At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what distance from the line will the field strength to be 2000 N/C

Answer:

Explanation:

initial speed, u = 19 m/s

(a) Let it rises upto height h.

Use third equation of motion

v² = u² - 2 gh

where, v is the final velocity and it is zero.

0 = 19 x 19 - 2 x 9.8 x h

h = 18.4 m

(c) Let the ball takes time t to reach to the maximum height.

use first equation of motion

v = u - gt

0 = 19 - 9.8 x t

t = 1.94 s

(c) The time taken by the ball to reach to the ground = 2 x time to reach to maximum height

T = 2 x t = 2 x 1.94 = 3.88 s

(d) When the ball reaches the ground, let the velocity is v.

Use third equation of motion

v² = u² - 2 gh

where, v is the final velocity

v² = 0 + 2 x 9.8 x 18.4

v = 19 m/s

(a) The ball reaches a height of approximately 18.68 meters. (b) It takes approximately 1.94 seconds to reach its highest point  (c) The time taken to hit the ground is 1.94 seconds. When the ball returns to the level from which it started, its velocity is -19.0 m/s.

(a) To find the height that the ball reaches, we can use the kinematic equation:
Δy = v2 / (2g)
where Δy is the change in height, v is the initial velocity, and g is the acceleration due to gravity. Plugging in the given values, we have:
Δy = (19.0 m/s)2 / (2 * 9.8 m/s2)
Calculating, we find that the ball rises to a height of approximately 18.68 meters.
(b) The time it takes for the ball to reach its highest point can be calculated using the equation:

t = v / g
where t is the time, v is the initial velocity, and g is the acceleration due to gravity. Substituting the given values, we get:
t = (19.0 m/s) / (9.8 m/s2)
Simplifying, we find that it takes approximately 1.94 seconds for the ball to reach its highest point.
(c) Since the time it takes for the ball to reach its highest point is the same as the time it takes for it to fall back down, the time it takes for the ball to hit the ground after reaching its highest point is also 1.94 seconds.
(d) When the ball returns to the level from which it started, its velocity is equal in magnitude but opposite in direction to its initial velocity. Therefore, the velocity when it returns is -19.0 m/s.

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To solve this problem we will apply the concepts related to the relationship between energy and frequency, from the latter we will obtain similar expressions that relate to the wavelength to find the two energy states between the given values. Finally we will make the comparative radius between the two. The relation between energy and frequency is given as,

[tex]E = hf[/tex]

Here,

E = Energy

h = Planck's constant

The relation between the speed of the electromagnetic waves (c), frequency (f) and wavelength ([tex]\lambda[/tex] ) is,

[tex]c = f\lambda[/tex]

Rearrange the above equation for frequency f as follows

[tex]f = \frac{c}{\lambda}[/tex]

Substitute,

[tex]E = h\frac{c}{\lambda}[/tex]

The wavelength x-ray or gamma ray photon is

[tex]\lambda = 1.0nm (\frac{1nm}{10^{9}nm})[/tex]

[tex]\lambda = 10^{-9} m[/tex]

Therefore the energy would be,

[tex]E_1 = \frac{hc}{\lambda}[/tex]

[tex]E_1 = \frac{(6.63*!0^{-34}J\cdo s)(3*10^{8}m/s)}{10^{-9}m}[/tex]

[tex]E_1 = 19.89*10^{-17} J[/tex]

The frequency is given as,

[tex]f = 10MHz (\frac{10^6z}{1.0MHz})[/tex]

[tex]f = 10^7Hz[/tex]

Now the second energy would be

[tex]E_2 = hf[/tex]

[tex]E_2 = (6.63*10^{-27}J\cdot s)(10^7Hz)[/tex]

[tex]E_2 = 6.63*10^{-27}J[/tex]

Therefore the ratio between them is

[tex]\frac{E_1}{E_2} = \frac{19.89*10^{-17}J}{6.63*10^{-27}J}[/tex]

[tex]\frac{E_1}{E_2} = 3*10^{20}[/tex]

Therefore the energy of 1nm x ray or gamma ray photon is [tex]3*10^{20}[/tex] times more than energy of 10MHz radio photon

The reaction produces -4.95 kJ of heat when 15.0 L of CO at 85°C and 112 kPa reacts with 14.4 L of H2 at 75°C and 744 torr.

