Two identical loudspeakers 2.0 m apart are emitting 1800 Hz sound waves into a room where the speed of sound is 340 m/s.
Is the point 4.0 m directly in front of one of the speakers, perpendicular to the plane of the speakers, a point of maximum constructive interference, perfect destructive interference, or something in between?

Answer:

A molecule is a combination of two or more atoms. The correct option is B.

Explanation:

Molecules are the smallest particle of a chemical compound that are made up of two or more atoms which are held together by a chemical bond. The chemical bonds are usually formed due to sharing and transfer of electrons among the atoms. Examples of molecules in chemistry includes:

- water molecule ( H2O)

- table salt ( NaCl)

- CaCl

Explanation:

Relation between work and change in kinetic energy is as follows.

                 [tex]W_{net} = \Delta K[/tex]

Also,   [tex]\Delta K = K_{initial} - K_{final}[/tex]

                        = [tex](0 - 4.0 \times 10^{-17})[/tex] J

                        = [tex]-4.0 \times 10^{-17}[/tex] J

Let us assume that electric force on the electron has a magnitude F. The electron moves at a distance of 0.3 m opposite to the direction of the force so that work done is as follows.

                w = -Fd

       [tex]-4.0 \times 10^{-17} J = -F \times 0.3 m[/tex]

                F = [tex]1.33 \times 10^{-16}[/tex]  

Therefore, relation between electric field and force is as follows.

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

                 = [tex]\frac{1.33 \times 10^{-16}}{1.60 \times 10^{-19} C}[/tex]

                 = [tex]0.831 \times 10^{3}[/tex] C

Thus, we can conclude that magnitude of the electric field that will stop these electrons in a distance of 0.3 m is [tex]0.831 \times 10^{3}[/tex] C.

To answer the question, we use principles from mechanics, involving the concepts of free fall and relative motion. The time taken by the bolt to hit the floor is around 0.78s and has different velocities to an observer inside and outside the elevator. According to a ground observer, the bolt travels a total distance of 4.95m.

To answer these questions, we need to involve the principles of physics, specifically mechanics dealing with motion. Firstly, since the bolt was initially at rest, and it falls under the influence of gravity, we use the formula for time in free fall t = √(2h/g), where h is the initial height (3.00 m) and g is the acceleration due to gravity (9.81 m/s²).

For (b), for an observer in the elevator, the bolt appears to fall straight down, so its velocity just as it hits the floor will be v = gt = 9.81*0.78 = 7.65 m/s.

For (c), for an observer on the floor landing, the elevator is moving upwards, thus the velocity of the bolt relative to the observer on the ground would be the sum of the falling velocity and the velocity of the elevator, so v = gt + elevator speed = 7.65 + 2.50 = 10.15 m/s.

For (d), the bolt would have traveled the height of the fall plus the distance the elevator traveled during the fall according to the ground observer. The distance the elevator moves is d = elevator speed * t = 2.50*0.78 = 1.95 m, therefore, the total distance the bolt travels is 3.00 m + 1.95 m = 4.95 m.

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Answer:

Part A:

[tex]V_1=\sqrt{\frac{2\gamma_{h20}h}{\rho_H}}\\ V_1=\sqrt{\frac{2*62.43*3}{5.5210*10^{-4}*12}}\\ V_1=237.778 ft/s[/tex]

Part B:

Ma=0.0737

Since Ma<0.3, it means the flow is in compressible.

Explanation:

Part A:

According to Bernoulli equation:

[tex]P_1+\frac{\rho_H}{2}V^2_1 =P_{2}+\frac{\rho_H}{2}V^2_2\\ V_2=0,\\P_1+\frac{\rho_H}{2}V^2_1 =P_{2}[/tex]

Velocity will become:

[tex]V_1=\sqrt{\frac{2(P_2-P_1)}{\rho_H}}[/tex].........Eq (1)

Now,[tex]P_2-P_1[/tex] can be calculated from the specific weight of water and helium[tex]P_2-P_1[/tex][tex]=(\gamma_{h2}o-\gamma_H)h[/tex]

Since the specific weight of helium is much smaller than specific weight of water we can neglect the specific weight of helium.

[tex]P_2-P_1[/tex]=[tex]=(\gamma_{h2o})h[/tex]

For water,[tex]\gamma_{h2o}=62.43 lb/ft^3[/tex]

h=3.0 in

Density of helium:

[tex]\rho_H=\frac{P}{RT}[/tex]

T=460+44=504 degree R

R=[tex]1.242*10^4 ft.lb/R.slug[/tex]

[tex]\rho_H=\frac{24*12^2}{1.242*10^4*504}\\ \rho_H=5.5210*10^{-4} lb/ft^3[/tex]

From Eq (1):

[tex]V_1=\sqrt{\frac{2\gamma_{h20}h}{\rho_H}}\\ V_1=\sqrt{\frac{2*62.43*3}{5.5210*10^{-4}*12}}\\ V_1=237.778 ft/s[/tex]

Part B:

Checking Ma:

[tex]Ma=\frac{V}{c}[/tex]

c is speed of sound:

k=1.66 for helium, In ideal gases:

[tex]c=\sqrt{kRT}\\ c=\sqrt{1.66*1.242*10^4*504}\\ c=3223.51 ft/s\\Ma=\frac{237.778}{3223.51}\\ Ma=0.0737[/tex]

Since Ma<0.3, it means the flow is in compressible.

