Three resistors are connected in series across a battery. The value of each resistance and its maximum power rating are as follows: 6.7Ω and 15.9 W, 30.4Ω and 9.12 W, and 16.3Ω and 12.3 W. (a) What is the greatest voltage that the battery can have without one of the resistors burning up? (b) How much power does the battery deliver to the circuit in (a)?

Answer:

a) vertical position at 2s: 50.68 m

b) horizontal position at 2s: 13.05 m

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the rock has two components: x-component and y-component. Being the equations to find the position as follows:

x-component:

[tex]x=V_{o}cos\theta t[/tex]   (1)

Where:

[tex]V_{o}=7.2 m/s[/tex] is the rock's initial speed

[tex]\theta=25\°[/tex] is the angle at which the rock was thrown

[tex]t=2 s[/tex] is the time

y-component:

[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex]   (2)

Where:

[tex]y_{o}=25 m[/tex]  is the initial height of the rock

[tex]y[/tex]  is the height of the rock at 2 s

[tex]g=9.8m/s^{2}[/tex]  is the acceleration due gravity

Knowing this, let's begin with the answers:

a) Vertical position at 2 s:

[tex]y=25 m+(7.2 m/s)sin(25\°) (2 s)-\frac{(9.8m/s^{2})(2)^{2}}{2}[/tex]   (3)

[tex]y=25 m+6.085 m-19. 6 m[/tex]   (4)

[tex]y=50.68 m[/tex]   (5)  This is the vertical position at 2 s

b) Horizontal position at 2 s:

[tex]x=(7.2 m/s) cos(25\°) (2 s)[/tex]    (5)

[tex]x=13.05 m[/tex]   (6)  This is the horizontal position at 2 s

Explanation:

It is given that,

Position of the man, h = 25 m

Initial velocity of the rock, u = 7.2 m/s

The rock is thrown upward at an angle of 25 degrees.

(a) Let y is the vertical position of the rock at t = 2 seconds. Using the equation of kinematics to find y as :

[tex]y=-\dfrac{1}{2}gt^2+u\ sin\theta\ t+h[/tex]

[tex]y=-\dfrac{1}{2}\times 9.8(2)^2+7.2\ sin(25)\times 2+25[/tex]

y = 11.48 meters

(b) Let x is the horizontal position of the rock at that time. It can be calculated as :

[tex]x=u\ cos\theta \times t[/tex]

[tex]x=7.2\ cos(25)\times 2[/tex]

x = 13.05 meters

Hence, this is the required solution.

Answer:

the height above the ground where  the board makes contact with the wall equals 5.297 meters above ground.

Explanation:

The wall and ladder are shown in the attached figure

In the figure we can see that

[tex]cos(28^{o})=\frac{H}{L}\\\\cos(28^{o})=\frac{H}{6}\\\\\therefore H=6.0\times cos(28^{o})==5.297meters[/tex]

Answer:

a)[tex] C_p=35.42\ \rm J/mol.K[/tex]

b)[tex] C_v=27.1\ \rm J/mol.K[/tex]

Explanation:

Given:

Let R be the gas constant which has value [tex]R=8.31432\times10^3\ \rm N\ m\ kmol^{-1}K^{-1}[/tex]

Work done in the process

[tex]W=PdV\\W=0.947\times(104-56.7)\times10^{-3}\ \rm L\ atm\\W=44.8\times10^{-3}\times101.33\ \rm J\\W=4.43\ \rm J[/tex]

Now Using First Law of thermodynamics

[tex]Q=W+\Delta U\\19.3=4.43+\Delta U\\\Delta U=14.77\ \rm J[/tex]

Let[tex]C_v[/tex] be the molar specific heat of the gas at constant volume given by

[tex]\Delta U=nC_v\Delta T\\14.77=\dfrac{C_v}{R}{PdV}\\14.77=\dfrac{C_v}{R}(0.987\times10^5(104-56.7)\times10^-6})\\C_v=3.26R\\C_v=27.1\ \rm J/mol.K[/tex]

Also we know that

Let[tex]C_p[/tex] be the molar specific heat of the gas at constant pressure given by

[tex]C_p-C_v=R\\C_p=R+C_v\\C_p=4.26R\\C_p=35.42\ \rm J/mol.K[/tex]

Answer:

The mass of the water on earth is [tex]5.4537\times10^{20}\ kg[/tex]

Explanation:

Given that,

Average depth h= 0.95 mi

[tex]h=0.95\times1.609[/tex]

[tex]h =1.528\ km[/tex]

[tex]h=1.528\times10^{3}\ m[/tex]

Radius of earth [tex]r= 6.37\times10^{6}\ m[/tex]

Density = 1000 kg/m³

We need to calculate the area of surface

Using formula of area

[tex]A =4\pi r^2[/tex]

Put the value into the formula

[tex]A=4\pi\times(6.37\times10^{6})^2[/tex]

