When static equilibrium is established for a charged conductor, the electric field just inside the surface of the conductor is

Answer:

2.52 rpm

Explanation:

given that

diameter of the wheel, d = 25 m

Mass of the person, m = 75 kg

Weight experienced, N = 810 N

Since diameter is 25, radius then is 25/2 = 12.5 m

We all know that,

v = rw

Also, the passengers weight is equal to the centripetal acceleration, and thus

mg = mv²/r

Substitute for v, we have

mg = m/r * (rw)²

mg = mr²w²/r

g = rw²

If we make w the subject of formula, we have

w² = g/r

w = √(g/r)

mg = 810

75 * g = 810

g = 810 / 75

g = 1.08 m/s²

w = √(g/r)

w = √(1.08 / 12.5)

w = √0.0864

w = 0.294 rad/s

Since the question asked us in rpm, we convert to rpm

0.294 * (60 / 2π)

2.52 revolution per minute.

Answer:

The object with the twice the area of the other object, will have the larger drag coefficient.

Explanation:

The equation for drag force is given as

[tex]F_{D} = \frac{1}{2}pu^{2} C_{D} A[/tex]

where [tex]F_{D}[/tex] IS the drag force on the object

p = density of the fluid through which the object moves

u = relative velocity of the object through the fluid

p = density of the fluid

[tex]C_{D}[/tex] = coefficient of drag

A = area of the object

Note that [tex]C_{D}[/tex] is a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag. The most interesting things is that it is dependent on the linear dimension, which means that it will vary directly with the change in diameter of the fluid

The above equation can also be broken down as

[tex]F_{D}[/tex] ∝ [tex]P_{D}[/tex] A

where [tex]P_{D}[/tex] is the pressure exerted by the fluid on the area A

Also note that [tex]P_{D}[/tex] = [tex]\frac{1}{2}pu^{2}[/tex]

which also clarifies that the drag force is approximately proportional to the abject's area.

In this case, the object with the twice the area of the other object, will have the larger drag coefficient.

Answer:

 v ’= 21.44 m / s

Explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

           f ’= f (v + v₀) / (v-[tex]v_{s}[/tex])

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

                v₀ = v_{s} = v ’

we substitute

               f ’= f (v + v’) / (v - v ’)

               f ’/ f (v-v’) = v + v ’

               v (f ’/ f -1) = v’ (1 + f ’/ f)

               v ’= (f’ / f-1) / (1 + f ’/ f) v

               v ’= (f’-f) / (f + f’) v

let's calculate

                v ’= (3400 -3000) / (3000 +3400) 343

                v ’= 400/6400 343

                v ’= 21.44 m / s

Answer:

Please see below as the answer is self-explanatory.

Explanation:

The low band of the VHF TV Spectrum, spans channels 2-6, from 54 to 88 Mhz.

In the analog TV, in the Americas, the total bandwidth of any channel is 6 Mhz, with the visual carrier modulated in VSS (Vestigial Side Band) at 1.25 Mhz from the lowest frequency of the channel.

The aural carrier is located at 4.5 Mhz from the visual carrier, and is FM modulated.

For Channel 6, which spans between 82 and  88 Mhz, the visual carrier is at 83.25 Mhz, so the aural carrier is at 87.75 Mhz, which falls within the FM Band, so it is possible to listen the audio part of this channel in a FM radio receiver, even at a lower volume, due to the FM radio has a greater deviation than TV aural carrier.

The reason why it is possible for TV station to sometimes pick up some of the audio portion on your FM radio receiver is because; TV waves can sometimes deviate into the FM radio frequency range.

Let us start with explaining the waves of TV and radio.

The frequency range utilized by TV stations is either the range 54 MHz to 88 MHz or 174 MHz to 222 MHz. In contrast, the frequency range utilized by FM Radio band is between 88 MHz and 174 MHz.

