1. Show that the inductive time constants RL has units of seconds. 2. If the inductance in the LR circuit is doubled, how is the half-life affected? 3. If the resistance in the LR circuit is doubled, how is the half-life affected?
4. If the charging voltage in the circuit is doubled, how is the half-life affected for the LR circuit? 5. To plot the equation V (1)=Vmax e^tR/L so the graph results in a straight line, what quantity do you have to plot vs, time? What is the expression for the slope of this straight line? Determine the expected self-inductance of a solenoid which has 1600 windings-each of enclosed cross- section radius 2.0 cm--and length 12 cm.

Answer:

Batteries are intended to store chemical energy and then it converts into electricidal energy. So overall, batteries produce energy.

Explanation:

Answer:

80 m or 81.3 m

Explanation:

Initially it had no velocity(as at rest). Let the height be h

Using conservation of energy:

Initial KE + PE = final KE + PE

(1/2) m(0)² + mgh = (1/2) m(40)² + mg(0)

mgh = (1/2) m (1600)

2gh = 16000

h = 1600/(2*10) {for g = 10}

h = 80

80 For g = 9.8, h = 81.3 m

Using Newton's eqⁿ of motion:

v² = u² + 2aS

40² = 0² + 2(g)S

1600/2g = S

For g = 10, S = 80 m

g = 9.8, S = 81.33 m

The angular momentum of the ball of clay before grazing the surface is (2/3)mrv.

The angular momentum of the ball of clay about the shaft before it grazes the surface is calculated as follows;

Moment of inertia of the ball;

I = (2/3) mr²

where;

Angular velocity is given as;

ω = v / r

where;

The angular momentum of the ball of clay before grazing the surface is calculated as;

L = Iω

L = (2/3)mr² x  (v / r)

L = (2/3)mrv

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Answer:

The correct option is (C).

Explanation:

The distance between two charges, [tex]d=10^{-8}\ m[/tex]

We need to find the magnitude of the electrostatic force between two electrons. The formula for the electric force between charges is given by :

[tex]F=\dfrac{kq^2}{r^2}\\\\=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(10^{-8})^2}\\\\=2.304\times 10^{-12}\ N[/tex]

So, the magnitude of force is equal to [tex]2.30\times 10^{-12}\ N[/tex].

A 10 kg mass slides down a flat hill that makes an angle of 10 degrees with the horizontal. Therefore, the resultant force on the sled is option (c) 97N.

If friction is negligible, the resultant force on the sled will be calculated below:

We know that gravitational force can be broken into two components - force parallel to the slope and force perpendicular to the slope.

The parallel component is given by

Fg * sin θ = 10*9.8*sin10 = 16.87 N.

The perpendicular component is given by

Fg * cos θ = 10*9.8*cos10 = 96.94 N.

The total force acting on the sled is the vector sum of the two components: Resultant force = √(16.87² + 96.94²) = 97 N.

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Kinetic energy is the energy that an object possesses due to its motion. Hence, the correct option is [109.5 J].

The formula to calculate kinetic energy is KE = 1/2 mv², where m is the mass of the object and v is its velocity or speed. In this question, the wagon wheel is a ring of mass 77.1 kg and radius 0.630 m. The wagon is moving at 1.22 m/s. We have to calculate the total kinetic energy of the wheel.The velocity of the wagon can be converted into the angular velocity of the wheel by using the formula v = wr. The angular velocity w is calculated as:w = v/rw = 1.22 m/s ÷ 0.630 m ≈ 1.936 rad/sNow that we know the angular velocity of the wheel, we can calculate its total kinetic energy using the formula KE = 1/2 Iw², where I is the moment of inertia of the wheel. For a ring-shaped object, the moment of inertia is given by I = mr².KE = 1/2 Iw²KE = 1/2 (mr²) (w²)KE = 1/2 (77.1 kg) (0.630 m)² (1.936 rad/s)²KE ≈ 109.5 J. Therefore, the total kinetic energy of the wagon wheel is approximately 109.5 J. Note: As per the question, the total kinetic energy of the wheel is required, not of the wagon.

