PHYSICS COLLEGE

A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surface of this cube?

Answers

Answer 1

Answer:

Answer:

The flux through the surface of the cube is $2.314\ Nm^(2)/C$

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = $8.0* 10^(- 2)\ m$

Volume Charge density, $\rho_(v) = 40 nC/m^(3) = 40* {- 9}\ C/m^(3)$

Now,

To calculate the electric flux:

$\phi = (q)/(\epsilon_(o))$ (1)

where

$\phi$ = electric flux

$\epsilon_(o) = 8.85* 10^(- 12)\ F/m$ = permittivity of free space

Volume Charge density for the given case is given by the formula:

$\rho_(v) = (Total\ charge, q)/(Volume of cube, V)$ (2)

Volume of cube, $V = a^(3)$

Thus

$V = (8.0* 10^(- 2))^(3) = 5.12* 10^(- 4)\ m^(3)$

Thus from eqn (2), the total charge is given by:

$q = \rho_(v)V = 40* {- 9}* 5.12* 10^(- 4)$

$q = 2.048* 10^(-11)\ F = 20.48\ pF$

Now, substitute the value of 'q' in eqn (1):

$\phi = (2.048* 10^(-11))/(8.85* 10^(- 12)) = 2.314\ Nm^(2)/C$

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A popular physics lab involves a hand generator and an assortment of wires with different values of resistance. In the lab, the leads of the generator are connected across each wire in turn. For each wire, students attempt to turn the generator handle at the same constant rate. Students must push harder on the handle when the leads of the generator are connected__________. This is because turning the handle at a given constant rate produces__________ , regardless of what is connected to the leads. So, when turning the handle at a constant rate, lab students must push harder in cases where there is________

Answers

Answer:

Explanation:

Students must push harder on the handle when the leads of the generator are connected across the wire with the lowest resistance.

This is because turning the handle at a given constant rate produces a constant voltage across the leads, regardless of what is connected to the leads.

So, when turning the handle at a constant rate, lab students must push harder in case where there is a greater current through the connected wire.

Which formula can be used to calculate the horizontal displacement of a horizontally launched projectile?x = vi(cos )
x = vi(cos )t
x = ayt
x = vxt (RIGHT ANSWER)

Answers

The formula for calculating the horizontal displacement of a horizontally launched projectile is $x=v*t$

A projectile launched horizontally with a velocity v, at a height y ,travels a horizontal distance x, while falling through a distance y. The horizontal velocity of a projectile remains constant throughout its motion, in the absence of air resistance. The vertical component of the velocity is under the action of the gravitational force and hence it increases in magnitude as it falls through the height.

The horizontal displacement of the projectile is a uniform motion and it occurs at a constant speed v.

Thus, the horizontal displacement of the projectile is given by the expression.

$x=v*t$

A 10.0kg object is moving at 1 m/s when a force is applied in the direction of the objects motion, causing it to speed up to 4 m/s. If the force was applied for 5s what is the magnitude of the force

Answers

Answer:

F = 6[N].

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.

$P=m*v\nor\nP=F*t$

where:

P = impulse or lineal momentum [kg*m/s]

m = mass = 10 [kg]

v = velocity [m/s]

F = force [N]

t = time = 5 [s]

Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.

$(m_(1)*v_(1))+F*t=(m_(1)*v_(2))$

where:

m₁ = mass of the object = 10 [kg]

v₁ = velocity of the object before the impulse = 1 [m/s]

v₂ = velocity of the object after the impulse = 4 [m/s]

$(10*1)+F*5=10*4\n10+5*F=40\n5*F=40-10\n5*F=30\nF=6[N]$

Hearing the siren of an approaching fire truck, you pull over to side of the road and stop. As the truck approaches, you hear a tone of 460 Hz; as the truck recedes, you hear a tone of 410 Hz. How much time will it take to jet from your position to the fire 5.00 km away, assuming it maintains a constant speed?