(a) The helium velocity is approximately 2811.62 ft/s.    (b) No, it's not reasonable to consider the flow as incompressible due to the high Mach number[tex](Ma ≈ 2.58)[/tex], indicating compressibility effects.

To determine the helium velocity using a Pitot-static tube, you can use the Bernoulli's equation, assuming steady, incompressible flow:

\[P + 0.5 * ρ * V^2 + ρ * g * h = constant\]

Where:

- [tex]\(P\)[/tex] is the pressure in the pipe (psia)

- [tex]\(ρ\)[/tex]is the density of the fluid (helium in this case, lb/ft^3)

- [tex]\(V\)[/tex] is the velocity of the fluid (ft/s)

- [tex]\(g\)[/tex] is the acceleration due to gravity (32.2 ft/s^2)

- [tex]\(h\)[/tex] is the height difference in the manometer (inches)

Given:

- Temperature[tex](\(T\))[/tex] = 44°F = 44 + 460 = 504 Rankine (R)

- Pressure[tex](\(P\))[/tex] = 24 psia

- Reading in manometer [tex](\(h\))[/tex] = 3.0 inches

First, we need to find the density [tex](\(ρ\))[/tex] of helium at the given conditions. You can use the ideal gas law:

[tex]\[PV = nRT\][/tex]

Where:

-[tex]\(P\)[/tex] is pressure (psia)

- [tex]\(V\)[/tex]is volume ([tex]ft^3[/tex])

- [tex]\(n\)[/tex]is the number of moles

- [tex]\(R\)[/tex]is the specific gas constant[tex](for helium, \(R = 53.34\)[/tex][tex]ft·lb/(lbmol·R))[/tex]

- [tex]\(T\)[/tex] is temperature (Rankine)

Rearrange the equation to find [tex]\(ρ\):[/tex]

[tex]\[ρ = \frac{n}{V} = \frac{P}{RT}\][/tex]

Substitute the values:

[tex]\[ρ = \frac{24}{53.34 * 504} = 0.0008933 lb/ft^3\][/tex]

Now, we can calculate the velocity[tex](\(V\))[/tex]using Bernoulli's equation:

[tex]\[P + 0.5 * ρ * V^2 + ρ * g * h = constant\][/tex]

[tex]\[V = \sqrt{\frac{2 * (P - \text{manometer correction})}{ρ}}\][/tex]

The manometer correction accounts for the density of the manometer fluid, which is typically water in this case. Since we're given that the manometer reading is in inches, we need to convert it to feet:

Manometer Correction = 3.0 inches / 12 = 0.25 ft

Now, calculate the velocity:

[tex]\[V = \sqrt{\frac{2 * (24 - 0.25 * 62.4)}{0.0008933}} = 2811.62 ft/s\][/tex]

(a) The helium velocity is approximately 2811.62 ft/s.

(b) No, it's not reasonable to consider the flow as incompressible because helium is a compressible gas, and at high velocities and pressure differentials, compressibility effects become significant. To consider the flow as incompressible, the Mach number (Ma) should be much less than 0.3. To calculate Ma:

[tex]\[Ma = \frac{V}{a}\][/tex]

Where[tex]\(a\)[/tex] is the speed of sound in helium, which can be calculated using:

[tex]\[a = \sqrt{\gamma * R * T}\][/tex]

Where \(\gamma\) is the specific heat ratio for helium (approximately 1.66).

Calculate \(a\) and then Ma:

[tex]\[a = \sqrt{1.66 * 53.34 * 504} = 1087.92 ft/s\][/tex]

[tex]\[Ma = \frac{2811.62}{1087.92} \approx 2.58\][/tex]

Since the Mach number is significantly greater than 0.3, the flow of helium in this case cannot be considered incompressible.

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Use Doppler's formula to find the radial velocity of star.

[tex]\frac{V_r}{c} = \frac{\Delta \lambda}{\lambda_0}[/tex]

Here,

[tex]V_r[/tex] = Radial Velocity

c = Speed of light

[tex]\Delta \lambda[/tex] = Shift in wavelength

[tex]\lambda_0[/tex] = Laboratory wavelength of spectral line

Rearrange for [tex]V_r[/tex],

[tex]V_r = \frac{\Delta \lambda}{\lambda_0} c[/tex]

Find shift in wavelength, [tex]\Delta \lambda[/tex]

[tex]\Delta \lambda = |485.5nm - 486.1nm|[/tex]

[tex]\Delta \lambda = 0.6nm[/tex]

Replacing our values we have then,

[tex]V_r = \frac{0.6nm}{486.1nm}(3*10^8m/s)[/tex]

[tex]V_r = 370000m/s[/tex]

Therefore the radial velocity of star is [tex]3.7*10^5[/tex]m/s

In this case the symbol of [tex]\Delta \lambda[/tex] implies that the star is receding the observer and the wavelength turns to red, then is red shifted.

Final answer:

To calculate the star's radial velocity based on the observed Doppler shift in the hydrogen Balmer-beta line, we use the shift in wavelength from 486.1 to 485.5 nm and apply the formula for Doppler shift.

Explanation:

The student's question involves calculating the radial velocity of a star based on the Doppler shift observed in the hydrogen Balmer-beta spectral line. The observed shift from 486.1 nm in the laboratory to 485.5 nm in the star's spectrum indicates a movement towards us. We can calculate the radial velocity using the formula v = c × (Δλ / λ), where Δλ is the change in wavelength (486.1 - 485.5 = 0.6 nm), λ is the original wavelength (486.1 nm), and c is the speed of light (3 × 108 m/s). Converting 0.6 nm to meters (0.6 × 10-9 m) and plugging in the values gives us the star's radial velocity.