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Answer:

Keyboard and Mobile devices

Explanation:

got it right on edge 2021

Answer:

Outlook allows users to insert symbols and characters not located on the  

keyboard

, but they can have shortcut keys.  

Inserting horizontal lines in the message body breaks up different sections and mostly benefits people using Outlook on  

mobile devices

.

Explanation:

The resistance (R) of the copper wire in the circuit is 1600 ohms, and the potential (V) at the other end of the wire is 21 volts.

To calculate the resistance of the copper wire (R), we use the formula R = (ρ * L) / A, where ρ represents the resistivity of the material, L is the length of the wire, and A is the cross-sectional area. In this case, the length of the copper wire is 8 mm (0.008 m), and the cross-sectional area is 0.3 mm^2 (0.3 * 10^(-6) m^2). With the resistivity of copper being 6 * 10^7 ohm^(-1) m^(-1), we can calculate the resistance as follows: R = (6 * 10^7 ohm^(-1) m^(-1) * 0.008 m) / (0.3 * 10^(-6) m^2), which gives us 1600 ohms. Since the copper wire and the carbon resistor are connected in series in the circuit, the potential difference (V) across each component is the same. In the case of a thick, high-conductivity copper wire, the small drift speed of electrons requires only a very small electric field, resulting in a negligible potential difference along the wire. As a result, the potential throughout the thick copper wire remains practically uniform.

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Answer:

V = f x λ

Explanation:

The options are confusing

Answer:

a. 0.947 m/s^2

b. 1304.54 N

c. 0.0966

Explanation:

mass of car = 13500 N = 13500/9.8 = 1377.55 kg

velocity = 50 km/h = 50,000 m/h = 13.9 m/s

raidus = 204 m

a. centripetal acceleartion = v^2/r = 13.9^2/204 = 0.947 m/s^2

b. centripetal force = m*centripetal acceleration = 1377.55 * 0.947 = 1304.54 N

c. In order for the car to round the curve safely, static friction = centripetal force

static friction = coefficient of friction (mu) * mg = mu* 1377.55*9.8 = 13500mu

13500mu = 1304.54

mu = 1304.54/13500 = 0.0966

The acceleration, force and coefficient of friction is required.

Centripetal acceleration is [tex]0.965\ \text{m/s}^2[/tex]

Centripetal force is [tex]1328\ \text{N}[/tex]

Coefficient of friction is [tex]0.1[/tex]

N = Weight of car = 13500 N

v = Velocity = [tex]50=\dfrac{50}{3.6}=13.89\ \text{m/s}[/tex]

r = Radius = [tex]2\times 10^2\ \text{m}[/tex]

m = Mass of car = [tex]\dfrac{N}{g}[/tex]

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

Centripetal acceleration is

[tex]a_c=\dfrac{v^2}{r}\\\Rightarrow a_c=\dfrac{13.89^2}{2\times 10^2}\\\Rightarrow a_c=0.965\ \text{m/s}^2[/tex]

Force is given by

[tex]F_c=ma_c\\\Rightarrow F_c=\dfrac{N}{g}a_c\\\Rightarrow F_c=\dfrac{13500}{9.81}\times 0.965\\\Rightarrow F_c=1328\ \text{N}[/tex]

Coefficient of friction is given by

[tex]\mu=\dfrac{F_c}{N}\\\Rightarrow \mu=\dfrac{1328}{13500}\\\Rightarrow \mu=0.098\approx 0.1[/tex]

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For accurate printing of longer storms, it is important to ensure that the events in the array are arranged in the correct order, with a 200ms delay between each event and a 400ms delay between each flash and boom.

When printing longer storms, it is crucial to maintain the correct order of events in the array. By arranging the events in the correct sequence, the storm will be printed accurately, providing a realistic representation. To achieve this, a delay of 200ms should be implemented between each event in the array. This delay ensures that each event is printed with the appropriate timing, creating a smooth and coherent storm simulation.

