A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle so that each side has a length of b. Find the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices.

Answer:

Explanation:

Intensity of the sunlight is expressed as I  = Power/cross sectional area. It is measured in W/m²

Given parameters

Power rating = 6.50Watts

Cross sectional area = 100cm²

Before we calculate the intensity, we need to convert the area to m² first.

100cm² = 10cm * 10cm

SInce 100cm = 1m

10cm = (10/100)m

10cm = 0.1m

100cm² = 0.1m * 0.1m = 0.01m²

Area (in m²) = 0.01m²

Required

Intensity of the sunlight I

I = P/A

I = 6.5/0.01

I = 650W/m²

Hence, the intensity of the sunlight in W/m² is 650W/m²

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:

0.0127m

Explanation:

Using

Ym= (1)(633x10^-9m)(2m) / (0.1x10^-3m) = 0.0127m

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:

4.9x10^-6T

Explanation:

See attached file

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: 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:

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:

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  induced current is [tex]I = 6.25*10^{-4} \ A[/tex]

Explanation:

From the question we are told that  

    The number of turns is  [tex]N = 1[/tex]

     The  cross-sectional area is  [tex]A = 8.20 cm^2 = 8.20 * 10^{-4} \ m^2[/tex]

    The  initial magnetic field is  [tex]B_i = 0.500 \ T[/tex]

     The  magnetic field at time =  1.02 s  is  [tex]B_t = 2.60 \ T[/tex]

     The  resistance is  [tex]R = 2.70\ \Omega[/tex]

The  induced emf is mathematically represented as

       [tex]\epsilon = - N * \frac{ d\phi }{dt}[/tex]

The  negative sign tells us that the induced emf is moving opposite to the change in magnetic flux

      Here  [tex]d\phi[/tex] is the change in magnetic flux which is mathematically represented as

        [tex]d \phi = dB * A[/tex]

Where  dB  is the change in magnetic field which is mathematically represented as

        [tex]dB = B_t - B_i[/tex]

substituting values

        [tex]dB = 2.60 - 0.500[/tex]

        [tex]dB = 2.1 \ T[/tex]

Thus  

      [tex]d \phi = 2.1 * 8.20 *10^{-4}[/tex]

     [tex]d \phi = 1.722*10^{-3} \ weber[/tex]

So  

     [tex]|\epsilon| = 1 * \frac{ 1.722*10^{-3}}{1.02}[/tex]

     [tex]|\epsilon| = 1.69 *10^{-3} \ V[/tex]

The  induced current i mathematically represented as

      [tex]I = \frac{\epsilon}{ R }[/tex]

  substituting values

       [tex]I = \frac{1.69*10^{-3}}{ 2.70 }[/tex]

       [tex]I = 6.25*10^{-4} \ A[/tex]

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:

2.55m

Explanation:

Using 1/do+1/di= 1/f

di= (1/f-1/do)^-1

( 1/0.0500-1/0.0510)^-1

= 2.55m

Answer:

The  voltage is [tex]V_c = 9.92 \ V[/tex]

Explanation:

From the question we are told that

     The voltage of the battery is  [tex]V_b = 24 \ V[/tex]

     The capacitance of the capacitor is  [tex]C = 3.0 mF = 3.0 *10^{-3} \ F[/tex]

     The  resistance of the resistor is [tex]R = 100\ \Omega[/tex]

     The time taken is  [tex]t = 0.16 \ s[/tex]  

Generally the voltage of a charging charging capacitor after time t is mathematically represented as

       [tex]V_c = V_o (1 - e^{- \frac{t}{RC} })[/tex]

Here [tex]V_o[/tex] is the voltage of the capacitor when it is fully charged which in the case of this question is equivalent to the voltage of the battery so  

      [tex]V_c = 24 (1 - e^{- \frac{0.16}{100 * 3.0 *10^{-1}} })[/tex]

      [tex]V_c = 9.92 \ V[/tex]

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:

constant horizontal force developed in the coupling C = 11.25KN

the friction force developed between the tires of the truck and the road during this time is 33.75KN

Explanation:

See attached file

The friction force between the tires of the truck and the road is 22500 N.

