The sound source of a ship’s sonar system operates at a frequency of 22.0 kHz . The speed of sound in water (assumed to be at a uniform 20∘C) is 1482 m/s . What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling straight toward the ship at 4.95 m/s ? Assume that the ship is at rest in the water.

Your target variable is Δf, the magnitude of the difference in frequency between the waves emitted from the sonar device and the waves received by the device after reflecting off the whale. Write an expression for Δf in terms of the relevant frequencies using the subscript notation introduced above.

Express your answer in terms of some or all of the variables fLe, fLr, fSe, and fSr.

Δf =

The new temperature is: d. 200 K

The Ideal Gas Law is given by:

[tex]PV = nRT[/tex]

where:
[tex]P[/tex] = pressure
[tex]V[/tex] = volume
[tex]n[/tex] = number of moles
[tex]R[/tex] = universal gas constant
[tex]T[/tex] = temperature

Given initial conditions:

[tex]P_1 = 2[/tex] atm (gauge pressure)
[tex]T_1 = 200[/tex] K
[tex]V_1 = V[/tex]
[tex]n = 10[/tex] moles

Final conditions:

[tex]P_2 = 1[/tex] atm (gauge pressure)
[tex]V_2 = 2V[/tex]
[tex]T_2 = ?[/tex]

We can use the combined gas law equation to relate the initial and final states of the gas:

[tex]\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}[/tex]

Substituting in the given values:

[tex]\frac{2V}{200} = \frac{1 \cdot 2V}{T_2}[/tex]

Solving for [tex]T_2[/tex]:

[tex]\frac{2V}{200} = \frac{2V}{T_2}[/tex]

By simplifying, we get:

[tex]\frac{2}{200} = \frac{2}{T_2}[/tex]

Cross-multiplying gives:

[tex]2 \cdot T_2 = 400[/tex]

Dividing both sides by 2 gives:

[tex]T_2 = 200[/tex] K

Therefore, the new temperature [tex](T_2)[/tex] after the volume is doubled and the pressure is halved is 200 K.

To solve this problem we will apply the geometric concepts of the Volume based on the consideration made of the radius measurement. The Volume must be written in differential terms of the radius and from the formula of the margin of error the respective response will be obtained.

The error in radius of sphere is not exceeding 2%

[tex]\frac{dr}{r} = \pm 0.02[/tex]

The objective is to find the percentage error in the volume.

The volume can be defined as

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

Differentiate with respect the radius we have,

[tex]\frac{dV}{dr} = 4\pi r^2[/tex]

[tex]dV = 4\pi r^2 \times dr[/tex]

[tex]dV = 4\pi r^2 (\pm 0.02r)[/tex]

[tex]dV = \pm 4\times 0.02 \times \pi r^3[/tex]

The percentage change in the volume is as follows

[tex]\% change = \frac{dV}{V} \times 100[/tex]

[tex]\% change = \frac{\pm 4 \times 0.02 \times \pi r^3 \times 3}{4\pi r^3}\times 100[/tex]

[tex]\% change = \pm 6\%[/tex]

Therefore the percentage change in volume is [tex]\pm 6\%[/tex]

Answer:

D and compound

Explanation:

because N2 is = to a compound

According to the US green building council, the US building account for 39% of world primary energy consumption . Electricity has approximately 78% of total building energy consumption and also contributes to GHG emissions

Answer:

40%

Explanation: United States Green Building Council is a body aimed at ensuring reduced green house gas emissions from activities taking place in building. they carry out surveys, carry out enlightenment activities and release the reports of and trending green house emission issues all these are to guarantee safe and healthy living for all. A total of 40% of Green house emissions are from buildings from the construction stage to it usage.

Answer:

a) [tex] W_B = F_B = 62.4 \frac{lb}{ft^3} (6ft*28ft*93ft)= 974937.6 lb[/tex]

b) [tex] W_g= 62.4 \frac{lb}{ft^3} * (9ft*28ft*93ft) -974937.6 lb =487468.8 lb[/tex]

Explanation:

Part a

For this case we have the situation illustrated on Figure 1.  We will have two forces involved in equilibrium the weight [tex] W_B[/tex] and the Bouyance force[tex] F_B[/tex], and since the system is on equilibrium we have:

[tex] \sum F_{vertical}=0[/tex]

So then we have:

[tex] W_B = F_B = \gamma_{w} V_s[/tex]

Where [tex] V_s[/tex] represent the submerged volume. [tex]\gamma_w[/tex] represent the specific weight for the fluid. So we can replace and we have:

[tex] W_B = F_B = 62.4 \frac{lb}{ft^3} (6ft*28ft*93ft)= 974937.6 lb[/tex]

Part b

As we can see on figure 2 attached we have the illustration for this case. We add the weight for the grain and now the depth is 9ft.