Additionally, it is necessary to introduce a 400ms delay between each "flash" and "boom" in the storm. This delay creates a distinct gap between these two elements, mimicking the natural occurrence of a lightning flash followed by the accompanying thunder. By allowing sufficient time between the flash and boom, the printed storm will convey a more realistic and immersive experience.

In summary, to accurately print longer storms, it is essential to maintain the correct order of events in the array and introduce appropriate delays. A 200ms delay between each event ensures accurate timing, while a 400ms delay between each flash and boom replicates the natural occurrence of lightning and thunder. Following these guidelines will result in a more realistic representation of storms when printing them.

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An automobile engine slows down from 4,087 rpm to 1,830 rpm in 1,419 revolutions the magnitude of the angular acceleration of the automobile engine is approximately 77.75 rad/s².

To calculate the magnitude of the angular acceleration, we can use the formula:

Angular acceleration (α) = (ω2 - ω1) / (t2 - t1)

where:

ω1 and ω2 are the initial and final angular velocities, respectively, and

t1 and t2 are the initial and final times, respectively.

Initial angular velocity ω1 = 4087 rpm

Final angular velocity ω2 = 1830 rpm

Number of revolutions (n) = 1419

First, we need to convert the angular velocities from rpm to radians per second (rad/s):

ω1 = (4087 rpm) * (2π rad/1 min) * (1 min/60 s) ≈ 426.97 rad/s

ω2 = (1830 rpm) * (2π rad/1 min) * (1 min/60 s) ≈ 191.46 rad/s

Next, we can calculate the time interval (t2 - t1) using the number of revolutions and the initial and final angular velocities:

t2 - t1 = (n / ω2) - (n / ω1)

t2 - t1 = (1419 / 191.46) - (1419 / 426.97) ≈ 3.3 s

Finally, we can calculate the magnitude of the angular acceleration:

α = (ω2 - ω1) / (t2 - t1)

α = (191.46 rad/s - 426.97 rad/s) / (3.3 s)

α ≈ -77.75 rad/s²

Therefore, the magnitude of the angular acceleration of the automobile engine is approximately 77.75 rad/s².

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Planet A exerts a force on planet B, the magnitude and direction of the gravitational force planet B exerts on planet A the gravitational force exerted by planet B on planet A is the same as the magnitude of the gravitational force exerted by planet A on planet B

Newton's third law states that if object A exerts a force on object B, then object B exerts an equal and opposite force on object A. Hence, if planet A exerts a gravitational force on planet B, then planet B exerts an equal and opposite gravitational force on planet A.The magnitude of the gravitational force exerted by planet B on planet A is the same as the magnitude of the gravitational force exerted by planet A on planet B, this is according to the law of universal gravitation,

This law states that the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The direction of the gravitational force exerted by planet B on planet A is towards planet B's center, just as the direction of the gravitational force exerted by planet A on planet B is towards planet A's center. Therefore, we can say that the magnitude and direction of the gravitational force planet B exerts on planet A is equal and opposite to the gravitational force planet A exerts on planet B

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Some diversification benefits can be achieved by combining securities in a portfolio as long as the correlation between the securities isThe correct answer is C. Between 0 and 1.

Diversification benefits can be achieved by combining securities in a portfolio as long as the correlation between the securities is between 0 and 1. Correlation measures the degree to which two securities move in relation to each other. A correlation of 1 indicates a perfect positive correlation, where the securities move in perfect tandem. A correlation of less than 1 indicates a less than perfect positive correlation, where the securities move somewhat together but not completely. When the correlation is between 0 and 1, it means that the securities have some degree of independence and tend to move differently from each other. This allows for diversification benefits as losses in one security may be offset by gains in another, reducing overall portfolio risk.

