Two infinite planes of charge lie parallel to each other and to the yz plane. One is at x = -5 m and has a surface charge density of σ = -2.6 µC/m^2. The other is at x = 3 m and has a surface charge density of σ = 5.8 µC/m^2. Find the electric field for the following locations. (a) x < -5 m
(b) -5 m < x < 3 m
(c) x > 3 m

Answer:

Explanation:

Let the charge on bead A be q nC  and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} = [tex]\frac{9\times10^9\times q\times(28-q)\times10^{-18}}{(5\times10^{-2})^2}[/tex]

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

The charges [tex]q_A[/tex] = 21.907nC and [tex]q_B[/tex] = 6.093nC are the charges on the beads.

We have to use Coulomb's Law and the principle of charge conservation.

1. Conservation of Charge:

  The total charge before and after they are touched must be the same. Initially, bead A has a charge of 28 nC, and bead B is neutral. After touching, let:

[tex]\[ q_A + q_B = 28 \, \text{nC} \][/tex]

2. Coulomb's Law:

  The force between two charges is given by Coulomb's Law:

[tex]\[ F = k \frac{q_A q_B}{r^2} \][/tex]

Let's set up the equation:

[tex]\[4.8 \times 10^{-4} = 8.99 \times 10^9 \frac{q_A q_B}{(0.05)^2}\][/tex]

Rearrange to solve for [tex]\( q_A q_B \)[/tex]:

[tex]\[q_A q_B = \frac{4.8 \times 10^{-4} \times (0.05)^2}{8.99 \times 10^9}\][/tex]

[tex]\[q_A q_B = \frac{4.8 \times 10^{-4} \times 0.0025}{8.99 \times 10^9}\][/tex]

[tex]\[q_A q_B = \frac{1.2 \times 10^{-6}}{8.99 \times 10^9}\][/tex]

[tex]\[q_A q_B = 1.3348 \times 10^{-16} \, \text{C}^2\][/tex]

Now, we have two equations:

1. [tex]\( q_A + q_B = 28 \times 10^{-9} \, \text{C} \)[/tex]

2. [tex]\( q_A q_B = 1.3348 \times 10^{-16} \, \text{C}^2 \)[/tex]

To solve these, we can substitute [tex]\( q_B = 28 \times 10^{-9} \, \text{C} - q_A \)[/tex] into the second equation:

[tex]\[q_A \left( 28 \times 10^{-9} - q_A \right) = 1.3348 \times 10^{-16}\][/tex]

[tex]\[28 \times 10^{-9} q_A - q_A^2 = 1.3348 \times 10^{-16}\][/tex]

[tex]\[q_A^2 - 28 \times 10^{-9} q_A + 1.3348 \times 10^{-16} = 0\][/tex]

This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where:

[tex]\[a = 1, \quad b = -28 \times 10^{-9}, \quad c = 1.3348 \times 10^{-16}\][/tex]

Solve using the quadratic formula [tex]\( q_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:

[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{(28 \times 10^{-9})^2 - 4 \times 1 \times 1.3348 \times 10^{-16}}}{2}\][/tex]

[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{7.84 \times 10^{-16} - 5.3392 \times 10^{-16}}}{2}\][/tex]

[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{2.5008 \times 10^{-16}}}{2}\][/tex]

[tex]\[q_A = \frac{28 \times 10^{-9} \pm 1.5814 \times 10^{-8}}{2}\][/tex]

This gives us two solutions for [tex]\( q_A \)[/tex]:

1. [tex]\( q_A = \frac{28 \times 10^{-9} + 1.5814 \times 10^{-8}}{2} \\q_A = \frac{4.3814 \times 10^{-8}}{2} \\q_A = 2.1907 \times 10^{-8} \, \text{C} \\q_A = 21.907 \, \text{nC} \)[/tex]

2. [tex]\( q_A = \frac{28 \times 10^{-9} - 1.5814 \times 10^{-8}}{2} \\q_A = \frac{1.2186 \times 10^{-8}}{2} \\q_A = 6.093 \times 10^{-9} \, \text{C} \\q_A = 6.093 \, \text{nC} \)[/tex]

Since bead A has the greater charge, we take:

[tex]\[q_A = 21.907 \, \text{nC}\][/tex]

And the charge on bead B:

[tex]\[q_B = 28 \, \text{nC} - 21.907 \, \text{nC} = 6.093 \, \text{nC}\][/tex]

So, the charges on the beads are:

[tex]\[q_A = 21.907 \, \text{nC}, \quad q_B = 6.093 \, \text{nC}\][/tex]

Answer:

[tex]Q=3.47\times 10^{-15}\ C[/tex]

Explanation:

Given that,

Number of extra electrons, n = 21749

We need to find the net charge on the metal ball. Let Q is the net charge.