Answers

Answer:

The truck will reach there in 250 seconds.

Explanation:

The frequency due to doppler effect, when the observer is stationary and the source is moving towards it is

$f_(obv)$ = $(v)/(v-v_(s) ) f$

where v= velocity of sound in air

$v_(s)$ = velocity of source of sound

f= frequency of sound and

$f_(obv)$ = frequency oberved due to Doppler effect

$(v)/(v-v_(0) ) f$ = 460------------------------------------------( 1 )

The frequency due to doppler effect, when the observer is stationary and the source is moving away from it

$f_(obv)$ = $(v)/(v+v_(s) ) f$

where v= velocity of sound in air

$v_(s)$ = velocity of source of sound

f= frequency of sound and

$f_(obv)$ = frequency oberved due to Doppler effect

$(v)/(v+v_(0) ) f$ = 410-------------------------------------------( 2 )

Dividing ( 1 ) by ( 2 )

$(v+v_(s) )/(v-v_(s) ) =(460)/(410)$

$(v+v_(s) )/(v-v_(s) ) =(46)/(41)$

41v + 41 $v_(s)$ = 46v - 46 $v_(s)$

87 $v_(s)$ = 5v

$v_(s)$ = $(5)/(87)v$

Velocity of Sound (v)= 348 m/s

$v_(s)$ =20 m/s

Therefore, the truck is moving at 20 m/s.

$Time=(Distance)/(Time)$

Distance= 5000 m

Time= $(5000)/(20)$

Time= 250 s

Time = 4 min 10 sec

24-gauge copper wire has a diameter of 0.51 mm. The speaker is located exactly 4.27 m away from the amplifier. What is the minimum resistance of the connecting speaker wire at 20°C? Hint: How many wires are required to connect a speaker!Compare the resistance of the wire to the resistance of the speaker (Rsp = 8 capital omega)

Answers

Answer:

R = 8.94 10⁻² Ω/m, R_sp / R_total = 44.8

Explanation:

The resistance of a metal cable is

R = ρ L / A

The area of a circle is

A = π R²

The resistivity of copper is

ρ = 1.71 10⁻⁸ ohm / m

Let's calculate

R = 1.71 10⁻⁸ 4.27 / (π (0.51 10⁻³)²)

R = 8.94 10⁻² Ω/m

Each bugle needs two wire, phase and ground

The total wire resistance is

R_total = 2 R

R_total = 17.87 10⁻² Ω

Let's look for the relationship between the resistance of the bugle and the wire

R_sp / R_total = 8 / 17.87 10⁻²

R_sp / R_total = 44.8

Final answer:

The resistance of the speaker wire can be calculated using the formula for the resistance of a wire, taking into account the resistivity of copper, the length and thickness of the wire, and whether a single or pair of wires is used.

Explanation:

The question is asking you to find the minimum resistance of a copper wire given its diameter and length, plus the resistance of the speaker it's connected to. Resistance of a wire is calculated using the formula R=ρL/A, where R is the resistance, ρ (rho) is the resistivity of the material (in this case, copper), L is the length of the wire, and A is the cross-sectional area of the wire.

First, you need to find the area of the 0.51 mm diameter wire. The area (A) of a wire is given by the formula π(d/2)^2 where d is the diameter of the wire. After calculating the area, use the formula R=ρL/A to calculate the resistance. For copper wire at 20°C, ρ is approximately 1.68 × 10^-8 Ω·m. Substituting these values into the formula will give you the resistance of the wire in ohms.

Note: you may need to consider whether you have just a single wire or a pair, since two wires are typically required to connect a speaker. If a pair is used, each wire will carry half the current, which affects the total resistance.

Learn more about Electric Resistance here:

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In an experiment, one of the forces exerted on a proton is F⃗ =−αx2i^, where α=12N/m2. What is the potential-energy function for F⃗ ? Let U=0 when x=0. Express your answer in terms of α and x.