[tex]A=5.099\times10^{14}\ m^2[/tex]

We need to calculate the volume of earth

[tex]V = Area\times height[/tex]

Put the value into the formula

[tex]V=5.099\times10^{14}\times1.528\times10^{3}[/tex]

[tex]V=7.791\times10^{17}\ m^3[/tex]

Now, 70 % volume of the total volume

[tex]V= 7.791\times10^{17}\times\dfrac{70}{100}[/tex]

[tex]V=5.4537\times10^{17}\ m^3[/tex]

We need to calculate the mass of the water on earth

Using formula of density

[tex]\rho = \dfrac{m}{V}[/tex]

[tex]m = \rho\times V[/tex]

Put the value into the formula

[tex]m=1000\times5.4537\times10^{17}[/tex]

[tex]m =5.4537\times10^{20}\ kg[/tex]

Hence, The mass of the water on earth is [tex]5.4537\times10^{20}\ kg[/tex]

Answer:

(A) - 0.273 kg m /s

(B) - 0.546 kg m /s

Explanation:

mass of paintball, m = 0.0032 kg

initial velocity, u = 85.3 m/s

(A) Momentum is defined as the product of mass of body and the velocity of body.

initial momentum, pi = mass x initial velocity

pi = 0.0032 x 85.3 = 0.273 kg m /s

Finally the ball stops, so final velocity, v = 0 m/s

final momentum, pf = mass x final velocity = 0.0032 x 0 = 0 m/s

change in momentum = pf - pi = 0 - 0.273 = - 0.273 kg m /s

(B) initial velocity, u = + 85.3 m/s

final velocity, v = - 85.3 m/s

initial momentum, pi = mass x initial velocity

pi = 0.0032 x 85.3 = 0.273 kg m /s

final momentum, pf = mass x final velocity = - 0.0032 x 85.3 = - 0.273 kgm/s

change in momentum = pf - pi = - 0.273 - 0.273 = - 0.546 kg m /s

(C) According to the Newton's second law, the rate of change of momentum is directly proportional to the force exerted.

As the change in momentum in case (B) is more than the change in momentum in case (A), so the force exerted on the skin is more in case (B).

Answer:

Explanation:

The force [tex]\vec{F}[/tex] on a charge q made by an electric field [tex]\vec{E}[/tex] its

[tex]\vec{F} = q \vec{E}[/tex]

The electric charge of the electron its

[tex]q \ = \ - \ 1.602 \ 10 ^{-19} \ C[/tex].

Taking the unit vector [tex]\hat{i}[/tex] pointing towards the east, the electric field will be:

[tex]\vec{E}= 3400 \ \frac{N}{C} \ \hat{i}[/tex].

So, the force will be:

[tex]\vec{F} =  \ - \ 1.602 \ 10 ^{-19} \ C \ * \ 3400 \ \frac{N}{C} \ \hat{i} [/tex]

[tex]\vec{F} =  \ - \ 5446.8 \ 10 ^{-19} \ N \ \hat{i} [/tex]

[tex]\vec{F} =  \ - \ 5.4468 \ 10 ^{-16} \ N \ \hat{i} [/tex]

[tex]\vec{F} =  \ - \ 5.4468 \ 10 ^{-16} \ N \ \hat{i} [/tex]

[tex]\vec{F} =  \ - \ 5.4468 \ 10 ^{-16} \ N \ \hat{i} [/tex]

So, the force its [tex] \ 5.4468 \ 10 ^{-16} [/tex] N to the west.

Answer:

(c) The forces have the same magnitude.

Explanation:

The electric Force between the two charges are proportional to the product of both charges:

[tex]F_{electric}=k*q_{1}q_{2}/r^{2}[/tex]

Both charges feels the same force but with opposite direction.

(c) The forces have the same magnitude.

In this exercise we have to have knowledge about the horizontal launch, so we have to use the known formulas to find that:

a) [tex]h*(1 - 1/2 g * h/v_0^2)[/tex]

b)[tex]h = v_0^2/ g[/tex]

c)[tex]h = v_0^2/ g[/tex]

So we have to remember some famous equations like the position and velocity of an object moving in a constant, like this:

[tex]y = y_0 + v_0*t + 1/2 * a * t^2\\v = v_0 + a * t[/tex]

where:

a) When the balls collide, h1 = h2. Then,

[tex]h_1 = h_2\\v_0 * t - 1/2 g * t^2 = h - 1/2 * g * t^2\\v_0 * t = h\\t = h / v_0[/tex]

Replacing in the equation of the height of the first ball:

[tex]h_1 = v_0 * h/v_0 - 1/2g * h^2/v_0^2\\h_1 = h - 1/2 g * h^2/ v_0^2\\h_1 = h*(1 - 1/2 g * h/v_0^2)[/tex]

b)  that the balls collide at t = h/v0. Then:

[tex]h/ v_0 = -v_0/-g\\h = v_0^2/ g[/tex]

c) The value of h for which the collision occurs at the instant when the first ball is at its highest point is the maximum value,Then:

[tex]h = v_0^2/ g[/tex]

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The pitch of the helical path followed by an electron in a magnetic field is given by the formula p = v * cos(theta) * T, where v is the velocity of the electron, theta is the angle between the direction of the magnetic field and the electron's velocity, and T is the period of the electron's motion in the magnetic field. By substituting the given values into this formula, we can calculate the pitch.

