A horizontal sheet of negative charge has a uniform electric field E = 3000N/C. Calculate the electric potential at a point 0.7m above the surface.

Answer:

T = 450 C

Explanation:

given,

pressure of rigid tank containing water = 200 kPa

water start condensing at 100 °C

vapor saturation pressure at 100 °C = 101421 Pa (from saturation table)

specific volume  = 1.6718 m³/kg

From super heated steam table

steam table is two set of table which is of energy transfer properties of water and steam.

at v = 1.6718 m³/kg and pressure = 200 kPa

temperature is equal to T = 450 C

Answer:

Car will take 5 sec to reach from 0 m/sec to 25 m/sec

Explanation:

We have given initial velocity u = 0 m/sec

And final velocity v = 25 m/sec

Acceleration [tex]a=5m/Sec^2[/tex]

From first equation of motion we know that [tex]v=u+at[/tex], here v is final velocity, u is initial velocity, a is acceleration and t is time

So [tex]25=0+5\times t[/tex]

[tex]t=4 sec[/tex]

So car will take 5 sec to reach from 0 m/sec to 25 m/sec

Answer:

[tex]V=\frac{4}{3}\pi R^{3}[/tex]

Explanation:

The volume of a body is defined as the capacity of the object. The amount of matter that object contains is called its volume.

All the three dimensional objects have volume.

The SI unit of volume is m^3. The volume of liquids is measured by teh unit litre or milli litre.

The volume of the sphere is given by

[tex]V=\frac{4}{3}\pi R^{3}[/tex]

where, R is the radius of the sphere.

Answer:

3.67 km

Explanation:

Joe distance towards coffee shop is,

[tex]OB=0.40 km[/tex]

And the Max distance towards bookstore is,

[tex]OA=3.65 km[/tex]

Now the distance between the Joy and Max will be,

By applying pythagorus theorem,

[tex]AB=\sqrt{OB^{2}+OA^{2}}[/tex]

Substitute 0.40 km for OB and 3.65 km for OA in the above equation.

[tex]AB=\sqrt{0.40^{2}+3.65^{2}}\\AB=\sqrt{13.4825} \\AB=3.67 km[/tex]

Therefore the distance between there destination is 3.67 km.

Answer:

Frequency will be 4.30 kHz

Explanation:

We have inductance [tex]L=37\mu H=37\times 10^{-6}H[/tex]

And capacitance [tex]C=37\mu H=37\times 10^{-6}F[/tex]

Inductive reactance is given by [tex]X_L=\omega L[/tex]

And capacitive reactance [tex]X_C=\frac{1}{\omega C}[/tex]

As in question it is given that inductive reactance and capacitive reactance is same

So [tex]X_L=X_C[/tex]

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

[tex]\omega ^2=\sqrt{\frac{1}{LC}}=\frac{1}{\sqrt{37\times 10^{-6}\times 37\times 10^{-6}}}=27027.027rad/sec[/tex]

We know that angular frequency [tex]\omega =2\pi f[/tex]

[tex]2\times 3.14\times f=27027.027[/tex]

f = 4303.66 Hz =4.30 kHz

Final answer:

To achieve an overall magnification of 200 with the given measurements, the eyepiece lens would need to have a focal length of approximately 5.68 cm.

Explanation:

To calculate the eyepiece focal length that will give an overall angular magnification of 200 in a microscope, we can use the formula of magnification for a compound microscope. Given that the distance between the objective and eyepiece lenses is 19 cm and the objective lens has a focal length of 5.5 mm, we can set up the equation as follows:

M = (D/fo) × (25/fe)

Where:

In the given problem, M = 200 and fo = 5.5 mm (which we should convert to centimeters to keep consistent units, thus fo = 0.55 cm). We are solving for fe.

Using the formula, we substitute the values:

200 = (25/0.55) × (25/fe)

Solving for fe, we find:

fe = (25/0.55) × (25/200)

fe = (45.45) × (0.125)

fe = 5.68 cm

Therefore, the eyepiece lens would need to have a focal length of approximately 5.68 cm to achieve an overall magnification of 200.