Now, in some cases, it is possible that the TV signal may deviate into the range of the FM Radio and as such in that case, the TV signal will pick the audio portion of an FM Radio. These TV waves are very high frequency waves.

Finally, it does not imply that the TV wave is broadcasting as an FM because it only deviated a bit from the TV range and not like that is where it is made to operate.

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Answer:

The 50N weight be hung  at 23 cm to maintain equilibrium

Explanation:

Given;

length of the uniform rod = 30 cm

center of the uniform rod = 15 cm

weight of 40N is hung at 5 cm mark

weight of 50 N will be hung at ?

     0------5cm-----------------15cm-------------P---------30cm

              ↓             10cm      Δ       xcm    ↓

             40N                                             50N

Take moment about the pivot point and apply the principle of moment

50N (x cm) = 40N (10 cm)

x = (400) / 50

x = 8cm

P = x cm + 15 cm

P = 8 cm + 15 cm

P = 23 cm

Therefore, the 50N weight be hung  at 23 cm to maintain equilibrium

Answer:

e)

Explanation:

In an RC series circuit, at any time, the sum of the voltages through the resistor and the capacitor must be constant and equal to the voltage of the DC voltage source, in order to be compliant with KVL.

At= 0, as the voltage through the capacitor can't change instantaneously, all the voltage appears through the resistor, which means that a current flows, that begins to charge the capacitor, up to a point that the voltage through the capacitor is exactly equal to the DC voltage, so no current flows in the circuit anymore, and the charge in the capacitor reaches to its maximum value.

Answer:

acting force is the answer

The direction of the magnetic force on the moving electron is upward.

The direction of the magnetic force on the electron can be determined by applying right hand rule.

This rule states that when the thumb is held perpendicular to the fingers, the thumb will point in the direction of the speed while the fingers will point in the direction of the field and the magnetic force will be perpendicular to the field.

Thus, we can conclude that, the direction of the magnetic force on the moving electron is upward.

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Answer:

The magnitude of the magnetic field is 55μT

Explanation:

Given;

number of turns of the solenoid per length, n = N/L = 1500 turns/m

current in the solenoid, I = 20 mA = 20 x 10⁻³ A

diameter of the solenoid, d = 4 cm = 0.04 m

The magnetic field at a point that is inside the solenoid;

B₁ = μ₀nI

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

B₁ = 4π x 10⁻⁷ x 1500 x 20 x 10⁻³

B₁ = 3.77 x 10⁻⁵ T

Given;

current in the wire, I = 2 A

distance of magnetic field from the wire, r = 1 cm = 0.01 m

The magnetic field at 1.0 cm from the wire;

[tex]B_2 = \frac{\mu_0I}{2\pi r} \\\\B_2 = \frac{4\pi*10^{-7}*2}{2\pi *0.01}\\\\B_2 = 4 *10^{-5} \ T[/tex]

The magnitude of the magnetic field;

[tex]B = \sqrt{B_1^2 +B_2^2} \\\\B = \sqrt{(3.77*10^{-5})^2 + (4*10^{-5})^2} \\\\B = 5.5 *10^{-5} \ T\\\\B = 55 \mu T[/tex]

Therefore, the magnitude of the magnetic field is 55μT

The magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire is [tex]5.5 \times 10^{-5}T[/tex]

Given the following parameters from the question  

The magnetic field at a point that is inside the solenoid is expressed according to the formula;  

Where;  

μ₀ is the permeability of free space = 4π x 10⁻⁷ m/A  

B₁ = 4π x 10⁻⁷ x 1500 x 20 x 10⁻³  

B₁ = 3.77 x 10⁻⁵ T

Next is to get the magnetic field strength in the second wire.