The equation of the reaction is;

CO(g) + H2(g) -------> CH2O(g)

The heat of reaction is obtained from;

Enthalpy of products - Enthalpy of reactants = (-116kJ/mol)  - (-110.5 kJ/mol)

= -5.5 kJ/mol

Number of moles of CO is obtained from;

PV = nRT

P =  112 kPa or 1.1 atm

T = 85°C + 273 = 358 K

n = ?

R = 0.082 atmLK-1mol-1

V = 15.0 L

n = PV/RT

= 1.1 atm × 15.0 L/ 0.082 atmLK-1mol-1 ×  358 K

= 0.56 moles

Number of moles of H2

n = PV/RT

P= 744 torr or 0.98 atm

V = 14.4 L

T = 75°C + 273 = 348 K

n =  0.98 atm ×  14.4 L/0.082 atmLK-1mol-1 ×  348 K

n = 0.49 moles

We can see that H2 is the limiting reactant here hence 0.49 moles of formaldehyde is produced.

If 1 mole of formaldehyde produces -5.5 kJ of heat

0.49 moles of formaldehyde produces -5.5 kJ ×  0.49 moles / 1 mole

= -4.95 kJ of heat

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Answer:

d. The length of the string is equal to one-half of a wavelength

Explanation:

A stretched string of length L, fixed at both ends, is vibrating in its third harmonic. How far from the end of the string can the blade of a screwdriver be placed against the string without disturbing the amplitude of the vibration

a. The length of the sting is equal to one-quarter of a wavelength.b. The length of the string is equal to the wavelength.c. The length of the string is equal to twice the wavelength.d. The length of the string is equal to one-half of a wavelength

e. The length of the string is equal to four times the wavelength

A stretched string of length L fixed at both ends is vibrating in its third harmonic H

How far from the end of the string can the blade of a screwdriver be placed against the string without disturbing the amplitude of the vibration

d. The length of the string is equal to one-half of a wavelength

There are two points during vibration , the node and the antinode

the node is the point where the amplitude is zero.

from the third harmonics, there are two nodes. The first node is half of the wavelength which is the closest to the fixed point.

for third harmonics=3/2lamda

Answer:

Power, [tex]P=3.93\times 10^{26}\ W[/tex]

Explanation:

Given that,

The distance from the earth to the sun is about, [tex]d=1.5\times 10^{11}\ m[/tex]

let us assume that the average intensity of solar radiation at the upper atmosphere of earth,. [tex]I=1390\ W/m^2[/tex]

We need to find the total power radiated by the sun. The intensity is defined as the total power divided by the area of a sphere of radius equal to the average distance between the earth and the sun. It is given by :

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

P is total power

[tex]P=4\pi r^2\times I[/tex]

[tex]P=3.93\times 10^{26}\ W[/tex]

So, the total power radiated by the sun is [tex]P=3.93\times 10^{26}\ W[/tex]. Hence, this is the required solution.

Final answer:

The total power radiated by the Sun is calculated using the area of the Sun and the power radiated per square meter at its surface. The Sun's total power output is found to be 3.82×1026 W. This value is important for understanding the energy Earth receives from the Sun, known as the solar constant (1360 W/m²).

Explanation:

The distance from the Earth to the Sun is about 1.50×1011 meters. The total power radiated by the Sun can be determined by using the following physics principle: The Sun, behaving as a perfect black body with an emissivity of exactly 1, radiates power uniformly across its surface area. The power radiated per square meter on the Sun's surface is found to be 6.3×107 W/m². The sun's radius is approximately 7.00×108 meters. Using the formula Power = Area × Intensity, we calculate the Sun's total power output.

The area of a sphere is given by 4πR2, where R is the radius of the sphere. Therefore, the total power output of the Sun is 4πR2σT4, which calculates to 3.82×1026 W. This immense amount of power is what we refer to as the solar luminosity.

At the distance of the Earth, this power is spread over a spherical area with a radius equal to the Earth-Sun distance. We can find the power per square meter at this distance, known as the solar constant, which is approximately 1360 W/m².

In order for the net effect at point A to be a downward force, the tension in cable AC (T) should be equal to the tension in cable AB (9.1 kN). The magnitude of the resulting downward force (R) would be the sum of the tensions in both cables, thus 2 * 9.1 kN = 18.2 kN.

To understand this scenario, it is essential to apply the principles of equilibrium and vector sum in Physics. The tension in the wires can be considered as forces experienced by point A. According to the question, the net effect of these forces should be a downward force, implying that they should negate the opposite upward force.

To find the tension T in cable AC, it's logical to assume that the force due to this tension needs to be equal and opposite to the force exerted by the tension in wire AB, which is 9.1 kN. Therefore, T should also be 9.1 kN for the net effect at point A to be a downward force.