We know that the charge on an electron is [tex]q=1.6\times 10^{-19}\ C[/tex]

To find the net charge if there are n number of extra electrons is :

Q = n × q

[tex]Q=21749\times 1.6\times 10^{-19}\ C[/tex]

[tex]Q=3.47\times 10^{-15}\ C[/tex]

So, the net charge on the metal ball is [tex]3.47\times 10^{-15}\ C[/tex]. Hence, this is the required solution.

Answer:

W = 289.70 kg

Explanation:

Given data:

Pressure in tank = 23 atm

Altitude 1000 ft

Air temperature  in tank T = 700 F

Volume of tank = 800 ft^3 = 22.654 m^3

from ideal gas equation we have

PV =n RT

Therefore number of mole inside the tank is

[tex]\frac{1}{n} = \frac{RT}{PV}[/tex]

              [tex] = \frac{8.206\times 10^{-5}} 644.261}{23\times 22.654}[/tex]

               [tex]= 1.02\times 10^{-4}[/tex]

               [tex]n = 10^4 mole[/tex]

we know that 1 mole of air weight is 28.97 g

therefore, tank air weight is [tex]W = 10^4\times 28.91 g = 289700 g[/tex]

               W = 289.70 kg

Answer:

The force on X Fx=0 N

The force on Y Fy=-2.18 N

Explanation:

We have an array of charges, we will use the coulomb's formula to solve this:

[tex]F=k*\frac{Q*Q'}{r^2}\\where:\\k=coulomb constant\\r=distance\\Q=charge[/tex]

but we first have to find the distance and the angule of the charge respect the charges on the Y axis:

[tex]r=\sqrt{(7*10^{-2}m)^2+ (3*10^{-2}m)^2} \\r=7.62cm=0.0762m[/tex]

we can notice that it is the same distance from both charges on Y axis.

we can find the angle with:

[tex]\alpha = arctg(\frac{7cm}{3cm})=66.80^o[/tex]

for the charge of 5µC [tex]\alpha =-66.80^o[/tex]

for the charge of -5µC [tex]\alpha =66.80^o[/tex]

the net force on the X axis will be:

[tex]F_{x5u}=9*10^9*\frac{5*10^{-6}*2*10^{-6}}{0.0762^2}*cos(-66.80)\\F_{x5u}=0.465N[/tex]

and

[tex]F_{x(-5u)}=9*10^9*\frac{-5*10^{-6}*2*10^{-6}}{0.0762^2}*cos(66.80)\\F_{x(-5u)}=-0.465N[/tex]

So the net force on X will be Zero.

for the force on Y we have:

[tex]F_{y5u}=9*10^9*\frac{5*10^{-6}*2*10^{-6}}{0.0762^2}*sin(-66.80)\\F_{y5u}=-1.09N[/tex]

and

[tex]F_{y(-5u)}=9*10^9*\frac{-5*10^{-6}*2*10^{-6}}{0.0762^2}*sin(66.80)\\F_{y(-5u)}=-1.09N[/tex]

Fy=[tex]F_{y5u}+F_{y(-5u)}[/tex]

So the net force on Y is Fy=-2.18N

To find the force on a charge of 2 µC on the x axis at x = 3 cm, we can use Coulomb's Law. The force is calculated to be 5.12 N.

To find the force on a charge of 2 µC on the x axis at x = 3 cm, we can use Coulomb's Law. Coulomb's Law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

The formula for Coulomb's Law is:

F = k * |q1 * q2| / r^2

Where F is the force, k is the electrostatic constant (k ≈ 8.99 * 10^9 N * m² / C²), q1 and q2 are the magnitudes of the charges, and r is the distance between them.

In this case, one charge is 5 µC and the other charge is -5 µC, with a distance of 14 cm between them (7 cm on the positive y-axis and 7 cm on the negative y-axis). The charge we're interested in is 2 µC at x = 3 cm. Plugging these values into the formula:

F = (8.99 * 10^9 N * m² / C²) * |(5 µC) * (2 µC)| / (0.14 m)²

F = 5.12 N

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Answer:

[tex]v=0.04m/s\\[/tex]

[tex]s=-28.592m\\[/tex]

Explanation:

[tex]a = 3t-4[/tex]

[tex]v(t)=\int\limits^t_0 {a(t)} \, dt =3/2*t^{2}-4t+v_0\\[/tex]

if t=3.6s and initial velocity, v0,  is -5m/s

[tex]v=0.04m/s\\[/tex]

