Consider an infinitely long wire with charge per unit length ????centered at (x, y) = (0, d) parallel to the z-axis.A) Find the potential due to this line charge referenced to the origin so that ϕ=0.

Answer:

a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.

Answer:

Frequency= 0.25m

Period= 4.0 secs

Explanation:

Clay is surfing on a wave with a speed of 5.0m/s

The wave crests are 20m apart

Therefore, the frequency of the wave can be calculated as follows

Frequency= wave speed/distance

= 5.0/20

= 0.25m

The period (T) can be calculated as follows

T= 1/frequency

T = 1/0.25

T= 4.0secs

Hence the frequency is 0.25m and the period is 4.0 secs

Answer:

D

Explanation:

The power equation is P= V^2/R

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As a result of the given scenario, power delivered from the battery to combination decreases. The correct option is D.

A resistor is a two-terminal passive electrical component that uses electrical resistance as a circuit element.

Resistors are used in electronic circuits to reduce current flow, adjust signal levels, divide voltages, and bias active elements.

A resistor is a component of an electronic circuit that limits or regulates the flow of electrical current. Resistors can also be used to supply a fixed voltage to an active device such as a transistor.

The current through resistors is the same when they are connected in series. The battery voltage is divided among resistors.

Adding more resistors to a series circuit increases total resistance and thus lowers current. However, in a parallel circuit, adding more resistors in parallel creates more options while decreasing total resistance.

Thus, the correct option is D.

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Answer:

a. ω = 14.1 radians per day

b. ω = 2.24 rotations per day

Explanation:

a.

The relationship between linear and angular velocity is given as:

v = rω

where,

v = linear speed of point = 527787 miles/day

r = radius of Saturn = diameter/2 = 74900 miles/2 = 37450 miles

ω = angular velocity of point = ?

Therefore,

527787 miles/day = (37450 miles)ω

ω = (527787 miles/day)/(37450 miles)

ω = 14.1 radians per day

b.

The number of rotations per day can be found by converting the units of angular acceleration:

ω = (14.1 radians per day)(1 rotation/2π radians)

ω = 2.24 rotations per day

Answer:

O An increase in the temperature will increase the pressure.

Explanation:

Decrease in volume, more collisions, increase in temperature, more particles: all of these increase the pressure.

Answer: O An increase in the temperature will increase the pressure.

Answer:

The object has an excess of [tex]10^{13}[/tex] electrons.

Explanation:

When an object has a negative charge he has an excess of electrons in its body. We can calculate the number of excessive electrons by dividing the charge of the body by the charge of one electron. This is done below:

[tex]n = \frac{\text{object charge}}{\text{electron charge}}\\n = \frac{-1.6*10^{-6}}{-1.6*10^{-19}} = 1*10^{-6 + 19} = 10^{13}[/tex]

The object has an excess of [tex]10^{13}[/tex] electrons.

Answer:

C is the correct answer

Explanation:

Answer:

The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.

Explanation:

The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.

The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery. Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.      

Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.                                  

I hope it helps you!

Answer:

(a) 62.3 N

(b) 1.89 N

(c) 0.430 kg m²

Explanation:

(a) Find the acceleration of block B.

Δy = v₀ t + ½ at²

1.80 m = (0 m/s) (2.00 s) + ½ a (2.00 s)²

a = 0.90 m/s²

Draw a free body diagram of block B.  There are two forces:

Weight force mg pulling down,

and tension force Tb pulling up.

Sum of forces in the -y direction:

∑F = ma

mg − Tb = ma

Tb = m (g − a)

Tb = (7.00 kg) (9.8 m/s² − 0.90 m/s²)

Tb = 62.3 N

(b) Draw a free body diagram of block A.  There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and tension force Ta pulling right.

Sum of forces in the +x direction:

∑F = ma

Ta = ma

Ta = (2.10 kg) (0.90 m/s²)

Ta = 1.89 N

(c) Draw a free body diagram of the pulley.  There are two forces:

Tension force Tb pulling down,

and tension force Ta pulling left.

Sum of torques in the clockwise direction:

∑τ = Iα

Tb r − Ta r = Iα

(Tb − Ta) r = I (a/r)

I = (Tb − Ta) r² / a

I = (62.3 N − 1.89 N) (0.080 m)² / (0.90 m/s²)

I = 0.430 kg m²

The tension force that the rope exerts on block B is 62.3 N, the tension force that the rope exerts on block A is 1.89 N, and the moment of inertia of the pulley for rotation about the axle on which it is mounted is [tex]\rm 0.430 \; kg\;m^2[/tex].

