Rank from largest to smallest the values of the magnetic field at the following distances from the axis of the conducting cylinder: Ra = 7.75 cm , Rb = 4.95 cm , r = 5.40 cm , and r>Ra.

Answer:

Explanation:

Balance point will be achieved as soon as the weight of the baby elephant creates torque equal to torque created by weight of woman about the pivot. torque by weight of woman

weight x distance from pivot

= 500x 5

= 2500 Nm

torque by weight of baby woman , d be distance of baby elephant from pivot at the time of balance

= 2500x d

for equilibrium

2500 d = 2500

d = 1 m

So elephant will have to walk up to 1 m close to pivot or middle point.

Answer:

A

Explanation:

Current, magnetic field and motion are mutually dependent and perpendicular to one another

Answer:

b) clockwise

Explanation:

The right hand rule states that to determine the direction of the magnetic force on a positive moving charge, point the thumb of the right hand in the direction of the potential v, the fingers in the direction of the magnetic field, B, and a perpendicular to the palm points in the direction of the force, F.

Since the south pole of the bar magnet is pointing downwards, the induced current developed in the wire repels it and moves in the clockwise direction.

Answer:

a

The speed of the particle is  [tex]v = 209485.71 m/s[/tex]

b

The potential difference is  [tex]\Delta V = 177.2 \ V[/tex]

Explanation:

From the question we are told that

    The mass of the particle is  [tex]m = 2.10 *10^{-16} kg[/tex]

     The charge on the particle is  [tex]q = 26.0 nC = 26.0 *10^{-9} C[/tex]

    The magnitude of the magnetic field is [tex]B = 0.600 T[/tex]

    The magnetic flux is  [tex]\O = 15.0 \mu Wb = 15.0 *10^{-6} Wb[/tex]

The magnetic flux is mathematically represented as

              [tex]\O = B *A[/tex]

Where A is the the area mathematically represented as  

           [tex]A = \pi r^2[/tex]

Substituting this into the equation w have

         [tex]\O = B (\pi r^2 )[/tex]

Making r the subject  of the formula  

         [tex]r = \sqrt{\frac{\O}{B \pi} }[/tex]

Substituting value  

        [tex]r = \sqrt{\frac{15 *10^{-6}}{3.142 * 0.6} }[/tex]

            [tex]r = 2.82 *10^{-3}m[/tex]

For the particle to form a circular path the magnetic force the partial experience inside the magnetic must be equal to the centripetal force of the particle and this is mathematically represented as

              [tex]F_q = F_c[/tex]

Where [tex]F_q = q B v[/tex]

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

Substituting this into the equation above

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

 making v the subject

       [tex]v = \frac{r q B}{m}[/tex]

substituting values

         [tex]v = \frac{2.82 * 10^{-3} * 26 *10^{-9} 0.6}{2.10*10^{-16}}[/tex]

           [tex]v = 209485.71 m/s[/tex]

The potential energy of the particle before entering the magnetic field is equal to the kinetic energy in the magnetic field

   This is mathematically represented as

                    [tex]PE = KE[/tex]

     Where [tex]PE = q * \Delta V[/tex]            

        and  [tex]KE = \frac{1}{2} mv^2[/tex]

Substituting into the equation above  

             [tex]q \Delta V = \frac{1}{2} mv^2[/tex]

Making the potential difference the subject

            [tex]\Delta V = \frac{mv^2}{2 q}[/tex]

              [tex]\Delta V = \frac{2.16 * 10^ {-16} * (209485.71)^2 }{2 * 26 *10^{-9}}[/tex]

              [tex]\Delta V = 177.2 \ V[/tex]

Final answer:

The speed of the particle in the magnetic field is found using the force equation for a moving charge in a magnetic field and the work-energy principle is used to determine the magnitude of the potential difference.

Explanation:

To answer the student's question:

The speed of the particle in the magnetic field can be calculated using the formula for the magnetic force acting on a moving charge, which is F = qvB. Since this force provides the centripetal force required for circular motion, F = mv2/r, where m is the mass of the particle, q is the charge of the particle, v is the speed of the particle, B is the magnetic field, and r is the radius of the circular path. Combining these equations, we can solve for the speed: v = qBr/m. To determine the radius, we can use the relationship between magnetic flux (ΦB), magnetic field strength (B), and area of the circular path (A): ΦB = BA or r = sqrt(ΦB / πB). The mass, charge, and magnetic field are given, and we can calculate the radius from the magnetic flux to then find the speed.

The magnitude of the potential difference the particle was initially accelerated through can be determined using the work-energy principle. The work done by the electric field on the charge is equal to the change in kinetic energy of the particle. Using the formula W = ΔKE = qΔV and knowing that the kinetic energy is KE = 1/2 mv2, we can solve for the potential difference: ΔV = KE/q, where KE is the kinetic energy of the particle right before entering the magnetic field.

Answer:

11.07Hz

Explanation:

Check the attachment for diagram of the standing wave in question.

Formula for calculating the fundamental frequency Fo in strings  is V/2L where;

V is the velocity of the wave in string

L is the length of the string which is expressed as a function of its wavelength.