W can do the balance of forces in the vertical and we got again:

[tex] W_B +W_g = F_B[/tex]

Where [tex] W_g[/tex] represent the weight for the grain.

And if we solve for [tex] W_g[/tex] we got:

[tex] W_g = F_B -W_B[/tex]

[tex] W_g =\gamma_w V_S -W_B[/tex]

Where [tex] \gamma_w[/tex] represent the specific weight of rthe water and [tex] V_s[/tex] the submerged volume. If we replace we got:

[tex] W_g= 62.4 \frac{lb}{ft^3} * (9ft*28ft*93ft) -974937.6 lb =487468.8 lb[/tex]

The weight of the unloaded barge is 975,769.6 lbs, and the weight of the grain is 486,572.8 lbs. These calculations were made based on Archimedes' principle, which states that the buoyant force (the weight of water displaced) is equal to the weight of the object.

This question is about calculating the weight of a river barge and its load based on the principles of fluid mechanics. Here, we are examining the fact that the weight of the water displaced by the barge equals the weight of the barge according to Archimedes' principle.

Firstly, we calculate the unloaded weight of the barge. The volume of water displaced by the barge when it is empty is the volume of a rectangular prism with dimensions 28ft x 93ft x 6ft, which gives us 15,624 cubic feet. Given that the density of water is 62.4 lbs/ft³, the weight of this water, which is equal to the weight of the empty barge, would be (Volume x Density) = 15,624ft³ x 62.4 lbs/ft³ = 975,769.6 lbs.

Secondly, let's calculate the weight of the grain. The volume of water displaced when the barge is loaded is the volume of a rectangular prism with dimensions 28ft x 93ft x 9ft, which equals 23,436 cubic feet. The weight of this water, which is equal to the weight of the loaded barge, is (Volume x Density) = 23,436ft³ x 62.4 lbs/ft³ = 1,462,342.4 lbs. Hence, the weight of the grain is the weight of the loaded barge minus the weight of the barge itself = 1,462,342.4 lbs - 975,769.6 lbs = 486,572.8 lbs.

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To solve this problem we will use the concepts given by Coulomb's law defined for force, said law is mathematically described as

[tex]F = \frac{kq_1 q_2}{r^2}[/tex]

Here,

k = Coulomb's constant

[tex]q_{1,2}[/tex]= Charge of two protons

r = Distance between them

F = Force

Our values are given as,

[tex]F =2.3*10^{-26} N[/tex]

[tex]q_1 =q_2 = 1.6*10^{-19} C[/tex]

[tex]k =9*10^9 Nm^2/C2[/tex]

Rearrenging to find the distance and replacing we have that

[tex]r^2=\frac{(9*10^9 )(1.6*10^{-19})^2 }{2.3*10^{-26} }[/tex]

[tex]r^2=10.01*10^{-3} m^2[/tex]

[tex]r =\sqrt{10.01*10^{-3} m^2}[/tex]

[tex]r = 0.100m[/tex]

Therefore the correct option is B.

Answer:

Can a room be gravitationally shielded? No, it can't.

Explanation:

the room cannot be gravitationally shielded because there is only one gravitational charge, in this case is mass. Mass can always be positive. the room can be electrically shielded because there are two type of charge, positive and negative charge than can cancel each other out.

The question pertains to physics, specifically the concept of density. The mass of the gold in the ring is computed to be 10.615 g based on the provided volume displacement and density of gold. To determine the cost of the ring, additional information such as the current market price for gold and labor costs would be necessary.

The subject matter of this question relates to the physical property of density and its application in determining the mass of gold in a wedding ring. Given that the density of gold is 19.3 g/cm³ and the ring displaced 0.55 mL of liquid, one can calculate the mass of the ring using the rule that 1 mL equals 1 cm³. Hence, if the ring displaces 0.55 mL of the liquid, it means the volume of the ring is 0.55 cm³.

To find mass, we use Density = Mass/Volume, so Mass = Density * Volume. Therefore, the mass of the ring will be 19.3 g/cm³ * 0.55 cm³ = 10.615 g.