It is given that a 2 Mg truck ( m = 2000 Kg) is initially moving with a speed of u = 15 m/s.

Distance traveled before coming to rest, s = 10m

The final velocity of the truck will be zero, v = 0

When the breaks are applied, only the frictional force is acting on the truck and it is opposite to the motion of the truck.

The frictional force is given by:

f = -ma

the acceleration of the truck = -a

The negative sign indicates that the acceleration is opposite to the motion.

Applying the third equation of motion we get:

v² = u² -2as

0 = 15² - 2×a×10

225 = 20a

a = 11.25 m/s²

So the magnitude of frictional force is:

f = ma = 2000 × 11.25 N

f = 22500 N

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Answer:

1248m

The time that wave moves from the wave source to the ocean floor is half the total travel time: t = 2.5/2 = 1.25s

The speed of sound in seawater is 1560 m/s

Therefore, s = vt = (1560 m/s)(1.25s) =1248 m = 1.2km

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The  wavelength is  [tex]\lambda = 622 nm[/tex]

Explanation:

  From the question we are told that

    The distance of the slit to the screen is  [tex]D = 5 \ m[/tex]

    The order of the fringe is m  =  6

     The distance between the slit is  [tex]d = 0.9 \ mm = 0.9 *10^{-3} \ m[/tex]

    The fringe distance is  [tex]Y = 1.9 \ cm = 0.019 \ m[/tex]

Generally the for a dark fringe the fringe distance is  mathematically represented as

        [tex]Y = \frac{[2m - 1 ] * \lambda * D }{2d}[/tex]

=>     [tex]\lambda = \frac{Y * 2 * d }{[2*m - 1] * D}[/tex]

substituting values

=>      [tex]\lambda = \frac{0.019 * 2 * 0.9*10^{-3} }{[2*6 - 1] * 5}[/tex]

=>     [tex]\lambda = 6.22 *10^{-7} \ m[/tex]

       [tex]\lambda = 622 nm[/tex]

Answer:

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

Explanation:

Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ([tex]P_{atm}[/tex]), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:

[tex]P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}[/tex]

Where:

[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Water total pressures inside the tank and at ground level, measured in pascals.

[tex]\rho[/tex] - Water density, measured in kilograms per cubic meter.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Water speeds inside the tank and at the ground level, measured in meters per second.

[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Heights of the tank and ground level, measured in meters.

Given that [tex]P_{1} = P_{2} = P_{atm}[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]z_{1} = 6.9\,m[/tex] and [tex]z_{2} = 4.9\,m[/tex], the expression is reduced to this:

[tex]\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)[/tex]

And final speed is now calculated after clearing it:

[tex]v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}[/tex]

[tex]v_{2} \approx 6.263\,\frac{m}{s}[/tex]

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

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:

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:

Explanation:

proton number + neutron number = atomic mass

30 + 35 = 65

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:

The distance is [tex]d = 1.747 *10^{-6} \ m[/tex]  

Explanation:

From the question we are told that  

       The order of maximum diffraction is  m =  2

         The wavelength is   [tex]\lambda = 775 nm = 775 * 10^{-9} \ m[/tex]

         The angle is  [tex]\theta = 62.5^o[/tex]

Generally the   condition for  constructive  interference for diffraction grating  is mathematically represented as

          [tex]dsin \theta = m * \lambda[/tex]

where  d is  the distance between the lines on a  diffraction grating

     So  

            [tex]d = \frac{m * \lambda }{sin (\theta )}[/tex]

substituting values  

           [tex]d = \frac{2 * 775 *1^{-9} }{sin ( 62.5 )}[/tex]

          [tex]d = 1.747 *10^{-6} \ m[/tex]

Answer:

I believe it's called rapid growth

Explanation:

that is my answer no matter what

Answer:

The unit of work is joules

Force and displacement are used to calculate the work done by an object. This work is measured in the units of Joules. Thus, the correct option is A.