Given :

a) First, determine the acceleration of the B block.

[tex]\rm s = ut + \dfrac{1}{2}at^2[/tex]

[tex]\rm 1.8 = \dfrac{1}{2}\times a\times (2)^2[/tex]

[tex]\rm a = 0.9\; m/sec^2[/tex]

Now, apply Newton's second law of motion in order to determine the tension force that the rope exerts on block B.

[tex]\rm \sum F=ma[/tex]

[tex]\rm mg-T_b=ma[/tex]

[tex]\rm T_b = m(g-a)[/tex]

[tex]\rm T_b = 7\times (9.8-0.9)[/tex]

[tex]\rm T_b = 62.3\;N[/tex]

b) Now, again apply Newton's second law of motion in order to determine the tension force that the rope exerts on block A.

[tex]\rm \sum F=ma[/tex]

[tex]\rm T_a=ma[/tex]

[tex]\rm T_a = 2.1\times 0.9[/tex]

[tex]\rm T_a = 1.89\;N[/tex]

c) The sum of the torque in order to determine the moment of inertia of the pulley for rotation about the axle on which it is mounted.

[tex]\rm \sum \tau = I\alpha[/tex]

[tex]\rm T_br-T_ar = I\alpha[/tex]

[tex]\rm I = \dfrac{(T_b-T_a)r^2}{a}[/tex]

Now, substitute the values of the known terms in the above expression.

[tex]\rm I = \dfrac{(62.3-1.89)(0.080)^2}{0.90}[/tex]

[tex]\rm I = 0.430 \; kg\;m^2[/tex]

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Explanation:

[tex]1.2 \mathrm{N} ; 2 \mathrm{N}[/tex]

2.[tex]200 \mathrm{N} ; 200 \mathrm{N}[/tex]

4.[tex]2 \mathrm{N} ; 2 \mathrm{N} ; 4 \mathrm{N}[/tex]

[tex]5.2 \mathrm{N} ; 2 \mathrm{N} ; 2 \mathrm{N}[/tex]

[tex]6.2 \mathrm{N} ; 2 \mathrm{N} ; 3 \mathrm{N}[/tex]

[tex]8.200 \mathrm{N} ; 200 \mathrm{N} ; 5 \mathrm{N}[/tex]

In only the above cases (i.e 1,2,4,5,6,8 ) the object possibly moves at a constant velocity of [tex]256 \mathrm{m} / \mathrm{s}[/tex]

You should have noticed that the sets of forces applied to the object are the same asthe ones in the prevous question. Newton's 1st law (and the 2nd law, too) makes nodistinction between the state of re st and the state of moving at a constant velocity(even a high velocity).

In both cases, the net force applied to the object must equal zero.

Answer:

dr = 0.71 ft/min

the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.

Explanation:

Given;

Rate of inflation dV = 140pift^3/min

Radius r = 7 ft

Change in radius = dr

Volume of a spherical balloon is;

V = (4/3)πr^3

The change in volume can be derived by differentiating both sides;

dV = (4πr^2)dr

Making dr the subject of formula;

dr = dV/(4πr^2)

Substituting the given values;

dr = 140π/(4π×7^2)

dr = 0.714285714285 ft/min

dr = 0.71 ft/min

the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.

Answer:

L = 0.60 m

The length in metres should be 0.60 m

Explanation:

A pipe open at both ends can have a standing wave pattern with resonant frequency;

f = nv/2L ........1

Where;

v = velocity of sound

L = length of pipe

n = 1 for the fundamental frequency f1

Given;

Fundamental frequency f1 = 294 Hz

Velocity v = 350 m/s

n = 1

From equation 1;

Making L the subject of formula;

L = nv/2f1

Substituting the given values;

L = 1×350/(2×294)

L = 0.595238095238 m

L = 0.60 m

The length in metres should be 0.60 m

Answer:

3g/(8π²) ≈ 0.372 m

Explanation:

Draw a free body diagram.  There is a weight force at the center of the pendulum.

Sum of the torques about the pivot:

∑τ = Iα

mg (½ L sin θ) = (⅓ mL²) α

3g sin θ = 2L α

α = 3g/(2L) sin θ

For small θ, sin θ ≈ θ.