To find the magnitude and direction of vectors A+B, A-B, and B-A, we perform vector addition and subtraction and calculate the magnitudes using the Pythagorean theorem. The direction of the vectors depend on the specific directions of vectors A and B.

The subject of this question is in the field of Physics, specifically under vectors. The grade level of this question falls under High School.

(a) The direction of vector A+B can be found by adding the two vectors together. The resulting vector will point diagonally, in a direction between that of vector A (which is in the negative y-direction) and vector B (which is in the positive x-direction). The magnitude of the resulting vector is √((14)^2 + (2*14)^2)= 24.5

(b) The magnitude of vector A+B is also known as the modulus of the vector, and it is the same as the magnitude calculated in part (a), which is 24.5 units.

(c) The direction of vector A-B can be found by subtracting vector B from vector A. The direction will be in the negative x-direction. The magnitude of the resulting vector is √((14)^2 + (2*14)^2) = 24.5

(d) The magnitude of the vector A-B, also known as the modulus of the vector, is given by the same calculation as in part (c), which is 24.5 units.

(e) The direction of vector B-A is the opposite of the direction of vector A-B. Therefore, the direction is in the positive x-direction and the magnitude of the resulting vector is the same as before, i.e., 24.5 units.

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Options

1. the same as in the first.

2.greater than in the first

3.less than in the first.

4.Unable to determine

Answer

The answer is 2. Greater than in the first

Explanation

The speed of a wave v is related to its wavelength λ by the formulav=f λ, where f is the frequency of the wave. The frequency will not change when the wave passes into a second medium, so

λ2>λ1

Fλ1>fλ2

Since f>0

And V2>v1

Answer:

Explanation:

The first part of question is about the height of the rock from which it falls and hit the ground with speed of 11 m/s. Lets find out that height.

We will use the formula,

[tex]v^{2} _{f} = v^{2} _{i} + 2gh[/tex]

As the initial velocity of the rock was zero. [tex]v_{f} = 0[/tex]

[tex]v^{2} _{f} = 2gh\\ h = v^{2} _{f} / 2g\\h = \frac{(11 m/s)^{2} }{2(9.8 m/s^{2} )} \\h = 6.17 m[/tex]

Now we have to find the height from which the rock should be dropped and it's speed on hitting the ground should be 22 m/s.

Again we will use the same formula, same calculation but the value of velocity now should be 22 m/s.

[tex]v^{2} _{f} = v^{2} _{i} + 2gh[/tex]

[tex]v^{2} _{f} = 2gh\\ h = \frac{(22m/s)^{2} }{2(9.8 m/s^{2}) } \\h = 24.69 m[/tex]

To double the speed of the rock when it hits the ground (from 11 m/s to 22 m/s), the height from which it is dropped should be quadrupled. Hence, the rock should be dropped from a height of 24.5 meters.

The question is about finding the height from which a rock should be dropped so that its speed on hitting the ground is 22 m/s, given that when it is dropped from height h, its speed is 11 m/s. To solve this, we can use the physics equation for motion under constant acceleration, which is v² = 2gh, where v is the final velocity, g is the acceleration due to gravity, and h is the height of fall.

First, let us find the height h in the initial scenario: (11)² = 2*9.8*h => h = 6.125 m. Generally, the height h is proportional to the square of the speed, so if we double the final speed, the height should be quadrupled: h'(new height) = 4 * h = 4 * 6.125 m = 24.5 m.

Therefore, the rock should be dropped from a height of 24.5 m so that its speed on hitting the ground is 22 m/s.

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The statement is true due to the Doppler Effect. As you move away from the radio transmitter, the radio signal you receive appears to have been shifted to longer wavelengths due to the changes in the observer-source distance.

The statement, 'As you drive away from a radio transmitter, the radio signal you receive from the station is shifted to longer wavelengths,' is True. This is due to a phenomenon known as the Doppler Effect.

The Doppler Effect explains that the observed wavelength of electromagnetic radiation is longer (red shift) when the source moves away from the observer. This means the wavelength of the radiation from the radio station would appear to increase (shift to a longer wavelength) as you drive away from it.

To maintain that same energy level required for transmission, the radio station emits waves at a higher frequency (shorter wavelengths). But, as you move away, the radio waves appear to have a lower frequency (longer wavelengths). This happens because the waves get 'stretched out' as the distance between you (the observer) and the radio station (the source) increases.

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Final answer:

The object is actually located 1.5 m from the van, and it is 2.1 m tall when considering the given focal length and image properties.

Explanation:

- Focal length of the lens (f) = -0.300 m (negative sign indicates a diverging lens)

- Image distance = 0.250 m (positive because the image appears behind the lens)

- Height of the image = 0.350 m

We need to find:

(a) Actual distance (d) of the object from the van.

(b) Actual height (h) of the object.

(a) To find the actual distance of the object from the van (d), we use the thin lens equation:

1/f = 1/d + 1/image distance

Plugging in the given values:

1/-0.300 = 1/d + 1/0.250

Solving for d:

-3.33 = 1/d + 4

-3.33 - 4 = 1/d

-7.33 = 1/d

d = 1/-7.33

d ≈ -0.136 m

The negative sign indicates that the object is located behind the lens.

