A uniformly charged thin rod of length L and positive charge Q lies along the x-axis with its left end at the origin as shown in Figure 1.

a. Set up a correct integral expression for the potential at point A,which lies a distance H above the right end of the rod. Point A has coordinates (L, H). You need to give appropriate limits of integration and expressions for r and dq

b. Set up a correct integral expression for the potential at point B on the x-axis, a distance D from the left end of the rod with the appropriate limits of integration.You need to give appropriate limits of integration and expressions for r and dq.

Explanation:

Given:

L = 0.02 H

C = [tex]2\:\mu \text{F}[/tex]

f = 200 Hz

The general form of the impedance Z is given by

[tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]

Since this is a purely inductive/capacitive circuit, R = 0 so Z reduces to

[tex]Z = \sqrt{(X_L - X_C)^2} = \sqrt{\left(\omega L - \dfrac{1}{\omega C} \right)^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{\left(2 \pi L - \dfrac{1}{2 \pi f C} \right)^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{\left[2 \pi (200\:\text{Hz})(0.02\:\text{H}) - \dfrac{1}{2 \pi (200\:\text{Hz})(2×10^{-6}\:\text{F})} \right]^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{(25.13\:\text{ohms} - 397.89\:\text{ohms})^2}[/tex]

[tex]\:\:\:\:\:\:\:=372.66\:\text{ohms}[/tex]

Answer:

219 ft

Explanation:

Here we can define the value t = 0s as the moment when the car starts decelerating.

At this point, the acceleration of the car is given by the equation:

A(t) = -10 ft/s^2

Where the negative sign is because the car is decelerating.

To get the velocity equation of the car, we integrate over time, to get:

V(t) = (-10 ft/s^2)*t + V0

Where V0 is the initial velocity of the car, we know that this is 88 ft/s

Then the velocity equation is:

V(t) = (-10 ft/s^2)*t + 88ft/s

To get the position equation we need to integrate again, this time we get:

P(t) = (1/2)*(-10 ft/s^2)*t^2 + (88ft/s)*t + P0

Where P0 is the initial position of the car, we do not know this, but it does not matter for now.

We want to find the total distance that the car traveled in a 3 seconds interval.

This will be equal to the difference in the position at t = 3s and the position at t = 0s

distance = P(3s) - P(0s)

 = ( (1/2)*(-10 ft/s^2)*(3s)^2+ (88ft/s)*3s + P0) - ( (1/2)*(-10 ft/s^2)*(0s)^2 + (88ft/s)*0s + P0)

=  ( (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s + P0) - ( P0)

=  (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s = 219ft

The car advanced a distance of 219 ft in the 3 seconds interval.

Answer:

W/7

Explanation:

By principle of moments,

Sum of clockwise moment = sum of anticlockwise moment

Weight × 7d = W × d

Weight = W/7

Since the two children are in rotational equilibrium, the weight of the second child is W/7.

The weight of the second child can be determined from the principle of moments.

The principle of moments states that for a body in equilibrium, the sum of the clockwise moments and anticlockwise moments about a point is zero.

Let the weight of the second child be X

From the principle of moments:

W × d = 7×d × X

X = W/7

Therefore, the weight of the second child is W/7.

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Answer:

  v = 2.75 10⁴ m / s

Explanation:

For this exercise we must use Kepler's third law which is an application of Newton's second law to the solar system

            F = ma

where force is the force of gravity

            F = [tex]G \frac{m M}{r^2}[/tex]

acceleration is centripetal

             a = [tex]\frac{v^2}{r}[/tex]

we substitute

           G m M / r² = m v² / r

           [tex]\frac{GM}{r}[/tex] = v²

           v = [tex]\sqrt{GM/r}[/tex]

indicate that the radius of the orbit is r = 1.17 AU, let's reduce to the SI system

            r = 1.17 AU (1.496 10¹¹ m / 1 AI) = 1.76 10¹¹ m

let's calculate

         v = [tex]\sqrt{\frac{6.67 \ 10^{-11} 1.991 \ 10^{30} }{ 1.76 \ 10^{11}} }[/tex]Ra (6.67 10-11 1.991 10 30 / 1.76 10 11

         v = [tex]\sqrt{7.5454 \ 10^8 }[/tex]ra 7.5454 10 8

         v = 2.75 10⁴ m / s

Answer:

[tex]X=0.0389m[/tex]

Explanation:

From the question we are told that:

Period of spring [tex]T_s=2.25s[/tex]

Initial Position of Mass [tex]x=0.0480m[/tex]

Final Mass period [tex]T_f=5.85s[/tex]

Generally the equation for the Mass location is mathematically given by

[tex]X=xcos*\frac{2\pi T_s}{T_f}[/tex]

[tex]X=0.048*cos*\frac{2\pi 5.85}{2.25}[/tex]

[tex]X=0.0389m[/tex]

Answer:

By the principle of corona discharge.