The wavelength of the string given is 1.5λ(one loop is equivalent to 0.5 wavelength)

Therefore L = 1.5λ

If L = 3.0m

1.5λ = 3.0m

λ = 3/1.5

λ = 2m

Also;

V = √T/m where;

T is the tension = 0.98N

m is the mass per unit length = 2.0g = 0.002kg

V = √0.98/0.002

V = √490

V = 22.14m/s

Fo = V/2L (for string)

Fo = 22.14/2(3)

Fo = 22.14/6

Fo = 3.69Hz

Harmonics are multiple integrals of the fundamental frequency. The string in question resonates in 2nd harmonics F2 = 3Fo

Frequency of the wave = 3×3.69

Frequency of the wave = 11.07Hz

0.46Ω

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir                      --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ                    ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

First, let's calculate the effective resistance in the circuit:

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P = [tex]\frac{V^{2} }{R}[/tex]

=> R = [tex]\frac{V^{2} }{P}[/tex]             -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = [tex]\frac{V^{2} }{P}[/tex]    --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = [tex]\frac{12.0^{2} }{4}[/tex]

R₁ = [tex]\frac{144}{4}[/tex]

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = [tex]\frac{V^{2} }{P}[/tex]    --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = [tex]\frac{12.0^{2} }{4}[/tex]

R₂ = [tex]\frac{144}{4}[/tex]

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

[tex]\frac{1}{R_{X} }[/tex] = [tex]\frac{1}{R_1}[/tex] + [tex]\frac{1}{R_2}[/tex]       -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

[tex]\frac{1}{R_X}[/tex] = [tex]\frac{1}{36}[/tex] + [tex]\frac{1}{36}[/tex]

[tex]\frac{1}{R_X}[/tex] = [tex]\frac{2}{36}[/tex]

Rₓ = [tex]\frac{36}{2}[/tex]

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

Now calculate the current I, flowing in the circuit:

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = [tex]\frac{11.7}{18}[/tex]

I = 0.65A

Now calculate the battery's internal resistance:

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A  into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = [tex]\frac{0.3}{0.65}[/tex]

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

Answer:

[tex]R_i_n_t=0.45 \Omega[/tex]

Explanation:

Internal resistance is a concept that helps model the electrical consequences of the complex chemical reactions that occur within a battery. When a charge is applied to a battery, the internal resistance can be calculated using the following equation:

[tex]R_i_n_t=(\frac{V_N_L}{V_F_L} -1)R_L[/tex]

Where:

[tex]V_F_L=Load\hspace{3}voltage=11.7V\\V_N_L= O pen\hspace{3}circuit\hspace{3}voltage=12V\\R_L=Load\hspace{3}resistance[/tex]

As you can see, we don't know the exactly value of the [tex]R_L[/tex]. However we can calculated that value using the next simple operations:

The problem tell us that the power of each lightbulb is 4.0 W at 12.0 V, hence let's calculated the power at 11.7V using Cross-multiplication:

[tex]\frac{12}{11.7} =\frac{4}{P}[/tex]

Solving for [tex]P[/tex] :

[tex]P=\frac{11.7*4}{12} =3.9W[/tex]

Now, the electric power is given by:

[tex]P=\frac{V^2}{R_b}[/tex]

Where:

[tex]R_b=Resistance\hspace{3}of\hspace{3}each\hspace{3}lightbulb[/tex]

So:

[tex]R_b=\frac{V^2}{P} =\frac{11.7^2}{3.9} =35.1\Omega[/tex]

Now, because of the lightbulbs are connected in parallel the equivalent resistance is given by:

[tex]\frac{1}{R_L} =\frac{1}{R_b} +\frac{1}{R_b} =\frac{2}{R_b} \\\\ R_L= \frac{R_b}{2} =\frac{35.1}{2}=17.55\Omega[/tex]

Finally, now we have all the data, let's replace it into the internal resistance equation:

[tex]R_i_n_t=(\frac{12}{11.7} -1)17.55=0.45\Omega[/tex]

Answer:

Average induced emf will be equal to 0.00156 volt

Explanation:

We have given diameter of the wire d = 16.1 cm

So radius [tex]r=\frac{16.1}{2}=8.05cm=0.08m[/tex]

Cross sectional area of the wire [tex]A=\pi r^2=3.14\times 0.08^2=0.02m^2[/tex]

Magnetic field is changing from 0.53 T to 0.19 T

So change in magnetic field dB = 0.19-0.53 = -0.34 T

Time taken to change in magnetic field dt = 0.23 sec

Induced emf is given by [tex]e=\frac{-d\Phi }{dt}=-A\frac{dB}{dt}=-0.02\times \frac{0.34}{0.23}=0.00156volt[/tex]

So average induced emf will be equal to 0.00156 volt

Answer: 6.29*10^-12 N

Explanation:

given,

Potential difference of the electron, v = 1950 V

Magnetic field of the electron, B = 1.50 T

If the electron is accelerated through 19500 V from rest its Potential Energy has to be

converted to Kinetic Energy. This allows us solve for the velocity.