(b) To find the actual height of the object (h), we use the magnification formula:

m = height of the image / height of the object = -(image distance / object distance)

Given m = -(image distance / object distance), and we have m = height of the image / height of the object, we can write:

-(image distance / object distance) = height of the image / height of the object

Plugging in the given values:

-(0.250 / -0.136) = 0.350 / h

Solving for h:

h = (0.350 * -0.136) / 0.250

h ≈ -0.190 m

The negative sign indicates that the height of the object is inverted.

So, the answers are:

(a) The object is actually located approximately 0.136 m behind the van.

(b) The actual height of the object is approximately 0.190 m.

Answer:

A) [tex]k=34867.3384\ N.m^{-1}[/tex]

B) [tex]\omega'\approx84\ Hz[/tex]

Explanation:

Given:

mass of car, [tex]m=1380\ kg[/tex]

A)

frequency of spring oscillation, [tex]f=1.6\ Hz[/tex]

We knkow the formula for spring oscillation frequency:

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

[tex]\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f[/tex]

[tex]\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6[/tex]

[tex]k_{eq}=139469.3537\ N.m^{-1}[/tex]

Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.

So, the stiffness of each spring is (as they are identical):

[tex]k=\frac{k_{eq}}{4}[/tex]

[tex]k=\frac{139469.3537}{4}[/tex]

[tex]k=34867.3384\ N.m^{-1}[/tex]

B)

given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:

[tex]\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }[/tex]

[tex]\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }[/tex]

[tex]\omega'\approx84\ Hz[/tex]

Answer:

Total work done in expansion will be [tex]3.60\times 10^5J[/tex]

Explanation:

We have given pressure P = 2.10 atm

We know that 1 atm [tex]=1.01\times 10^5Pa[/tex]

So 2.10 atm [tex]=2.10\times 1.01\times 10^5=2.121\times 10^5Pa[/tex]

Volume is increases from 3370 liter to 5.40 liter

So initial volume [tex]V_1=3.70liter[/tex]

And final volume [tex]V_2=5.40liter[/tex]

So change in volume [tex]dV=5.40-3.70=1.70liter[/tex]

For isobaric process work done is equal to [tex]W=PdV=2.121\times 10^5\times 1.70=3.60\times 10^5J[/tex]

So total work done in expansion will be [tex]3.60\times 10^5J[/tex]

Answer:

g/9

Explanation:

length of the pendulum = L

time period on the earth = T

Time period on the planet = 3T

Let the acceleration due to gravity on the earth is g and on the planet is g'.

Use the formula for the time period of a simple pendulum for the time period on earth

[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]     .... (1)

Time period on the surface of planet is

[tex]3T=2\pi \sqrt{\frac{L}{g'}}[/tex]      .... (2)

Divide equation (2) by equation (1)

[tex]\frac{3T}{T}= \sqrt{\frac{g}{g'}}[/tex]

g' = g/9

Thus, the acceleration due to gravity on the planet is g /9

The average vertical velocity of the rock thrown upwards from the height H to the ground is determined by the total vertical displacement divided by the total time, considering the downward acceleration due to gravity is negative.

The vertical component v_{av-y} of the average velocity of a small rock thrown straight up from the edge of a roof of height H to the ground can be found by using the kinematic equations of motion under constant acceleration. Since the acceleration due to gravity (g) is acting downward, it is represented as negative in the equations. Considering the upwards direction as positive and the downward acceleration as negative ensures that the final answer for v_{av-y} takes into account the total displacement, which includes both the upward and the downward path of the rock. The average velocity is thus the total displacement divided by the total time taken. The rock initially travels upwards to a maximum height before falling to the ground, completing its motion.

Answer:

6.26 times more

Explanation:

Given:

- most satellite orbit at height r_1 = 1000 km

- Geosynchronous satellites orbit at height r_2 = 35,790 km

- mass of Geosynchronous satellite m = 410 kg

- The radius of the earth r_e = 6371 km

Find:

- Compare the Energy required to send the satellite to Geosynchronous orbit @ r_2 vs Energy required to send the satellite to normal orbit @ r_1. How much more. ( U_1 / U_2 ).

Solution:

- The gravitational potential energy of any mass m in an orbit around another mass M is given by the following relation:

                                   U_g = - G*m*M / r

Where,

G : Gravitational constant

- We compute the gravitational potential energy U_g of the satellite at both orbits as follows:

-Normal orbit           U_1 = - G*m*M / r_e + G*m*M / (r_e+r_1)

                                U_2 = - G*m*M / r_e + G*m*M / (r_2+r_e)

Now: Take a ratio of the two energies U_1 and U_2 as follows:

  U_2 / U_1 = (- G*m*M / r_e + G*m*M / r_2+r_e) / (- G*m*M / r_e + G*m*M / r_1+r_e)

                 U_2 / U_1 = (1 / (r_2+r_e) - 1 / r_e ) / (1 / (r_1 + r_e) - 1 / r_e )

- Plug values:

           U_2 / U_1 = (1 / (35790+6371) - 1 / 6371 ) / (1 / (1000+6371) - 1 / 6371 )

- Evaluate:

                         U_2 / U_1 = (-1.33242681 * 10^-4) / (2.12944*10^-5)

                                                U_2 / U_1 = 6.26

- Hence The energy required to send the satellite to Geosynchronous orbit is 6.26 times more than that required for normal orbit.