The pitch of a helical path can be determined by considering how the charged particle reacts in a magnetic field. The helical path of an electron in a magnetic field is caused by the perpendicular and parallel components of the electron's velocity. In particular, the parallel component of the velocity, v(cos(theta)), is responsible for the linear movement of the electron along the field lines. This leads to the corkscrew-like helical path, and the pitch of this helix is equivalent to the linear distance moved by the electron in one revolution.

The pitch (p) can be calculated by the formula p = v * cos(theta) * T, where v is the velocity of the electron, theta is the angle between the direction of the magnetic field and the velocity of the electron, and T is the period of circular motion of the electron in the magnetic field. From the provided data, we can first calculate the velocity of the electron using its kinetic energy, 22.5 eV: v = sqrt((2 * KE) / m), where KE is the kinetic energy and m is the mass of the electron (9.11 * 10^-31 kg). Next, we calculate the period using the formula T = 2 * pi * m / (q * B), where q is the charge of the electron (-1.6 * 10^-19 C) and B is the magnetic field intensity (4.55 * 10^4 T).

Finally, we substitute these values into the pitch formula and calculate the pitch of the helical path taken by the electron in the magnetic field.

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The pitch of the helical path taken by the electron is approximately [tex]\( 9.20 \times 10^{-11} \)[/tex] meters.

To find the pitch of the helical path taken by the electron, we need to follow these steps:

1. Calculate the speed of the electron:

  The kinetic energy (K.E.) of the electron is given by:

  [tex]\[ \text{K.E.} = \frac{1}{2}mv^2 \][/tex]

  where [tex]\( m \)[/tex] is the mass of the electron [tex](\( 9.11 \times 10^{-31} \) kg)[/tex] and [tex]\( v \)[/tex] is the speed of the electron.

First, we need to convert the kinetic energy from electron volts (eV) to joules (J):

  [tex]\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \][/tex]

  [tex]\[ 22.5 \, \text{eV} = 22.5 \times 1.602 \times 10^{-19} \, \text{J} = 3.6045 \times 10^{-18} \, \text{J} \][/tex]

  Now, using the kinetic energy formula:

 [tex]\[ \text{K.E.} = \frac{1}{2}mv^[/tex]

To find the pitch of the helical path taken by the electron, we need to determine the components of the electron's velocity and the motion along the magnetic field.

1. Calculate the speed of the electron:

The kinetic energy (KE) of the electron is given by:

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

  where m is the mass of the electron [tex](\( 9.11 \times 10^{-31} \) kg)[/tex] and v is the speed of the electron.

  First, convert the kinetic energy from electron volts (eV) to joules (J):

 [tex]\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \][/tex]

 [tex]\[ 22.5 \, \text{eV} = 22.5 \times 1.602 \times 10^{-19} \, \text{J} = 3.6045 \times 10^{-18} \, \text{J} \][/tex]

  Now, solve for the speed \( v \) using the kinetic energy formula:

  [tex]\[ \frac{1}{2} mv^2 = 3.6045 \times 10^{-18} \, \text{J} \][/tex]

 [tex]\[ v^2 = \frac{2 \times 3.6045 \times 10^{-18}}{9.11 \times 10^{-31}} \][/tex]

  [tex]\[ v^2 = 7.9137 \times 10^{12} \][/tex]

 [tex]\[ v = \sqrt{7.9137 \times 10^{12}} \approx 2.81 \times 10^6 \, \text{m/s} \][/tex]

2. Determine the components of the velocity:

The angle between the velocity and the magnetic field is [tex]\( 65.5^\circ \)[/tex]. The component of the velocity parallel to the magnetic field [tex](\( v_{\parallel} \))[/tex] and the component perpendicular to the magnetic field[tex](\( v_{\perp} \))[/tex] can be found using trigonometry:

 [tex]\[ v_{\parallel} = v \cos(65.5^\circ) \][/tex]

  [tex]\[ v_{\perp} = v \sin(65.5^\circ) \][/tex]

  Calculate these components:

  [tex]\[ v_{\parallel} = 2.81 \times 10^6 \, \text{m/s} \times \cos(65.5^\circ) \approx 2.81 \times 10^6 \times 0.417 \approx 1.17 \times 10^6 \, \text{m/s} \][/tex]

 [tex]\[ v_{\perp} = 2.81 \times 10^6 \, \text{m/s} \times \sin(65.5^\circ) \approx 2.81 \times 10^6 \times 0.909 \approx 2.55 \times 10^6 \, \text{m/s} \][/tex]

3. Find the cyclotron frequency [tex](\( f \))[/tex] and the period [tex](\( T \))[/tex] of the electron's circular motion:

  The cyclotron frequency is given by:

  [tex]\[ f = \frac{qB}{2\pi m} \][/tex]

  where q is the charge of the electron [tex](\( 1.602 \times 10^{-19} \) C)[/tex], B is the magnetic field [tex](\( 4.55 \times 10^4 \) T)[/tex], and m is the mass of the electron.