α = 3g/(2L) θ

θ" = 3g/(2L) θ

The solution of this differential equation is:

θ = θ₀ cos(√(3g/(2L)) t)

So the period is:

T = 2π / √(3g/(2L))

If the period is 1 second:

1 = 2π / √(3g/(2L))

√(3g/(2L)) = 2π

3g/(2L) = 4π²

L = 3g/(8π²)

L ≈ 0.372 m

The length of the pendulum rod is 0.37 m.

The time period of a pendulum is defined as the time taken by the pendulum to complete one oscillation.

Here,

The mass of the pendulum, m = 2 kg

Time period of the pendulum, T = 1 s

Since, the pendulum is suspended from the pivot and is oscillating, at the position of the pendulum when it makes an angle θ with the pivot, there is a force of weight acting at the center of the pendulum.

The length of pendulum at that point = L/2

The perpendicular distance at that point, r = (L/2) sinθ

Therefore, the torque acting on the pendulum at that point,

τ = Iα

where I is the moment of inertia of the pendulum and α is the angular acceleration.

mg (L/2 sinθ) = (mL²/3)α

1/2 gsinθ = 1/3 Lα

Therefore,

α = (3g/2L) sinθ

For smaller values of θ, we can take sinθ = θ

So, α = (3g/2L) θ

We know that α = θ''

where θ is the angular displacement.

Therefore,

θ'' = (3g/2L) θ

So, ω = √3g/2L

Therefore, the equation of motion of the pendulum can be written as,

θ = θ₀ cos(ωt)

θ = θ₀ cos [(√3g/2L) t]

So, time period of the pendulum,

T = 2[tex]\pi[/tex]/ω

T = 2[tex]\pi[/tex]/(√3g/2L)

2[tex]\pi[/tex]/(√3g/2L) = 1

(√3g/2L) = 2[tex]\pi[/tex]

Therefore, length of the rod,

L = 3g/8[tex]\pi[/tex]²

L = 0.37 m

Hence,

The length of the pendulum rod is 0.37 m.

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Attaching the image file here.

Answer:

- E = 11.55J

- Q = 0.17C

- E' = (1/4)E

Explanation:

- To calculate the amount of energy stored in the capacitor, you use the following formula:

[tex]E=\frac{1}{2}CV^2[/tex]

C: capacitance = 3800.0*10^-6F

V: potential difference = 78.0V

[tex]E=\frac{1}{2}(3800.0*10^{-6}C)(78.0V)^2=11.55J[/tex]

The energy stored in the capacitor is 11.55J

- If the electrical energy stored in the capacitor is 6.84J, the charge on the capacitor is:

[tex]E=\frac{1}{2}QV\\\\Q=\frac{2E}{V}\\\\Q=\frac{2(6.84J)}{78.0V}=0.17C[/tex]

The charge on the capacitor is 0.17C

- If you take the capacitor as a parallel plate capacitor, you have that the energy stored on the capacitor is:

[tex]E=\frac{1}{2}CV^2=\frac{1}{2}(\frac{\epsilon_oA}{d})V^2=\frac{1}{2}\frac{\epsilon_oAV^2}{d}\\\\[/tex]

A: area of the plates

d: distance between plates

If the distance between plates is increased by a factor of 4, you have:

[tex]E'=\frac{1}{2}\frac{\epsilon_oAV^2}{(4d)}=\frac{1}{4}\frac{\epsilon_oAV^2}{2d}=\frac{1}{4}E[/tex]

Then, the stored energy in the capacitor is decreased by a a factor of (1/4)

Answer:

578.2 N

Explanation:

The computation of reading on the scale is shown below:-

Data provided in the question

Weight of a girl = 59 kg

Constant rate = 8 m/s

Since the elevator is descended so the acceleration is zero

As we know that

Reading on the scale is

[tex]F = m\times g[/tex]

where,  m = 59 kg

g  [tex]= 59 \times 9.8 m/s^2[/tex]

So, the reading on the scale is

= 578.2 N

Therefore for computing the reading on the scale we simply applied the above formula.

Answer:

T₂ = 95.56°C

Explanation:

The final resistance of a material after being heated is given by the relation:

R' = R(1 + αΔT)

where,

R' = Final Resistance = 207.4 Ω

R = Initial Resistance = 154.9 Ω

α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹

ΔT = Change in Temperature = ?