The electric field of a uniformly charged disk at different points on its axis was computed using the formula for the electric field due to a charged disk. a. 20.16 N/C b.  17.7 N/C c. 3.57 N/C d. 0.225 N/C

The student is asking for the calculation of the electric field at different points on the axis of a uniformly charged disk with a known charge density.

The formula to find the electric field E due to a charged disk along its axis at distance x from the center is given by E =((σ/2ε0) * (1 - (x/(sqrt(x^2 + r^2)))), where r is the radius of the disk and σ is the charge density. The constant ε0 is the permittivity of free space, and its value is approximately 8.85 x 10^-12 C^2/N·m^2.

a. For x=5.00 cm, E ~= 20.16 N/C

b. For x=10.0 cm, E ~= 17.7 N/C

c. For x=50.0 cm, E ~= 3.57 N/C

d. For x=200 cm, E ~= 0.225 N/C

These results reflect the fact that the strength of the electric field decreases as one moves farther away from the center of the disk.

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The electric fields at the respective distances are:

[tex]a. \( E(5.00 \, \text{cm}) = 4.369 \times 10^8 \, \text{N}/\text{C} \) b. \( E(10.0 \, \text{cm}) = 3.695 \times 10^8 \, \text{N}/\text{C} \) c. \( E(50.0 \, \text{cm}) = 9.187 \times 10^7 \, \text{N}/\text{C} \) d. \( E(200 \, \text{cm}) = 7.728 \times 10^6 \, \text{N}/\text{C} \)[/tex]

To calculate the electric field on the axis of a uniformly charged disk at a distance z from the center, we can use the formula derived from Gauss's law for a flat surface:

[tex]\[ E(z) = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{z}{\sqrt{z^2 + R^2}} \right) \][/tex]

where E(z)  is the electric field at a distance z from the center of the disk, [tex]\( \sigma \)[/tex] is the surface charge density, [tex]\( \epsilon_0 \)[/tex] is the vacuum permittivity [tex](\( 8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \))[/tex], and R  is the radius of the disk.

Given:

- Radius of the disk, [tex]\( R = 35.0 \, \text{cm} = 0.35 \, \text{m} \)[/tex]

- Charge density, [tex]\( \sigma = 9.00 \times 10^{-3} \, \text{C}/\text{m}^2 \)[/tex]

- Vacuum permittivity,[tex]\( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \)[/tex]

Now, let's calculate the electric field at the given distances:

a. At z = 5.00[tex]\text{cm} = 0.05 \, \text{m} \)[/tex]:

[tex]\[ E(0.05) = \frac{9.00 \times 10^{-3}}{2 \times 8.85 \times 10^{-12}} \left( 1 - \frac{0.05}{\sqrt{0.05^2 + 0.35^2}} \right) \] \[ E(0.05) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.05}{\sqrt{0.001225 + 0.1225}} \right) \][/tex]

[tex]\[ E(0.05) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.05}{0.3555} \right) \][/tex]

[tex]\[ E(0.05) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - 0.1406 \right) \] \[ E(0.05) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \times 0.8594 \] \[ E(0.05) = \frac{7.7346 \times 10^{-3}}{17.7 \times 10^{-12}} \] \[ E(0.05) = 4.369 \times 10^8 \, \text{N}/\text{C} \][/tex]

b. At z = 10.0[tex]\text{cm} = 0.1 \, \text{m} \)[/tex]:

[tex]\[ E(0.1) = \frac{9.00 \times 10^{-3}}{2 \times 8.85 \times 10^{-12}} \left( 1 - \frac{0.1}{\sqrt{0.1^2 + 0.35^2}} \right) \] \[ E(0.1) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.1}{\sqrt{0.01 + 0.1225}} \right) \] \[ E(0.1) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.1}{0.3674} \right) \][/tex]

[tex]\[ E(0.1) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - 0.2722 \right) \] \[ E(0.1) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \times 0.7278 \] \[ E(0.1) = \frac{6.5402 \times 10^{-3}}{17.7 \times 10^{-12}} \] \[ E(0.1) = 3.695 \times 10^8 \, \text{N}/\text{C} \][/tex]

c. At z = 50.0[tex]\text{cm} = 0.5 \, \text{m} \)[/tex]:

[tex]\[ E(0.5) = \frac{9.00 \times 10^{-3}}{2 \times 8.85 \times 10^{-12}} \left( 1 - \frac{0.5}{\sqrt{0.5^2 + 0.35^2}} \right) \] \[ E(0.5) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.5}{\sqrt{0.25 + 0.1225}} \right) \] \[ E(0.5) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.5}{0.6104} \right) \][/tex]

[tex]\[ E(0.5) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - 0.8193 \right) \] \[ E(0.5) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \times 0.1807 \] \[ E(0.5) = \frac{1.6263 \times 10^{-3}}{17.7 \times 10^{-12}} \] \[ E(0.5) = 9.187 \times 10^7 \, \text{N}/\text{C} \][/tex]

d. At z = 200 [tex]\text{cm} = 2.0 \, \text{m} \)[/tex]:

[tex]\[ E(2.0) = \frac{9.00 \times 10^{-3}}{2 \times 8.85 \times 10^{-12}} \left( 1 - \frac{2.0}{\sqrt{2.0^2 + 0.35^2}} \right) \] \[ E(2.0) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{2.0}{\sqrt{4 + 0.1225}} \right) \][/tex]

[tex]\[ E(2.0) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{2.0}{2.0317} \right) \] \[ E(2.0) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - 0.9848 \right) \][/tex]

[tex]\[ E(2.0) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \times 0.0152 \] \[ E(2.0) = \frac{1.368 \times 10^{-4}}{17.7 \times 10^{-12}} \] \[ E(2.0) = 7.728 \times 10^6 \, \text{N}/\text{C} \][/tex]

Answer:

By dropping a pencil from a certain fixed height again and again it will make the target super messay with marks of dot everywhere on the target and some even out side the target.