[tex]\[ f = \frac{1.602 \times 10^{-19} \times 4.55 \times 10^4}{2\pi \times 9.11 \times 10^{-31}} \][/tex]

[tex]\[ f \approx \frac{7.29 \times 10^{-15}}{5.72 \times 10^{-31}} \approx 1.27 \times 10^{16} \, \text{Hz} \][/tex]

  The period [tex](\( T \))[/tex] is the inverse of the frequency:

  [tex]\[ T = \frac{1}{f} \approx \frac{1}{1.27 \times 10^{16}} \approx 7.87 \times 10^{-17} \, \text{s} \][/tex]

4. Calculate the pitch of the helical path:

The pitch is the distance the electron travels parallel to the magnetic field in one period:

 [tex]\[ \text{Pitch} = v_{\parallel} \times T \][/tex]

 [tex]\[ \text{Pitch} \approx 1.17 \times 10^6 \, \text{m/s} \times 7.87 \times 10^{-17} \, \text{s} \approx 9.20 \times 10^{-11} \, \text{m} \][/tex]

The temperature of the boiling chlorine is calculated using the resistance change of a platinum wire thermometer and the temperature coefficient of resistivity. After calculations, the boiling point of chlorine is found to be -34.6°C.

The temperature of the boiling chlorine can be deduced using the relation between temperature change and resistance change for a platinum wire thermometer. The equation we use is:

R = R0 (1 + αΔT)

Where R represents the resistance at the new temperature, R0 the resistance at a reference temperature (20°C in this case), α the temperature coefficient of resistivity, and ΔT the change in temperature. We can rearrange this equation to solve for the new temperature:

ΔT = ∂(T₂ - T₁)

Giving us:

(99.6 ohms / 125 ohms - 1) = 3.72 × 10-3°C-1 ∂T

Simplifying:

ΔT = (0.7968 - 1) / 3.72 × 10-3°C-1

ΔT = -0.2032 / 3.72 × 10-3°C-1

ΔT = -54.6°C

Thus, the boiling point of chlorine is:

20°C – 54.6°C = -34.6°C

Answer:

Electric field at a distance of 1.45 cm will be [tex]172.41\times 10^4N/C[/tex]

Explanation:

We have given the distance d = 1.45 cm = 0.0145 m

And the potential difference [tex]V=2.5\times 10^4volt[/tex]

There is a relation between potential difference and electric field

Electric field at a distance d due to a potential difference is given by

[tex]E=\frac{V}{d}[/tex], here E is electric field, V is potential difference and d is distance

So [tex]E=\frac{V}{d}=\frac{2.5\times 10^4}{0.0145}=172.41\times 10^4N/C[/tex]

The magnitude of the uniform electric field between the plates is approximately [tex]\( 1.724 \times 10^4 \) V/m.[/tex]

The magnitude of the uniform electric field between the plates is [tex]\( E = \frac{\Delta V}{d} \)[/tex], where [tex]\( \Delta V \)[/tex] is the potential difference and ( d ) is the distance between the plates.

Now, we can calculate the electric field ( E ):

[tex]\[ E = \frac{\Delta V}{d} = \frac{2.50 \times 10^4 \text{ V}}{1.45 \times 10^{-2} \text{ m}} \][/tex]

[tex]\[ E = \frac{2.50 \times 10^4}{1.45} \times 10^2 \text{ V/m} \][/tex]

[tex]\[ E = 1.724 \times 10^4 \text{ V/m} \][/tex]

Answer:

The electric field intensity at the aircraft is [tex]7.18\times10^{5}\ N/C[/tex]

Explanation:

Given that,

Altitude = 2000 m

Charge = +40 C

At the position of the airplane, the electric field will be due to both the positive and negative charges and will be downwards due to both the charges.

The position is equidistant from both the charges.