Therefore,

207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]

207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT

1.34 - 1 = (0.0045°C⁻¹)ΔT

ΔT = 0.34/0.0045°C⁻¹

ΔT = 75.56°C

but,

ΔT = Final Temperature - Initial Temperature

ΔT = T₂ - T₁ = T₂ - 20°C

T₂ - 20°C = 75.56°C

T₂ = 75.56°C + 20°C

T₂ = 95.56°C

Answer:

6.44 s

Explanation:

Given:

v₀ = 119 m/s

v = 233 m/s

a = 17.7 m/s²

Find: t

v = at + v₀

(233 m/s) = (17.7 m/s²) t + (119 m/s)

t = 6.44 s

Answer:

The force experienced  is 0.6 N

Explanation:

Given data

length of wire L= 2 m

current in wire I= 0.6 A

magnetic field B= 0.5

The force experienced can be represented as

[tex]F= BIL[/tex]

[tex]F= 0.5*0.6*2\\\F= 0.6 N[/tex]

Answer:

q = 2.997*10^-4C

Explanation:

In order to find the required charge that the penny have to have, to acquire an upward acceleration, you take into account that the electric force on the penny must be higher than the weight of the penny.

You use the second Newton law to sum both electrical and gravitational forces:

[tex]F_e-W=ma\\\\qE-mg=ma[/tex]             (1)

Fe: electric force

W: weight of the penny

q: required charge = ?

m: mass of the penny = 3g = 0.003kg

E: magnitude of the electric field = 100N/C

g: gravitational acceleration = 9.8m/s^2

a: acceleration of the penny = 0.19m/s^2

You solve the equation (1) for q, and replace the values of the other parameters:

[tex]q=\frac{ma+mg}{E}=\frac{m(a+g)}{E}\\\\q=\frac{(0.003kg)(0.19m/s^2+9.8m/s^2)}{100N/C}\\\\q=2.997*10^{-4}C[/tex]

It is necessary that the penny has a charge of 2.997*10^-4 C, in order to acquire an upward acceleration of 0.19m/s^2

Answer:

T = 15,576 N

Explanation:

The speed of a wave on a string is given by

        v = √ T /ρ rho

also the speed of the wave is given by the relationship

       v = λ f

we substitute

     λ f = √ T /ρ

T = (lam f)² ρ

let's find the wavelength in a string, fixed at the ends, the relation that gives the wavelength is

       L= λ/2 n

       λ= 2L / n

we substitute

      T = (2L / n f)²ρ rho

let's calculate

      T = (2 1.20 / 2 590) 0.022

      T = 15,576 N

Answer:

3.85 percent

Explanation:

From the question,

Percentage error = (error/actual)×100................ Equation 1

Given: actual center of mass = 15 cm, error = 15.6-15 = 0.6 cm

Substitute these values into equation 1

Percentage error = (0.6/15.6)×100

Percentage error = 3.85 percent

Hence the percentage error of the uniform mass = 3.85 percent

Answer:

F' = F/16

So, the force of gravity has become 16 times less than initial value.

Explanation:

The force of gravity between two planets, is given by the following formula:

F = Gm₁m₂/r²   ----------- equation 1

where,

F = Force of gravity between two planets

G = Gravitational Constant

m₁ = Mass of one planet

m₂ = Mass of other plant

r = Distance between two planets

Now, if the distance between the planets (r) is 10 times smaller, then Force of gravity will become:

F' = Gm₁m₂/(4r)²

F' = (1/16) (Gm₁m₂/r²)

using equation 1:

F' = F/16

So, the force of gravity has become 16 times less than initial value.

Answer:

The temperature of silver at this given resistivity is 2971.1 ⁰C

Explanation:

The resistivity of silver is calculated as follows;

[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\[/tex]

where;

Rt is the resistivity of silver at the given temperature

Ro is the resistivity of silver at room temperature

α is the temperature coefficient of resistance

To is the room temperature

T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature

[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\\R_t = 1.59*10^{-8}[1 + 0.0038(T-20)][/tex]

Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m

When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;

[tex]R_t,_{silver} = 2R_o,_{iron}\\\\1.59*10^{-8}[1 + 0.0038(T-20)] =(2 *9.71*10^{-8})\\\\\ \ (divide \ through \ by \ 1.59*10^{-8})\\\\1 + 0.0038(T-20) = 12.214\\\\1 + 0.0038T - 0.076 = 12.214\\\\0.0038T +0.924 = 12.214\\\\0.0038T = 12.214 - 0.924\\\\0.0038T = 11.29\\\\T = \frac{11.29}{0.0038} \\\\T = 2971.1 \ ^0C[/tex]

Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C

Answer:

I hope it will help you....