Explanation:

The charge on each ball will decreases over time due to the electrical discharge in air.

According to the principle of corona discharge, when the curvature is small, the discharge of the charge takes placed form the pointed ends.

Answer:

1.67 N

Explanation:

Applying,

F = u(dm/dt)+m(du/dt)................ Equation 1

Where F = force, m = mass of the vehicle, u = speed.

Since u is constant,

Therefore, du/dt = 0

F = u(dm/dt)............... Equation 2

From the question,

Given: u = 10 m/s, dm/dt = 10 kg/min = (10/60) kg/s

Substitute these values into equation 2

F = 10(10/60)

F = 100/60

F = 1.67 N

Answer:

v₂ = 53.23 m/s

Explanation:

Given that,

The mass of a golf club, m₁ = 158 g = 0.158 kg

The initial speed of a golf club, u₁  =  48.2 m/s

The mass of a golf ball, m₂ = 46 g = 0.046 kg

It was at rest, u₂ = 0

Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s, v₁ = 32.7 m/s

We use the conservation of energy to find the speed of the golf ball just after impact as follows :

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.158(48.2)-0.158(32.7)}{0.046}\\\\=53.23\ m/s[/tex]

So, the speed of the golf ball just after the impact is equal to 53.23 m/s.

Answer:

The other angle is 75⁰

Explanation:

Given;

velocity of the projectile, v = 10 m/s

range of the projectile, R = 5.1 m

angle of projection, 15⁰

The range of a projectile is given as;

[tex]R = \frac{u^2sin(2\theta)}{g}[/tex]

To find another angle of projection to give the same range;

[tex]5.1 = \frac{10^2 sin(2\theta)}{9.81} \\\\100sin(2\theta) = 50\\\\sin(2\theta) = 0.5\\\\2\theta = sin^{-1}(0.5)\\\\2\theta = 30^0\\\\\theta = 15^0\\\\since \ the \ angle \ occurs \ in \ \ the \ first \ quadrant,\ the \ equivalent \ angle \\ is \ calculated \ as;\\\\90- \theta = 15^0\\\\\theta = 90 - 15^0\\\\\theta = 75^0[/tex]

Check:

sin(2θ) = sin(2 x 75) = sin(150) = 0.5

sin(2θ) = sin(2 x 15) = sin(30) = 0.5

Answer:

Test 1

1.True

2.True

3.True

4.False

5.True

6.True

7.False

8.True

9.True

10.True

yung iba nasa pic

Answer:

pretty sure its 6.2 x 10^-13

Explanation:

I looked it up I'm not a bigbrain but want to help

Answer:

Due to the applied filed the electrons move in a particular direction.

Explanation:

Initially when the switch is off, the free electrons move here and there in any random directions in the conductor with the random speeds called thermal velocity.  So, tat the net flow is almost zero.

When the battery is connected is switch is ON, the random motion of the electrons aligned in a particular direction due to the force applied by the electric filed, so the net flow is not zero it increases and thus the current flow.

Answer:

P =40.69 atm

Explanation:

We need to find the approximate pressure at a depth of 400 m.

It can be calculated as follows :

P = Patm + ρgh

Put all the values,

[tex]P=1\ atm+1025 \times 9.81\times 400\times 9.8692\times 10^{-6}\ atm/Pa\\\\P=40.69\ atm[/tex]

So, the approximate pressure is equal to 40.69 atm.

Answer:

When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.

Explanation:

As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.

Blueshifts happens when the source of the wave and the observer are moving closer towards one another.

Assume that the wave is of frequency [tex]f\; {\rm Hz}[/tex] at the source. In other words, the source of the wave sends out a peak after every [tex](1/f)\; {\text{seconds}}[/tex].

Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.

The source of this wave sends out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. It would appear to the observer that consecutive peaks arrive every [tex](1/f)\; {\text{seconds}}\![/tex]. That would correspond to a frequency of [tex]f\; {\rm Hz}[/tex].

On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of [tex]v \; {\rm m \cdot s^{-1}}[/tex].