[tex]s(t)=\int\limits^t_0 {v(t)} \, dt =1/2*t^{3}-2t^{2}+v_0*t+s_0\\[/tex]

if t=3.6s and the initial displacement, s0, is -8m:

[tex]s=-28.592m\\[/tex]

Answer:

a)a=2.63m/s^2

b)x=475.25m

Explanation:

To solve the first part of this problem we use the following equation

Vf=final speed=50m/s

Vo=initial speed=0

t=time

a=aceleration

a=(Vf-Vo)/t

a=(50-0)/19=2.63m/s^2

B )

For the second part of this problem we use the following equation

X=(Vf^2-Vo^2)/2a

X=(50^2-0^2)/(2*2.63)=475.25m

Answer:

(a) 0.9 s

(b) 7.78 m/s

Explanation:

height, h = 4 m

Horizontal distance, d = 7 m

Let it takes time t to reach the ground and u be the initial velocity of the jet.

(a) Use second equation of motion in vertical direction

[tex]s = ut + \frac{1}{2}at^{2}[/tex]

In vertical direction, uy = 0 m/s, a = g = - 9.8 m/s^2, h = - 4 m

By substituting the values, we get

[tex]-4 = 0 - \frac{1}{2}\times 9.8\times t^{2}[/tex]

t = 0.9 second

Thus, the time taken by water jet in air is 0.9 second.

(b) Use

Horizontal distance = horizontal velocity x time

d = u t

7 = u x 0.9

u = 7.78 m/s

Thus, the initial velocity of water jet is 7.78 m/s.

The duration the water is in the air is found using the formula for the motion under gravity, which depends on the vertical distance and the acceleration due to gravity. After finding the time, the initial velocity of the water is calculated using the horizontal distance and the fraction of time the water was in motion.

To determine how long the water is in the air (a), and the initial velocity of the water (b) when a water gun is fired horizontally from a hill, we can use the principles of projectile motion. The time a projectile is in the air is solely determined by its vertical motion. Since the water gun is fired horizontally, it has an initial vertical velocity of 0 m/s.

For part (a), the time (t) it takes for the water to reach the ground can be calculated using the formula for the motion under gravity, which is y = 0.5 * g * t2, where y is the vertical distance (4 meters in this case) and g is the acceleration due to gravity (approximated to 9.81 m/s2). Solving for t gives us the time the water is in the air.

For part (b), once we have the time, we can use the horizontal distance (7 meters) to find the initial velocity (v0) using the formula x = v0 * t. This provides the initial horizontal velocity of the water gun's jet. The overall process involves solving for time first and then using that time to find the initial velocity.

Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

[tex]F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}[/tex]

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

[tex]acceleration=\frac{30}{5}=6m/s^{2}[/tex]

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

[tex]v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds[/tex]

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.

Answer:

a)[tex]v=6.19m/s[/tex]

b)[tex]v=5.51m/s[/tex]

c)[tex]a=3.3*10^{3}m/s^{2}[/tex]

d)[tex]x=5.78*10^{-3}m[/tex]

Explanation:

h1=195m

h2=1.55

a) Velocity just before the ball strikes the floor:

Conservation of the energy law

[tex]E_{o}=E_{f}[/tex]

[tex]E_{o}=mgh_{1}[/tex]

[tex]E_{f}=1/2*mv^{2}[/tex]

so:

[tex]v=\sqrt{2gh_{1}}=\sqrt{2*9.81*1.95}=6.19m/s[/tex]

b) Velocity just after the ball leaves the floor:

[tex]E_{o}=E_{f}[/tex]

[tex]E_{o}=1/2*mv^{2}[/tex]

[tex]E_{f}=mgh_{2}[/tex]

so:

[tex]v=\sqrt{2gh_{2}}=\sqrt{2*9.81*1.55}=5.51m/s[/tex]

c) Relation between Impulse, I, and momentum, p:

[tex]I=\Delta p\\ F*t=m(v_{f}-v{o})\\ (ma)*t=m(v_{f}-v{o})\\\\ a=\frac{ v_{f}-v{o}}{t}=\frac{ 5.51- (-6.19)}{3.5*10^{-3}}=3.3*10^{3}m/s^{2}[/tex]

d) The compression of the ball:

The time elapsed between the ball touching the ground and it is fully compressed, is half the time the ball is in contact with the ground.