The magnitude R of the downward force can be determined by considering the combined effect of tensions in cables AB and AC. Since point A is in equilibrium, R will be the result of the total upward forces. Hence, R is equal to 2 times the tension in any one cable (as they are equal), which gives us R = 2 * 9.1 kN = 18.2 kN.

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A) Force of the wall on the ladder: 186.3 N

B) Normal force of the ground on the ladder: 725.2 N

C) Minimum value of the coefficient of friction: 0.257

D) Minimum absolute value of the coefficient of friction: 0.332

Explanation:

a)

The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:

[tex]W=mg[/tex]: weight of the ladder, with m = 20 kg (mass) and [tex]g=9.8 m/s^2[/tex] (acceleration of gravity)

[tex]W_M=Mg[/tex]: weight of the person, with M = 54 kg (mass)

[tex]N_1[/tex]: normal reaction exerted by the wall on the ladder

[tex]N_2[/tex]: normal reaction exerted by the floor on the ladder

[tex]F_f = \mu N_2[/tex]: force of friction between the floor and the ladder, with [tex]\mu[/tex] (coefficient of friction)

Also we have:

L = 4.1 m (length of the ladder)

d = 3.0 m (distance of the man from point A)

Taking the equilibrium of moments about point A:

[tex]W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}[/tex]

where

[tex]Wsin 21^{\circ}[/tex] is the component of the weight of the ladder perpendicular to the ladder

[tex]W_M sin 21^{\circ}[/tex] is the component of the weight of the man perpendicular to the ladder

[tex]N_1 sin 69^{\circ}[/tex] is the component of the normal  force perpendicular to the ladder

And solving for [tex]N_1[/tex], we find the force exerted by the wall on the ladder:

[tex]N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N[/tex]

B)

Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of [tex]N_2[/tex].

We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.

Therefore, we have:

[tex]\sum F_y = 0\\N_2 - W - W_M =0[/tex]

And substituting and solving for N2, we find:

[tex]N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N[/tex]

C)

Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.

The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.

Therefore, we can write:

[tex]\sum F_x = 0\\F_f - N_1 = 0[/tex]

And re-writing the equation,

[tex]\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257[/tex]

So, the minimum value of the coefficient of friction is 0.257.

D)

Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.

From part C), we saw that the coefficient of friction can be written as

[tex]\mu = \frac{N_1}{N_2}[/tex]

This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was

[tex]N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}[/tex]

We see that this quantity is maximum when d is maximum, so when

d = L

Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:

[tex]N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)[/tex]

And substituting, we get

[tex]N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N[/tex]

And therefore, the minimum coefficient of friction in order for the ladder not to slip is

[tex]\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332[/tex]

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Answer:

1.2 cm

Explanation:

Thermal Expansion

It's the tendency that materials have to change its size and/or shape under changes of temperature. It can be in one (linear), two (surface) or three (volume) dimensions.

The formula to compute the expansion of a material under a change of temperature from [tex]T_o[/tex] to [tex]T_f[/tex] is given by.

[tex]\Delta L=L_o.\alpha .(T_f-T_o)[/tex]

Where Lo is the initial length and [tex]\alpha[/tex] is the linear temperature expansion coefficient, which value is specific for each material. The data provided in the problem is as follows:

[tex]L_o=30\ m,\ T_f-T_o=30^oC,\ \alpha=13\times 10^{-6}\ ^oC^{-1}[/tex]

Computing the expansion we have

[tex]\Delta L=30\times 13\times 10^{-6}(30)=0.0117=1.17\ cm[/tex]

The expansion gap should be approximately 1.2 cm

Answer:

Explanation:

Given

speed of Electron [tex]u=2\times 10^7\ m/s[/tex]

final speed of Electron [tex]v=4\times 10^7\ m/s[/tex]

distance traveled [tex]d=1.2\ cm[/tex]

using equation of motion

[tex]v^2-u^2=2as[/tex]

where v=Final velocity

u=initial velocity

a=acceleration

s=displacement

[tex](4\times 10^7)^2-(2\times 10^7)^2=2\times a\times 1.2\times 10^{-2}[/tex]

[tex]a=5\times 10^{16}\ m/s^2[/tex]

acceleration is given by [tex]a=\frac{qE}{m}[/tex]

where q=charge of electron

m=mass of electron

E=electric Field strength

[tex]5\times 10^{16}=\frac{1.6\times 10^{-19}\cdot E}{9.1\times 10^{-31}}[/tex]