PE = Vq

PE = 1950 * 1.6*10^-19

PE = 3.12*10^-16 J

Also, PE = 1/2mv²

3.12*10^-16 = 1/2mv²

v = 2.62*10^7 m/s

to get F(max), we use,

F(max) = qvB

F(max) = 1.6*10^-19 * 2.62*10^7 * 1.5

F(max) = 6.29*10^-12 N

6.3 x 10⁻¹²N

As stated by Lorentz Force law, the magnitude of a magnetic force, F, can be expressed in terms of a fixed amount of charge, q, which is moving at a constant velocity, v, in a uniform magnetic field, B, as follows;

F =  qvB sin θ              ------------(i)

Where;

θ = angle between the velocity and the magnetic field vectors

When the electron passes through a potential difference, V, it is made to accelerate as it gains some potential energy ([tex]P_{E}[/tex]) which is then converted to kinetic energy ([tex]K_{E}[/tex]) as it moves. i.e

[tex]P_{E}[/tex] = [tex]K_{E}[/tex]                 ----------------(ii)

But;

[tex]P_{E}[/tex] = qV

And;

[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex] x m x v²

Therefore substitute these into equation (ii) as follows;

qV = [tex]\frac{1}{2}[/tex] x m x v²

Make v subject of the formula;

2qV = mv²

v² = [tex]\frac{2qV}{m}[/tex]

v = [tex]\sqrt{\frac{2qV}{m} }[/tex]      ---------------(iii)

From the question;

q = 1.6 x 10⁻¹⁹C       (charge on an electron)

V = 1.95 x 10³V

m = 9.1 x 10⁻³¹kg

Substitute these values into equation (iii) as follows;

v = [tex]\sqrt{\frac{2*1.6*10^{-19} * 1.95*10^{3}}{9.1*10^{-31}} }[/tex]

v =  [tex]\sqrt{\frac{6.24*10^{-16}}{9.1*10^{-31}} }[/tex]

v = 2.63 x 10⁷m/s

Now, from equation (i), the magnitude of the magnetic force will be maximum when the angle between the velocity and the magnetic field is 90°. i.e when θ = 90°

Substitute the values of θ, q v and B = 1.50T into equation (i) as follows;

F =  qvB sin θ

F = 1.6 x 10⁻¹⁹ x 2.63 x 10⁷ x 1.50 x sin 90°

F = 6.3 x 10⁻¹²N

Therefore the maximum magnitude of the magnetic force this particle can experience is 6.3 x 10⁻¹²N

Answer:

[tex]386.2^{\circ}F[/tex]

Explanation:

We are given that

[tex]P_1=200lbf/in^2[/tex]

[tex]P_2=60lbf/in^2[/tex]

[tex]v_1=200ft/s[/tex]

[tex]v_2=1700ft/s[/tex]

[tex]T_1=500^{\circ}F[/tex]

[tex]Q=0[/tex]

[tex]C_p=1BTU/lb^{\circ}F[/tex]

We have to find the exit temperature.

By steady energy flow equation

[tex]h_1+v^2_1+Q=h_2+v^2_2[/tex]

[tex]C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}[/tex]

[tex]1BTU/lb=25037ft^2/s^2[/tex]

Substitute the values

[tex]1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}[/tex]

[tex]500+1.598=T_2+115.4[/tex]

[tex]T_2=500+1.598-115.4[/tex]

[tex]T_2=386.2^{\circ}F[/tex]

We can use the First Law of Thermodynamics and the provided initial conditions from the problem statement to calculate the exit temperature of steam in a steady-flow system, such as a nozzle. The calculation requires knowledge of thermal dynamics principles and the use of steam tables.

This question is an application of the First Law of Thermodynamics (also known as the Law of Energy Conservation) and it pertains specifically to the flow of steam. In this case, we're working with a steady-flow system (a nozzle) where only the kinetic energy changes are important. There is no mention of the incoming steam doing work on the nozzle, nor of any heat is transferred to or from the nozzle, so, in this case, the work done (W) and heat transfer (Q) are both zero.

Furthermore, the exit pressure of the steam is given, as is the exit velocity; what we need to ascertain is the exit temperature, T2. To do this, we can use the enthalpies for steam at the given initial state, and then use the first law of thermodynamics to determine the exit enthalpy. The corresponding temperature can then be found in the steam tables associated with the exit enthalpy and pressure.

Therefore, the calculation requires an understanding of Thermal Dynamics principles and the use of Steam Tables.

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Answer: C. Chemical Weathering

Explanation:

Answer:

Chemical Weathering

Explanation:

Answer:

[tex]\frac {f_i}{\sqrt 2}[/tex]

Explanation:

Frequency is given by [tex]\frac {1}{2\pi}\sqrt{{\frac {k}{m}}[/tex]

Let the initial frequency be denoted by [tex]f_i[/tex]

[tex]f_i=\frac {1}{2\pi}\sqrt{{\frac {k}{M}}[/tex]

When M is 2M then

[tex]f_f=\frac {1}{2\pi}\sqrt{{\frac {k}{2M}}=\frac {f_i}{\sqrt 2}[/tex]

Therefore, the final frequency is

[tex]\frac {f_i}{\sqrt 2}[/tex]

Answer:

0.44 V

Explanation:

Parameters given:

Radius of loop, r = 10.7 cm = 0.107 m

Magnetic field strength, B = 0.803 T

Rate of shrinkage of Radius, dr/dt = 81 cm/s = 0.81m/s

EMF induced is given in terms of magnetic Flux, Φ, as:

EMF = dΦ/dt

Magnetic Flux, Φ, is given as:

Φ = B * A (where A is area of loop)

Therefore, EMF is:

EMF = d(B*A)/dt

The area of the loop is given as A = πr²

EMF = d(Bπr²) / dt = Bπ*d(r²)/dt

=> EMF = Bπ*2r*(dr/dt)

EMF = 0.803 * π * 2 * 0.107 * 0.81

EMF = 0.44 V

Answer:

120 m

Explanation:

Given:

wavelength 'λ' = 2.4m

pulse width 'τ'= 100T ('T' is the time of one oscillation)

The below inequality express the range of distances to an object that radar can detect

τc/2 < x < Tc/2 ---->eq(1)

Where, τc/2 is the shortest distance

First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'

f = c/λ (c= speed of light i.e 3 x [tex]10^{8}[/tex] m/s)

f= 3 x [tex]10^{8}[/tex] / 2.4

f=1.25 x  [tex]10^{8}[/tex] hz.

As, T= 1/f

time of one oscillation T= 1/1.25 x  [tex]10^{8}[/tex]

T= 8 x [tex]10^{-9}[/tex] s

It was given that pulse width 'τ'= 100T

τ= 100 x 8 x [tex]10^{-9}[/tex] => 800 x [tex]10^{-9}[/tex] s

From eq(1), we can conclude that the shortest distance to an object that this radar can detect:

[tex]x_{min}[/tex]= τc/2 =>  (800 x [tex]10^{-9}[/tex] x 3 x [tex]10^{8}[/tex])/2

[tex]x_{min}[/tex]=120m

Answer:

60833 loops

Explanation:

Given parameters:

Input voltage  = 120V

Output voltage  = 100kV

Primary loops  = 73 loops

Current = 14.1A

Unknown:

Number of loops in the secondary coil  = ?

Solution:

A step up transformer of this nature increases the output voltage. Here, the voltage is raised on the output side.

One thing about a transformer is that the output and input power are always the same.

To find the number of loops in the secondary coil;

             [tex]\frac{N_{p} }{N_{s} } = \frac{V_{p} }{V_{s} }[/tex]

Where N is the number of loops and V is the voltage

            s and p are the secondary and primary parts

Note: 100kV  = 100000V

 Now input the parameters;

          [tex]\frac{73}{N_{s} } = \frac{120}{100000}[/tex]

     120N[tex]_{s}[/tex]   = 73 x 100000

            N[tex]_{s}[/tex]  = 60833 loops

Answer:

Explanation:

Given that,

We have a projectile that is shot from the earth

Then, it's escape velocity is

Ve = √(2GM / R)

Ve² = 2GM / R

M is the mass of earth

R is the radius of Earth

Using conservation of energy

Ki + Ui = Kf + Uf

The final kinetic energy is zero since the body is finally at rest

Then,

Ki + Ui = Uf

Kinetic energy can be determine using

K = ½mv²

Potential energy can be determine using

U = -GmM / R

Where m is mass of the body

Then,

Ki + Ui = Uf

½mv² - GmM / R = -GmM / r

r is the maximum height reached

m cancel out

½ v² - GM / R = -GmM / r

A. We want to find the maximum height reached when initial speed (v) is equal to 0.785 escape velocity

So, v = 0.785•Ve

So,

½v² - GM / R = -GM / r

½(0.785•Ve)² - GM / R = -GM / r

0.308•Ve² - GM / R = -GM / r

0.308 × 2GM / R - GM / R = -GM / r

Divide through by GM

0.616 / R - 1 / R = -1 / r

-0.384 / R = -1 / r

Cross multiply

-0.384r = -R

r = -R / -0.384

r = 2.6 R

B. When it initial kinetic energy is is 0.785 of the kinetic energy required to escape Earth

Ki = 0.785 Ke

Ki = 0.785 ½mVe²

Ki = 0.785 × ½m× 2GM / R

Ki = 0.785•G•m•M / R

So,

Ki + Ui = Uf

0.785•G•m•M / R - GmM / R = -GmM / r

Let G•m•M cancels out

0.785 / R - 1 / R = -1 / r

-0.215 / R = -1 / r

Cross multiply.