Again, the source of this wave would send out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. However, by the time the source sends out the second peak, the source would have been [tex]v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m}[/tex] closer to the observer then when the source sent out the first peak.

When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than [tex](1/f)\; {\text{seconds}}[/tex]. The observed frequency of this wave would be larger than the original [tex]f\; {\rm Hz}[/tex].

Answer:

The intensity of 4 lawn movers is 86 dB.

Explanation:

Intensity of one lawnmower = 80 dB

Let the intensity is I.

Use the formula of intensity

[tex]dB = 10 log\left ( \frac{I}{Io} \right )\\\\80=10log\left ( \frac{I}{Io} \right )\\\\10^8 = \frac{I}{10^{-12}}\\\\I = 10^{-4} W/m^2[/tex]

Now the intensity of 4 lawn movers is

[tex]dB = 10 log\left ( \frac{4I}{Io} \right )\\\\dB=10log\left ( \frac{4\times10^{-4}}{10^{-12}} \right )\\\\dB = 86 dB\\[/tex]

Answer:

h = 2755102 m = 2755.102 km

Explanation:

According to the given condition:

Potential Energy = Energy Consumed by Bulb

[tex]mgh = Pt\\\\h = \frac{Pt}{mg}[/tex]

where,

h = height = ?

P = Power of bulb = 75 W

t = time = (2 h)(3600 s/1 h) = 7200 s

m = mass of bulb = 20 g = 0.02 kg

g = acceleration due to gravity = 9.8 m/s²

Therefore,

[tex]h = \frac{(75\ W)(7200\ s)}{(0.02\ kg)(9.8\ m/s^2)}[/tex]

h = 2755102 m = 2755.102 km

Answer:

have an increased resistance

Answer: hello tables and data related to your question is missing attached below are the missing data

answer:

a) I = I₁ = I₂ = I₃ = 0.484 mA

b) I₁ =  0.016 amps

   I₂ =  0.0016 amps

   I₃ = 7.27 * 10^-4 amps

c)  I₁ = 1.43 * 10^-3 amp

    I₂ =  0.65 * 10^-3 amps

Explanation:

A) magnitude of current for Part 1

Resistors are connected in series

Req = r1 + r2 + r3

       = 3300 Ω  ( value gotten from table 1 ) ,

          V = 1.6 V ( value gotten from table )

hence I ( current ) = V / Req = 1.6 / 3300 = 0.484 mA

The magnitude of current is the same in the circuit

Vi = I * Ri

B) magnitude of current for part 2

Resistors are connected in parallel

V = 1.6 volts

Req = [ ( R1 * R2 / R1 + R2 ) * R3 / ( R1 * R2 / R1 + R2 ) +  R3 ]

      = [ ( 100 * 1000 / 100 + 1000) * 2200 / ( 100 * 1000 / 100 + 1000 ) + 2200]

      = 87.30 Ω

For a parallel circuit the current flow through each resistor is different

hence the magnitude of the currents are

I₁ = V / R1 = 1.6 / 100 = 0.016 amps

I₂ = V / R2 = 1.6 / 1000 = 0.0016 amps

I₃ = V / R3 = 1.6 / 2200 = 7.27 * 10^-4 amps

C) magnitude of current for part 3

Resistors are connected in combination

V = 1.6 volts

Req = R1 + ( R2 * R3 / R2 + R3 )

       = 766.66 Ω

Total current ( I ) = V / Req = 1.6 / 766.66 = 2.08 * 10^-3 amps

magnitude of currents

I₁ = ( I * R3 ) / ( R2 + R3 ) = 1.43 * 10^-3 amps

I₂ = ( I * R2 ) / ( R2 + R3 ) = 0.65 * 10^-3 amps

Answer:

true

Explanation:

im not sure please dont attack me

Solution :

Part A .

Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, d = [tex]\text{4 km 29}[/tex] degree left of [tex]y[/tex]-axis.

So the [tex]x[/tex] component is = -4 x sin (29°) = -1.939 km

           [tex]y[/tex] component is = 4 x cos (29°) = 3.498 km

Part B

Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{v = 2 cm/s}[/tex] , [tex]\text{-x direction}[/tex]

So the [tex]x[/tex] component is = -2 cm/s

           [tex]y[/tex] component is = 0

Part C

Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{a = 13 m/s, 36 degree}[/tex] left of [tex]y[/tex]-axis.