We need to calculate the resultant electric field

The contribution due to the positive charge at 3000 m altitude

Using formula of electric field

[tex]E_{+}=k\dfrac{|q|}{r^2}[/tex]

Put the value into the formula

[tex]E_{+}=8.99\times10^{9}\times\dfrac{40}{(3000-2000)^2}[/tex]

[tex]E_{+}=3.59\times10^{5}\ N/C[/tex]

The contribution due to the negative charge at 1000 m altitude

Using formula of electric field

[tex]E_{-}=8.99\times10^{9}\times\dfrac{40}{(2000-1000)^2}[/tex]

[tex]E_{-}=3.59\times10^{5}\ N/C[/tex]

We need to calculate the electric field intensity at the aircraft

The resultant electric field is

[tex]E=E_{+}+E_{-}[/tex]

Put the value into the formula

[tex]E=3.59\times10^{5}+3.59\times10^{5}[/tex]

[tex]E=7.18\times10^{5}\ N/C[/tex]

Hence, The electric field intensity at the aircraft is [tex]7.18\times10^{5}\ N/C[/tex]

Answer:

Explanation:

The digits which are reliable and first uncertain digit is called significant figure.

In any measurement, if the significant figure is more than the accuracy of the measurement is also more.

If there are some numbers in a calculations and they having different numbers of significant figures, then in the final result the number of significant figures is the least number of significant figures which are in the individual number.

When performing calculations with measurements that have different numbers of significant figures, the final answer will be rounded to the same number of significant figures as the measurement with the fewest significant figures. This applies to both addition and subtraction, as well as multiplication and division.

For addition and subtraction, the result should be rounded to the same decimal place as the least precise measured value. For example, if one measurement has two decimal places and another measurement has three decimal places, the final answer should be rounded to two decimal places.

For multiplication and division, the final answer should have the same number of significant figures as the measurement with the fewest significant figures. For example, if one measurement has three significant figures and another measurement has four significant figures, the final answer should have three significant figures.

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Answer:

a) The average velocity is 32 m/s

b) See the attached figure. Slope of the line = 32.

Graphic: a line that passes through the points (0; 50) and (5; 210)

Explanation:

a) The average velocity is calculated as the displacement over time:

v = ΔX / Δt

where

ΔX : displacement ( final position - initial position)

Δt : time (final time - initial time)

Considering the origin of the reference system as the position where the observer is:

ΔX = 210 m - 50 m = 160 m

Δt  = 5 s

v = 160 m / 5 s = 32 m/s

b) In the graphic position vs time, plot a line that passes through the points (0, 50) (because at time 0, the car is 50 m away from you, the center of the reference system) and (5, 210). The x-axis, time, is in seconds and the y-axis, position, in meters. The slope of the line is calculated as:

slope = (X₂ - X₁) /(t₂ - t₁) = (210 - 50) / (5 - 0) = 32. Then, the velocity is equal to the slope of the line. See the attached figure.

Answer:

None of the above

It should be position is changing and acceleration is constant.

Explanation:

Since the velocity is changing, this means the object is moving, so the position must also be changing.

Acceleration is the change in velocity in time, if this change of velocity happens at a constant rate, the acceleration must be constant too.

So, for example, if the velocity were to stay the same (not changing), acceleration would be zero, because there wouldn't be a change in time on the velocity.

So in this case the answer sould be position is changing and acceleration is constant. But this isn't in the options so the correct answer is "None of the above"

In straight line motion, if velocity changes at a constant rate, then the position is changing and the acceleration is constant and non-zero. This is defined under the principles of kinematics and implies that as the velocity alters constantly, the object is in motion, hence its position is changing.

In straight line motion, if the velocity of an object is changing at a constant rate, then its position is changing and its acceleration is constant and non-zero. This condition is defined under the laws of physics, more specifically, under the study of kinematics.

The acceleration is constant because you're considering a situation where velocity is changing at a constant rate. In this case, the change in velocity is the acceleration, which is a constant and not zero. This situation is described by the kinematic equations for constant acceleration.

The position is changing because the object is moving. A change in position over time constitutes motion, and in this case, because the velocity (the rate of change of position) is changing, the object's position cannot be constant.

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Answer:

B. children have too much energy and play will rid them of that energy.

Explanation:

The surplus energy theory of play suggests that human being have excess of energy that must be released through active play.

This theory is given by Friedreich Schiller. Therefore, the correct answer is children have too much energy and play will rid them of that energy. rest of the option are examples of other theories, whereas Option B is the example of surplus energy theory.