Answer:

This is a conceptual problem so I will try my best to explain the impossible scenario. First of all the two dust particles ara virtually exempt from any external forces and at rest with respect to each other. This could theoretically happen even if it's difficult for that to happen. The problem is that each of the particles have an electric charge which are equal in magnitude and sign. Thus each particle should feel the presence of the other via a force. The forces felt by the particles are equal and opposite facing away from each other so both charges have a net acceleration according to Newton's second law because of the presence of a force in each particle:

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

Having seen Newton's second law it should be clear that the particles are actually moving away from each other and will not remain at rest with respect to each other. This is in contradiction with the last statement in the problem.

The situation is impossible because the gravitational force between equally charged particles is much weaker than the electric force, making it impossible for the two forces to have the same magnitude, resulting in non-zero net force and movement.

The scenario described by the student is impossible because the gravitational force and electric force between two particles with identical mass and charge are not equal in magnitude. According to physics, specifically Coulomb's Law and Newton's Law of Universal Gravitation, the gravitational force is significantly weaker than electric force when the particles have the same magnitude of charge as that of their mass in standard units.

The electric force (Coulomb's Law) is known to be much stronger than gravitational force. For example, the electric repulsion between two electrons is approximately 1042 times stronger than their gravitational attraction. Therefore, if two dust particles are floating in empty space and are at rest with respect to each other, with identical charges and mass, the electric forces would cause them to repel each other with much greater force than the gravitational forces would attract them. Hence, they cannot remain at a constant distance with zero net force on each other.

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy  5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant [tex]\gamma =1.4[/tex]

specific heat is given as [tex]=\frac{\gamma R}{\gamma -1}[/tex]

gas constant =287 J⋅kg−1⋅K−1

[tex]Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}[/tex]

specific heat at constant volume

[tex]Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}[/tex]

change in internal energy [tex]= Cv(T_2 -T_1)[/tex]

                            [tex]  \Delta U = 717.5 (800-295)  = 3.62*10^5 J kg^{-1}[/tex]

change in enthalapy [tex]\Delta H = Cp(T_2 -T_1)[/tex]

                                 [tex] \Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}[/tex]

change in entropy

[tex]\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})[/tex]

[tex]\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})[/tex]

[tex]\Delta S = 382.79 J kg^{-1} K^{-1}[/tex]

Final answer:

To calculate the distance the firefighters should position their water cannon from the building, analyze the vertical motion of the water stream. Therefore, the firefighters should position their water cannon approximately 11.40 meters from the building.

Explanation:

To find the horizontal distance from the building where the firefighters should position their water cannon, we can use the projectile motion equations.

Given:

[tex]- Initial \ velocity \ of \ the \water \ cannon, \( v_0 = 25.0 \, \text{m/s} \)\\- Launch angle, \( \theta = 52.0^\circ \)\\- Vertical displacement, \( y = 12.0 \, \text{m} \)\\- Vertical acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \)\\[/tex]

First, we'll find the time it takes for the water to reach a height of 12.0 m. We'll use the kinematic equation for vertical motion:

[tex]\[ y = v_{0y}t + \frac{1}{2}gt^2 \][/tex]

Where:

[tex]- \( v_{0y} \) is the initial vertical component of the velocity\\- \( t \) is the time[/tex]

Since the initial velocity is at an angle, we need to find its vertical component:

[tex]\[ v_{0y} = v_0 \sin(\theta) \][/tex]

Substitute the given values:

[tex]\[ v_{0y} = 25.0 \, \text{m/s} \times \sin(52.0^\circ) \]\[ v_{0y} \approx 25.0 \, \text{m/s} \times 0.788 \]\[ v_{0y} \approx 19.7 \, \text{m/s} \][/tex]

Now, let's use the equation for vertical motion to solve for \( t \):

[tex]\[ 12.0 \, \text{m} = (19.7 \, \text{m/s})t - \frac{1}{2}(9.81 \, \text{m/s}^2)t^2 \][/tex]

This is a quadratic equation, we can solve it to find \( t \). Let's denote [tex]\( a = -4.905 \, \text{m/s}^2 \) and \( b = 19.7 \, \text{m/s} \):[/tex]

[tex]\[ -4.905t^2 + 19.7t - 12.0 = 0 \][/tex]

Now, we can use the quadratic formula to solve for \( t \):[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

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

Let's calculate \( t \).