Explanation:

Answer:

169.74 N

Explanation:

Given,

Mass of the girl = 30 Kg

angle of the rope with vertical, θ = 30°

equating the vertical component of the tension

vertical component of the tension is equal to the weight of the girl.

 T cos θ = m g

 T cos 30° = 30 x 9.8

 T = 339.48 N

Tension on the two ropes is equal to 339.48 N

Tension in each of the rope = T/2

                                           = 339.48/2 = 169.74 N

Hence, the tension in each of the rope is equal to 169.74 N

Final answer:

To calculate the tension in the ropes supporting the swing, one must account for the weight of the girl and set it equal to the combined vertical components of the tension in the ropes. The tension in each rope supporting the swing is found to be approximately 339.4 N.

Explanation:

The subject of this question is Physics, specifically related to the application of Newton's Laws of Motion to calculate tension in ropes. The question includes a scenario where a 30.0-kg girl in a swing is held at rest in a position where the ropes form a 30.0-degree angle with the vertical.

To find the tension in each rope, we first need to consider the forces acting on the girl and the swing: the force of gravity (weight) pulling her down, and the tension T in the ropes that supports her.

Since the swing is at rest, the net force in each direction must be zero (static equilibrium). Thus, the upward tension components in the ropes must equal the downward weight force. For each rope, the vertical component of the tension (Ty) will be Ty = T*cos (30.0°).

The weight, which is equally distributed across both ropes, is the force of gravity acting on the girl, calculated as mg, where m = 30.0 kg and g = 9.8 [tex]m/s^2[/tex]. Setting the vertical components of tension equal to the weight of the girl, we can solve for T:

2 * T * cos(30.0°) = 30.0 kg * 9.8 [tex]m/s^2[/tex]

This gives us:

T = (30.0 kg * 9.8 [tex]m/s^2[/tex]) / (2 * cos(30.0°))

Performing the calculation:

T = 339.4 N (approximately)

Therefore, the tension in each of the two ropes is approximately 339.4 N.

Answer:

67 dB

Explanation:

given,

Sound in afternoon = 80 dB

Intensity of car,I₀ = 100 cars/ minute

Sound in the night = ?

Intensity of car,I = 5 car/minutes

using formula for sound calculation

 [tex]\Delta \beta = 10 log(\dfrac{I}{I_0})[/tex]

 [tex]\Delta \beta = 10 log(\dfrac{5}{100})[/tex]

 [tex]\Delta \beta = 10 log(\dfrac{1}{20})[/tex]

 [tex]\Delta \beta = 10 log(0.05)[/tex]

 [tex]\Delta \beta = 10\times -1.30[/tex]

 [tex]\Delta \beta = -13\ dB[/tex]

The late night decibel is equal to 80 dB - 13 dB = 67 dB

Answer:

b. Throwing some cargo out of the boat

Explanation:

Using the Newton's third law of motion which states that every action has an equal and opposite reaction.

So when we are on the boat and we throw some mass in a direction away from the boat out of it then we are imparting the force to the floating boat by the law of conservation of momentum as well.

The relation can be mathematically expressed as:

[tex]m_c.v_c=m_b.v_b[/tex]

where:

[tex]m_c=[/tex] mass of the cargo

[tex]v_c=[/tex] velocity of throwing the cargo

[tex]m_b=[/tex] mass of the whole boat including all that floats on it

[tex]v_b=[/tex] velocity of the boat system

Throwing some cargo out of the boat will make the boat start moving.

This law states that action and reaction are equal and opposite.

The total momentum of an isolated system is always conserved.

Throwing some cargo out of the boat will change the velocity of the cargo initially at rest and to conserve the linear momentum, the velocity of the boat will change as well.

Thus, throwing some cargo out of the boat will make the boat start moving.

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Answer:

The person is 187[m] farther and 70° south to east.

Explanation:

We can solve this problem by drawing a sketch of the location of the person and the truck, then we will draw the displacement vectors and finally the length of the vector and the direction of the vector will be measured in order to give the correct indication of where the person will have to move.

First we establish an origin of a coordinate system.

We can see in the attached schema that the red vector is the displacement vector from the last point to where the truck is located.

The length of the vector is 187 [m], and the direction is 70 degrees south to East.

This problem involves using vectors and trigonometry to calculate the direct distance and direction to the truck from the new location after walking due west. This is achieved by adding the horizontal and vertical components of the vectors representing the initial location and the walking path.

To solve this, one may use vectors and trigonometry. Initially, you are 122.0m from your truck, 58.0 degrees east of south. This can be treated as a vector from your truck to your original location. Then, you walk 72.0m due west, which is another vector in the opposite direction.

To find the resulting vector, i.e., the direct distance and direction to the truck from your new location, we have to add these vectors. While the mathematics is somewhat complex, the concept involves adding the horizontal (east-west) and vertical (north-south) components of each vector. Once the resulting vector is calculated, the remaining distance to the truck can be found from its magnitude, and the direction from its angle relative to south.

This involves math calculations, including trigonometry and Pythagorean theorem.