Answer:

Explanation:

Given

ambient Pressure =98.10 kPa

(a)gauge pressure 152 kPa

we know

Absolute pressure=gauge pressure+Vacuum  Pressure

[tex]P_{abs}[/tex]=152+98.10=250.1 kPa or 36.27 psi

(b)[tex]P_{gauge}[/tex]=67.5 Torr or 8.99 kpa

as 1 Torr is 0.133 kPa

[tex]P_{abs}[/tex]=8.99+98.10=107.09 kPa or 15.53 psi

(c)[tex]P_{vaccum}[/tex]=0.1 bar or 10 kPa

Thus absolute pressure=98.10-10=88.10 kPa or 12.77 psi

as 1 kPa is equal to 0.145 psi

(d)[tex]P_{vaccum}[/tex]=0.84 atm  or 85.113 kPa

as 1 atm is equal to 101.325 kPa

[tex]P_{abs}[/tex]=98.10-85.11=12.99 kPa or 1.88 psi

Answer:

[tex]F=1.07\times 10^{-3}\ N[/tex]

Explanation:

Given that,

Charge on two balls, [tex]q_1=q_2=-38\ nC=-38\times 10^{-9}\ C[/tex]

Distance between charges, r = 11 cm = 0.11 m

We need to find the repulsive force between balls. Mathematically, it is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

[tex]F=9\times 10^9\times \dfrac{(38\times 10^{-9})^2}{(0.11)^2}[/tex]

F = 0.001074 N

or

[tex]F=1.07\times 10^{-3}\ N[/tex]

So, the magnitude of repulsive force between the balls is [tex]1.07\times 10^{-3}\ N[/tex].

Answer:

[tex]N=1.375*10^{11}[/tex] electrons

Explanation:

The total charge Q+ at the coin is equal, but with opposite sign, to the charge negative Q- removed of it. Q- is the sum of the charge of the N electrons removed from the coin:

[tex]Q_{-}=N*q_{e}[/tex]

[tex]Q_{-}=-Q_{+}[/tex]

q_{e}=-1.6*10^{-19}C    charge of a electron

We solve to find N:

[tex]N=-Q_{+}/q_{e}=-2.2*10^{-8}/(-1.6*10^{-19})=1.375*10^{11}[/tex]

Answer:

[tex]50.91 \mu C[/tex]

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for [tex]q_{1}[/tex] and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by[tex]q_{1}[/tex] on q. Then we can know the magnitude of the force exerted by [tex]q_{2}[/tex] about q, finally this will allow us to know the magnitude of [tex]q_{2}[/tex]

[tex]q_{1}[/tex] exerts a force on q in +y direction, and [tex]q_{2}[/tex] exerts a force on q in -y direction.

[tex]F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\[/tex]

The net force on q is:

[tex]F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}[/tex]

Rewriting for [tex]q_{2}[/tex]:

[tex]q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C[/tex]

Answer:

W = 0

Explanation:

given data:

starting position is ( 2cm , 7 cm)

ending position is  (10 cm, 7 cm)

we know thta work done is given as

[tex] W = q\Delta v[/tex]

here , q is particle charge

[tex]\Delta v[/tex] is change in potential

as we know that intial and final position are having uniform eleectric potential, therefore it have same potential i.e. V_1 = V_2, thus

one more reason of being work done equal to zero is,

Since the distance covered by the item and the direction of motion is always perpendicular to each other during the round journey, no work is therefore carried out.

W = 0

Answer:

Current through circuit will be [tex]0.2706\times 10^6A[/tex]

Explanation:

We have given source voltage v = 50 volt

Resistance [tex]R=25Mohm=25\times 10^6ohm[/tex]

Capacitance [tex]C=0.1\mu F=0.1\times 10^{-6}F[/tex]

Time t = 5 sec

Time constant of RC circuit is given by [tex]\tau =RC=25\times 10^6\times 0.1\times 10^{-6}=2.5sec[/tex]

We know that voltage across capacitor is given by [tex]v_c=v_s(1-e^{\frac{-t}{\tau }})[/tex]

[tex]v_c=50(1-e^{\frac{-5}{2.5 }})=43.2332v[/tex]

So current will be [tex]=\frac{v_s-v_c}{R}=\frac{50-43.2332}{25\times 10^{-6}}=0.2706\times 10^6A[/tex]

So current through circuit will be [tex]0.2706\times 10^6A[/tex]

Answer:

the potential at a distance of 16 cm from the charge [tex]4\mu C[/tex] will be 225000 volt

Explanation:

We have given charge [tex]q=4\mu C=4\times 10^{-6}C[/tex]

Distance between the charge r = 16 cm = 0.16 m

Electric potential is given by [tex]V=\frac{1}{4\pi \varepsilon _0}\frac{q}{r}=\frac{Kq}{r}[/tex], here K  is constant which value is [tex]9\times 10^9Nm^2/C^2[/tex]

So potential [tex]V=\frac{9\times 10^9\times 4\times 10^{-6}}{0.16}=225000volt[/tex]

So the potential at a distance of 16 cm from the charge [tex]4\mu C[/tex] will be 225000 volt

The electric potential, V, of a point charge, is found using the formula V = kq/r. Given the charge of 4.00 μC and the distance of 16.0 cm, the electric potential comes out to approximately 2.24 x 10⁵ V.