We have the quadratic equation:

[tex]\[ -4.905t^2 + 19.7t - 12.0 = 0 \][/tex]

Using the quadratic formula:

[tex]\[ t = \frac{-19.7 \pm \sqrt{(19.7)^2 - 4(-4.905)(-12.0)}}{2(-4.905)} \]\[ t = \frac{-19.7 \pm \sqrt{389.21 - 235.2}}{-9.81} \]\[ t = \frac{-19.7 \pm \sqrt{154.01}}{-9.81} \]\[ t = \frac{-19.7 \pm 12.41}{-9.81} \][/tex]

Now, we have two possible values for \( t \):

[tex]1. \( t_1 = \frac{-19.7 + 12.41}{-9.81} \)\\2. \( t_2 = \frac{-19.7 - 12.41}{-9.81} \)[/tex]

Calculating each value:

[tex]1. \( t_1 = \frac{-7.29}{-9.81} \) \( t_1 \approx 0.742 \) seconds\\2. \( t_2 = \frac{-32.11}{-9.81} \) \( t_2 \approx 3.27 \) seconds[/tex]

Since the time \( t \) cannot be negative, we'll take [tex]\( t = t_1 \approx 0.742 \) seconds[/tex].

Now, to find the horizontal distance \( x \) from the building, we can use the equation for horizontal motion:

[tex]\[ x = v_{0x} \cdot t \][/tex]

where \( v_{0x} \) is the initial horizontal component of the velocity, given by:

[tex]\[ v_{0x} = v_0 \cdot \cos(\theta) \][/tex]

Substituting the given values:

[tex]\[ v_{0x} = 25.0 \, \text{m/s} \cdot \cos(52.0^\circ) \]\[ v_{0x} \approx 25.0 \, \text{m/s} \cdot 0.615 \]\[ v_{0x} \approx 15.375 \, \text{m/s} \][/tex]

Now, we can find \( x \):

[tex]\[ x = 15.375 \, \text{m/s} \cdot 0.742 \, \text{s} \]\[ x \approx 11.40 \, \text{m} \][/tex]

Therefore, the firefighters should position their water cannon approximately 11.40 meters from the building.

Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, [tex]d=24.44\times 2=48.88\ m[/tex]

(a) Acceleration, [tex]a=-4\ m/s^2[/tex]

We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

[tex]s=\dfrac{v^2-u^2}{2a}[/tex]

[tex]s=\dfrac{-(24.44)^2}{2\times -4}[/tex]

s = 74.66 meters

s = 74.66 + 48.88 = 123.54 meters

(b) Acceleration, [tex]a=-8\ m/s^2[/tex]

[tex]s=\dfrac{v^2-u^2}{2a}[/tex]

[tex]s=\dfrac{-(24.44)^2}{2\times -8}[/tex]

s = 37.33 meters

s = 37.33 + 48.88 = 86.21 meters

Hence, this is the required solution.

Answer:

(e) The particles move apart with a velocity that increases for a while and then becomes constant.

Explanation:

Each particle feels a repulsive (because they have same sign of charge) electric force from the each other:

[tex]F=\frac{kq_{1}q{2}}{d^{2}}[/tex]

and

[tex]F=ma\\[/tex]

So each particle feels a repulsive force proportional to the quadratic inverse of the distance.that means that the charges begin to move away, and the further away they are from each other, the force (and therefore the acceleration) decreases, at a rate inversely proportional to the square of the distance. Theoretically this acceleration will never be zero, but in practice it will at some point reach a value very close to zero. Then the speed will grow for a while and when the acceleration has reached almost zero, the speed will practically remain constant.