[tex]E=248.3\ kN/C[/tex]                

Answer:

Total weight of the hot tub and water is 5676.6 pounds-force

Explanation:

rho = 62.4lbm/ft^3 × 1ft^3/7.481gal = 8.34lbm/gal

Mass of water = rho × volume = 8.34lbm/gal × 600 gallons = 5004lbm = 5004×0.45359kg = 2269.8kg

Total mass of hot tub and water = 320kg + 2269.8kg = 2589.8kg

Local acceleration due to gravity = 32ft/s^2 = 32ft/s^2 × 1m/3.2808ft = 9.75m/s^2

Total weight of hot tub and water = 2589.8kg × 9.75m/s^2 = 25250.55N = 25250.55/4.4482 lbf = 5676.6 lbf

Answer:

Explanation:

Given

Electric Field Strength [tex]E=4.5\times 10^{3}\ V/m[/tex]

Potential Difference between Plates is given by [tex]V=12.5\ kV[/tex]

In conducting plates a Potential difference exist between two plate which accelerate the charge when put between the conducting plates

The potential difference is given by

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

where E=Electric Field strength

d=distance between Plates

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

[tex]d=\frac{12.5\times 10^3}{4.5\times 10^{3}}[/tex]

[tex]d=2.78\ m[/tex]

Answer:

(a) 20.84 ft/s

(b) 11.24 ft

(c) -68160 N

Explanation:

Parameters given:

Mass of Bus, Mb = 12000 lb

Mass of car, Mc = 2800 lb

Initial speed of bus before collision, u(b) = 18 ft/s

Initial speed of car before collision, u(c) = 33 ft/s

Coefficient of friction, μk = 0.6

(a)Combined velocity after collision.

Since the bus and car are entangled and move together after the collision, they have the same velocity after collision.

Using the law of conservation of momentum, we have:

Mb*u(b) + Mc*u(c) = Mb*v(b) + Mc*v(c)

Where v(b) = velocity of the bus,L after collision,

v(c) = velocity of the car after collision.

Since v(b) = v(c) = v,

Mb*u(b) + Mc*u(c) = (Mb + Mc)*v

=> (12000 * 18) + (2800 * 33) = (12000 + 2800) * v

=> 308400 = 14800 * v

=> v = 20.84 ft/s

Combined velocity after collision is 20.84 ft/s

(b) The force acting on the bus and car after collision is given as:

F = m*a

Where F = force,

m = combined mass of bus and car,

a = acceleration of the bus and car after collision.

We know that the only force acting on the combined mass of the bus and car is the Frictional force, Fr and it is given as:

Fr = μk*m*g

Where g = acceleration due to gravity

Hence,

Fr = ma

=> - μk*m*g = m*a

The negative sign signifies that the fictional force is acting in the opposite direction to the lotion.

=> a = - μk*g

a = - 0.6 * 32.2

a = - 19.32 ft/s^2

Using one of the equations of linear motion, we can find the distance moved by the car and bus after collision:

v*v = u*u + 2*a*s

Where s = distance moved

The final velocity, v, of the car and bus is 0 because they come to rest and the initial velocity, u, is 20.84 ft/s, the velocity of the car and bus after collision.

Hence,

0 = (20.84*20.84) + (-2*19.32*s)

=> s = -434.3056/-38.64

s = 11.24 ft

(c) Impulsive force is the force that two bodies which are colliding exert on one another. It is given mathematically as

I. F. = (momentum change)/time

Momentum change of the bus is:

Momentum change = final momentum - initial momentum

Momentum change = Mb*v(b) - Mb*u(b)

Momentum change = (12000*20.84) - (12000*18)

Momentum change = 250080 - 216000

Momentum change = - 34080 kgft/s

=> I. F. = - 34080/0.5

I. F. = - 68160 N

The Impulsive force of the bus on the car is - 68160N

The negative sign means the force is acting opposite to the motion of the car.

Answer:

a.Insulators can be used to increase the amount of current that can flow through a resistor without increasing its temperature.

Explanation:

A conductor has low resistance, while an insulator has much higher resistance. Devices called resistors control amounts of resistance into an electrical circuits. Electric charges inside an insulator are bound to the individual atoms or molecules, not being able to move inside the material.

When you turn up the resistance, the electric current flowing through the circuit is reduced

Ohm's law:

V = I * R

R = V/I

An Ohmic conductor would have a linear relationship between the current and the voltage. With non-Ohmic conductors, the relationship is not linear. Most metals are good conductors example silver, copper etc.