The electric potential, V, of a point charge, can be calculated by the formula V = kq/r where:

In the question, q is given as 4.00 μC, which is equivalent to 4.00 x 10⁻⁶ C. The distance r is given as 16.0 cm, which is equivalent to 0.16 m. We get by substituting the values into the formula:

V = (8.99 × 10⁹ Nm²/C² x 4.00 x 10⁻⁶ C)/0.16 m

Through calculation, the electric potential 16.0 cm from a 4.00 μC point charge is therefore approximately 2.24 x 10⁵ V.

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Answer:

The second statement is true: If the concentration of Y is increased by a factor of 1.5, the rate will increase by a factor of 2.25.

Explanation:

Hi there!

Let´s write the rate law for the original reaction and the reaction with X increased by 1.5:

rate 1 =k [X][Y]²

rate 2 = k[1.5 X][Y]²

Now we have to demonstrate if rate 2 = 2.25 rate 1

Let´s do the cocient between the two rates:

rate 2/ rate 1

if rate 2 = 2.25 rate 1

Then,

rate 2 / rate 1 = 2.25 rate 1 / rate 1 = 2.25

Let´s see if this is true using the expressions for the rate law:

rate 2 / rate 1

k[1.5 X][Y]² / k [X][Y]² = 1.5 k [X][Y]² / k[X][Y]² = 1.5

2.25 ≠ 1.5

Then the first statement is false.

Now let´s write the two expressions of the rate law, but this time Y will be increased by 1.5:

rate 1 = k[X][Y]²

rate 2 = k[X][1.5Y]²

Again let´s divide both expressions to see if the result is 2.25

rate 2 / rate 1

k[X][1.5Y]²/ k [X][Y]²  

(distributing the exponent)

(1.5)²k [X][Y]² / k [X][Y]² = (1.5)² = 2.25

Then the second statement is true!

Answer:

a) True

b)False

c) True

Explanation:

The order of reactants decides the exponents of respective reactant concentrations.

Since X is first order, exponent is = 1

Overall third order, Exponent X + Exponent Y = 3

Hence, exponent Y = 2.

Hence,

Rate = k[X][Y]^2

b)

If X conc is increased by 1.5 Rate should increase by 1.5 because k proportional to [X]. Hence, statement is False

c)

If Y conc is increased by 1.5 Rate should increase by 2.25 because k proportional to [Y]^2 = (1.5)^2 = 2.25. Hence, statement is True

Answer:

6

Explanation:

Number of lines emanate from + 5 micro coulomb is 15 .

They terminates at negative charges that means at - 3 micro coulomb and - 2 micro Coulomb.

the electric field lines terminates at - 3 micro Coulomb and - 2 micro Coulomb is in the ratio of 3 : 2.

So the lines terminating at - 3 micro coulomb

                                    = [tex]\frac{3}{5}\times 15 = 9[/tex]

So the lines terminating at - 2 micro coulomb

                                    = [tex]\frac{2}{5}\times 15 = 6[/tex]

So, the number of filed lines terminates at - 2 micro Coulomb are 6.

The number of lines emanating from the negative charged particles are 9.09 and 6.06 respectively.

In an electric field, the number of lines emanating from a charged particle is directly proportional to the magnitude of the charge. Thus, this is given by this mathematical expression:

[tex]q \;\alpha \;L\\\\q=kL\\\\5=k15\\\\k=\frac{5}{15}[/tex]

k = 0.33.

For the -3.0 μC particle:

[tex]L=\frac{q}{k} \\\\L=\frac{3.0}{0.33}[/tex]

L = 9.09.

For the -2.0 μC particle:

[tex]L=\frac{q}{k} \\\\L=\frac{2.0}{0.33}[/tex]

L = 6.06.

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Answer:

10.22 cm

Explanation:

linear charge density, λ = 7.5 x 10^-12 C/m

distance from line, r = 14.5 cm = 0.145 m

initial speed, u = 3000 m/s

final speed, v = 0 m/s

charge on proton, q = 1.6 x 10^-19 C

mass of proton, m = 1.67 x 10^-27 kg

Let the closest distance of proton is r'.

The potential due t a line charge at a distance r' is given  by

[tex]V=-2K\lambda ln\left (\frac{r'}{r}  \right )[/tex]

where, K = 9 x 10^9 Nm^2/C^2

W = q V

By use of work energy theorem

Work = change in kinetic energy

[tex]qV = 0.5m(u^{2}-v^{2})[/tex]

By substituting the values, we get

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

[tex]-2K\lambda ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{2q}[/tex]

[tex]- ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{4Kq\lambda }[/tex]

[tex]- ln\left ( \frac{r'}{r} \right )=\frac{1.67 \times 10^{-27}\times 3000\times 3000}{4\times 9\times 10^{9}\times 1.6\times 10^{-19}\times 7.5\times 10^{-12} }[/tex]

[tex]- ln\left ( \frac{r'}{r} \right )=0.35[/tex]

[tex]\frac{r'}{r} =e^{-0.35}[/tex]

[tex]\frac{r'}{r} =0.7047[/tex]

r' = 14.5 x 0.7047 = 10.22 cm

Answer:

56.7°

Explanation:

Imagine a rectangle triangle.