Answer:

Explanation:

The rock will decelerate at the rate of g or 9.8 m s⁻²while on its way up because of gravitational attraction towards the earth.  . In other words , it will accelerate with the magnitude - 9.8 m s⁻² while on its way up.

On its way down , it will accelerate due to gravitational attraction. In this case acceleration will be positive and equal to 9.8 m s⁻².

At the highest point in its trajectory , rock  becomes stationary or comes to  rest momentarily. So at the peak position , velocity is zero.

At the highest point , as the gravitational force continues to act on the body undiminished, its acceleration will remain the same ie 9.8 m s⁻² at the highest point as well. It will act in the downward direction.

Answer:

[tex]t=4.06s[/tex]

Explanation:

From the exercise we know

[tex]y_{o}=1.2m\\v_{o}=20m/s\\y=1.6m\\g=-9.8m/s^2[/tex]

To find how long does it takes the marker to reach the highest point we need to use the equation of position:

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

[tex]1.6m=1.2m+(20m/s)t-\frac{1}{2}(9.8m/s^2)t^2[/tex]

[tex]0=-0.4+20t-4.9t^2[/tex]

Now, we need to use the quadratic formula:

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

[tex]a=-4.9\\b=20\\c=-0.4[/tex]

Solving for t

[tex]t=0.020s[/tex] or [tex]t=4.06s[/tex]

So, the answer is t=4.06s because the other option is almost 0 and doesn't make any sense for the motion of the marker

Answer:

2.551 m/s

Explanation:

Given:

Boulder is dropped from the cliff

thus,

Initial speed of the boulder, u = 0 m/s

Final speed to be obtained, v = 90 km/h =[tex]90\times\frac{5}{18}[/tex] =25 m/s

also, in the case of free fall the acceleration of the boulder will be equal to the acceleration due to the gravity i.e g = 9.8 m/s²

Now, from the Newton's equation of motion

v = u + at

where, a is the acceleration = g = 9.8 m/s²

t is the time

on substituting the respective values, we get

25 = 0 + 9.8 × t

or

t = [tex]\frac{\textup{25}}{\textup{9.8}}[/tex]

or

t = 2.551 m/s

The statement that heuristics allow a priori theoretical concepts is true. Heuristics are mental shortcuts that can definitely be informed by a priori theoretical concepts, while empiricism emphasizes the role of experience in the formation of ideas.

Regarding the question on the statements about heuristics and empiricism, the best choice would be 'A'. Heuristics allow a priori theoretical concepts'. Heuristics are mental shortcuts or rules of thumb that simplify decisions, particularly under conditions of uncertainty. They are not necessarily based on theoretical concepts but rather on practical, personal experience or common sense. However, they can certainly be informed by a priori theoretical concepts in the sense that our theoretical understanding can shape the rules of thumb we use.

On the other hand, empiricism is a philosophical system that emphasizes the role of experience, especially sensory perception, in the formation of ideas, while discounting a priori reasoning, intuitiveness, and innate ideas. Hence, options B, C, and D are incorrect. Option E is also incorrect as heuristics can definitely be based on a trial and error approach.

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Answer:

213 ft/s

Explanation:

Given:

Δx = 330 ft

v₀ = 0 ft/s

t = 3.1 s

Find: Δv

x = x₀ + ½ (v + v₀)t

330 ft = ½ (v + 0 ft/s) (3.1 s)

v = 213 ft/s

The change in velocity is:

Δv = 213 ft/s − 0 ft/s

Δv = 213 ft/s

Answer:

55.66 m

Explanation:

While falling by 50 m , initial velocity u = 0

final velocity = v , height h = 50 , acceleration g = 9.8

v² = u² + 2gh

= 0 + 2 x 9.8 x 50

v = 31.3 m /s

After that deceleration comes into effect

In this case final velocity v = 17 m/s

initial velocity u = 31.3 m/s

acceleration a = - 61 m/s²

distance traveled h = ?

