy of a single photon of this emission. To find the energy of a mole of photons, we can multiply this value by Avogadro's number (6.02 x 10^23 photons/mol).

Therefore, the energy of a mole of photons of this emission is (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m) x (6.02 x 10^23 photons/mol).

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A fully loaded, slow-moving freight elevator has a cab with a total mass of 1400 kg, which is required to travel upward 37 m in 3.6 min, starting and ending at rest. The elevator's counterweight has a mass of only 930 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable?

Answers

Answer:

789.8 W

Explanation:

mass of the cab = 1400 kg, the counter weight of the elevator = 930 kg

weight of the cab = 1400 × 9.81 where weight = mg and m is mass and g is acceleration due to gravity.

weight of the cab = 13734 N

counter weight of the elevator = 930 × 9.81 = 9123.3 N

the exerted force of the elevator = weight of the cab - counter weight of the elevator = 13734 - 9123.3 = 4610.7 N

Average power by the motor P = F × v = F × distance / time

where v is speed in m/s, and time is in seconds

P = 4610.7 × 37 / ( 3.6 × 60) = 789.80 W

where (3.6 × 60 ) is the time in seconds

Last night, Shirley worked on her accounting homework for one and one half hours. During that time, she completed 6 problems. What is the velocity in problems per hour? a. 6 per hour
b. 4 per hour
c. 10 per hour
d. 0.67 per hour
e. 15 per hour