Answer:

a) They will hit the ground with a speed of 19.6 m/s.

b) They are at a height of 20 m.

c) It is not a safe jump.

Explanation:

Hi there!

a) The equations of height and velocity in function of time of a free falling body are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the object at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering downward as negative direction).

v = velocity of the object at time t.

Using the equation of velocity, let's find the velocity at which they will hit the ground. The pebble is dropped (initial velocity = 0) and it takes 2 s to reach the ground:

v = v0 + g · t     (v0 = 0)

v = g · t

v = -9.8 m/s² · 2.0 s

v = -19.6 m/s

They will hit the ground with a speed of 19.6 m/s.

b)Now, we have to use the equation of height:

h = h0 + v0 · t + 1/2 · g · t²

If we place the origin of the frame of reference on the ground, we have to find the initial height (h0) knowing that at t = 2.0 s, h = 0 m

0 m = h0 - 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 20 m

They are at a height of 20 m.

c)According to a NASA paper (Issues on Human Acceleration Tolerance After Long-Duration Space Flights, figure 10), if you fall with a vertical velocity greater than 17 m/s it is unlikely that you will survive. So, it is not a safe jump.  

Answer:

11995200 N/m²

Explanation:

Pressure: The is the ratio of force to the surface area in contact. The S.I unit of pressure is N/m².

Generally pressure in fluid can be expressed as

P = ρgh......................... Equation 1

Where P = maximum pressure on the submarine hull, ρ = Density of ocean, h = depth of ocean, g = acceleration due to gravity.

Given: h = 1200 m, ρ = 1020 kg/m³

Constant: g = 9.8 m/s²

Substitute into equation 1

P = 1200(1020)(9.8)

P = 11995200 N/m²

Hence the maximum pressure on the submarine hull = 11995200 N/m²

Final answer:

The maximum pressure on a submarine hull that descends 1200 m in the ocean, with ocean density of 1020 kg/m³, is calculated using the formula P = [tex]P_{o}[/tex] + ρgh. With an atmospheric pressure of 101325 Pa, the final pressure would be 12116485 Pa at that depth.

Explanation:

To calculate the maximum pressure on a submarine hull that can descend 1200 m in the ocean, we use the formula for the pressure exerted by a fluid at a certain depth. The formula is P = [tex]P_{o}[/tex] + ρgh, where P is the pressure at depth, [tex]P_{o}[/tex] is the atmospheric pressure on the surface, ρ (rho) is the density of the fluid, g is the acceleration due to gravity, and h is the depth.

Atmospheric pressure [tex]P_{o}[/tex] is approximately 101325 Pa (or 1 atm). The density of sea water is given as 1020 kg/m³ and the depth is 1200 m. Taking g as 9.8 m/s², the standard acceleration due to gravity, we can substitute the values into the formula:

P = 101325 Pa + (1020 kg/m³)×(9.8 m/s²)×(1200 m)

P = 101325 Pa + 12003360 Pa

P = 12116485 Pa

Therefore, the maximum pressure on the submarine hull at a depth of 1200 m would be 12116485 Pascal (Pa).

Answer:

0.033 A

Explanation:

Current: This can be defined as the rate of flow of electric charge in a circuit.

The S.I unit of current is Ampere (A)

From Ohm's law.

V = IR ............................ Equation 1

Where V = Potential difference, I = current, R = resistance.

Making I the subject of the equation,

I = V/R................... Equation 2

Given: V = 52.3 V, R = 1570 Ω

Substitute into equation 2

I = 52.3/1570

I = 0.033 A.

Hence the current in the resistor = 0.033 A

Answer:

C

Explanation:

Correct answer: C. Fly the approach to runway 26. The winds from the storm could suddenly gust up and you don’t want to be in a short final or even in the flare when a tailwind from a storm gusts up. No other traffic and calm winds currently means that you can land on any runway you want. Go for the safe bet and land on 26.

In this scenario, the pilot should fly a left hand approach to runway 08.

In this scenario, with a calm wind, the pilot should fly a left hand approach to runway 08.

Since the wind is calm, there is no need for the pilot to adjust the approach. Flying in a clockwise pattern to runway 08 is the normal procedure.

Choosing any other option would not be appropriate or necessary.