v² = u² + 2gh

(17)² = (31.3)² - 2x 61xh

h = 690.69 / 2 x 61

= 5.66 m

Total height during which he was in air

= 50 + 5.66

= 55.66 m

Explanation:

Given that,

Wavelength of the laser light, [tex]\lambda=1062\ nm=1062\times 10^{-9}\ m[/tex]

The laser pulse lasts for, [tex]t=34\ ps=34\times 10^{-12}\ s[/tex]

(a) Let d is the distance covered by laser in the given by, [tex]d=c\times t[/tex]

[tex]d=3\times 10^8\times 34\times 10^{-12}[/tex]

d = 0.0102 meters

Let n is the number of wavelengths found within the laser pulse. So,

[tex]n=\dfrac{d}{\lambda}[/tex]

[tex]n=\dfrac{0.0102}{1062\times 10^{-9}}[/tex]

n = 9604.51

(b) Let t is the time need to be fit only in one wavelength. So,

[tex]t=\dfrac{\lambda}{c}[/tex]

[tex]t=\dfrac{1062\times 10^{-9}}{3\times 10^8}[/tex]

[tex]t=3.54\times 10^{-15}\ s[/tex]

Hence, this is the required solution.

Answer:

a) Linear equation

Explanation:

Definition of acceleration

[tex]a=\frac{dv}{dt}\\[/tex]

if a=constant and we integrate the last equation

[tex]v(t)=v_{o}+a*t[/tex]

So the relation between the time and the velocity is linear. If we plot the velocity in function of time, the plot is a line, and the acceleration is the slope of this line.

Answer:

[tex]a)9.5\frac{ft}{s^2}\\ b) 12.66\frac{ft}{s^2}[/tex]

Explanation:

A body has acceleration when there is a change in the velocity vector, either in magnitude or direction. In this case we only have a change in magnitude. The average acceleration represents the speed variation that takes place in a given time interval.

a)

[tex]a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{38\frac{ft}{s}-0}{4 s- 0}=9.5\frac{ft}{s^2}\\[/tex]

b)

[tex]a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{76\frac{ft}{s}-38\frac{ft}{s}}{7 s- 4s}\\a_{avg}=\frac{38\frac{ft}{s}}{3s}=12.66\frac{ft}{s^2}[/tex]

Final answer:

The car's average acceleration for part (a) is 9.5 [tex]ft/s^2[/tex], and for part (b) it is 12.67 [tex]ft/s^2[/tex], calculated using the change in velocity over the time period.

Explanation:

Car's Average Acceleration

To find the car's average acceleration in part (a), we use the formula for average acceleration, which is the change in velocity (deltaV) divided by the time (t).

Average Acceleration (a) = (deltaV) / (t)For part (a), the initial velocity (V0) is 0 ft/s, the final velocity (V) is 38 ft/s, and the time (t) is 4.0 s.

Therefore, a = (38 ft/s - 0 ft/s) / 4.0 s = 9.5 [tex]ft/s^2[/tex]

For part (b), the final velocity (V) is now 76 ft/s and the time (t) is 3.0 s (the additional time).

Average Acceleration (a) = (76 ft/s - 38 ft/s) / 3.0 s = 12.67 [tex]ft/s^2[/tex]

Answer:

The weight of Earth's atmosphere exert is [tex]516.6\times10^{17}\ N[/tex]

Explanation:

Given that,

Average pressure [tex]P=1.01\times10^{5}\ Pa[/tex]

Radius of earth [tex]R_{E}=6.38\times10^{6}\ m[/tex]

Pressure :

Pressure is equal to the force upon area.

