A battery charges a parallel-plate capacitor fully and then is removed. The plates are then slowly pulled apart. What happens to the potential difference between the plates as they are being separated?

Complete Question

A proton of mass m​p​​= 1.67×10​−27​​ kg and a charge of q​p​​= 1.60×10​−19​​ C is moving through vacuum at a constant velocity of 10,000 m/s directly to the east when it enters a region of uniform electric field that points to the south with a magnitude of E = 3.62e+3 N/C . The region of uniform electric field is 5 mm wide in the east-west direction. How far (in meters) will the proton have been deflected towards the south by the time it exits the region of uniform electric field. You may neglect the effects of friction and gravity, and assume that the electric field is zero outside the specified region. Answer is to be in units of meters

Answer:

    [tex]s = 0.039 \ m[/tex]

Explanation:

From the question we are told that

    The  mass of the proton is  [tex]m = 1.67 *10^{-27} \ g[/tex]

    The charge of on the proton is [tex]q = 1.60 *10^{-19} \ C[/tex]

      The speed of the proton is  [tex]v = 10000 \ m/s[/tex]

     The magnitude of the electric field is  [tex]E = 3.62*10^{3 } \ N/C[/tex]

       The width covered by the electric field    [tex]d = 5mm = 5 *10^{-3} \ m[/tex]      

Generally the acceleration of the proton due to the electric toward the south  (at the point where the force on the proton is equal to the electric force due to the electric field) is  mathematically represented as

       [tex]a = \frac{q* E}{m}[/tex]  

Substituting values

       [tex]a = \frac{1.60*10^{-19 } * 3.26 *10^{3}}{ 1.67*10^{-27}}[/tex]

      [tex]a = 3.12*10^{11} \ m/s^2[/tex]

Generally the time it will take the proton to cross the electric field is  mathematically represented as

      [tex]t = \frac{d}{v}[/tex]

Substituting values

      [tex]t = \frac{5 *10^{-3}}{10000}[/tex]

     [tex]t = 5 *10^{-7} \ s[/tex]

Generally the the distance covered by the proton toward the south is  

       [tex]s = ut + \frac{1}{2} * a*t^2[/tex]

   Here  u = 0  m/s  this  because before the proton entered the electric field region the it velocity towards the south is  zero

     So

       [tex]s = \frac{1}{2} * a*t^2[/tex]

Substituting values

      [tex]s = \frac{1}{2} * 3.12 *10^{11}*(5 *10^{-7})^2[/tex]

      [tex]s = 0.039 \ m[/tex]

Answer:

W = 14700 J

Explanation:

This is an exercise on Newton's second law.

To solve it we must fix a coordinate system, the most common is an axis parallel to the ramp and the other perpendicular axis, we write Newton's second law

Y Axis . Perpendicular to the ramp

       N - Wy = 0

X axis. Parallel to the ramp, we assume it is positive when the ramp is going up

        F - Wx = m a

in this case F = 1470 N and it is parallel to the plane.

Work is defined by

      W = F .d

boldface indicates vectors

      W = F d cos θ

     let's calculate

      W = 1470 10 cos 0

       W = 14700 J

Answer:

The number of bright spot is  m =4

Explanation:

From the question we are told that

    The number of lines is  [tex]s = 500 \ lines / mm = 500 \ lines / 10^{-3} m[/tex]

     The wavelength of the laser is  [tex]\lambda = 480 nm = 480 *10^{-9} \ m[/tex]

Now the the slit is mathematically evaluated as

        [tex]d = \frac{1}{s} = \frac{1}{500} * 10^{-3} \ m[/tex]

Generally the diffraction grating is mathematically represented as

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

Here m is the order of fringes (bright fringes) and at maximum m  [tex]\theta = 90^o[/tex]

    So

          [tex]\frac{1}{500} * sin (90) = m * (480 *10^{-3})[/tex]

=>        [tex]m = 4[/tex]

This  implies that the number of bright spot is  m =4

Answer:

B: I2>I1

Explanation:

See attached file

Answer:

An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about  2 years.

Explanation:

Given;

orbital period of 3 years, P = 3 years

To calculate the years of an orbital with a semi-major axis, we apply Kepler's third law.

Kepler's third law;

P² = a³

where;

P is the orbital period

a is the orbital semi-major axis

(3)² = a³

9 = a³

a = [tex]a = \sqrt[3]{9} \\\\a = 2.08 \ years[/tex]

Therefore, An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about  2 years.