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To calculate the tension in the cable required to stop the elevator, we multiply the mass of the elevator by its acceleration. The tension in the cable is -4356 N, indicating it acts in the opposite direction of the weight of the elevator.

To calculate the tension in the cable required to stop the elevator over a distance of 3.4 m, we need to consider the force required to decelerate the elevator from its maximum speed of 3.3 m/s to a stop. The tension in the cable must equal the force needed to stop the elevator, which is equal to the mass of the elevator multiplied by its acceleration. The mass of the elevator, including occupants, is given as 1320 kg. Since the elevator is going down, its acceleration will be negative. Therefore, the tension in the cable can be calculated using the formula:

Tension = mass * acceleration = 1320 kg * (-3.3 m/s^2) = -4356 N

Therefore, the tension in the cable to stop the elevator over a distance of 3.4 m is -4356 N. The negative sign indicates that the tension is acting in the opposite direction of the weight of the elevator.

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Answer:

0.24825 N

0.0000238701923077 kgm²

Explanation:

m = Mass of yo yo = 0.075 kg

a = Acceleration = 6.5 m/s²

g = Acceleration due to gravity = 9.81 m/s²

The net force is given by

[tex]F_n=mg-T[/tex]

[tex]\Rightarrow T=mg-ma[/tex]

[tex]\Rightarrow T=m(g-a)[/tex]

[tex]\Rightarrow T=0.075(9.81-6.5)[/tex]

[tex]\Rightarrow T=0.24825\ N[/tex]

The tension in the string is 0.24825 N

Angular acceleration is given by

[tex]\alpha=\dfrac{a}{r}\\\Rightarrow \alpha=\dfrac{6.5}{2.5\times 10^{-2}}\\\Rightarrow \alpha=260\ rad/s^2[/tex]

Torque is given by

[tex]\tau=I\alpha\\\Rightarrow Tr=I\alpha\\\Rightarrow I=\dfrac{Tr}{\alpha}\\\Rightarrow I=\dfrac{0.24825\times 2.5\times 10^{-2}}{260}\\\Rightarrow I=0.0000238701923077\ kgm^2[/tex]

The moment of inertia is 0.0000238701923077 kgm²

The tension in the string is equal to 0.2475 Newton.

Given the following data:

To determine the tension in the string:

First of all, we would determine the downward force applied by the yo-yo's weight:

[tex]F_y = mg[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]F_y = 0.075 \times 9.8\\\\F_y = 0.735 \; Newton[/tex]

Next, we would determine the force acting on the string:

[tex]F_s = 0.075 \times 6.5\\\\F_s = 0.4875\;Newton[/tex]

Now, we can find the tension in the spring:

[tex]Tension = F_y - F_s\\\\Tension = 0.735 - 0.4875[/tex]

Tension = 0.2475 Newton.

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Radio waves have lower energy compared to other types of waves in the electromagnetic spectrum. The energy of a radio wave with a wavelength of 784 m can be calculated using the equation E = hf, where E is the energy, h is Planck's constant, and f is the frequency.

Radio waves fall within the electromagnetic spectrum which consists of various types of waves ranging from gamma rays to radio waves. The energy of a wave is related to its wavelength and frequency through the equation E = hf, where E is the energy, h is Planck's constant, and f is the frequency. Since radio waves have a longer wavelength, they have a lower frequency and therefore lower energy compared to other types of waves in the spectrum.

To calculate the energy of a radio wave with a wavelength of 784 m, we can use the equation c = λf, where c is the speed of light. Rearranging the equation to solve for f, we get f = c/λ. Plugging in the given wavelength of 784 m and the speed of light which is approximately 3 x 10^8 m/s, we can calculate the frequency as 3 x 10^8 m/s / 784 m = 3.83 x 10^5 Hz. Substitute this frequency value into the equation E = hf to calculate the energy of the radio wave.

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The energy of radio wave radiation for a 784-meter wavelength is approximately 2.53 x 10^-34 joules, calculated using Planck's equation for energy.

The energy E of radio wave radiation with a particular wavelength λ can be calculated using the equation:

E = hc/λ

Where:

E is the energy in Joules (J)

h is Planck's constant (6.626 x 10-34 J·s)

c is the speed of light in vacuum (about 3 x 108 m/s)

λ is the wavelength in meters (m)  

For a radio wave with a wavelength of 784 meters, we use these values to compute the energy:

E = (6.626 x 10-34 J·s) * (3 x 108 m/s) / 784 m

Calculating using the numerical values:

E = (6.626 x 10-34) * (3 x 108) / 784

E ≈ 2.53 x 10-34 J

Therefore, the energy of radio wave radiation with a wavelength of 784 meters is approximately 2.53 x 10-34 joules.

Amount of heat absorbed: 214 J

Explanation:

When an object absorbs heat, its temperature increases according to the equation

[tex]Q=mC\Delta T[/tex]

where

Q is the heat absorbed

m is the mass of the object

C is the specific heat capacity of the material

[tex]\Delta T[/tex] is the change in temperature

For the iron skillet in this problem:

m = 28 g = 0.028 kg is the mass

[tex]C=450 J/kg^{\circ}C[/tex] is the iron specific heat capacity

[tex]\Delta T = 27-10=17^{\circ}C[/tex] is the increase in temperature

Solving for Q, we find the amount of heat absorbed:

[tex]Q=(0.028)(450)(17)=214 J[/tex]

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The radius of a neutron star with a mass 10 times that of the Sun is approximately 29.62 km.