[tex]t_{2}=t/2=3.5/2=1.75ms[/tex]

Kinematics equation:

[tex]x(t)=v_{o}t+1/2*a*t_{2}^{2}[/tex]

Vo is the velocity when the ball strike the floor, we found it at a) 6.19m/s.

a, is the acceleration found at c) but we should to use it with a negative sense, because its direction is negative a Vo, a=-3.3*10^3

So:

[tex]x=6.19*1.75*10^{-3}-1/2*3.3*10^{3}*(1.75*10^{-3})^2=5.78*10^{-3}m[/tex]

Answer and Explanation:

The inertial reference frame is one with constant velocity or non-accelerated frame of reference.

The value of acceleration and velocity change will vary in the two frames and will not be same.

As in case, we observe the acceleration and velocity of a moving train from the platform and the one observed in the train itself will be different.

In case of energy, it is dependent on the frame of reference but the energy change is independent of the frame of reference.

Answer:

Explanation:

a) Hardness is a measure of the resistance of a material to permanent deformation (plastic) on its surface,

Hardness tests play an important role in material testing, quality control and component acceptance.

Hardness test are needed to be perform as a quality assurance procedure, to validate materials are according to the specific hardness required,

We depend on the data to verify the quality of the components to determine if a material has the necessary properties for its intended use.

Through the years, the establishment of increasingly productive and effective means of testing, has given way to new cutting-edge methods that perform and interpret hardness tests more effectively than ever. The result is a greater capacity and dependence on "letting the instrument do the work", contributing to substantial increases in performance and consistency and continuing to make hardness tests very useful in industrial and R&D applications.

b)

"The hardness test is a critical aspect of engineering practice for several reasons:

1. Material Performance: Hardness is a measure of a material's resistance to deformation, which is directly related to its wear resistance, strength, and durability. By performing hardness tests, engineers can predict how a material will perform under service conditions.

2. Quality Control: Hardness testing is used to ensure that materials meet the required specifications. It is a non-destructive (in some cases) or semi-destructive method to verify the quality of the material without compromising its integrity.

3. Material Selection: The results of hardness tests help engineers to select appropriate materials for different applications. A material that is too soft or too hard for a particular application can lead to premature failure or unnecessary costs.

4. Heat Treatment Verification: Heat treatment processes such as quenching, annealing, and tempering are used to alter the mechanical properties of materials. Hardness testing is used to confirm that the desired properties have been achieved after such treatments.

5. Failure Analysis: In the event of a material failure, hardness testing can provide valuable information about the cause of the failure. It can indicate if the material was too brittle or too soft for the intended application.

6. Comparative Analysis: Hardness tests provide a simple way to compare the properties of different materials or different batches of the same material.

7. Research and Development: In the development of new materials, hardness testing is an essential tool for characterizing materials and understanding their properties.

B) Two possible sources of error in hardness testing, other than parallax error, and their effects on the results are:

1. Indenter Geometry Errors: The geometry of the indenter (e.g., ball, cone, or pyramid) must be precise according to the test standard being used. Wear, damage, or incorrect shape of the indenter can lead to inaccurate measurements. For instance, a worn or blunt indenter will give lower hardness values than the actual hardness of the material.

2. Loading Rate and Duration Errors: The rate at which the load is applied and the duration for which it is held can significantly affect the hardness measurement. Some materials exhibit time-dependent plasticity (creep), which can affect the size of the indentation. If the loading rate or duration is not consistent with the standards or previous tests, the results may not be comparable or may not accurately represent the material's hardness. For example, applying the load too quickly may result in an underestimation of hardness due to the material's inability to fully resist the indenter.

To minimize these errors, it is essential to follow standardized testing procedures, maintain and calibrate equipment regularly, and ensure that the test conditions are consistent. Additionally, operators should be well-trained and aware of the potential sources of error to ensure the accuracy and reliability of the hardness test results."

Final answer:

By applying the conservation of energy principle, accounting for the work done by the electric field as change in potential energy, we calculate the change in kinetic energy to find the speed of the charged particle at a different electric potential.

Explanation:

To calculate the speed of the particle at point B, we need to consider the change in electric potential energy and convert it to kinetic energy. The work done on the particle by the electric field is equal to the change in potential energy, which is given by W = q(VB - VA), where q is the charge of the particle, VA and VB are the electric potentials at points A and B respectively. Since no other forces act on the particle, the work done by the electric field causes a change in kinetic energy, which is defined by ½mv22 - ½mv12 = q(VB - VA), where m is the mass of the particle and v1 and v2 are the speeds at points A and B.

Rearranging the equation to solve for v2 gives us v2 = √(2q(VB - VA)/m + v12). Substituting in the given values, we can find the speed at point B.

Answer:

The speed of the train is 7.75 m/s towards station.

The speed of the train is 8.12 m/s away from the station.