Answer:

An object lying on the surface of the earth has energy like it maybe kinetic or potential energy

Explanation:

Hope it will help you :)

Answer:

density d = 1.59 g/cm^3

The density of the rock is 1.59 g/cm^3

Explanation:

The density of an object can be derived by measuring its mass and then measuring its volume by submerging it in a graduated cylinder.

Density = mass/volume of water displaced

d = m/v ........1

Given;

mass m = 344 g

Volume of water displaced v = 216 cm^3

from equation 1, we can calculate the value of the density;

Substituting the given values;

d = 344/216 g/cm^3

d = 1.592592592592 g/cm^3

d = 1.59 g/cm^3

The density of the rock is 1.59 g/cm^3

Answer:

The  tension on the clotheslines is  [tex]T = 8.83 \ N[/tex]

Explanation:

The  diagram illustrating this  question is  shown on the first uploaded image

From the question we are told that  

    The distance between the two poles is  [tex]d = 12 \ m[/tex]

     The mass tie to the middle of the clotheslines [tex]m = 1 \ kg[/tex]

     The length at which the clotheslines sags is  [tex]l = 4 \ m[/tex]

Generally the weight due to gravity at the middle of the  clotheslines is mathematically represented as

          [tex]W = mg[/tex]

let the angle which the tension on the  clotheslines makes with the horizontal be  [tex]\theta[/tex] which mathematically evaluated using the SOHCAHTOA as follows

        [tex]Tan \theta = \frac{ 4}{6}[/tex]

=>     [tex]\theta = tan^{-1}[\frac{4}{6} ][/tex]

=>     [tex]\theta = 33.70^o[/tex]

   So the vertical component of this  tension is  mathematically represented a  

      [tex]T_y = 2* Tsin \theta[/tex]

Now at equilibrium the  net horizontal force is  zero which implies that

          [tex]T_y - mg = 0[/tex]

=>       [tex]T sin \theta - mg = 0[/tex]

substituting values

          [tex]T = \frac{m*g}{sin (\theta )}[/tex]

substituting values

           [tex]T = \frac{1 *9.8}{2 * sin (33.70 )}[/tex]

           [tex]T = 8.83 \ N[/tex]

Answer:

B = 0.025T

Explanation:

In order to calculate the strength of the magnetic field at the center of the solenoid, you use the following formula:

[tex]B=\frac{\mu N i}{L}[/tex]         (1)

μ: magnetic permeability of vacuum = 4π*10^-7 T/A

N: turns of the solenoid = 500

i: current = 4.0A

L: length of the solenoid = 0.10m

You replace the values of the parameters in the equation (1):

[tex]B=\frac{(4\pi*10^{-7}T/A)(500)(4.0A)}{0.10m}=0.025T[/tex]

The strength of the magnetic field at the center of the solenoid = 0.025T

Answer:

Magnetic field strength at the center is 2.51x10^-2T

Explanation:

Pls see attached file for step by step calculation

Answer:

2005 J

Explanation:

500 cal × (4.18 J/cal) = 2090 J

ΔU = q + w

ΔU = 2090 J − 85 J

ΔU = 2005 J

Given that,

Output power = 87 W

Decibel reading = 120 dB

We need to calculate the intensity of sound

Using formula of intensity of sound

[tex]dB=10\log(\dfrac{I}{I_{0}})[/tex]

Put the value into the formula

[tex]120=10\log(\dfrac{I}{1\times10^{-12}})[/tex]

[tex]12=log(\dfrac{I}{1\times10^{-12}})[/tex]

[tex]10^{12}=\dfrac{I}{1\times10^{-12}}[/tex]

[tex]I=10^{12}\times1\times10^{-12}[/tex]

[tex]I=1\ W/m^2[/tex]

We need to calculate the distance

Using relation of power out[ut and intensity

[tex]I=\dfrac{P}{4\pi r^2}[/tex]

Put the value into the formula

[tex]1=\dfrac{87}{4\pi r^2}[/tex]

[tex]r=\sqrt{\dfrac{87}{4\pi}}[/tex]

[tex]r=2.63\ m[/tex]

Hence, The distance is 2.63 m

Complete Question

 Now if you look at the equation for acceleration  given in the question i.e

        [tex]a = a_1 + F * m[/tex]