-0.215r = -R

r = -R / -0.215

r = 4.65 R

Answer:

The value of path integral is 2.2 [tex]\mu T.m[/tex]

Explanation:

Given:

Current carrying by long wire [tex]I = 2.5[/tex] A

Area of rectangle [tex]A = 18.75[/tex] [tex]m^{2}[/tex]

Angle with surface normal [tex]\theta =[/tex] 45°

According to the ampere's circuital law,

     [tex]\int\limits {B} \, ds = \mu _{o} I_{net}[/tex]

Where [tex]\mu _{o} = 4\pi \times 10^{-7}[/tex] [tex]ds=[/tex] area element

Here, [tex]I _{net} = I \cos 45[/tex]

 [tex]I_{net} = 2.5 \times \frac{1}{\sqrt{2} }[/tex]

Put value of current in above equation,

  [tex]\int\limits {B} \, ds = 4\pi \times 10^{-7} \times 2.5 \times \frac{1}{\sqrt{2} }[/tex]

  [tex]\int\limits {B} \, ds = 22.2 \times 10^{-7}[/tex]

  [tex]\int\limits {B} \, ds = 2.2 \times 10^{-6}[/tex]

  [tex]\int\limits {B} \, ds = 2.2[/tex] [tex]\mu T.m[/tex]

Therefore, the value of path integral is 2.2 [tex]\mu T.m[/tex]

Answer:

B=1.89*10^{-4} T

Explanation:

First we have to calculate the electric force on the electron, and then we have to take into account that this force is equal to the force generated by the magnetic field.

The formula is

[tex]F=q_eE\\\\E=\frac{V}{d}[/tex]

q: charge of the electron = 1.6*10^{-19}C

V: potential

d: separation between plates

[tex]E=\frac{100V}{17*10^{-3}m}=5882.35N/C[/tex]

[tex]F=(1.6*10^{-19}C)(5882.35N/C)=9.4*10^{-16}N[/tex]

This force must equal the Lorentz's force

[tex]F_E=F_B\\\\F_B=qvB\\\\B=\frac{F_B}{q_ev}[/tex]

But before we have to calculate the speed of the electron by using  (me=9.1*10^{-31}kg)

[tex]E_e=\frac{1}{2}m_ev^2=q_eV\\\\v=\sqrt{\frac{2q_eV}{m_e}}=\sqrt{\frac{2(1.6*10^{-19}C)(2.8*10^3V)}{9.1*10^{-31}kg}}\\\\v=3.13*10^7\frac{m}{s}[/tex]

Hence, we have

[tex]B=\frac{9.4*10^{-16}N}{(1.6*10^{-19}C)(3.1*10^{7})\frac{m}{s}}=1.89*10^{-4}T[/tex]

hope this helps!!

Answer:

The magnetic field that is necessary is equal to 1.88x10⁻⁴ T

Explanation:

If the electron is accelerated:

[tex]e*V=\frac{1}{2} mv^{2} \\v=\sqrt{\frac{2eV}{m} }[/tex]

Where

e = 1.6x10⁻¹⁹C

V = 2.8 kV = 2800 V

m = 9.1x10⁻³¹kg

Replacing:

[tex]v=\sqrt{\frac{2*1.6x10^{-19}*2800 }{9.1x10^{-31} } } =3.13x10^{7} m/s[/tex]

When the electron is moving in straight line, the magnetic force is balanced with the electric force, thus:

V = E * d

Where V = 100 V

d = 17 mm = 0.017 m

E = V/d = 100/0.017 = 5882.35 N/C

The magnetic field that is necessary is equal to:

B = E/v = 5882.35/3.13x10⁷ = 1.88x10⁻⁴ T

Answer:

F=30N

a= 3m/s^2

m=?

F=ma

30=m(3)

30/3=m

m=10kg

The mass of the ball is 10kg

Final answer:

The mass of an object thrown with a force of 30 Newtons resulting in an acceleration of 3 m/s² is calculated using the formula m = F/a, resulting in a mass of 10 kg.

Explanation:

The question is regarding the calculation of the mass of an object when a known force is applied to it and an acceleration is observed as a result. According to Newton's Second Law of Motion, which is typically expressed as F = ma, where F is the force applied to an object in newtons (N), m is the mass of the object in kilograms (kg), and a is the acceleration of the object in meters per second squared (m/s²). To find the mass of the object, we rearrange this formula to get m = F / a. Given the applied force of 30 N and the acceleration of 3 m/s² due to that force, we can calculate the mass of the object.

Mass (m) = F / a = 30 N / 3 m/s² = 10 kg

Therefore, the mass of the object that was thrown with a force of 30 Newtons, resulting in an acceleration of 3 m/s², is 10 kg.

Answer:

(a) So range of frequency [tex]f > 43.27[/tex] Hz

(b) the reactance is 89.75 Ω

Explanation:

Given:

(a)

Capacitance of a capacitor [tex]C= 23 \times 10^{-6}[/tex] F

Reactance of capacitive circuit [tex]X_{C} =[/tex] 160 Ω

From the formula of reactance,

[tex]X_{C} = \frac{1}{\omega C}[/tex]

[tex]X_{C} = \frac{1}{2\pi fC}[/tex]

   [tex]f = \frac{1}{2\pi X_{C} C }[/tex]

   [tex]f = \frac{1}{6.28 \times 160 \times 23 \times 10^{-6} }[/tex]

   [tex]f = 43.27[/tex] Hz

So range of frequency [tex]f > 43.27[/tex] Hz

(b)

Capacitance [tex]C = 41 \times 10^{-6}[/tex] F

Frequency [tex]f = 43.27[/tex] Hz

From the formula of reactance,

[tex]X_{C} = \frac{1}{2\pi fC}[/tex]

[tex]X_{C} = \frac{1}{6.28 \times 43.27 \times 41 \times 10^{-6} }[/tex]

[tex]X_{C} =[/tex] 89.75 Ω

Therefore, the reactance is 89.75 Ω

Answer:

Explanation:

wavelength of sound = velocity / frequency

= 340 / 780

= .4359 m

Let the observer be at equal distance d from in phase source .

let it moves by distance x for destructive interference .

path difference from source

= d + x - (d - x )

=  2x

for destructive interference

path difference = wave length / 2

2x = .4359 / 2 m

x = .4359 / 4

= .108975 m

10.9 cm

so observer must move by distance 10.9 cm towards on of the centers.