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Answer:

[tex]-43\hat{k}[/tex]

Explanation:

given,

[tex]\vec{A} = 4 \hat{i} + 7 \hat{j}[/tex]

[tex]\vec{B} = 5 \hat{i} - 2 \hat{j}[/tex]

vector product [tex] \vec{A} \times \vec{B} = ?[/tex]

[tex]\vec{A} \times \vec{B}[/tex] = [tex]\begin{bmatrix}i & j & k\\ 4 & 7 &0 \\ 5 & -2 & 0\end{bmatrix}[/tex]

now, expanding the vector

[tex]\vec{A} \times \vec{B}= \hat{k}(-2\times 4 - 7\times 5)[/tex]

[tex]\vec{A} \times \vec{B}= -43\hat{k}[/tex]

the vector product is equal to [tex]-43\hat{k}[/tex]

Answer:

Electric potential, [tex]V=-4.01\times 10^4\ volts[/tex]

Explanation:

Given that,

Charge 1, [tex]q_1=2.2\ \mu C[/tex]

Point charge 2, [tex]q_2=-8\ \mu C[/tex]

Distance between charges, d = 2.6 m

We need to find the electric potential midway between them. The electric potential is given by :

[tex]V=\dfrac{kq}{r}[/tex]

In this case, r = 1.3 m (midway between charges)

[tex]V=\dfrac{kq_1}{r}-\dfrac{kq_2}{r}[/tex]

[tex]V=\dfrac{k}{r}(q_1-q_2)[/tex]

[tex]V=\dfrac{9\times 10^9}{1.3}(2.2\times 10^{-6}-8\times 10^{-6})[/tex]

[tex]V=-40153.84\ volts[/tex]

or

[tex]V=-4.01\times 10^4\ volts[/tex]

So, the electric potential midway between the charges is [tex]V=-4.01\times 10^4\ volts[/tex]. Hence, this is the required solution.

Final answer:

The electric potential midway between two point charges of +2.20 μC and -8.00 μC, separated by 2.60 m, is calculated separately for each charge using the formula V = kq/r and summed up. The total electric potential at the midpoint is -4.00 × 10⁴ V.

Explanation:

To find the electric potential midway between two point charges, we need to consider the contribution from each charge separately and then sum them up.

The electric potential V due to a single point charge q at a distance r is given by the formula:

V = kq/r

where k is the Coulomb's constant (k ≈ 8.99 × 109 Nm²/C²).

In this case, we have two charges, +2.20 μC and -8.00 μC, and they are separated by 2.60 m. So the distance from the midpoint to each charge is 1.30 m (half of 2.60 m).

Calculating the potential due to the +2.20 μC charge:

V₁ = (8.99 × 109)(+2.20 × 10⁻⁶) / 1.30 = 1.53 × 10⁴ V

And for the -8.00 μC charge:

V₂ = (8.99 × 10⁹)(-8.00 × 10⁻⁶) / 1.30 = -5.53 × 10⁴ V

The total electric potential at the midpoint is the sum of V₁ and V₂:

Vtotal = V₁ + V₂ = 1.53 × 10⁴ V - 5.53 × 10⁴ V = -4.00 × 10⁴V

The electric potential midway between the two charges is -4.00 × 10⁴V.

Explanation:

Implicit techniques for the discovery of extra-solar planets are used by astronomers. Evidence from the radial velocity of a change in the rotation of the star indicates the presence of a planet. If, with reference to Earth, the inclination of the planet's orbit is viewed as an edge-on, the detection of light originating from the star throughout its planetary transit would then be a verification.

Answer:

P = 0.533 W

Explanation:

given,

Resistance of the bulb, R = 270 Ω

Potential of the battery, V=  12 V

Power output of the bulb = ?

we know,

P = I² R

also, V = IR

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

[tex]P =\dfrac{12^2}{270}[/tex]

[tex]P =\dfrac{144}{270}[/tex]

     P = 0.533 W

Hence, the Power delivered by the bulb is equal to 0.533 W.