We need to calculate the weight of earth's atmosphere

Using formula of pressure

[tex]P=\dfrac{F}{A}[/tex]  

[tex]F=PA[/tex]

[tex]F=P\times 4\pi\times R_{E}^2[/tex]

Where, P = pressure

A = area

Put the value into the formula

[tex]F=1.01\times10^{5}\times4\times\pi\times(6.38\times10^{6})^2[/tex]

[tex]F=516.6\times10^{17}\ N[/tex]

Hence, The weight of Earth's atmosphere exert is [tex]516.6\times10^{17}\ N[/tex]

Answer:

5.164 x 10^19 N

Explanation:

P = 1.01 x 10^5 Pa

R = 6.38 x 10^6 m

Area = 4 π R²

[tex]A= 4\times 3.14\times6.38\times10^{6}\times6.38\times10^{6}[/tex]

A = 5.112 x 10^14 m^2

Pressure is defined as the force exerted per unit area.

The formula for the pressure is

P = F / A

Where, F is the force, A be the area

here force is the weight of atmosphere.

F = P x A  

F = 1.01 x 10^5 x 5.112 x 10^14

F = 5.164 x 10^19 N

The minimum coefficient of kinetic friction required to stop the automobile before it hits the barrier is approximately 0.2041.

To stop the automobile before it hits the barrier, the minimum coefficient of kinetic friction needed can be found using the equation F = μk Mcg. The force of friction is equal to the product of the coefficient of kinetic friction (μk), the mass of the car (Mc), and the acceleration due to gravity (g). Since the car is stopping, its acceleration is negative, equal in magnitude to the square of its velocity divided by twice the stopping distance.

Using the given information, we can solve for the coefficient of kinetic friction:

F = μk Mcg

μk Mcg = (Mc)((-v^2)/(2d))

Substituting the values, we get:

μk = ((-v^2)/(2d))/g

where v = 20 m/s and d = 50 m.

Substituting the values in the equation, we get:

μk = ((-20^2)/(2*50))/9.8 ≈ -0.2041 ≈ 0.2041

Therefore, the minimum coefficient of kinetic friction required to stop the automobile before it hits the barrier is approximately 0.2041.

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Answer:7 cm

Explanation:

Given

one sphere has a radius(r) of 4.85 cm

Also another sphere has a mass 3 times the previous one

since the density remains same therefore

second sphere has a volume 3 times of former.

Let R be the radius of bigger Sphere

[tex]\frac{4\pi R^3}{3}=3\times \frac{4\pi r^3}{3}[/tex]

[tex]R=\left ( 3\right )^\frac{1}{3}r=1.44\times 4.85=6.99 \approx 7 cm[/tex]

Answer:

Voltage, V = 0.0524 volts

Explanation:

Thickness of the membrane, [tex]d=7.95\ nm=7.95\times 10^{-9}\ m[/tex]

Electric field strength, [tex]E=6.6\ MV/m=6.6\times 10^6\ V/m[/tex]

We need to find the voltage across it. The relationship between the voltage, electric field and the distance is given by :

[tex]V=E\times d[/tex]

[tex]V=6.6\times 10^6\ V/m\times 7.95\times 10^{-9}[/tex]

V = 0.0524 volts

So, the voltage across the thick membrane is 0.0524 volts. Hence, this is the required solution.

The voltage across a 7.95 nm thick membrane with an electric field strength of 6.60 MV/m, is found to be 52.47 mV.

To find the voltage across a membrane, we can use the relationship between electric field strength (E) and voltage (V) given by the formula:

V = E * d

where:

V is the voltage

E is the electric field strength

d is the thickness of the membrane

In this case, the electric field strength (E) is given as 6.60 MV/m and the thickness of the membrane (d) is 7.95 nm.

First, convert the thickness from nanometers to meters:

d = 7.95 nm = 7.95 * 10⁻⁹ m

Now use the formula to calculate the voltage:

V = 6.60 * 10⁶ V/m * 7.95 * 10⁻⁹ m

V = 5.247 * 10⁻² V = 52.47 mV

Therefore, the voltage across the 7.95 nm thick membrane is 52.47 mV.