Explanation:

Given that,

Frequency of the whistles f= 175 Hz

Beat frequency [tex]\Delta f= 4.05 Hz[/tex]

Speed of observer = 0

We need to calculate the frequency

Using formula of beat frequency

[tex]\Delta f=f'-f[/tex]

[tex]f'=\Delta f+f[/tex]

[tex]f'=4.05+175[/tex]

[tex]f'=179.05\ Hz[/tex]

When the train moving towards station, then the frequency heard is more than the actual

Using Doppler effect

[tex]f'=f(\dfrac{v-v_{o}}{v-v_{s}})[/tex]

[tex]v=v-\dfrac{vf}{f'}[/tex]

Put the value into the formula

[tex]v=343-\dfrac{343\times175}{179.05}[/tex]

[tex]v=7.75\ m/s[/tex]

The speed of the train is 7.75 m/s towards station.

When the train moving away form the station

Again beat frequency

[tex]\Delta f=f-f'[/tex]

[tex]f'=f-\Delta [/tex]

[tex]f'=175-4.05[/tex]

[tex]f'=170.95\ Hz[/tex]

We need to calculate the speed

Using Doppler effect

[tex]f'=f(\dfrac{v-v_{o}}{v+v_{s}})[/tex]

[tex]v=\dfrac{vf}{f'}-v[/tex]

Put the value into the formula

[tex]v=\dfrac{343\times175}{170.95}-343[/tex]

[tex]v=8.12\ m/s[/tex]

The speed of the train is 8.12 m/s away from the station.

Hence, This is the required solution.

Answer:

mechanical energy per unit mass is 887.4 J/kg

power generated is 443.7 MW

Explanation:

given data

average velocity = 3 m/s

rate = 500 m³/s

height h = 90 m

to find out

total mechanical energy and power generation potential

solution

we know that mechanical energy is sum of potential energy and kinetic energy

so

E = [tex]\frac{1}{2}[/tex]×m×v² + m×g×h    .............1

and energy per mass unit is

E/m =  [tex]\frac{1}{2}[/tex]×v² + g×h

put here value

E/m =  [tex]\frac{1}{2}[/tex]×3² + 9.81×90

E/m = 887.4 J/kg

so mechanical energy per unit mass is 887.4 J/kg

and

power generated is express as

power generated = energy per unit mass ×rate×density

power generated = 887.4× 500× 1000

power generated = 443700000

so power generated is 443.7 MW

The total mechanical energy per unit mass of the river water is the sum of the potential and kinetic energy. The potential energy is calculated using the height and the gravitational constant, while kinetic energy is calculated using velocity. The power generation potential of the river is the total mechanical energy multiplied by the volume flow rate.

The total mechanical energy per unit mass of the river water at this location can be determined using the principle of mechanical energy conservation and the knowledge of potential and kinetic energy. The total mechanical energy (E) equals the potential energy (PE) plus the kinetic energy (KE) per unit mass.

The potential energy is given by: PE = mgh, where m is the mass, g is the acceleration due to gravity (~9.81 m/s^2), and h is the height (90 m above lake surface). But since we are looking for energy per unit mass, m cancels out and we get PE = gh.

The kinetic energy is given by: KE = 0.5mv^2, where v is the velocity (3 m/s). In per unit mass terms, this simplifies to KE = 0.5v^2.

Therefore, E = PE + KE = gh + 0.5v^2.

The power generation potential can be calculated using the equation: P = E * volume flow rate (in our case, 500 m^3/s).

So,​​ P = (gh + 0.5v^2) * 500.

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To determine the distance to the epicenter of an earthquake, we can use the time difference between the arrival of P-waves and S-waves. In this case, the P-wave arrives 2.00 minutes before the S-wave. The distance of the earthquake center can be calculated by multiplying the time difference by the speed difference between the two waves.

To determine the distance to the epicenter of an earthquake, we can use the time difference between the arrival of P-waves and S-waves. In this case, the P-wave arrives 2.00 minutes before the S-wave. We know that the speed of the P-wave is 9000 m/s and the speed of the S-wave is 5000 m/s.

We can calculate the distance using the formula: distance = speed × time.

So, the distance of the earthquake center can be calculated as follows:

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Answer:

Heat required to melt 1 lb of ice is 151.469 KJ

Explanation:

We have given mass of ice = 1 lb

We know that 1 lb = 0.4535 kg

Latent heat of fusion for ice =334 KJ/kg

Amount if heat for fusion of ice is given by

[tex]Q=mL[/tex], here m is mass of ice and L is latent heat of fusion

So heat [tex]Q=mL=0.4535\times 334=151.469kj[/tex]

So heat required to melt 1 lb of ice is equal to 151.469 KJ

Answer:

The diameter is 0.378 ft.