Substitute into the formula:

[tex]B_2=\dfrac{\mu_0 I}{2 \pi r} \\B_2=\frac{4\pi \times 10^{-7}\times 2}{2 \times 3.14\times 0.01} \\B_2 =4.0 \times 10^{-5}T[/tex]

Get the resultant magnetic field:

[tex]B = \sqrt{(0.00003771)^2+(0.00004)^2} \\B =5.5 \times 10^{-7}T[/tex]

Therefore the magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire is [tex]5.5 \times 10^{-5}T[/tex]

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Answer:

The 1/2 inch barrel will burst at the same height of 20 ft

Explanation:

The pressure on a column of fluid increases with depth, and decreases with height. This means that if you increase the height of the fluid in the column, the pressure at the bottom will increase.

From the equation of fluid pressure,

P = ρgh

where

P is the pressure at the bottom of the fluid due to its height

ρ is the density of the fluid in question

h is the height to which the water stand.

You notice how apart from the height 'h' in the equation, all the other parts of the right hand side of the equation cannot be varied; they are a fixed property of the fluid and gravity. And there is no consideration for the horizontal diameter of the water's cross section area.

We can also think of the pressure at the bottom of the fluid to be as a result of an incremental weight of an infinitesimally small vertical section of the water down.

That been said, we can then say that if the barrel with the 1 ft diameter dimension bursts when filled with water up to 20 ft, then, the barrel with the reduced diameter will still burst at the same height as the former pipe.

NB: The only way to stop the pipe from bursting is to increase the thickness of the barrel wall to counteract the pressure forces due to the height.

Answer:

2.64 x 10⁻⁶T

Explanation:

The magnitude of the magnetic field produced by a long straight wire carrying current is given by Biot-Savart law as follows: "The magnetic field strength is directly proportional to the current on the wire and inversely proportional to the distance from the wire". This can be written mathematically as;

B = (μ₀ I) / (2π r)                ----------------(i)

B is magnetic field

I is current through the wire

r is the distance from the wire

μ₀  is the magnetic constant = 4π x 10⁻⁷Hm⁻¹

From the question;

I = 0.7A

r = 0.053m

Substitute these values into equation (i) as follows;

B =  (4π x 10⁻⁷ x 0.7) / (2π x 0.053)

B = 2.64 x 10⁻⁶T

Therefore the approximate magnitude of the magnetic field at that location is 2.64 x 10⁻⁶T

Answer:

So, according to Einstein's special relativity a person on Mars observe the light to be traveling at c = 3 x 10⁸ m/s.

Explanation:

The special theory of relativity has two main postulates:

1- VALIDITY OF PHYSICAL LAWS

The laws of physics such as Newton's Laws and Maxwell's Equations are valid in all inertial frame of references.

2- CONSTANCY OF SPEED OF LIGHT

The speed of light in vacuum is the same for all observers in uniform translational relative motion, and it is independent of the motion of the source or the observer. Thus, speed of light is a universal constant and its value is c = 3 x 10⁸ m/s.

So, according to Einstein's special relativity a person on Mars observe the light to be traveling at c = 3 x 10⁸ m/s.

Answer:

Explanation:

For circular path in magnetic field

mv² / R = Bqv ,

m is mass , v is velocity , R is radius of circular path , B is magnetic field , q is charge on the particle .

a )

R = mv / Bq

If v is changed  to 2v , keeping other factors unchanged , R will be doubled

b )

magnitude of acceleration inside field

= v² / R

= Bqv / m

As v is doubled , acceleration will also be doubled

c )

If T be the time inside the magnetic field

T = π R / v

=  π  / v x  mv / Bq

= π m / Bq

As is does not contain v that means T  remains unchanged .

d )

Net force acting on electron

= m v² / R = Bqv

Net force = Bqv

As v becomes twice force too becomes twice .

So a . b , d are correct answer.

Answer:

Mass of shot (m) = 4 kg

Explanation:

Given:

Velocity (v) = 15 m/s

Mechanical kinetic energy (K.E) = 450 J

Find:

Mass of shot (m) = ?