The distances traveled by the child on the toboggan are [tex]0.9, 3.6, and\ 8.1[/tex] respectively.

The distance traveled by the child on the toboggan can be calculated using the equations of motion for uniformly accelerated motion. The equation that relates distance (d), initial velocity (u), acceleration (a), and time (t) is given by:

[tex]d = ut + (1/2)at^2[/tex]

In this case, the child starts at rest [tex](u = 0)[/tex] and has an acceleration of a[tex]= 1.8 m/s^2[/tex].

Calculate the distances for the given times:

(a) For t = 1.0 s:

[tex]d = (0) \times (1.0 ) + (1/2) \times (1.8 ) \times (1.0 )^2\\d = 0 + 0.9 \\d = 0.9\ m[/tex]

(b) For t = 2.0 s:

[tex]d = (0) \times (2.0 ) + (1/2) \times (1.8 ) \times (2.0 )^2\\d = 0 + 3.6\\d = 3.6\ m[/tex]

(c) For t = 3.0 s:

[tex]d = (0) \times (3.0 ) + (1/2) \times (1.8 ) \times (3.0)^2\\d = 0 + 8.1\\d = 8.1\ m[/tex]

So, the distances traveled by the child on the toboggan are [tex]0.9, 3.6, and\ 8.1[/tex] respectively.

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Final answer:

Using the kinematic equation for motion with constant acceleration, the distance traveled by the child on the toboggan is 0.9 m after 1.0 s, 3.6 m after 2.0 s, and 8.1 m after 3.0 s.

Explanation:

To solve for the distance traveled by the child on the toboggan in each case, we'll use the kinematic equation for motion with constant acceleration: d = v_i * t + \frac{1}{2} * a * t^2, where d is the distance traveled, v_i is the initial velocity, t is the time, and a is the acceleration.

Since the child starts at rest, the initial velocity v_i is 0 m/s. The acceleration a is given as 1.8 m/s^2.

Answer:

Explanation:

We know that the differential work made by the gas  its defined as:

[tex]dW =  P \ dv[/tex]

We can solve this by integration:

[tex]\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv[/tex]

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

[tex]P \ V = \ n \ R \ T[/tex]

[tex]P = \frac{\ n \ R \ T}{V}[/tex]

This give us

[tex] \int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv [/tex]

As n, R and T are constants

[tex] \int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv [/tex]

[tex] \Delta W= \ n \ R \ T  \left [ ln (V) \right ]^{v_2}_{v_1} [/tex]

[tex] \Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 ) [/tex]

[tex] \Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 ) [/tex]

[tex] \Delta W = \ n \ R \ T  ln (\frac{v_2}{v_1})[/tex]

But the volume is:

[tex]V = \frac{\ n \ R \ T}{P}[/tex]

[tex] \Delta W = \ n \ R \ T  ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )[/tex]

[tex] \Delta W = \ n \ R \ T  ln(\frac{P_1}{P_2})[/tex]

Now, lets use the value from the problem.

The temperature its:

[tex]T = 27 \° C = 300.15 \ K[/tex]

The ideal gas constant:

[tex]R = 8.314 \frac{m^3 \ Pa}{K \ mol}[/tex]

So:

[tex] \Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K  ln (\frac{20 atm}{1 atm}) [/tex]

[tex] \Delta W = 7475.69 joules[/tex]

We know that, for an ideal gas, the energy is:

[tex]E= c_v n R T[/tex]

where [tex]c_v[/tex] its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

[tex]\Delta E = \Delta Q - \Delta W[/tex]

where [tex]\Delta W[/tex] is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

[tex]\Delta E = 0[/tex]

[tex]\Delta Q = \Delta W[/tex]

The car lands approximately 45.67 meters away from the base of the cliff, and its impact speed is 36.1 m/s.

The situation described involves typing a projectile motion. Ignore any effect of air resistance for simplicity's sake. The stunt man's car exits the ramp at an angle, and continues its journey downwards under the force of gravity.