We see that evaluating it in terms of dimension it is  incorrect  instead the equation should be  

       [tex]a = a_1 + \frac{F}{m}[/tex]

So in the solution below we will be making use of [tex]a = a_1 + \frac{F}{m}[/tex]

Answer:

The values of a is [tex]a = 4.714 \ m/s^2[/tex]

Explanation:

From the question we are told that

    The  expression for the acceleration is  [tex]a = a_1 + \frac{F}{m}[/tex]

     The value of  [tex]a_ 1 = 3.0 \ m/s^2[/tex]

     The values of  [tex]F = 12.0 \ kg \cdot m/s^2[/tex]  

      The values of m is  [tex]m = 7.0 \ kg[/tex]

substituting values

        [tex]a = 3 + \frac{12}{7}[/tex]

       [tex]a = 4.714 \ m/s^2[/tex]

Answer:

a. When the total displacement is -(A + B)

b. A + B = 1 m or -(A + B) = -11 m

c. 0 m

Explanation:

a. Under what circumstances can you end up back at your starting point?

If we have the displacement A and displacement B. The total displacement is A + B. We would end up at the starting point if we take a displacement -(A + B) from point B

b. What is the magnitude of the largest displacement you can end up from the starting point?

The maximum displacement we can obtain is when A and B are in the same direction. So A + B = 5 m + 6 m = 11 m or -A - B = -(A + B) = -11 m.

c. When A and B are perpendicular, what is the component of B in the direction of A?

Since A is perpendicular to B, the angle between A and B is 90°

So the component of B in A,s direction is Bcos90° = B × 0 = 0 m

Answer:

The two methods will yield different results as one is subject to experimental errors that us the Archimedes method of measurement, the the density measurement method will be more accurate

Explanation:

This is because the density method using the calculated volume will huve room for less errors that's occur in practical method i.e Archimedes method due to human error

Answer:

a) 1.94 m/s

b) 120 m

Explanation:

Convert km/h to m/s:

6 km/h = 1.67 m/s

a) The final speed is found with Pythagorean theorem:

v = √((1.67 m/s)² + (1 m/s)²)

v = 1.94 m/s

b) The time it takes the boat to cross the river is:

t = (200 m) / (1.67 m/s)

t = 120 s

The displacement in the direction of the current is:

x = (1 m/s) (120 s)

x = 120 m

q = 40 C

e = 1.6×10^-19 C

n = ?

n = q/e

n = 40/1.6×10^-19 C

= 2.6×10^20

Answer:

(a) L =  128.75 m

(b) L =  182.56 m

(c) L =  114.28 m

(d) Mass of Gold = 7.68 kg = 7680 gram

(e) Cost of Gold Wire = $ 307040

Explanation:

The resistance of the wire is given as:

R = ρL/A

where,

R = Resistance

ρ = resistivity

L = Length

A = cross-sectional area

(a)

For Gold Wire:

ρ = 2.44 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.44 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.44 x 10⁻⁸ Ω.m)

L =  128.75 m

(b)

For Copper Wire:

ρ = 1.72 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(1.72 x 10⁻⁸ Ω.m)

L =  182.56 m

(c)

For Aluminum Wire:

ρ = 2.75 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.75 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.75 x 10⁻⁸ Ω.m)

L =  114.28 m

(d)

Density = Mass/Volume

Mass = (Density)(Volume)

Volume of Gold = AL = (3.14 x 10⁻⁶ m²)(128.75 m) = 4.04 x 10⁻⁴ m³

Therefore,

Mass of Gold = (1.9 x 10⁴ kg/m³)(4.04 x 10⁻⁴ m³)

Mass of Gold = 7.68 kg = 7680 gram

(e)

Cost of Gold Wire = (Unit Price of Gold)(Mass of Gold)

Cost of Gold Wire = ($ 40/gram)(7680 grams)

Cost of Gold Wire = $ 307040

(a) L is = 128.75 m

(b) L is = 182.56 m

(c) L is = 114.28 m

(d) Mass of Gold is = 7.68 kg = 7680 gram

(e) Cost of Gold Wire is = $307040

When The resistance of the wire is given as:

R is = ρL/A

Now, where

R is = Resistance

ρ is = resistivity

L is = Length

A is = cross-sectional area

(a) For Gold Wire is:

ρ is = 2.44 x 10⁻⁸ Ω.m

A is = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R is = 1 Ω

Thus,

1 Ω is = (2.44 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L is = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.44 x 10⁻⁸ Ω.m)

L is =  128.75 m

(b) For Copper Wire is:

ρ is = 1.72 x 10⁻⁸ Ω.m

Then A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R is = 1 Ω

Thus,

After that 1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

Now, L = (1 Ω)(3.14 x 10⁻⁶ m²)/(1.72 x 10⁻⁸ Ω.m)

Therefore, L =  182.56 m

(c) For Aluminum Wire is:

ρ is = 2.75 x 10⁻⁸ Ω.m

A is = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R is = 1 Ω

Thus,

Then 1 Ω = (2.75 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

After that L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.75 x 10⁻⁸ Ω.m)

L =  114.28 m

(d) Density is = Mass/Volume

Mass is = (Density)(Volume)

Then Volume of Gold = AL = (3.14 x 10⁻⁶ m²)(128.75 m) = 4.04 x 10⁻⁴ m³

Therefore,

Now, Mass of Gold = (1.9 x 10⁴ kg/m³)(4.04 x 10⁻⁴ m³)

Then Mass of Gold = 7.68 kg = 7680 gram

(e) The Cost of Gold Wire is = (Unit Price of Gold)(Mass of Gold)

Than Cost of Gold Wire = ($ 40/gram)(7680 grams)

Therefore, The Cost of Gold Wire is = $ 307040

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Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with  due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

[tex]K_{1} = K_{2} + W_{f}[/tex]

Where:

[tex]K_{1}[/tex], [tex]K_{2}[/tex] are the initial and final translational kinetic energies of the tobbogan, measured in joules.

[tex]W_{f}[/tex] - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

[tex]f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})[/tex]

Where:

[tex]f[/tex] - Friction force, measured in newtons.

[tex]\Delta s[/tex] - Distance travelled by the toboggan in the rough region, measured in meters.

[tex]m[/tex] - Mass of the toboggan, measured in kilograms.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

[tex]f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}[/tex]

If [tex]m = 375\,kg[/tex], [tex]v_{1} = 4.50\,\frac{m}{s}[/tex], [tex]v_{2} = 1.20\,\frac{m}{s}[/tex] and [tex]\Delta s = 5.40 \,m[/tex], then:

[tex]f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}[/tex]

[tex]f = 653.125\,N[/tex]

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

[tex]\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%[/tex]

[tex]\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%[/tex]

[tex]\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%[/tex]

[tex]\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%[/tex]

[tex]\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%[/tex]

If [tex]v_{1} = 4.50\,\frac{m}{s}[/tex] and [tex]v_{2} = 1.20\,\frac{m}{s}[/tex], then:

[tex]\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%[/tex]

[tex]\%K_{loss} = 92.889\,\%[/tex]

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

[tex]\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%[/tex]

[tex]\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%[/tex]

If [tex]v_{1} = 4.50\,\frac{m}{s}[/tex] and [tex]v_{2} = 1.20\,\frac{m}{s}[/tex], then:

[tex]\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%[/tex]

[tex]\%v_{loss} = 73.333\,\%[/tex]

The speed of the toboggan is reduced in 73.333 %.

The average frictional force exerted on the toboggan by the rough surface is 661.5 N.

The percentage of the toboggan kinetic energy reduction is 7.11%.

The percentage of the toboggan speed reduction is 26.67%.

The given parameters;

The normal force on the toboggan is calculated as follows;

Fₙ = mg

Fₙ = 375 x 9.8 = 3675 N

The acceleration of the toboggan is calculated as follows;

[tex]v^2 = u^2 + 2as\\\\2as = v^2 - u^2\\\\a = \frac{v^2 - u^2 }{2s} \\\\a = \frac{(1.2)^2 - (4.5)^2 }{2(5.4)}\\\\a = -1.74 \ m/s^2[/tex]

The coefficient of friction is calculated as follows;

[tex]\mu_k = \frac{a}{g} \\\\\mu_k = \frac{1.74}{9.8} \\\\\mu_k = 0.18[/tex]

The average frictional force exerted on the toboggan by the rough surface;

[tex]F_k = \mu_k F_n\\\\F_k = 0.18 \times 3675\\\\F_k = 661.5 \ N[/tex]