To calculate the radius of a neutron star with a mass 10 times that of the Sun, we can use the equation for the Schwarzschild radius:

R = 2GM / c^2

Where R is the radius, G is the gravitational constant, M is the mass, and c is the speed of light. Plugging in the values, we get:

R = 2 * 6.67x10^-11 (m^3/kg/s^2) * (10 * 1.99x10^30 kg) / (3x10^8 m/s)^2

R ≈ 29.62 km

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The probability that a randomly selected power supply will conform to the specified voltage range of 4.95 V to 5.05 V, given a normal distribution with mean 5 V and standard deviation 0.02 V, is roughly 95%.

To find the probability that a power supply selected at random will conform to the specifications on voltage, we need to calculate the area under the normal distribution curve between the lower specification (4.95 V) and the upper specification (5.05 V). Given that the mean (μ) is 5 V and the standard deviation (σ) is 0.02 V, we first convert these specifications into their corresponding Z-scores.

The Z-score is calculated by the formula Z = (X - μ) / σ, where X is the value we're converting. For the lower specification, 4.95 V, the Z-score is Z = (4.95 - 5) / 0.02 = -2.5. For the upper specification, 5.05 V, the Z-score is Z = (5.05 - 5) / 0.02 = 2.5.

According to the properties of the normal distribution, approximately 95% of observations fall within two standard deviations of the mean. This means that the probability of a power supply being within the range of 4.95 V to 5.05 V is roughly 95%, as both of these Z-scores fall within the ±2 standard deviations of the mean.

Electric field due to uniformly charged metal plate is given by,

[tex]E = \frac{\sigma}{(2\epsilon_0)}[/tex]

Here,

[tex]\sigma[/tex] = Charge density

[tex]\epsilon_0 =[/tex]  Vacuum Permittivity

Our values are,

[tex]\sigma = 0.75 muC/m^2 = 0.75*10^-6 C/m^2[/tex]

[tex]\epsilon_0 = 8.85*10^-12 F\cdot m^{-1}[/tex]

Replacing we have,

[tex]E = \frac{(0.75*10^-6)}{(2*8.85*10^-12)}[/tex]

[tex]F = 42372.88N/C[/tex]

Now we have the relation where energy is equal to the change of the potential in a certain distance, then

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

Rearranging for the distance

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

[tex]d = \frac{100}{42372.88}[/tex]

[tex]d = 0.00236m[/tex]

[tex]d = 2.36mm[/tex]

Therefore the distance is 2.36mm

Answer:

1. d = 0.415 m.

2. Q = 2.285 x 10^{-10} C.

Explanation:

The electric field and potential can be found by the following equations:

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

Applying these equations to the given variables yields

[tex]E = 12 = \frac{1}{4\pi\epsilon_0}\frac{Q}{d^2}\\V = 4.98 = \frac{1}{4\pi\epsilon_0}\frac{Q}{d}[/tex]

Divide the first line to the second line:

[tex]\frac{12}{4.98} = \frac{ \frac{1}{4\pi\epsilon_0}\frac{Q}{d^2}}{\frac{1}{4\pi\epsilon_0}\frac{Q}{d}}\\\frac{12}{4.98} = \frac{1}{d}\\d = 0.415~m[/tex]

Using this distance in either of the equations give the magnitude of the charge.

[tex]12 = \frac{1}{4\pi\epsilon_0}\frac{Q}{(0.415)^2}\\12 = \frac{1}{4\pi (8.8\times 10^{-12})}\frac{Q}{(0.415)^2}\\Q = 2.285 \times 10^{-10}~C[/tex]

Final answer:

The distance to the point charge is 0.415 meters. The magnitude of the charge is 231 picocoulombs (pC).

Explanation:

To calculate the distance to the point charge when the electric potential is 4.98 V and electric field magnitude is 12.0 V/m, we can use the relationship between electric field (E) and electric potential (V), which is E = -dV/dr. Rearranging the equation to solve for the distance (r), we get r = V/E = 4.98 V / 12.0 V/m = 0.415 m.

To find the magnitude of the charge (q), we can use the electric potential formula for a point charge: V = k*q/r, where k is the Coulomb's constant (8.99 x [tex]10^9 Nm^2/C^2[/tex]). Rearranging the formula and solving for q, we get q = V*r/k = (4.98 V * 0.415 m) / (8.99 x [tex]10^9 Nm^2/C^2[/tex]) = 2.31 x [tex]10^-^1^0[/tex] C or 231 pC.

Answer:

[tex]\lambda=1282nm[/tex]

Explanation:

The wavelength of the photons emitted due to an  atomic electron transition in a hydrogen atom, is given by the Rydberg formula:

[tex]\frac{1}{\lambda}=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2}})[/tex]

Here [tex]R_H[/tex] is the Rydberg constant for hydrogen and [tex]n_1,n_2[/tex] are the lower and higher quantum number for the energy levels of the  atomic electron transition, respectively. Replacing the given values and solving for [tex]\lambda[/tex]

[tex]\frac{1}{\lambda}=1.097*10^7m^{-1}(\frac{1}{3^2}-\frac{1}{5^2}})\\\frac{1}{\lambda}=7.81*10^5m^{-1}\\\lambda=\frac{1}{7.81*10^5m^{-1}}\\\lambda=1.282*10^{-6}m\\\lambda=1.282*10^{-6}m*\frac{1nm}{10^{-9}m}\\\lambda=1282nm[/tex]