Firstly, we need to break down the initial velocity of the car into its horizontal and vertical components. Horizontal component (Ux) is governed by the formula Ux = U cos θ which results to 20 cos 20° = 18.79 m/s. Vertical component (Uy) is calculated by Uy = U sin θ which gives us 20 sin 20° = 6.84 m/s.

Following this, we need to calculate the total time the car is in the air (t), governed by the equation h = Uy * t - (1/2) * g * t², where g represents gravity (9.8 m/s²) and h is the height of the cliff (28m). Solving for t gives us approximately 2.43 seconds.

Finally, we find the horizontal distance (X) the car covers during the time it's in the air: X = Ux * t  = 18.79 m/s * 2.43 s ≈ 45.67 m.

To calculate the impact speed, we take the square root of the sum of the squares of the final horizontal and vertical velocities. The final vertical velocity (Vy) = Uy + g * t = 6.84 m/s + 9.8 m/s² * 2.43 s = 30.95 m/s. The impact speed then equals √((18.79 m/s)² + (30.95 m/s)²) = 36.1 m/s.

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Answer:

Answered

Explanation:

impacts of the physics of matter on aviation operations.

Thrust, drag and lift.

Thrust:

It is the force developed by airplane engines that cause it to pull forward. With the help of huge propellers of course attached to the wings.

Drag:

It is the resistive force on the plane caused by the friction between air and plane. Its magnitude depends upon surface area, speed and viscosity of the air.

Lift:

The drag produced is utilized such that one of its component acts opposite to the weight. This causes the plane to take flight and stay in air. Lift can be deduced using Bernoulli's principle.

Bernoulli's principle is equivalent to law of conservation of energy. Meaning it tries to keep the energy of a system constant. In doing so, it produces low pressure zone above the wing. Which causes a net upward force, lift.

Answer:

The wavelength of the light is [tex]7200\ \AA[/tex].

Explanation:

Given that,

Distance between the slit centers d= 1.2 mm

Distance between constructive fringes [tex]\beta= 0.3\ cm[/tex]

Distance between fringe and screen D= 5 m

We need to calculate the wavelength

Using formula of width

[tex]\beta=\dfrac{D\lambda}{d}[/tex]

Put the value into the formula

[tex]0.3\times10^{-2}=\dfrac{5\times\lambda}{1.2\times10^{-3}}[/tex]

[tex]\lambda=\dfrac{0.3\times10^{-2}\times1.2\times10^{-3}}{5}[/tex]

[tex]\lambda=7.2\times10^{-7}\ m[/tex]

[tex]\lambda=7200\ \AA[/tex]

Hence, The wavelength of the light is [tex]7200\ \AA[/tex].

Answer:

(A) 476 kPa

Explanation:

If the volume remains constant, the ideal gas law says:

P/T=constant

so: P1/T1=P2/T2

P2=P1*T2/T1=460*298/288=476KPa

Answer:

A. The final pressure at the tire would be 476 kPa

Explanation:

Since the volume is constant the ideal gas equation would be used to obtain the final pressure at the tire.

Given

the initial temperature [tex]T_{1}[/tex] = 288 K

the initial pressure [tex]P_{1}[/tex] = 460 k Pa

the final temperature [tex]T_{2}[/tex] = 298 K

the final pressure [tex]P_{2}[/tex] = ?

Using the ideal gas equation;

PV = nRT

[tex]P_{1}[/tex] / [tex]T_{1}[/tex]  = [tex]P_{2}[/tex] / [tex]T_{2}[/tex]

Making [tex]P_{2}[/tex] the subject formula

[tex]P_{2}[/tex]  = ([tex]P_{1}[/tex] x [tex]T_{2}[/tex] ) / [tex]T_{1}[/tex]

[tex]P_{2}[/tex]  = (460 x 298) / 288

[tex]P_{2}[/tex]  = 475.972 kPa

[tex]P_{2}[/tex]  ≈ 476 kPa

Therefore the final pressure at the tire would be 476 kPa