The triangle has 3 sides.

One side is the height of the tower, let's name it A.

Another side is the distance from the base of the tower to the point where the waire touches the ground. Let's name that B.

Sides A and B are perpendicular.

The other side is the length of the wire. Let's name it C.

From trigonometry we know that:

cos(a) = B / C

Where a is the angle between B anc C, between the wire and the ground.

Therefore

a = arccos(B/C)

a = arccos(552/1005) = 56.7°

[tex]56.7^\circ[/tex] angle the wire makes with the ground.

Given :

A guy wire 1005 feet long is attached to the top of a tower.

When pulled taut, it touches level ground 552 feet from the base of the tower.

Solution :

We know that,

[tex]\rm cos \theta = \dfrac{Base}{Hypotanuse}[/tex]

Given that,

base = 552 ft

hypotanuse = 1005 ft

Therefore,

[tex]\rm cos\theta = \dfrac{552}{1005}[/tex]

[tex]\rm \theta = cos^-^1 \dfrac{552}{1005}[/tex]

[tex]\theta = 56.7^\circ[/tex]

[tex]56.7^\circ[/tex] angle the wire makes with the ground.

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Answer:

[tex]t=6.96s[/tex]

Explanation:

From this exercise, our knowable variables are hight and initial velocity

[tex]v_{oy}=96ft/s[/tex]

[tex]y_{o}=112ft[/tex]

To find how much time does the ball strike the ground, we need to know that the final position of the ball is y=0ft

[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]

[tex]0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}[/tex]

Solving for t using quadratic formula

[tex]t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]

[tex]a=-\frac{1}{2} (32.2)\\b=96\\c=112[/tex]

[tex]t=-0.999s[/tex] or [tex]t=6.96s[/tex]

Since time can't be negative the answer is t=6.96s

Answer:

Current density [tex]j=1.44\times 10^7A/m^2[/tex]

Electron density [tex]=5.55\times 10^{18}electron/sec[/tex]

Explanation:

We have given power = 100 watt

Current = 0.89 A

Diameter d = 0.280 mm

So radius [tex]r=\frac{d}{2}=\frac{0.28}{2}=0.14mm=0.14\times 10^{-3}m[/tex]

Area [tex]A=\pi r^2=3.14\times (0.14\times 10^{-3})^2=0.016\times 10^{-6}m^2[/tex]

We know that current density [tex]J=\frac{I}{A}=\frac{0.89}{0.016\times 10^{-6}}=1.44\times 10^7A/m^2[/tex]

Now we have to calculate the electron density

We have current i = 0.89 A = 0.89 J/sec

Charge on 1 electron [tex]1.6\times 10^{-19}C/electron[/tex]

So electron density [tex]=\frac{0.89j/sec}{1.6\times 10^{-19}C/electron}=5.55\times 10^{18}electron/sec[/tex]

Final answer:

The current density in the filament is 14.4 x 10^6 A/m^2. The electron current in the filament is 1.42 x 10^-19 C/s.

Explanation:

The current density in a filament can be calculated by dividing the current in the lightbulb by the cross-sectional area of the filament. To find the current density, we first need to find the cross-sectional area of the filament. The diameter of the filament is given as 0.280 mm, so the radius is 0.140 mm (0.140 mm = 0.280 mm / 2).

Using the formula for the area of a circle (A = πr^2), we can find the cross-sectional area of the filament:

A = π(0.140 mm)^2 = 0.0616 mm^2

Now, we can calculate the current density by dividing the current (0.890 A) by the cross-sectional area (0.0616 mm^2), and converting mm to m:

Current density = 0.890 A / (0.0616 mm^2) * (1 m / 1000 mm)^2 = 14.4 x 10^6 A/m^2

The electron current in the filament can be calculated using the formula:
Electron current = charge * number of charges passing per second

The charge of an electron is approximately 1.6 x 10^-19 C, and the current is given as 0.890 A, so:

Electron current = (1.6 x 10^-19 C) * (0.890 A) = 1.42 x 10^-19 C/s

Answer:371.564 mi

Explanation:

Given

Airplane flies northwest for 250 mi and then travels west 150 mi

That is first it travels 250cos45 in - ve x direction and simultaneously 250sin45 in y direction

after that it travels 150 mi in -ve x direction

So its position vector is given by

[tex]r=-250cos45\hat{i}-150\hat{i}+250sin45\hat{j}[/tex]

[tex]r=-\left ( 250cos45+150\right )\hat{i}+250sin45\hat{j}[/tex]

so magnitude of displacement is

[tex]|r|=\sqrt{\left ( \frac{250}{\sqrt{2}}+150\right )^2+\left ( \frac{250}{\sqrt{2}}\right )^2}[/tex]

|r|=371.564 mi