Answer:

Explanation:

mass of rocket m = .5 kg

height of wall h = 40 m

initial horizontal velocity u = .5 m /s

horizontal acceleration = force / mass

= 20 / .5

a = 40 m /s²

Let rocket falls or covers 40 m vertically downwards in time t

h = 1/2 gt² , initial vertical velocity = 0

40 = 1/2 x 9.8 x t²

t = 2.8566 s

During this period it will cover horizontal distance with initial velocity of .5 m /s and acceleration a = 40m /s²

horizontal distance = ut + 1/2 at²

= .5 x 2.8566 + .5 x 40 x 2.8566²

= 1.4283 + 163.2033

= 164.63 m .

The distance from the base of the wall does the rocket land is 164.63 m .

A 500 g model rocket is resting horizontally at the top edge of a 40-m-high wall when it is accidentally bumped. The bump pushes it off the edge with a horizontal speed of 0.5 m/s

We know that

horizontal acceleration = force / mass

a = 20 / .5

a = 40 m /s²

Now

h = 1/2 gt² ,

Here initial vertical velocity = 0

So,

40 = 1/2 x 9.8 x t²

t = 2.8566 s

Now

horizontal distance = ut + 1/2 at²

= .5 x 2.8566 + .5 x 40 x 2.8566²

= 1.4283 + 163.2033

= 164.63 m .

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Complete Question:

An object oscillates back and forth on the end of a spring.

Which of the following statements are true at some time during the course of the motion? Check all that apply. Check all that apply. The object can have zero acceleration and, simultaneously, nonzero velocity. The object can have zero velocity and, simultaneously, nonzero acceleration. The object can have zero velocity and, simultaneously, zero acceleration. The object can have nonzero velocity and nonzero acceleration simultaneously.

Answer:

a) the object can have zero acceleration and, simultaneously, nonzero velocity

b) the object can have zero velocity and, simultaneously, nonzero acceleration

d) the object can have nonzero velocity and nonzero acceleration simultaneously

Explanation:

For an object oscillating back and forth on the end of a spring, when the object swings to the extremes, it momentarily stops and the velocity becomes zero. Rather than being zero, acceleration is maximum at these points because the net force at these extremes is maximum. This justifies option B

At the middle, which is the equilibrium position, the net force is zero because the object is motionless, and hence the acceleration is zero. At this point, the velocity is maximum and not zero. This justifies option A

When the object is swinging and it is neither at the middle nor at the extremes, both acceleration and velocity are not zero. This justifies option D

It is not possible for both the acceleration and the velocity of the swinging object to be simultaneously zero. Option C is wrong

An object can have zero acceleration and non-zero velocity or have non-zero velocity and acceleration simultaneously at some point during its motion. However, an object can also have zero velocity and zero acceleration at all times during its motion.

Both the statements "The object can have zero acceleration and, simultaneously, nonzero velocity" and "The object can have nonzero velocity and nonzero acceleration simultaneously" are true at some time during the course of motion.

For example, when an object is thrown upwards, at the highest point of its trajectory, its velocity becomes zero, but it still experiences the acceleration due to gravity, which is nonzero. This satisfies the first statement. Additionally, when an object moves in a circular path at a constant speed, it has a nonzero velocity and nonzero acceleration directed towards the center of the circle. This satisfies the second statement.

The statement "The object can have zero velocity and, simultaneously, zero acceleration." is true at all times during the course of motion. For instance, when an object is at rest, both its velocity and acceleration are zero.

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The charge on the capacitor after 2 s is 0.43 C.

Explanation:

The formula for finding the voltage while charging a capacitor is

[tex]V = V_{0}(1-e^{\frac{-t}{RC} })[/tex]

Here , V₀ is the initial potential before charging and t is the time at which we have to determine the voltage, R is the resistance and C is the capacitance for the given circuit.

The given problem have given us the values for V₀ = 50 V and maximum current I is given as 500 mA.

Then, resistance can be determined using Ohm's law: [tex]R = \frac{V}{I} =\frac{50}{500*10^{-3} } = 100 Ohm[/tex]

The capacitance is defined as the ratio of charge to the unit voltage.

[tex]C = \frac{Q}{V} = \frac{IT}{IR} = \frac{T}{R}[/tex]

Here T is the time constant which is given as 1 s and R is found to be 100 ohm, then capacitance will be [tex]\frac{1}{100} = 10 mF[/tex]

So, the values for parameters like V₀ = 50 V, R = 100 Ω, C = 10 mF and t = 2 s.

Then, [tex]V =50*(1-e^{\frac{-2}{100*10*10^{-3} } }) =50*(1-e^{-2}) = 50*(1-0.1353)[/tex]

V= 43 V.