This calculation gives us the mass of the gold in the ring.

However, to determine the cost of the ring, we need more information such as the market price of gold per gram and the cost of workmanship and other potential elements in the ring, which are not provided in the question.

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Answer:

Explanation:

q = - 4.25 nC = - 4.5 x 10^-9 C

(A) d = 0.250 m

The formula for the electric field is given by

[tex]E = \frac{1}{4\pi \epsilon _{0}}\frac{q}{d^{2}}[/tex]

By substituting the values

[tex]E = \frac{9\times 10^{9}\times 4.5\times10^{-9}}{0.25\times 0.25}[/tex]

E = 648 N/C

(B) As the charge is negative in nature so the direction of electric field is towards the charge and downwards.

(a) The magnitude of the electric field is 612 N/C.

(b) The direction of the electric field will be up, away from the particle.

(c) The distance from the particle is 1.71 m.

The magnitude of the electric field is calculated as follows;

E = (kq)/r²

where;

k is Coulomb's constant

q is the charge

r is distance

E = ( 9 x 10⁹ x 4.25 x 10⁻⁹)/(0.25 x 0.25)

E =  612 N/C

The direction of the electric field is always opposite to the direction of the negative charge.

Thus, the direction of the electric field will be up, away from the particle.

The distance from the particle is determined using the following formula;

E = (kq)/r²

r² = kq/E

r² = (9 x 10⁹ x 4.25 x 10⁻⁹) / 13

r² = 2.94

r = √2.94

r = 1.71 m

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To solve this problem we will make a graph that allows us to understand the components acting on the body. In this way we will have the centripetal Force and the Force by gravity generating a total component. If we take both forces and get the trigonometric ratio of the tangent we would have the angle is,

[tex]T_x = nsinA = \frac{mv^2}{r}[/tex]

[tex]T_y = ncosA = mg[/tex]

Dividing both.

[tex]tan A = \frac{v^2}{rg}[/tex]

[tex]tan A = \frac{11.7^2}{50*9.8}[/tex]

[tex]A = tan^{-1} (0.279367)[/tex]

[tex]A = 15.608\°[/tex]

Therefore the angle that should the curve be banked is 15.608°

Answer:

23.0760769 %

Explanation:

[tex]m_1[/tex] = Mass of freight car = 15000 kg

[tex]m_2[/tex] = Mass of second car = 50000 kg

[tex]v_1[/tex] = Velocity of freight car = 2 m/s

[tex]v_2[/tex] = Velocity of second car = 0

v = Combined mass velocity

As the linear momentum of the system is conserved we have

[tex]m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}\\\Rightarrow v=\dfrac{15000\times 2+50000\times 0}{15000+50000}\\\Rightarrow v=0.46153\ m/s[/tex]

The initial kinetic energy

[tex]K_i=\dfrac{1}{2}15000\times 2^2\\\Rightarrow K_i=30000\ J[/tex]

Final kinetic energy

[tex]K_f=\dfrac{1}{2}(15000+50000)\times 0.46153^2\\\Rightarrow K_f=6922.82307\ J[/tex]

The percentage is given by

[tex]\dfrac{6922.82307}{30000}\times 100=23.0760769\ \%[/tex]

The change in percentage of initial kinetic energy is 23.0760769 %

Answer:

The distance is 641.8207 light years, and the star Betelgeuse is further away when compared to other stars

Historical event: Benedict XI succeeds Boniface VIII as pope (1302)

Explanation:

the solution is in the attached Word file

Explanation:

Technology-based:

1.As the name implies technology, no technology will be clarified about it. It only depends on the variables, which describes them.

2. That is based on a single facility's findings.

3. It takes into account the contaminants type and volume, and their equations to monitor them.

4. This is reserved for city or urban wastewater treatment plants only.

5. It takes into account the pH, need for oxygen and the suspended solids.

Water quality based on:

1.This is enforced if there is a need to apply stricter limits to pollutants that are not pleased with the limits of technology.

2. All basing on water quality was risk-based.

3. They placed some less importance on the technologies which is used in the technology based limit.

Water quality dependent restrictions are used as cool water fishery for streams.

The act on clean water will also include bodies of water belonging to wildlife, agriculture and others. The law also included that the physical chemical and biological variables of all the state water bodies must be controlled by these water quality based limits.