The initial kinetic energy of the toboggan is calculated as follows;

[tex]K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2} \times 375\times 4.5^2\\\\K.E_i = 3,796.88 \ J[/tex]

The final kinetic energy of the toboggan is calculated as follows;

[tex]K.E_f = \frac{1}{2} mv^2\\\\K.E_f = \frac{1}{2} \times 375\times 1.2^2\\\\K.E_f = 270 \ J[/tex]

The percentage of the toboggan kinetic energy reduction is calculated as follows;

[tex]\frac{K.E_f}{K.E_i} \times 100\% = \frac{270}{3796.88} \times 100\% = 7.11 \%[/tex]

The percentage of the toboggan speed reduction is calculated as follows;

[tex]\frac{1.2}{4.5} \times 100\% = 26.67 \%[/tex]

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Answer:

v = 16 m/s

Explanation:

It is given that,

Acceleration of a particle along x -axis is [tex]2.5\ m/s^2[/tex]

At t = 0s, its velocity is 6 m/s

We need to find the velocity at t = 4 s

It means that the initial velocity of the particle is 6 m/s

Let v is the velocity at t = 4 s

So,

v = u + at

[tex]v=6+2.5\times 4\\\\v=16\ m/s[/tex]

So, the velocity at t = 4 s is 16 m/s.

Answer:

v = 16 m/s

Explanation:

It is given that,

Acceleration of a particle along x -axis is  

At t = 0s, its velocity is 6 m/s

We need to find the velocity at t = 4 s

It means that the initial velocity of the particle is 6 m/s

Let v is the velocity at t = 4 s

So,

v = u + at

So, the velocity at t = 4 s is 16 m/s.

Answer:

Length of a tube = 1.574 m

Explanation:

Given:

Fundamental frequency (f1) = 108 Hz

First overtone (f2) = 216 Hz

Speed of sound (v) = 340 m/s

Find:

Length of a tube

Computation:

We know that,

f = v / λ

f = nv / 2L  [n = number 1,2,3]

So,

f1 = 1(340) / 2L

f1 = 170 / L

L = 170 / 108 = 1.574 m

f2 = 2(340) / 2L

L = 340 / 216

L = 1.574 m

Answer: Amplitude

Explanation: The intensity or loudness of a sound depends upon the extent to which the sounding body vibrates (i.e the amplitude of vibration).

Loudness is measured in units called decibels and amplitude in metre

Answer:

[tex]V(30g)=1.4\frac{cm}{s} \\V(13g)=5.9\frac{cm}{s}[/tex]

Explanation:

So my calculations may be off (the final step is plugging in a bunch of things and getting a value, which creates room for error [and they honestly seem too small]) but I'm confident that the process is correct. I'll upload my work shortly but here is the method:

And with that, you've got it! This whole process is kinda long and involved so I would try practicing it a lot before any tests/quizzes so it doesn't eat up your time.

Edit: In my work I made u to be the initial velocities and v to be the final velocities because it was easier to keep track of

Answer:

The final velocity of the 30 g object is 16.8  cm/s

The final velocity of the 13 g object is 21.3 cm/s

Explanation:

Let's study the elastic collision with conservation of linear momentum, assigning object 1 to the 30 g object, and object 2 to the 13 gr object:

[tex]p_{1\,i}+p_{2\,i}=p_{1\.f}+p_{2\,f}\\(30)\,(19.5)+(13)\,(15) = (30)\,v_{1\,f} +(13)\,v_{2\,f}\\780 = (30)\,v_{1\,f} +(13)\,v_{2\,f}[/tex]

so we can write one of the unknowns in terms of the other one:

[tex]v_{1\,f}=(780-13\,v_{2\,f})/30[/tex]

Now we analyze the equation for conservation of kinetic energy that verifies in elastic collisions:

[tex]\frac{30}{2} (19.5)^2+\frac{13}{2} \,(15)^2=\frac{30}{2} (v_{1\,f})^2+\frac{13}{2} \,(v_{2\,f})^2\\7166.25=15\, (v_{1\,f})^2+6.5\,(v_{2\,f})^2[/tex]

now we can write this quadratic equation replacing [tex]v_{1\,f}[/tex] with its expression in terms of [tex]v_{2\,f}[/tex] and solve it (with the help of a graphing calculator is simpler by looking for the roots).