Then, [tex]Q = CV = 10*10^{-3} * 43 V = 0.43 C[/tex]

Thus, the charge on the capacitor after 2 s is 0.43 C.

The unit of charge is Coulomb.

The charge on the capacitor after 2 second is 0.43 Coulomb.

To finding the voltage while charging a capacitor is given as,

      [tex]V=V_{0}(1-e^{-\frac{t}{RC} } )[/tex]

Here , V₀ is the initial potential, R is the resistance and C is the capacitance

It is given that V₀ = 50 V and maximum current I is given as 500 mA.

So,  [tex]R=\frac{V}{I}=\frac{50}{500*10^{-3} } =100ohm[/tex]

Time constant = RC = 1

 So,     [tex]C=\frac{1}{R} =\frac{1}{100}F[/tex]

Now, we have  V₀ = 50 V, R = 100 Ω, C = 1/100 F and t = 2 s.

 [tex]V=50(1-e^{-\frac{2}{100*\frac{1}{100} } } )=50(1-e^{-2} )=50(1-0.1353)=43V[/tex]

We know that,  [tex]Q=CV=\frac{1}{100}*43=0.43C[/tex]

Thus, the charge on the capacitor after 2 s is 0.43 C.

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Answer:

a) [tex]\Delta U_{g} = 12.945\,J[/tex], b) [tex]\Delta U_{k} = 12.945\,J[/tex], c) [tex]k = 2930.059\,\frac{N}{m}[/tex]

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

[tex]\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)[/tex]

[tex]\Delta U_{g} = 12.945\,J[/tex]

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

[tex]\Delta U_{k} = 12.945\,J[/tex]

c) The spring constant of the gun is:

[tex]\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}[/tex]

[tex]k = \frac{2\cdot \Delta U_{k}}{x^{2}}[/tex]

[tex]k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}[/tex]

[tex]k = 2930.059\,\frac{N}{m}[/tex]

Answer:

[tex]1.6026299569\times 10^{-11}\ m[/tex]

Explanation:

Grating constant

[tex]d=\dfrac{1}{6200}=0.000161\ cm=0.000161\times 10^{-2}\ m[/tex]

Number of slits

[tex]N=3.14\times 6200=19468[/tex]

Order

[tex]m=\dfrac{d}{\lambda}\\\Rightarrow m=\dfrac{0.000161\times 10^{-2}}{624\times 10^{-9}}\\\Rightarrow m\approx 2[/tex]

At m = 1

[tex]\Delta\lambda=\dfrac{\lambda}{mN}\\\Rightarrow \Delta\lambda=\dfrac{624\times 10^{-9}}{1\times 19468}\\\Rightarrow \Delta\lambda=3.2052599137\times 10^{-11}\ m[/tex]

At m = 2

[tex]\Delta\lambda=\dfrac{\lambda}{mN}\\\Rightarrow \Delta\lambda=\dfrac{624\times 10^{-9}}{2\times 19468}\\\Rightarrow \Delta\lambda=1.6026299569\times 10^{-11}\ m[/tex]

The wavelengths can be close by [tex]1.6026299569\times 10^{-11}\ m[/tex]

Answer:

It has no effect on the amplitude.

Explanation:

When the sandbag is dropped, then the cart is at its maximum speed. Dropping the sand bag does not affect the speed instantly, this is because the energy remains within the system after the bag as been dropped. The cart will always return to its equilibrium point with the same amount of kinetic energy, as a result the same maximum speed is maintained.

Answer:

It decreases the amplitude.

Explanation:

If it is located at the end, its kinetic energy will be equal to zero, so the system as a whole would only have potential energy, which is defined as:

Ep = (1/2) * k * A²

Where A is the amplitude of the oscillation.

According to the energy principle, said energy must remain conserved, therefore, the total energy is equal to:

Etot = (1/2) * k * A²

Since the system is in equilibrium, according to the expression of conservation of energy, the energy in the position located at the end would be equal to the energy in equilibrium. As the sandbag falls, its kinetic energy decreases, which is why the total energy of the system also decreases.

According to the total energy expression, it is directly proportional to the square of the oscillation amplitude. If the total energy is decreased, the amplitude of the oscillation would also decrease.

Answer:

a)0.48 m/s

b) 0.583 m/s

Explanation:

As the wagon rolls,

momentum'p'= m x v => 95.8 x 0.530 = 50.774 Kgm/s

(a)Rock is thrown forward,

momentum of rock = 0.325 x 15.1 =  4.9075 Kgm/s

Conservation of momentum says momentum of wagon is given by

50.774 - 4.9075 = 45.8665

Therefore, Speed of wagon = 45.8665 / (95.8-0.325) = 0.48 m/s

(b) Rock is thrown backward,

momentum of wagon = 50.774  + 4.9075 = 55.68  Kgm/s

Therefore, speed of wagon = 55.68 / (95.8-0.325) = 0.583 m/s

Final answer:

The speed of the wagon after the rock is thrown directly forward is 0.529 m/s, and after the rock is thrown directly backward is 1.63 m/s.