Work can be defined as the force that is applied on an object which shows some displacement. Examples of work done include lifting an object against the Earth's gravitational force, and driving a car up on a hill. Work is a form of energy. It is a vector quantity as it has both the direction as well as the magnitude. The standard unit of work done is the joule (J). This unit is equivalent to a newton-meter (N·m).

The nature of work done by an object can be categorized into three different classes. These classes are positive work, negative work and zero work. The nature of work done depends on the angle between the force and displacement of the object. Positive work is done if the applied force displaces the object in its direction, then the work done is known as positive work. Negative work is opposite of positive work as in this work, the applied force and displacement of the object are in opposite directions to each other and zero work is done when there is no displacement.

Therefore, the correct option is A.

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Answer:

work done in pumping the entire fuel is 1399761 J

Explanation:

weight per volume of the gasoline = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 6 m

The height of the tractor tank above the top of the tank = 5 m

The total volume of the fuel is gotten below

we know that the tank is cylindrical.

we assume that the fuel completely fills the tank.

therefore, the volume of a cylinder =  

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x  x 6 = 42.417 m^3

we then proceed to find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 42.417 = 279952.2 N

therefore,

the work done to pump the fuel through to the 5 m height = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 279952.2 x 5 = 1399761 J

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:

1.03A

Explanation:

For computing the magnitude of the current in the circuit we need to do the following calculations

LCR circuit impedance

[tex]Z = \sqrt{R^2 + (X_L - X_c)^2} \\\\ = \sqrt{110^2 + (210 - 110)^2}[/tex]

= 148.7Ω

Now the phase angle is

[tex]\phi = tan^{-1} (\frac{X_L - X_C}{R}) \\\\ = tan^{-1} (\frac{210 - 110}{110})\\\\ = 42.3^{\circ}[/tex]

Now the rms current flowing in the circuit is

[tex]I_{rms} = \frac{V_{rms}}{Z} \\\\ = \frac{146}{148.7}[/tex]

= 0.98 A

The current flowing in the circuit is

[tex]I = I_{rms}\sqrt{2} \\\\ = (0.98) (1.414)[/tex]

= 1.39 A

And, finally, the current across the generator is

[tex]I'= I cos \phi[/tex]

[tex]= (1.39) cos 42.3^{\circ}[/tex]

= 1.03A

Hence, the magnitude of the circuit current is 1.03A

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:

a

    [tex]\alpha = 2327.7 \ rev/s^2[/tex]

b

   [tex]\theta = 9124.5 \ rev[/tex]

Explanation:

From the question we are told that

    The maximum  angular   speed is  [tex]w_{max} = 391000 \ rpm = \frac{2 \pi * 391000}{60} = 40950.73 \ rad/s[/tex]

     The  time  taken is  [tex]t = 2.8 \ s[/tex]

     The  minimum angular speed is  [tex]w_{min}= 0 \ rad/s[/tex] this is because it started from rest

Apply the first equation of motion to solve for acceleration we have that

       [tex]w_{max} = w_{mini} + \alpha * t[/tex]

=>     [tex]\alpha = \frac{ w_{max}}{t}[/tex]

substituting values

       [tex]\alpha = \frac{40950.73}{2.8}[/tex]

       [tex]\alpha = 14625 .3 \ rad/s^2[/tex]

converting to [tex]rev/s^2[/tex]

  We have

           [tex]\alpha = 14625 .3 * 0.159155 \ rev/s^2[/tex]

           [tex]\alpha = 2327.7 \ rev/s^2[/tex]

According to the first equation of motion the angular displacement is  mathematically represented as

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

substituting values

      [tex]\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2[/tex]

      [tex]\theta = 57331.2 \ radian[/tex]

converting to revolutions  

        [tex]revolution = 57331.2 * 0.159155[/tex]

        [tex]\theta = 9124.5 \ rev[/tex]