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Answer:

0.05 m/s

Explanation:

We start by finding the average velocity of water in the pipe. This is done by saying

R(e) = ρv(avg)d/μ

Where,

R(e) = Reynolds number, and that's 2000

ρ = Density of water, 1000 kg/m³

μ = Viscosity of water, 10^-3

d = diameter of pipe

v(avg) = average velocity

Since we're interested in average velocity, we make v(avg) the subject of formula. So that

V(avg) = R(e).μ/ρ.d

V(avg) = 2000 * 10^-3 / 1000 * 0.08

V(avg) = 2 / 80

V(avg) = 0.025 m/s

The maximum flow rate of water in the pipe usually is twice the average velocity, and as such

V(max) = 2 * V(avg)

V(max) = 2 * 0.025

V(max) = 0.05 m/s

The Sun would have to shine for approximately 2 billion years for one percent of its mass to be converted into energy.

To calculate the time required for the Sun to convert one percent of its mass into energy, we need to determine the total energy the Sun can produce and then calculate the amount of mass that would be converted.

Energy produced per second by the Sun via the proton-proton chain = 400 trillion trillion joules (4 x 10²⁶joules)

Mass of the Sun = 2 million trillion trillion kilograms (2 x 10³⁰ kilograms)

First, we need to calculate the total energy the Sun can produce per year:

Energy produced per year = Energy produced per second × Number of seconds in a year

Number of seconds in a year = 365 days × 24 hours × 60 minutes × 60 seconds

Number of seconds in a year = 31,536,000 seconds

Energy produced per year = 4 x 10²⁶joules/second × 31,536,000 seconds/year

Now, we can calculate the mass that would be converted to energy:

Mass converted to energy per year = Energy produced per year / (Speed of light)²

The speed of light (c) is approximately 3 x 10⁸meters per second.

Mass converted to energy per year = (4 x 10²⁶ joules/second × 31,536,000 seconds/year) / (3 x 10^8 meters/second)²

Now, let's calculate the mass converted to energy as a percentage of the Sun's total mass:

Mass percentage = (Mass converted to energy per year / Mass of the Sun) × 100

Mass percentage = ((4 x 10²⁶ joules/second × 31,536,000 seconds/year) / (3 x 10⁸ meters/second)²) / (2 x 10³⁰ kilograms) × 100

Finally, we can calculate the number of years required for one percent of the Sun's mass to be converted into energy:

Years = 1% / Mass percentage

Years = 1% / ((4 x 10²⁶ joules/second × 31,536,000 seconds/year) / (3 x 10⁸ meters/second)²) / (2 x 10³⁰kilograms) × 100

After performing the calculations, the result is approximately 2 billion years.

The Sun would have to shine for approximately 2 billion years for one percent of its mass to be converted into energy.

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Answer:

Yes

Explanation:

sorry if im wrong

Answer:

For which the reasoning of the boy is correct for small heights, but as height increases his analysis is not correct.

Explanation:

The force of gravity comes from Newton's second law with the force the universal attraction

         F = ma

         F = [tex]G \frac{m_1 M}{(R_e +h)^2}[/tex]

we substitute

          [tex]G \frac{m_1 M}{ (R_e+ h)^2}[/tex] = m₁ a

where Re is the radius of the Earth 6.37 106 m

          a = [tex]G\frac{M}{R_e^2} \ ( 1 + \frac{h}{R_e})^{-2}[/tex]

In general, the height is much less than the radius of the earth, therefore the term ha / Re is very small and we can use a series expansion leaving only the first fears.

             (1 + x)⁻² = 1 -2x + [tex]\frac{2 \ 1}{2!}[/tex]  x²

we substitute

          a = g₀ ([tex]1 - 2 \frac{h}{R_e}[/tex] )

with

         g₀ = [tex]G \frac{M}{R_e^2}[/tex]

let's launch the expression.

* For small height compared to the radius of the earth we can neglect the last term

          g = g₀

* For height comparable to the radius of the Earth

          g = g₀  [tex](1 - \frac{2h}{Re} )[/tex]

We see that the acceleration of gravity is decreasing.

For which the reasoning of the boy is correct for small heights, but as height increases his analysis is not correct.