Explanation:

To find the speed of the wagon after the rock is thrown, we can use the principle of conservation of momentum. Since there is no external force acting on the system, the total momentum before and after the rock is thrown will be the same.

For part (a) when the rock is thrown directly forward, the momentum of the wagon and rider will remain the same, but the momentum of the rock will change to zero. So, the final momentum of the system will be the same as the initial momentum.

Using the equation: initial momentum = final momentum

(mass of wagon + rider + rock) x initial velocity of wagon = (mass of wagon + rider) x final velocity of wagon

(95.8 kg) x (0.530 m/s) = (95.8 kg + 0.325 kg) x vf

vf = 0.529 m/s

For part (b) when the rock is thrown directly backward, the momentum of the wagon and rider will change to zero, but the momentum of the rock will remain the same. So, the final momentum of the system will be the same as the initial momentum.

Using the equation: initial momentum = final momentum

(mass of wagon + rider + rock) x initial velocity of wagon = (mass of rock) x final velocity of rock

(95.8 kg) x (0.530 m/s) = (0.325 kg) x vf

vf = 1.63 m/s

Answer:force equals to rate of change of momentum

Explanation:

F=force

t=time

m=mass

v=final velocity

u=initial velocity

(mv-mu)/t=rate of change of momentum

Force=rate of change of momentum

F=(mv-mu)/t

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the netforce, and inversely proportional to the mass of the object.

Answer:

So energy is not conserved and inelastic shock

Explanation:

In the collision between a bullet and a ballistic pendulum, characterized in that the bullet is embedded in the block, if the kinetic energy is conserved the shock is elastic and if it is not inelastic.

Let's find the kinetic energy just before the crash

         K₀ = ½ m vₓ₀²

After the crash we can use the law of conservation of energy

Starting point. Right after the crash, before starting to climb

        Em₀ = K = ½ (m + M) v₂²

Final point. At the maximum height of the pendulum

       [tex]Em_{f}[/tex] = U = (m + M) g h

Where m is the mass of the bullet and M is the mass of the pendulum

        Em₀ = Em_{f}

        ½ (m + M) v₂² = (m + M) g h

        v₂ = √ 2g h

Now we can calculate the final kinetic energy

       K_{f} = ½ (m + M) v₂²²

       K_{f} = ½ (m + M) (2gh)

The relationship between these two kinetic energies is

      K₀ / K_{f} = ½ m vₓ₀² / (½ (m + M) 2 g h)

      K₀ / K_{f} = m / (m + M)  vₓ₀² / 2 g h

We can see that in this relationship the Ko> Kf

So energy is not conserved and inelastic shock

To determine if the collision between the ball and the pendulum is elastic or inelastic, we need to calculate the kinetic energy of the system before and after the collision.

An elastic collision is one that conserves kinetic energy, while an inelastic collision does not conserve kinetic energy. To determine if the collision between the ball and the pendulum is elastic or inelastic, we need to calculate the kinetic energy of the system before and after the collision.

In this case, we can calculate the kinetic energy of the system before the collision using the value of vxo and the kinetic energy just after the collision using the experimental value of h in Eq. 9.2.

If the kinetic energy of the system is the same before and after the collision, then it is an elastic collision. If the kinetic energy decreases after the collision, then it is an inelastic collision.

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Answer:

0.48 W

Explanation:

Given that:

the mass of the steel ball = 0.0367 kg

C = 0.270 N

g (acceleration due to gravity) = 9.8

Now;

At Terminal Velocity Weight is balance by drag force

mg =Cv

Making v the subject of the formula:we have:

[tex]v = \frac {mg}{C}[/tex]

[tex]v = \frac {0.0367*9.8}{0.270}[/tex]

v = 1.332 m/s

Thus, the Rate at which gravitational energy is converted to thermal energy once the ball reaches terminal velocity is:

P=Fv

P = mgv

[tex]P = (0.0367*9.8)*1.332[/tex]

P=0.48 W

Final answer:

To find the rate at which gravitational energy is converted to thermal energy at terminal velocity, equate the drag force to the gravitational force on the ball and solve for power as the product of force and velocity.

Explanation:

The question asks for the rate at which gravitational energy is converted to thermal energy once the ball reaches terminal velocity in a vat of corn syrup. To find this rate, we must recognize that at terminal velocity, all the gravitational force acting on the ball (minus any buoyant force, which is ignored here) is converted into thermal energy due to viscous drag. The viscous drag force [tex]F_{drag}[/tex] is given by [tex]F_{drag[/tex] = -Cv, where C is the drag coefficient, and v is the velocity of the ball. The gravitational force acting on the ball can be calculated using [tex]F_{gravity[/tex] = mg, where m is the mass of the ball, and g is the acceleration due to gravity. At terminal velocity, [tex]F_{drag[/tex] = [tex]F_{gravity[/tex], hence the power (rate of energy conversion) P can be found as P = Fv = mgv. Assuming standard gravity (9.81 m/s2), the mass of the ball (0.0367 kg), and solving for v using [tex]F_{drag[/tex], one can determine the rate of gravitational energy conversion to thermal energy.

Answer:

the coefficient of lift is 3.5561

the coefficient of drag is 2.3153

Explanation:

the solution is in the attached Word file