Computation:

Mechanical kinetic energy (K.E) = 1/2mv²

Mechanical kinetic energy (K.E) = [1/2](m)(15)²

450 = [1/2](m)(15)²

900 = 225 m

Mass of shot (m) = 4 kg

Answer:

Option 3 = both spheres are at the same potential.

Explanation:

So, let us complete or fill the missing gap in the question above;

" A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not equal to r1) by a conducting wire. After the charges on the spheres are in equilibrium BOTH SPHERES ARE AT THE SAME POTENTIAL"

The reason both spheres are at the same potential after the charges on the spheres are in equilibrium is given below:

=> So, if we take a look at the Question again, the kind of connection described in the question above (that is a charged sphere, say X is connected another charged sphere, say Y by a conducting wire) will eventually cause the movement of charges(which initially are not of the same potential) from X to Y and from Y to X and this will continue until both spheres are at the same potential.

Answer:

 x = 0.006 m

Explanation:

The potential at one point is given by

          V = k ∑ [tex]q_{i} / r_{i}[/tex]

remember that the potential is to scale, let's apply to our case

          V = k (q₁ / x₁ + q₂ / x₂ + q₃ / x)

in this case they indicate that the potential is zero

          0 = k (2 10⁻⁶ / (- 1 10⁻²) + (-6 10⁻⁶) / 2 10⁻² + ​​3 10⁻⁶ / x)

         3 / x = + 2 / 10⁻² + ​​3 / 10⁻²

         3 / x = 500

          x = 3/500

          x = 0.006 m

Answer:

V = 451.47 volts

Explanation:

Given that,

Distance, d = 0.21 m

Initial speed, u = 0

Time, t = 33.3 ns

Let v is the final velocity. Using second equation of motion as :

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

a is acceleration, [tex]a=\dfrac{v-u}{t}[/tex] and u = 0

So,

[tex]d=\dfrac{1}{2}(v-u)t[/tex]

[tex]v=\dfrac{2d}{t}\\\\v=\dfrac{2\times 0.21}{33.3\times 10^{-9}}\\\\v=1.26\times 10^7\ m/s[/tex]

Now applying the conservation of energy i.e.

[tex]\dfrac{1}{2}mv^2=qV[/tex]

V is voltage

[tex]V=\dfrac{mv^2}{2q}\\\\V=\dfrac{9.1\times 10^{-31}\times (1.26\times 10^7)^2}{2\times 1.6\times 10^{-19}}\\\\V=451.47\ V[/tex]

So, the voltage is 451.47 V.

Answer:

The  time interval is  [tex]t = 3 \ s[/tex]

Explanation:

From the question we are told that

    The angular acceleration is  [tex]\alpha = 4.0 \ rad/s^2[/tex]

     The  time taken is  [tex]t = 4.0 \ s[/tex]

      The angular displacement is  [tex]\theta = 80 \ radians[/tex]

The angular displacement can be represented by the second equation of motion as shown below

          [tex]\theta = w_i t + \frac{1}{2} \alpha t^2[/tex]

where  [tex]w_i[/tex] is the initial velocity at the start of the 4 second interval

So substituting values

        [tex]80 = w_i * 4 + 0.5 * 4.0 * (4^2)[/tex]

=>    [tex]w_i = 12 \ rad/s[/tex]

Now considering this motion starting from the start point (that is rest ) we have

       [tex]w__{4.0 }} = w__{0}} + \alpha * t[/tex]

Where  [tex]w__{0}}[/tex] is the angular velocity at rest which is zero  and  [tex]w__{4}}[/tex] is the angular velocity after 4.0 second which is calculated as 12 rad/s s

        [tex]12 = 0 + 4 t[/tex]

=>       [tex]t = 3 \ s[/tex]

Following are the response to the given question:

Given:

[tex]\to \alpha = 4.0 \ \frac{rad}{s^2}\\\\[/tex]

[tex]\to \theta= 80\ radians\\\\\to t= 4.0 \ s\\\\ \to \theta_0=0\\[/tex]