So the [tex]x[/tex] component is = -13 x sin (36°) = -7.6412 [tex]m/S^2[/tex]

           [tex]y[/tex] component is = -13 x cos (36°) = -10.517 [tex]m/S^2[/tex]

The x- and y-components of the vectors  is mathematically given as as follows for each Part respectively

x= -1.939 km, y= 3.498 km

x= -2 cm/s, 0

y=, x= -7.6412m/s^2, -10.517m/s^2

What are the x- and y-components of the vectors?

Question Parameters:

Generally, we follow a basic principle where

x component= Fsin\theta

y component= Fcos\theta

Therefore

For A

x component is

x= -4 x sin (29°)

x= -1.939 km

 y component is

y= 4 x cos (29°)

y= 3.498 km

For B

x component is

x= -2 cm/s            

y component is

y= 0

For C

x component is

x= -13 x sin (36°)

x= -7.6412m/s^2      

y component is

y= -13 x cos (36°)

y= -10.517m/s^2  

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Answer:

Pressure is more in the open container than the closed one.

Explanation:

The pressure due to the fluid at a depth is given by

Pressure = depth x density of fluid x gravity

So, when the container is open, the atmospheric pressure is also add  up but when  the container is closed only the pressure due to the fluid is there.

So, when the container is open, the pressure is atmospheric pressure + pressure due to the fluid.

hen the container is closed only the pressure due to the fluid is there.

Answer:

Following are the solution to the given question:

Explanation:

Please find the complete question in the attached file.

The cost after 30 days is 60 dollars. As energy remains constant, the cost per hour over 30 days will be decreased.

[tex]\to \frac{\$60}{\frac{30 \ days}{24\ hours}} = \$0.08 / kwh.[/tex]

Thus, [tex]\frac{\$0.08}{\$0.12} = 0.694 \ kW \times 0.694 \ kW \times 1000 = 694 \ W.[/tex]

The electricity used is continuously 694W over 30 days.

If just resistor loads (no reagents) were assumed,

[tex]\to I = \frac{P}{V}= \frac{694\ W}{120\ V} = 5.78\ A[/tex]

Energy usage reduction percentage = [tex](\frac{60\ W}{694\ W} \times 100\%)[/tex]

This bulb accounts for [tex]8.64\%[/tex] of the energy used, hence it saves when you switch it off.

Solids cannot diffuse.

Answer:

d

Explanation:

Ya gon find the Kenitic Energy first

K=½mv²===> K=½×0.2×(0.2)²===> 0.1(0.04)===> 0.004

and now the replacement:

0.004=½×0.4V²====> v²=0.02===> V=0.14m/s

Answer:

[tex]K.E_{max}=0.8973J[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=1.64kg[/tex]

Equation of Mass

[tex]X=0.33cos(3.17t)[/tex]...1

Generally equation for distance X is

[tex]X=Acos(\omega t)[/tex]...2

Therefore comparing equation

Angular Velocity [tex]\omega=3.17rad/s[/tex]

Amplitude A=0.33

Generally the equation for Max speed is mathematically given by

[tex]V_{max}=A\omega[/tex]

[tex]V_{max}=0.33*3.17[/tex]

[tex]V_{max}=1.0461m/s[/tex]

Therefore

[tex]K.E_{max}=0.5mv^2[/tex]

[tex]K.E_{max}=0.5*1.64*(1.0461)^2[/tex]

[tex]K.E_{max}=0.8973J[/tex]

Answer:

1. CaO is not an organic compound because it doesn’t contain a carbon molecule.

2.

Name General Structure Properties/Uses

Alcohol R-OH (contains a hydroxyl group)  Can be poisonous, can be made from fermentation or distillation

Aldehyde R-COH (contains a carbon atom double-bonded to an oxygen and single-bonded to a hydrogen)  Makes up formaldehyde and acetaldehyde

Ketone R-CO-R (contains a carbon atom double-bonded to an oxygen atom and then connected to carbon chains through the other two single bonds)   Makes up acetone

Fatty acid R-COOH (contains a carbon atom double-bonded to an oxygen atom, single-bonded to a hydroxyl, and single-bonded to the carbon chain) Makes up fatty acids like acetic acid and stearic acid; used to form esters

Ether R-O-R (contains double carbon chains connected to an oxygen atom through single bonds)   Ethyl ether is very volatile and flammable, used in veterinary medicine

3. Carbon is able to make four covalent bonds with other elements. This gives it a lot of diversity and the ability to form differently shaped molecules that perform specific functions or fit specific cell receptors in the body. H can form only one bond, and oxygen forms only two bonds, so they don't have as much potential to form a good starting point for organic molecules.