We get two answers for  [tex]v_{2\,f}[/tex] : one 15 cm/s, and the other one 21.28 cm/s.

We select the 21.28 cm/s answer since otherwise, the situation is the same as the initial one at which the second object was moving at 15 cm/s.

This velocity can be rounded to one decimal to:  21.3 cm/s

Given the value 21,28 for   [tex]v_{2\,f}[/tex] , then:

[tex]v_{1\,f}=(780-13\,v_{2\,f})/30\\v_{1\,f}=(780-13\,(21.28))/30=16.78 \,\,cm/s[/tex]

which can be rounded to 16.8 cm/s

Answer:

The average current will be "17.69 nA".

Explanation:

The given values are:

Charge,

q = 9.2 pC

Time,

t = 0.52ms

The equivalent circuit of the cell surface is provided by:

⇒  [tex]i_{avg}=\frac{charge}{t}[/tex]

Or,

⇒  [tex]i_{avg}=\frac{q}{t}[/tex]

On substituting the given values, we get

⇒         [tex]=\frac{9.2\times 10^{-12}}{0.52\times 10^{-3}}[/tex]

⇒         [tex]=17.69^{-9}[/tex]

⇒         [tex]=17.69 \ nA[/tex]

Answer:

The horizontal component is  [tex]v_h = 1.7096 \ m/s[/tex]

Explanation:

A diagram illustrating the projection is  shown on the first uploaded image  (from  IB Maths Resources from British international school Phuket )

From the question we are told that  

    The initial velocity is  [tex]v_o = 7.6 \ m/s[/tex]

      The angle of projection is  [tex]\theta = 1.27 \ rad = 72.77^o[/tex]

The  horizontal component of this  projectile  velocity is  mathematically represented as

          [tex]v_h = v_o * cos (\theta )[/tex]

substituting values  

         [tex]v_h = 7.6 * cos (72.77 )[/tex]

        [tex]v_h = 1.7096 \ m/s[/tex]

Complete Question

A very large sheet of a conductor carries a uniform charge density of [tex]4.00\ pC/mm^2[/tex]  on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?

Answer:

The electric field is  [tex]E = 4.5198 *10^{5} \ N/C[/tex]

Explanation:

From the question we are told that

    The charge density is  [tex]\sigma = 4.00pC /mm^2 = 4.00 * 10^{-12 } * 10^{6} = 4.00 *10^{-6}C/m[/tex]

    The position outside the surface is  [tex]a = 3.00 \ mm = 0.003 \ m[/tex]

Generally the electric field is mathematically represented as

          [tex]E = \frac{\sigma}{\epsilon _o }[/tex]

Where  [tex]\epsilon_o[/tex] is  the permitivity of free space with values  [tex]\epsilon _o = 8.85 *10^{-12} F/m[/tex]

substituting values  

           [tex]E = \frac{4.0*10^{-6}}{8.85 *10^{-12} }[/tex]

           [tex]E = 4.5198 *10^{5} \ N/C[/tex]

Explanation:

u = 2.5 m/s

v = 0

t = 3sec

s = ?

s = (u+v)/t

s = (0+2.5)/3

s = 2.5/3 = 0.83 m

Answer:

2.20°

Explanation:

For the central bright spot, we will use the constructive pattern for a double slit interference,

[tex]m\times w = d \times Sin\beta[/tex]

where w indicates the wavelength

and [tex]\beta[/tex] indicates the angle between the bright spot and center line.

now we will use the given values,

1 × 500 × 10^-9 = 1.3 × 10^-5 × Sin [tex]\beta[/tex]

Solving for [tex]\beta[/tex],

[tex]\beta[/tex] = 2.204° ~ 2.20°

Therefore the correct answer is 2.20°

Answer:

From the image, the force as shown in option A will exert the biggest torque on the cylinder about its central axes.

Explanation:

The image is shown below.

Torque is the product of a force about the center of rotation of a body, and the radius through which the force acts. For a given case such as this, in which the cylinders are identical, and the forces are of equal magnitude, the torque at the maximum radius away from the center will exert the maximum torque. Also, the direction of the force also matters. To generate the maximum torque, the force must be directed tangentially away from the circle formed by the radius through which the force acts away from the center. Option A satisfies both condition and hence will exert the most torque on the cylinder.