The student's reasoning gone wrong when the analysis is undertaken for the increasing heights, to drop the object.

The given problem is based on the concept of gravity and gravitational force. The force of gravity comes from Newton's second law with the force the universal attraction as,

F = ma

[tex]F=G\dfrac{mM}{(R+h)^{2}}\\\\\\ma = G\dfrac{mM}{(R+h)^{2}}[/tex]

Here, a is the linear acceleration, m is the mass of object, M is the mass of Earth, R is the radius of Earth and h is the height from where the objects will be dropped. Then,

[tex]a = \dfrac{GM}{R^{2}} \times(1+h/R)^{-2}[/tex]

In general, the height is much less than the radius of the earth, therefore the term h/ R is very small, hence can be neglected.

[tex]a = \dfrac{GM}{R^{2}}\\\\a=g = \dfrac{GM}{R^{2}}[/tex]

g is the gravitational acceleration.

For small height compared to the radius of the earth we can neglect the last term as,

a = g

And for the height comparable to radius of Earth,

[tex]a = \dfrac{GM}{R^{2}} \times(1+h/R)^{-2}\\\\a=g \times(1+h/R)^{-2}[/tex]

Clearly, the acceleration of gravity is decreasing, for which the reasoning of the boy is correct for small heights, but as height increases his analysis is not correct.

Thus, we can conclude that the student's reasoning gone wrong when the analysis is undertaken for the increasing heights, to drop the object.

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The linear homogeneous recurrence relation with constant coefficients that the sequence is: An = -(-1)n + (3)n or An = (1)n + (3)n

The sequence An = (n-3)(2n) + n(2n) satisfies the linear homogeneous recurrence relation with constant coefficients. In order to prove this, we need to first find the general formula of the sequence.

The formula can be found by replacing n with n+1 as follows:

An+1 = (n+1-3)(2n+2) + (n+1)(2n+2)

An+1 = n(2n+4) + (n+1)(2n+2)

An+1 = 2n² + 4n + 2n² + 4n + n + 2n + 2

An+1 = 4n² + 7n + 2

The characteristic equation of the linear homogeneous recurrence relation is given by:

r² - 2r - 3 = 0

Solving this quadratic equation, we get:r = -1 or r = 3

Hence, the general formula of the sequence is given by:An = A(-1)n + B(3)n

Now, we need to find the values of A and B using the initial values of the sequence. The first two terms of the sequence are given by:

A0 = -6 and A1 = 2

Substituting these values in the general formula, we get:

-6 = A + B2 = -A + 3B

From the above two equations, we can solve for A and B to get:

A = -1 and B = 1

Hence, the linear homogeneous recurrence relation with constant coefficients that the sequence An = (n-3)(2n) + n(2n) satisfies is given by:

An = -(-1)n + (3)n or An = (1)n + (3)n

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a. The electric field is  E = (kρa^3/r^2) r

b. The magnitude of the electric field inside the hole of the sphere is (kρa^3/b^2 - kρ(4/3)πa^3/(b^2 - a^2)) j

a)

The electric field inside an insulating sphere with radius a and uniform charge density rho is,

E = (kq/r^2) r

Where

q = volume charge density * volume of the sphere

  = rho*(4/3)πa^3

q = (4/3)πa^3ρ

The electric field is:,

E = (kq/r^2) r

  = (kρa^3/r^2) r

b)

        E = E_solid_sphere - E_hollow_sphere

           E_solid_sphere = (kρa^3/b^2) j

           E_hollow_sphere = -(kρ(4/3)πa^3/(b^2 - a^2)) j

         E = E_solid_sphere - E_hollow_sphere

            = (kρa^3/b^2 - kρ(4/3)πa^3/(b^2 - a^2)) j

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Given, Initial speed of the helicopter = 25 m/s. Final speed of the helicopter = 60 m/s, Time taken by the helicopter to change its speed from 25 m/s to 60 m/s is 5 seconds. Therefore, the acceleration of the helicopter is 7 m/s².Hence, option (A) is correct.