Answer:

Explanation:

Given

mass of vehicle [tex]m=1500\ kg[/tex]

Speed of vehicle [tex]u=50\ km/hr\approx 13.89\ m/s[/tex]

Kinetic Energy Possessed by mass

[tex]K_i=\frac{1}{2}mu^2[/tex]

[tex]K_i=\frac{1}{2}\times 1500\times (13.89)^2[/tex]

[tex]K_i=144.69\ kJ[/tex]

when vehicle is slowed down to speed of [tex]v=10\ km/hr\approx 2.78\ m/s[/tex]

Kinetic Energy at this speed

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

[tex]K_f=\frac{1}{2}\times 1500\times (2.78)^2[/tex]

[tex]K_f=5.78\ kJ[/tex]

Energy Recovered [tex]=K_i-K_f[/tex]

Energy Recovered[tex]=144.69-5.78=138.9\ kJ[/tex]        

Answer:

[tex]F=0.26N[/tex]

Explanation:

Assuming that thet act like point charges, the attractive force is given by Coulomb's law:

[tex]F=\frac{kq_1q_2}{d^2}[/tex]

Where k is the Coulomb constant, [tex]q_1[/tex] and [tex]q_2[/tex] are the magnitudes of the point charges and d is the distance of separation between them. Thus, we replace the given values and get how strong is the attractive force between them:

[tex]F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(0.7*10^{-6}C)(-0.6*10^{-6}C)}{(12*10^{-2}m)^2}\\F=0.26N[/tex]

Incomplete question as many data is missing.I have assumed value of charge and electric field.The complete question is here

A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 5.00×10⁴ V/m.

What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0 degrees  downward from the horizontal?

Answer:

[tex]W_{work}=2.67*10^{-3}J[/tex]

Explanation:

Given data

Charge q=28 nC

Electric field E=5.00×10⁴ V/m.

Distance d=2.70 m

Angle α=45°

To find

Work done by electric force

Solution

[tex]W_{work}=F_{force}*D_{distance}Cos\alpha \\where\\F_{force}=q_{charge}*E_{Electric-Field}\\So\\W_{work}=qE*D*Cos\alpha \\W_{work}=(28*10^{-9}C )(5.00*10^{4}V/m )(2.70m)Cos(45)\\W_{work}=2.67*10^{-3}J[/tex]

The spring does 0.1 Joules of work when the object is released and travels back to its initial position.

We have,

The work done by the spring can be calculated using the formula for the potential energy stored in a spring:

Potential Energy (PE) = 0.5 * k * x²,

where k is the spring constant and x is the displacement from the equilibrium position.

Given:

Spring constant (k) = 20 N/m,

Displacement (x) = 0.10 m (10 cm).

Calculate the potential energy when the spring is stretched to 10 cm:

PE = 0.5 * k * x²

= 0.5 * 20 N/m * (0.10 m)²

= 0.5 * 20 * 0.01 J

= 0.1 J.

The spring stores 0.1 Joules of potential energy when stretched to 10 cm.

When the mass is released and returns to its initial position, this potential energy is converted back into kinetic energy as the mass accelerates.

Therefore,

The spring does 0.1 Joules of work when the object is released and travels back to its initial position.

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Final answer:

The spring does 0.1 Joules of work when the mass travels back to its initial position.

Explanation:

To find the work done by a spring when an object is released and travels back to its initial position, we can use the formula:

Work = (1/2) * k * x^2

Where k is the spring constant and x is the displacement from the equilibrium position. In this case, the spring constant is 20 N/m and the displacement is 10 cm (which is 0.1 m). Plugging these values into the formula, we get:

Work = (1/2) * 20 N/m * (0.1 m)^2 = 0.1 J

So, the spring does 0.1 Joules of work when the mass travels back to its initial position.

Answer:

0.044 V

Explanation:

E = Electric field = [tex]5.5\times 10^6\ V/m[/tex]

d = Thickness of membrane = 8 nm

When the electric field strength is multiplied by the membrane thickness we get the voltage

Voltage across a gap is given by

[tex]V=Ed\\\Rightarrow V=5.5\times 10^6\times 8\times 10^{-9}\\\Rightarrow V=0.044\ V[/tex]

The voltage across the membrane is 0.044 V

Final answer:

The voltage across an 8.00 nm-thick membrane with an electric field strength of 5.50 MV/m is calculated using the formula V = Ed, resulting in a voltage of 44.0 mV.

Explanation:

The voltage across a membrane can be determined by using the relationship between electric field strength (E), voltage (V), and the distance (d) the electric field spans. The electric field is uniform and the formula to use is V = Ed. In this case, the electric field (E) is given as 5.50 MV/m or 5.50 x 10⁶V/m, and the thickness of the membrane (d) is 8.00 nm or 8.00 x 10⁻⁹ m.

To find the voltage across the membrane, we simply multiply the electric field by the thickness of the membrane:

V = (5.50 x 10⁶ V/m) x (8.00 x 10⁻⁹ m)

Therefore:

V = 44 x 10⁻³ V

V = 44.0 mV

So, the voltage across an 8.00 nm-thick membrane with the given electric field strength is 44.0 mV.