Explanation:

Given that,

Mass of shot = 12 lb

Density of fresh water = 62.4 lb/ft

Specific gravity = 6.8

We need to calculate the volume of shot

[tex]V = \dfrac{4}{3}\pi r^3\ ft^3[/tex]

The density of shot is

Using formula of density

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

Put the value into the formula

[tex]\rho =\dfrac{12}{ \dfrac{4}{3}\pi r^3}[/tex]

We need to calculate the radius

Using formula of specific gravity

[tex]specific\ gravity =\dfrac{density\ of\ shot}{dnsity\ of\ water}[/tex]

Put the value into the formula

[tex]6.8=\dfrac{\dfrac{12}{\dfrac{4}{3}\pi r^3}}{62.4}[/tex]

[tex]r^3=\dfrac{12}{\dfrac{4}{3}\pi\times6.8\times62.4}[/tex]

[tex]r^3=0.0067514[/tex]

[tex]r =(0.0067514)^{\frac{1}{3}}[/tex]

[tex]r=0.1890\ ft[/tex]

The diameter will be

[tex]d = 2\times r[/tex]

[tex]d =2\times0.1890[/tex]

[tex]d =0.378\ ft[/tex]

Hence, The diameter is 0.378 ft.

Answer:

[tex]\eta = 0.411[/tex]

Explanation:

As we know that efficiency is defined as the ratio of output useful work and the input energy to the engine

So here we know that the

input energy = 441.3 kJ

energy rejected = 259.8 kJ

so we have

[tex]Q_1 - Q_2 = W[/tex]

[tex]W = 441.3 kJ - 259.8 kJ = 181.5 kJ[/tex]

now efficiency is defined as

[tex]\eta = \frac{W}{Q_1}[/tex]

[tex]\eta = \frac{181.5}{441.3}[/tex]

[tex]\eta = 0.411[/tex]

Answer:

[tex]K_e_q=22.75878093\frac{N}{m}[/tex]

[tex]f=1.363684118Hz[/tex]

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

[tex]\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}[/tex]

Replacing the data provided:

[tex]\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}[/tex]

[tex]K_e_q=22.75878093\frac{N}{m}[/tex]

Finally, to calculate the frequency of oscillation we use this:

[tex]f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }[/tex]

Replacing m and k:

[tex]f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz[/tex]

Answer:

28.5 m/s

Explanation:

The speed has two orthogonal components, horizontal and vertical. To decompose the speed into these components we can use these trigonometric equations:

Vh = V * cos(a)

Vv = V * sin(a)

The forwards velocity is the horizontal component, so we use

Vh = V * cos(a)

Vh = 50 * cos(55)

Vh = 50 * 0.57

Vh = 28.5 m/s

Answer:

Navier Stokes equation

( 1 )  it is a partial differential equation that is describe the flow of incompressible fluids

Reynolds equation

(1) it is partial differential equation that governs the pressure distribution of thin viscous fluid in lubrication

Explanation:

Navier Stokes equation

( 1 )  it is a partial differential equation that is describe the flow of incompressible fluids

(2) Navier Stokes equation is used to model weather and ocean current and water flow in the pipe and air flow around wing

( 3) equation is

[tex]\nabla .\overrightarrow{v} = 0[/tex]   momentum equation

[tex]\rho \frac{d\overrightarrow{v}}{dt} = \nabla p + \rho \overrightarrow{g} + \mu \nabla ^2 v^2[/tex]

here [tex]\nabla p[/tex] is pressure gradient and [tex]\rho \overrightarrow{g}[/tex] is body force and [tex]\mu \nabla ^2 v^2[/tex] is diffusion term

and

Reynolds equation

(1) it is partial differential equation that governs the pressure distribution of thin viscous fluid in lubrication

(2) it is drive in 1886 from Navier Stokes law

(3) equation is attach

here

Answer:

(a) d = 20 m

(b) d' = 80 m

(c) x = 28 m

(d) x' = 96 m

Solution:

As per the question:

Initial velocity of the object, v = 0

Constant acceleration of the object, [tex]a_{c} = 10 m/s^{2}[/tex]

(a) Distance traveled, d in t = 2.0 s is given by the second eqn of motion:

[tex]d = vt + \frac{1}{2}a_{c}t^{2}[/tex]

[tex]d = 0.t + \frac{1}{2}\times 10\times 2^{2} = 20 m[/tex]

(b) Distance traveled, d' in t = 4.0 s is given by the second eqn of motion:

[tex]d' = vt + \frac{1}{2}a_{c}t^{2}[/tex]