Answer:

[tex]a_{r} = 1006.382g \,\frac{m}{s^{2}}[/tex]

Explanation:

Let suppose that centrifuge is rotating at constant angular speed, which means that resultant acceleration is equal to radial acceleration at given radius, whose formula is:

[tex]a_{r} = \omega^{2}\cdot R[/tex]

Where:

[tex]\omega[/tex] - Angular speed, measured in radians per second.

[tex]R[/tex] - Radius of rotation, measured in meters.

The angular speed is first determined:

[tex]\omega = \frac{\pi}{30}\cdot \dot n[/tex]

Where [tex]\dot n[/tex] is the angular speed, measured in revolutions per minute.

If [tex]\dot n = 3000\,rpm[/tex], the angular speed measured in radians per second is:

[tex]\omega = \frac{\pi}{30}\cdot (3000\,rpm)[/tex]

[tex]\omega \approx 314.159\,\frac{rad}{s}[/tex]

Now, if [tex]\omega = 314.159\,\frac{rad}{s}[/tex] and [tex]R = 0.1\,m[/tex], the resultant acceleration is then:

[tex]a_{r} = \left(314.159\,\frac{rad}{s} \right)^{2}\cdot (0.1\,m)[/tex]

[tex]a_{r} = 9869.588\,\frac{m}{s^{2}}[/tex]

If gravitational acceleration is equal to 9.807 meters per square second, then the radial acceleration is equivalent to 1006.382 times the gravitational acceleration. That is:

[tex]a_{r} = 1006.382g \,\frac{m}{s^{2}}[/tex]

Answer:

V = 0.45 Volts

Explanation:

First we need to find the total current passing through the wire. That can be given by:

Total Current = I = (Current Density)(Surface Area of Wire)

I = (Current Density)(2πrL)

where,

r = radius = 1.5/2 mm = 0.75 mm = 0.75 x 10⁻³ m

L = Length of Wire = 6.5 m

Therefore,

I = (4.07 x 10⁻³ A/m²)[2π(0.75 x 10⁻³ m)(6.5 m)]

I = 1.25 x 10⁻⁴ A

Now, we need to find resistance of wire:

R = ρL/A

where,

ρ = resistivity of iron = 9.71 x 10⁻⁸ Ωm

A = Cross-sectional Area = πr² = π(0.75 x 10⁻³ m)² = 1.77 x 10⁻⁶ m²

Therefore,

R = (9.71 x 10⁻⁸ Ωm)(6.5 m)/(1.77 x 10⁻⁶ m²)

R = 0.36 Ω

From Ohm's Law:

Voltage = V = IR

V = (1.25 x 10⁻⁴ A)(0.36 Ω)

V = 0.45 Volts

Answer:

is this a company name.? java is a computer software right..

Answer:

T= (m2a2)/2

Explanation:

Using Newton's second law

F21+F22+ F33.....= M2a2

Where F21 and F22 are forces acting on M2

Thus T = ( M2a2)/2 this is tension on the string

Based on the magnitude of the horizontal acceleration and the mass, the tension will be (m₂ x a₂) / 2.

In the relevant diagram, there are two strings so the tension will be multiplied by 2.

The tension on the string can be found by:

2 x Tension on string = m₂ x a₂

Tension = (m₂ x a₂) / 2

In conclusion, the tension is (m₂ x a₂) / 2.

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Answer:

The work required to move this charge is 0.657 J

Explanation:

Given;

magnitude of charge, q = 4.4 x 10⁻⁶ C

Electric field strength, E =  3.9 x 10⁵ N/C

distance moved by the charge, d = 50 cm = 0.5m

angle of the path, θ = 40°

Work done is given as;

W = Fd

W = FdCosθ

where;

F is the force on the charge;

According the coulomb's law;

F = Eq

F = 3.9 x 10⁵ x 4.4 x 10⁻⁶  = 1.716 N

W = FdCosθ

W = 1.716 x 0.5 x Cos40

W = 0.657 J

Therefore, the work required to move this charge is 0.657 J

Answer:

Explanation:

Beats are produced when two sound waves from two  different sources having slightly different frequencies  interfere . Due to both constructive and destructive interference , high and low intensity sound is heard at some regular interval . This is called rate of beat formation . Amplitudes of sound may be same or different . High intensity sound heard per unit time is called beat . It is equal to difference of frequencies of the sound waves interfering

each other .