Answer:

f = 614.28 Hz

Explanation:

Given that, the length of the air column in the test tube is 14.0 cm. It can be assumed that the speed of sound in air is 344 m/s. The test tube is a kind of tube which has a closed end. The frequency in of standing wave in a closed end tube is given by :

[tex]f=\dfrac{nv}{4l}[/tex]

[tex]f=\dfrac{1\times 344}{4\times 0.14}[/tex]

f = 614.28 Hz

So, the frequency of the this standing wave is 614.28 Hz. Hence, this is the required solution.

To solve this problem we will first find the distance between each of the points 'x' and then use it as the variable of the distance in the function of the electric field. According to the graph the value of 'x' is,

[tex]x = \frac{\sqrt{3}}{2}a[/tex]

[tex]x = \frac{\sqrt{3}}{2}(0.5)[/tex]

[tex]x = 0.43301m[/tex]

The magnitude of the electric field is

E=\frac{kQ}{x^2}

Here,

k = Coulomb's Constant

Q = Charge

x = Distance

[tex]E=\frac{(9*10^9)(5*10^{-9})}{(0.43301)^2}[/tex]

[tex]E = 240.002N/C \approx 240N/C[/tex]

Therefore the correct answer is A.

For both cases we will use the proportional values of the distance referring to the amplitude and intensity. Theoretically we know that the intensity is inversely proportional to the square of the distance, while the amplitude is inversely proportional to the distance, therefore,

PART A )  Intensity is inversely proportional to the square of the distance

[tex]Intensity \propto \frac{1}{distance^2}[/tex]

Therefore the intensity of the two values would be

[tex]\frac{I_{27}}{I_{13}} = \frac{(13km)^2}{(27km)^2}[/tex]

[tex]\mathbf{\therefore \frac{I_{27}}{I_{13}} = 0.232 }[/tex]

PART B) Amplitude is inversely proportional to the distance

[tex]Amplitude \propto \frac{1}{distance}[/tex]

[tex]\frac{A_{27}}{A_{13}}= \frac{(13km)}{(27km)}[/tex]

[tex]\mathbf{\therefore\frac{A_{27}}{A_{13}}= 0.4815}[/tex]

The intensity ratio of an earthquake P wave passing through the Earth and detected at two points is equal to the square of the amplitude ratio.

The ratio of the intensities of an earthquake P wave passing through the Earth and detected at two points is equal to the square of the ratio of their amplitudes.

Let's assume the amplitudes of the earthquake P wave at the two points are given by A27 and A13.

The intensity of a wave is given by the square of its amplitude. Therefore, the ratio of the intensities I27/I13 is equal to the square of the ratio of the amplitudes A27/A13.

So, the answer is:

(a) I27/I13 = (A27/A13)2

(b) A27/A13

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Final answer:

Each consecutive squirt from the water gun is separated by 3 meters, calculated by multiplying the speed of the squirts (15 m/s) by the interval between each squirt (0.2 seconds).

Explanation:

The question is asking to calculate the distance at which each consecutive squirt from a water gun is separated when the water gun fires squirts at a certain rate and speed. Given that the water gun fires 5 squirts per second and the speed of the squirts is 15 m/s, we can use the formula distance = speed × time to find the separation distance between squirts.

Since there are 5 squirts per second, each squirt is separated by 1/5 of a second, or 0.2 seconds. To find the separation distance, we multiply the speed of the squirts by the time interval between each squirt:

Distance = Speed × Time
Distance = 15 m/s × 0.2 s
Distance = 3 meters

Therefore, each consecutive squirt is separated by 3 meters.

Answer:[tex]A_x[/tex] = Acos[tex]\theta[/tex]

Explanation:

A vector in this situation have two components

1) The X-component

2) The Y-component

So as we put [tex]cos\theta[/tex] with the x-axis while [tex]sin\theta[/tex] with the y- axis and this case our answer should be

[tex]A_x[/tex] = Acos[tex]\theta[/tex]

I hope this will answer your question,

An image is also provided please have a look at that.

Thank you.

Answer: Ax = Acos(theta)

Therefore, the x component of A: Ax = Acos(theta)

Explanation:

The attached image is a pictorial representation of the question:

From the attached image,

Cos(theta) = adjacent/hypothenus

Cos(theta) = Ax/A

Making Ax the subject of formula,

Ax = Acos(theta)

Therefore, the x component of A: Ax = Acos(theta)

Answer:

[tex]F_H_n=230.04 N[/tex]

The Required  horizontal force is 230.04N

Explanation:

Since the velocity is constant so acceleration is zero; a=0

Now the horizontal force required to move the pickup is equal to the frictional force.