[tex]d' = 0.t + \frac{1}{2}\times 10\times 4^{2} = 80 m[/tex]

Now, when initial velocity, v = 4 m/s, then

(c) Distance traveled, x in t = 2.0 s is given by the second eqn of motion:

[tex]x = vt + \frac{1}{2}a_{c}t^{2}[/tex]

[tex]x = 4\times 2.0 + \frac{1}{2}\times 10\times 2^{2} = 28 m[/tex]

(d) Distance traveled, x' in t = 4.0 s is given by the second eqn of motion:

[tex]x = vt + \frac{1}{2}a_{c}t^{2}[/tex]

[tex]x = 4\times 4.0 + \frac{1}{2}\times 10\times 4^{2} = 96 m[/tex]

Answer:

Part a)

H = 26.8 m

Part b)

error = 7.18 %

Explanation:

Part a)

As the stone is dropped from height H then time taken by it to hit the floor is given as

[tex]t_1 = \sqrt{\frac{2H}{g}}[/tex]

now the sound will come back to the observer in the time

[tex]t_2 = \frac{H}{v}[/tex]

so we will have

[tex]t_1 + t_2 = 2.42[/tex]

[tex]\sqrt{\frac{2H}{g}} + \frac{H}{v} = 2.42[/tex]

so we have

[tex]\sqrt{\frac{2H}{9.81}} + \frac{H}{336} = 2.42[/tex]

solve above equation for H

[tex]H = 26.8 m[/tex]

Part b)

If sound reflection part is ignored then in that case

[tex]H = \frac{1}{2}gt^2[/tex]

[tex]H = \frac{1}{2}(9.81)(2.42^2)[/tex]

[tex]H = 28.7 m[/tex]

so here percentage error in height calculation is given as

[tex]percentage = \frac{28.7 - 26.8}{26.8} \times 100[/tex]

[tex]percentage = 7.18 [/tex]

The janitor can determine the depth of the elevator shaft by timing the drop of the rock and using the displacement for free-falling objects formula, considering the time for the sound to travel up the shaft. The percent error if the sound travel time is ignored can be calculated by comparing the depth calculated with and without the sound travel time.

The janitor wants to determine how far it is from the point they dropped the rock to the bottom of the elevator shaft by measuring the time it takes for the rock to drop and hit the bottom. Here we can use the formula for the displacement of items in free-fall: d = 0.5gt², where d is the displacement (depth of the shaft), g is the acceleration due to gravity (9.8 m/s²), and t is the time it takes for the rock to fall (2.42s).

So, the depth of the shaft is d = 0.5 x 9.80 m/s² x (2.42 s)² = 28.63 m. However, we need to account for the time it takes for the sound to travel back up the shaft. The speed of sound is given as 336 m/s, and the time it takes to travel a certain distance is the distance divided by the speed, so in this case, it’s 28.63 m / 336 m/s = 0.085 s. Subtracting this from the total time gives us the true fall time of the rock, which we can plug back into the displacement formula to get the corrected depth of the shaft.

The percent error introduced if we ignore the travel time for the sound is then the depth obtained with sound travel time subtracted from the depth obtained without it, divided by the depth obtained with sound travel time, times 100.

https://brainly.com/question/34634691

#SPJ3

Answer:

Explanation:

Area of plane = .35 x .7 = 0.245 m²

a) When plane is perpendicular to field ( plane is parallel to yz plane. )

Flux = field x area

3000 x .245 = 735 weber

b ) When plane is parallel to xy plane , the plane also becomes parallel to electric field so no flux will pass though the given plane.

Flux through the plane = 0

c ) Since normal to the plane makes 30 degree with x axis, it will also make 30 degree with direction of the field.

Flux through the plane

= electric field x area x cos 30

3000 x .245 x .866

636.51 weber.

Final answer:

To find the time the football is in the air before hitting the ground, we can analyze the vertical motion using the given initial velocity and launch angle.

Explanation:

To find the time it takes for the football to hit the ground, we need to analyze the vertical motion of the football. We can use the formula:
t = (2 * vy) / g
where t is the time, vy is the vertical component of the initial velocity, and g is the acceleration due to gravity.

Given that the initial velocity is 22.0 m/s and the launch angle is 58.5°, we can find the vertical component of the velocity using the formula:
vy = v * sin(θ)
where v is the initial velocity and θ is the launch angle.

Using this information, we can calculate the time it takes for the football to hit the ground.