To find the acceleration of the helicopter, we can use the formula:

Acceleration= (Final velocity - Initial velocity)/Time,

Acceleration = (60 m/s - 25 m/s)/5 s,

Acceleration = 35 m/s/5 s,

Acceleration = 7 m/s²

Therefore, the acceleration of the helicopter is 7 m/s².

We know that,

Acceleration = change in velocity / time taken,

Acceleration can be defined as the rate at which an object changes its velocity.

It can be measured in units such as m/s², cm/s², etc.

Here, initial velocity = 25 m/s, Final velocity = 60 m/s time = 5 s.

Hence,

Acceleration= (Final velocity - Initial velocity)/Time

Acceleration = (60 m/s - 25 m/s)/5 s

Acceleration= 35/5

Acceleration= 7 m/s²

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Answer:

A malleable material is one in which a thin sheet can be easily formed by hammering or rolling. ... In contrast, ductility is the ability of a solid material to deform under tensile stress.

Answer:

Stars i think

Explanation:

The total work performed around the thermodynamic cycle in Figure 7 is 350 Joules.
In a thermodynamic cycle, the total work performed is equal to the net area enclosed by the cycle on a pressure-volume (PV) diagram. From the given figure, we can observe that the cycle consists of two quasi-static processes: Process 1-2 and Process 2-3.

In Process 1-2, the system undergoes an expansion at constant pressure, represented by the horizontal line between points 1 and 2 on the PV diagram. As the volume increases, work is done by the system, and the work done in this process is given by the equation W_1-2 = PΔV = 2 * 50 = 100 Joules (where P is the constant pressure and ΔV is the change in volume).

In Process 2-3, the system undergoes compression at constant volume, represented by the vertical line between points 2 and 3 on the PV diagram. Work is done on the system during this process, and the work done in this process is given by the equation W_2-3 = -PΔV = -3 * 50 = -150 Joules (where P is the constant pressure and ΔV is the change in volume).

The total work performed around the cycle is the sum of the work done in each process, i.e., 100 Joules + (-150 Joules) = -50 Joules.
The total work performed around the thermodynamic cycle in Figure 7 is -50 Joules.

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Answer:

When focused light is projected onto the retina, it stimulates the rods and cones. The retina then sends nerve signals are sent through the back of the eye to the optic nerve. The optic nerve carries these signals to the brain, which interprets them as visual images.

Explanation:

Hope it will help u

Answer:

C hope it helps

call the client and inform her that she was incorrectly charged.

Answer:

The answer is C. call the client and inform her that she was incorrectky charged

Explanation:

Answer:

the final velocity of the skater after throwing the snowball is 3.17 m/s.

Explanation:

Given;

mass of the ice skater, m₁ = 72 kg

initial velocity of the ice skater, u₁ = 3.1 m/s

mass of the snowball, m₂ = 0.21 kg

initial speed of the snowball, u₂ = 28.0 m/s

Let the final velocity of the skater after throwing the snowball = v

Apply the principle of conservation of linear momentum to determine v;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

72 x 3.1   +  0.21 x 28  = v(72 + 0.21)

229.08 = v(72.21)

v = 229.08 / 72.21

v = 3.17 m/s

Therefore, the final velocity of the skater after throwing the snowball is 3.17 m/s.

Light is an electromagnetic wave that consists of oscillating electric and magnetic fields. It exhibits various properties, including wavelength, frequency, speed, and polarization.

Polarization refers to the orientation of the electric field vector of a light wave. Unpolarized light consists of electric field vectors oscillating in all possible directions perpendicular to the direction of propagation. Polarized light, on the other hand, has its electric field vectors confined to a specific orientation. Polarization can be achieved through various mechanisms, such as reflection, scattering, or passing light through certain materials. Polarizers, such as polarizing filters, can selectively transmit or block light waves based on their polarization orientation. The study of light and its properties, including polarization, has contributed to numerous advancements in various fields, such as optics, telecommunications, and technology.

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