Answer:

E = 7.77 N/C

Explanation:

The charge of a single electron is 1.6 x 10^{-19} C. The net charge of the object is therefore the multiplication of the number of excess electrons and the charge of a single electron:

[tex]Q = (2.6\times 10^4) \times 1.6\times 10^{-19} = 4.16 \times 10^{-15}~C[/tex]

The electric field can be found by the following formula

[tex]E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}[/tex]

where 'r' can be calculated as

[tex]r = \sqrt{(2\times 10^{-3})^2 + (1\times 10^{-3})^2} = 0.0022~m\\r^2 = 4.84\times 10^{-6}[/tex]

Finally, the electric field at the position (2.00 mm, 1.00 mm) is

[tex]E = \frac{1}{4\pi(8.8\times 10^{-12})}\frac{4.16\times 10^{-15}}{4.84\times 10^{-6}} = 7.77~N/C[/tex]

The magnitude of the electric field it produces at the position is 7.5 N/C.

The given parameters:

The position of the charged object is calculated as follows;

[tex]r^2 = (2.0 \times 10^{-3})^2 + (1.0 \times 10^{-3})^2\\\\r^2 = 5\times 10^{-6} \ m^2[/tex]

The charge of the electron is calculated as follows;

[tex]Q = nq\\\\Q = 2.6 \times 10^4 \times 1.6\times 10^{-19}\\\\Q =4.16 \times 10^{-15} \ C[/tex]

The magnitude of the electric field it produces at the position is calculated as follows;

[tex]E = \frac{F}{Q}= \frac{kQ}{r^2} = \frac{9\times 10^9 \times 4.16 \times 10^{-15}}{5\times 10^{-6}} \\\\E = 7.5 \ N/C[/tex]

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Answer:

1/2 Hz

Explanation:

A simple harmonic motion has an equation in the form of

[tex]x(t) = Acos(\omega t - \phi)[/tex]

where A is the amplitude, [tex]\omega = 2\pi f[/tex] is the angular frequency and [tex]\phi[/tex] is the initial phase.

Since our body has an equation of  x = 5cos(π t + π/3) we can equate [tex]\omega = \pi[/tex] and solve for frequency f

[tex]2\pi f = \pi[/tex]

f = 1/2 Hz

0.5Hz

The general equation of the displacement, x, of a body undergoing simple harmonic motion at a given point in time (t) is given by;

x = A cos (ωt ± ∅)  --------------------------(i)

where;

A = amplitude of the wave

ω = angular velocity of the wave

∅ = phase constant of the wave

From the question;

x = 5cos(π t + π/3)      -----------------------------(ii)

Comparing equations (i) and (ii), the following deductions among others can be made;

A = 5cm

ω = π

But the angular velocity (ω) of the wave is related to its frequency (f) as follows;

ω = 2 π f        --------------------(iii)

Substitute the value of ω = π  into equation (iii) as follows;

π = 2 π f

Divide through by π;

1 = 2f

Solve for f;

f = 1/2

f = 0.5

Frequency (f) is measured in Hz. Therefore, the frequency of the oscillation is 0.5Hz

Answer:

T= 89.25 K

Explanation:

Given that

V= 590 mph

We know that

1 mph  = 0.44 m/s

That is why ,V= 263.75 m/s

We know that speed of the gas molecule is given as

[tex]V=\sqrt{\dfrac{3RT}{M}}[/tex]

R= 8.314 J/mol.k

M= 32 g/mol = 0.032 kg/mol

T=Temperature in Kelvin unit

[tex]V^2=\dfrac{3RT}{M}[/tex]

[tex]T=\dfrac{V^2\times M}{3R}[/tex]

[tex]T=\dfrac{263.76^2\times 0.032}{3\times 8.314}\ K[/tex]

T= 89.25 K

Therefore the average temperature ,T = 89.25 K

Answer:

temperature does the average speed of an oxygen molecule equal that of an airplane moving at 590 mph  = 89.24 K

Explanation:

Average speed of oxygen molecule is given by

[tex]v= \sqrt{\frac{3RT}{M} }[/tex]

[tex]590\times0.44704 = 263.75[/tex] m/s

R= 8.314 J/mol K = universal gas constant

M= molecular weight of oxygen = 32 g/mol =0.032 Kg/mol

now plugging these values to find T we get

[tex]263.75=\sqrt{\frac{3(8.314)(T)}{0.032}}[/tex]

solving the above equation we get

T= 89.24 K