To find:

[tex]\to \omega=?\\\\\to t=?\\\\[/tex]

Solution:

Using formula:

[tex]\to \theta- \theta_0 = w_{0} t+ \frac{1}{2} \alpha t^2\\\\ \to 80-0= \omega_{0}(4) + \frac{1}{2} (4)(4^2)\\\\ \to 80= \omega_{0}(4) + \frac{1}{2} (4)(16)\\\\\\to 80= \omega_{0}(4) + (4)(8)\\\\\to 80= \omega_{0}(4) + 32\\\\\to 80-32 = \omega_{0}(4) \\\\\to \omega_{0}(4)= 48 \\\\\to \omega_{0}= \frac{48}{4} \\\\ \to \omega_{0} = 12 \frac{rad}{ s} \\\\[/tex]  

It would be the angle for rotation at the start of the 4-second interval.

This duration can be estimated by leveraging the fact that the wheel begins from rest.  

[tex]\to \omega = \omega_{0} + \alpha t\\\\\to 12 = 0 +4(t) \\\\\to 12 = 4(t) \\\\ \to t=\frac{12}{4}\\\\\to t= 3\ s[/tex]

Therefore, the answer is "[tex]12\ \frac{rad}{s}[/tex] and [tex]3 \ s[/tex]".

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Answer: Translational Kinetic Energy

Rotational Kinetic Energy

Explanation:

An object has translational kinetic energy when it is undergoing through a linear displacement.

Rotational energy is kinetic energy due to the rotation of an object .

Here the wheel of bicycle undergoes both translational and rotational kinetic energy has it moves with linear displacement with rotation in it.

Hence, the tires have been two kinds of energy : translational and rotational kinetic energy

Answer:

4.9x10^-6T

Explanation:

See attached file

Answer:

The speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s

Explanation:

Given;

speed of the ball thrown inside boxcar, [tex]V_B[/tex] = 23.2 m/s

speed of the boxcar moving along the tracks, [tex]V_T[/tex] = 34.9 m/s

Determine the speed of the ball relative to the ground if the ball is thrown forward.

If the ball is thrown forward, the speed of the ball relative to the ground will be sum of the ball's speed plus speed of the boxcar.

[tex]V_{relative \ speed} = V_B + V_T\\\\V_{relative \ speed} = 23.2 + 34.9\\\\V_{relative \ speed} = 58.1 \ m/s[/tex]

Therefore,  the speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s.

Answer:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

Explanation:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

Answer:

Heat Flow Rate : ( About ) 87 W

Explanation:

The heat flowing out of the system each minute, will be represented by the following equation,

Q( cup ) + Q( water ) = m( cup ) [tex]*[/tex] c( al ) [tex]*[/tex] ΔT + m( w ) [tex]*[/tex] c( w ) [tex]*[/tex] ΔT

So as you can see, the mass of the aluminum cup is 150 grams. For convenience, let us convert that into kilograms,

150 grams = .15 kilograms - respectively let us convert the mass of water to kilograms,

800 grams = .8 kilograms

Now remember that the specific heat of aluminum is 900 J / kg [tex]*[/tex] K, and the specific heat of water = 4186 J / kg [tex]*[/tex] K. Therefore let us solve for " the heat flowing out of the system per minute, "

Q( cup ) + Q( water ) = .15 [tex]*[/tex] ( 900 J / kg [tex]*[/tex] K )  [tex]*[/tex] 1.5 + .8 [tex]*[/tex] ( 4186 J / kg [tex]*[/tex] K ) [tex]*[/tex] 1.5,

Q( cup ) + Q( water ) = 5225.7 Joules

And the heat flow rate should be Joules per minute,

5225.7 Joules / 60 seconds = ( About ) 87 W

Answer:

   s_400 = 16.5 m , s_700 = 29.4 m

Explanation:

The limit of the human eye's solution is determined by the diffraction limit that is given by the expression

                   θ = 1.22 λ / D

where you lick the wavelength and D the mediator of the circular aperture.