Answer:

3.85 nm

Explanation:

The volume of a sphere is:

V = 4/3 * π * r^3

The volume of a shell is the volume of the big sphere minus the volume of the small sphere

Vs = 4/3 * π * r2^3 - 4/3 * π * r1^3

Vs = 4/3 * π * (r2^3 - r1^3)

If the difference between the radii is 1 nm

r2 = r1 + 1 nm

Vs = 4/3 * π * ((r1 + 1)^3 - r1^3)

Vs = 4/3 * π * ((r1 + 1)^3 - r1^3)

Vs = 4/3 * π * (r1^3 + 3*r1^2 + 3*r1 + 1 - r1^3)

Vs = 4/3 * π * (3*r1^2 + 3*r1 + 1)

The volme of the shell is equall to the volume of the inner shell:

4/3 * π * (r1^3) = 4/3 * π * (3*r1^2 + 3*r1 + 1)

r1^3 = 3*r1^2 + 3*r1 + 1

0 = -r1^3 + 3*r1^2 + 3*r1 + 1

Solving this equation electronically:

r1 = 3.85 nm

Answer:

Explanation:

Given

Velocity of helicopter relative to air 12.5 m/s to west

In vector form

[tex]V_{ha}=-12.5\hat{i}[/tex]

where [tex]V_h[/tex]= velocity of helicopter relative to ground

Also the velocity of air

[tex]V_a=4.55\hat{j}[/tex]

[tex]V_{ha}=V_h-V_a[/tex]

[tex]V_h=V_{ha}+V_a[/tex]

[tex]V_h=-12.5\hat{i}+4.55\hat{j}[/tex]

Speed relative to east[tex]=\sqrt{12.5^2+4.55^2}[/tex]

=13.302

For Direction

[tex]tan\theta =\frac{4.55}{-12.5}[/tex]

[tex]180-\theta =20[/tex]

[tex]\theta =160^{\circ}[/tex] relative to east

For 1 km travel it takes

[tex]t=\frac{1000}{13.302}=75.17 s[/tex]

Answer:

L =4166.66 N   and  D = 46296.29 N and T = 96047.34 N

Explanation:

given,

flight speed = 200 ft/s

turn rate = 5 deg/s

weight of aircraft = 50000 lb

L/D = 10

for equilibrium in the horizontal position

w - L cos ∅ - D sin ∅ = 0 .............(1)

D cos ∅ - L sin ∅ = 0................(2)

L/D = tan ∅ = 10

∅ = 84.29°

50000 - L cos  84.29°- D sin  84.29°= 0

D cos  84.29° - L sin 84.29° = 0

on solving the above equation we get

L =4166.66 N   and  D = 46296.29 N

thrust force calculation:

T = W sin ∅ + D

  = 50000×sin 84.29 + 46296.29

T = 96047.34 N

Answer:

B. a conversion factor

Explanation:

The expression "1 in. = 2.54 cm" is called a conversion factor.

With this expression inches can be converted to centimeters.

Inversely the expression can also be used to convert centimeters to inches

By rearranging the equation we get

[tex]\frac{1}{2.54}\ inches=1\ cm\\\Rightarrow 1\ cm=0.394\ inches[/tex]

The statement which shows the equal nature of two different expressions is called an equation

Rule of thumb is a general approximation of a result of a test or experiment.

Hi your answer is B on edge

Answer:

The distance is 2 cm

Solution:

According to the question:

Magnetic field of Earth, B_{E} = [tex]5.0\times 10^{- 5} T[/tex]

Current, I = 5.0 A

We know that the formula of magnetic field is given by:

[tex]B = \farc{\mu_{o}I}{2\pi d}[/tex]

where

d = distance from current carrying wire

Now,

[tex]d = \frac{\mu_{o}I}{2\pi B}[/tex]

[tex]d = \frac{4\pi\times 10^{- 7}\times 5.0}{2\pi\times 5.0\times 10^{- 5}}[/tex]

d = 0.02 m 2 cm

The distance from the wire where the magnitude of the magnetic field equals the Earth's magnetic field is approximately 0.4 meters.

To find the distance from the wire where the magnitude of the magnetic field is equal to the strength of the Earth's magnetic field, we can use the equation:

B = μ0 * I / (2π * r)

Where B is the magnetic field, μ0 is the permeability of free space (4π x [tex]10^-7[/tex]A), I is the current, and r is the distance from the wire.

Plugging in the given values, we have:

B_wire = μ0 * 5.0A / (2π * r) and B_earth = 5.0 x [tex]10^-5 T[/tex]

Setting B_wire equal to B_earth and solving for r:

5.0 x[tex]10^-5[/tex]* 5.0A / (2π * r)

Solving for r, we find that the distance from the wire where the magnitude of the magnetic field is equal to the Earth's is approximately 0.4 meters.