Explanation:

pf

CaO is not an organic compound because it doesn’t contain a carbon molecule.

(which contains a hydroxyl group)  Can be poisonous, can be made from fermentation or distillation

Aldehyde R-COH (contains a carbon atom double-bonded to oxygen and single-bonded to hydrogen)  Makes up formaldehyde and acetaldehyde

Ketone R-CO-1R (contains a carbon atom double-bonded to an oxygen atom and then connected to carbon chains through the other two single bonds)   Makes up acetone

Fatty acid R-COOH (contains a carbon atom double-bonded to an oxygen atom, single-bonded to a hydroxyl, and single-bonded to the carbon chain) Makes up fatty acids like acetic acid and stearic acid; used to form esters11

Ether -O-R (contains double carbon chains connected to an oxygen atom through single bonds)   Ethyl ether is very volatile and flammable, used in veterinary medicine

Carbon can make four covalent bonds with other elements. This gives it a lot of diversity and the ability to form differently shaped molecules that perform specific functions or fit specific cell receptors in the body. H can form only one bond, and oxygen forms only two bonds, so they don't have as much potential to form a good starting point for organic molecules.

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#SPJ2

Answer:

5N

Explanation:

We have a simple problem of momentum here.

ΔMomentum= mΔv= FΔt

Solve for F

mΔv/Δt=F

Plug in givens

1*(2-1.5)/0.1=F

F=5N

The amount of force that the wall imparts on the ball is 5.0N

According to Newton's second law, the formula for calculating the force applied is expressed as:

[tex]F=ma[/tex]

m is the mass of the object

a is the acceleration of the object

Since acceleration is the change in velocity of an object, hence [tex]a=\frac{\triangle v}{t}[/tex]

The applied force formula becomes [tex]F=\frac{m\triangle v}{t}[/tex]

Given the following parameters

m = 1.0kg

[tex]\triangle v=2.0-1.5\\\triangle v=0.5m/s[/tex]

t = 0.1sec

Substitute the given parameter into the formula

[tex]F=\frac{1.0\times 0.5}{0.1}\\F=\frac{0.5}{0.1}\\F=5N[/tex]

Hence the amount of force that the wall imparts on the ball is 5.0N

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Answer:

0.857 cm

Explanation:

We are given that:

The focal length for a convex lens to be (f) = 1.5cm

The object distance (u) = - 2.0 cm

We are to determine the image distance (v) = ??? cm

By applying the lens formula:

[tex]\dfrac{1}{f} = \dfrac{1}{u}+\dfrac{1}{v}[/tex]

By rearrangement and making (v) the subject of the above formula:

[tex]v = \dfrac{uf}{u-f}[/tex]

replacing the given values:

[tex]v = \dfrac{(-2.0)(1.5)}{(-2.0 -1.5)}[/tex]

[tex]v = \dfrac{-3.0}{(-3.5)}[/tex]

v = 0.857 cm

Answer:

7.8% of the original volume.

Explanation:

From the given information:

Temperature [tex]T_1[/tex] = 22° C = 273 + 22 = 295° C

Pressure [tex]P_1[/tex] = 240 kPa

Temperature [tex]T_2[/tex] = 45° C

At initial temperature and pressure:

Using the ideal gas equation:

[tex]P_1V_1 =nRT_1[/tex]

making V_1 (initial volume) the subject:

[tex]V_1 = \dfrac{nRT_1}{P_1}[/tex]

[tex]V_1 = \dfrac{nR*295}{240}[/tex]

Provided the pressure maintained its rate at 240 kPa, when the temperature reached 45° C, then:

the final volume [tex]V_2[/tex] can be computed as:

[tex]V_2 = \dfrac{nR*318}{240}[/tex]

Now, the change in the volume ΔV =  V₂ - V₁

[tex]\Delta V = \dfrac{nR*318}{240}- \dfrac{nR*295}{240}[/tex]

[tex]\Delta V = \dfrac{23nR}{240}[/tex]

∴

The required fraction of the volume of air to keep up the pressure at (240) kPa can be computed as:

[tex]= \dfrac{\dfrac{23nR}{240}}{ \dfrac{295nR}{240}}[/tex]

[tex]= {\dfrac{23nR}{240}} \times { \dfrac{240}{295nR}}[/tex]

[tex]= 0.078[/tex]

= 7.8% of the original volume.