In our case, the dilated pupil has a diameter of approximately 8 mm = 8 10-3 m and the eye responds to a wavelength between 400 nm and 700 nm.

by introducing these values ​​into the formula

λ = 400 nm      θ = 1.22 400 10⁻⁹ / 8 10⁻³ = 6 10⁻⁵ rad

λ = 700 nm     θ = 1.22 700 10⁻⁹ / 8 10⁻³-3 = 1.07 10⁻⁴ rad

Now we can use the definition radians

          θ= s / R

where s is the supported arc and R is the radius. Let's find the sarcos for each case

λ = 400 nm       s_400 = θ R

                         S_400 = 6 10⁻⁵ 275 10³

                         s_400 = 16.5 m

λ = 700 nm s_ 700 = 1.07 10⁻⁴ 275 10³

                          s_700 = 29.4 m

Answer:

I believe it's called rapid growth

Explanation:

that is my answer no matter what

The change in internal energy (ΔE) of the system is equal to -18823 Kilojoules.

Given the following data:

Conversion:

1 kcal = 4.184 kJ

[tex]4.50 \times 10^2 \;kcal[/tex] = [tex]4.50 \times 10^2 \times 4.184 = 18828 \; kJ[/tex]

To determine the change in internal energy (ΔE) of the system, we would apply the first law of thermodynamics.​

Mathematically, the first law of thermodynamics is given by the formula:

[tex]\Delta E = Q - W[/tex]

Where;

Substituting the given parameters into the formula, we have;

[tex]\Delta E = 5 - 18828\\\\\Delta E = -18823[/tex]

Change in internal energy, E = -18823 Kilojoules

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Answer:

Weight is 83 385 N

Explanation:

Weight is calculated by multiplying the mass by the gravitational acceleration constant

       Weight = mass* gravity

Assuming that the gravitational constant is 9.81 m/s^2

        Weight = mass* gravity

        Weight of crane = (8500 kg)*(9.81 m/s^2)

        Weight = 83 385 kg*m/s^2 or 83 385 N

Answer:

9.42*10^-4 m^3/s

Explanation:

d=3/100 m

=0.03 m

A=3.14*0.03^2/6

=4.71*10^-4 m^2

Volume flow rate V = A * s

= 4.71*10^-4 * 2

= 9.42*10^-4

So, the volume flw rate of fluid is 9.42*10^-4 m^3/s

Answer:

Since their wings and body develop the drag. When there is warm air then they expand their wings. Since,soaring birds and glider pilots have no engine, they always maintain their high speed to lift their weight in air for hours without expending power by convection

Explanation:

Answer:

The maximum speed of the motorcycle should be 21 m/s

Explanation:

Since the hill is considered to be a circular arc, the motorcycle will experience centripetal force that tends to flip it away from the center of the hill.

Since the motorcycle does not lose contact with the ground, it means that the weight of the motorcycle downwards just balances the centripetal force on the motorcycle.

we know that the centripetal force on the motorcycle is equal to

centripetal force = [tex]\frac{mv^{2} }{r}[/tex]

where m is the mass of the motorcycle,

v is the velocity of the motorcycle,

and r is the radius of the hill = 45.0 m

Also we now that the weight of the motorcycle is equal to

weight = mg

where m is still the mass of the motorcycle,

and g is the acceleration due to gravity = 9.81 m/s

Equating the both forces since they are equal, we'll have

[tex]\frac{mv^{2} }{r}[/tex] = mg

the mass of the motorcycle will cancel out, and we'll be left with

[tex]v^{2} = gr[/tex]

[tex]v = \sqrt{gr}[/tex]

[tex]v = \sqrt{9.81*45}[/tex]

[tex]v = \sqrt{441.45}[/tex]

[tex]v[/tex] = 21 m/s