[tex]F_H_n=F_f\\F_h=mg*u[/tex]

where:

F_{Hn} is the required Force

u is the friction coefficient

m is the mass

g is gravitational acceleration=9.8m/s^2

[tex]200=mg*u[/tex]                         Eq (1)

Now, weight increases by 42% and friction coefficient decreases by 19%

New weight=(1.42*m*g) and new friction coefficient=0.81u

[tex]F_H=(1.42m*g*.81u)[/tex]          Eq (2)

Divide Eq(2) and Eq (1)

[tex]\frac{F_H_n}{200}=\frac{1.42m*g*0.81u}{m*g*u}\\F_H_n=1.42*0.81*200\\F_H_n=230.04 N[/tex]

The Required  horizontal force is 230.04N

Answer: %(∆W) = 0.37%

Explanation:

According to Newton's law of gravitation which states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers

F = Gm1m2/r^2

Where

F = force between the masses

G universal gravitational constant

m1 and m2 = mass of the two particles

r = distance between the centre of the two mass

Therefore, weigh of an object on earth is inversely proportional to the square of its distance from the centre of the earth

W₁/W₂ = r₂²/r₁² .....1

W₂ = W₁r₁²/r₂²

At sea level the weight of the plane is W1 and at distance r₁ from the centre of the earth which is equal to the radius of the earth.

The radius of the earth is = 6378.1km

r₁ = radius of the earth = 6 378.1km = 6,378,100m

r₂ = r₁ + 11,719.342m = 6,378,100m + 11,719.342m

r₂ = 6,389,819.342m

W₂ = W₁r₁²/r₂²

W₂ = W₁[(6378100)²/(6,389,819.342)²]

W₂ = W₁[0.996335234422]

W₂/W₁ = 0.9963

fraction reduction of the weight is

ΔW/W₁ = 1 - W₂/W₁ = 1 - 0.9963 = 0.0037

percentage change :

%(∆W) = 0.0037 × 100% = 0.37%

Therefore, the percentage weight loss is 0.37%

Answer:

A) At 2.2 s the position of the particle is 9.4 m.

B) At t =2.2 s the velocity is 9.7 m/s.

C) At t = 2.2 s the acceleration of the particle is 8.8 m/s²

Explanation:

Hi there!

A)The velocity of the particle is given by the variation of the position over time. If the time interval is very small, we get the instantaneous velocity that can be expressed as follows:

dx/dt = 2 · t²

Separating variables, we can find the equation of position as a function of time:

dx = 2 · t² · dt

Integrating both sides between x0 = 2.3 m and x and from t0 = 0 and t:

∫ dx = 2 ∫ t² · dt

x - 2.3 m = 2/3 · t³

x = 2.3 m + 2/3 m/s³ · t³

Replacing t = 2.2 s:

x = 2.3 m + 2/3 m/s³ · (2.2 s)³

x = 9.4 m

At 2.2 s the position of the particle is 9.4 m

B) Now, let´s evaluate the velocity function at t = 2.2 s:

v = 2 · t²

v = 2 m/s³ · (2.2 s)²

v = 9.7 m/s

At t =2.2 s the velocity is 9.68 m/s

C) The acceleration is the variation of the velocity over time (the derivative of the velocity):

dv/dt = a

a = 4 · t

At t = 2.2 s:

a = 4 m/s³ · 2.2 s

a = 8.8 m/s²

At t = 2.2 s the acceleration of the particle is 8.8 m/s²

(A) The particle's position at time, t = 2.2 s is 7.1 m.

(B) The velocity of the particle at 2.2 s is 9.68 m/s.

(C) The acceleration of the particle at 2.2 s is 8.8 m/s².

The given parameters:

The particle's position at time, t = 2.2 s is calculated as follows;

[tex]x = \int\limits^{t_1}_{t_0} {v} \, dt\\\\ x = \int\limits^{t_1}_{t_0} {2t^2}\\\\ x = [\frac{2t^3}{3} ]^{2.2}_0\\\\ x = \frac{2(2.2)^3}{3} \\\\ x = 7.1 \ m[/tex]

The velocity of the particle at 2.2 s is calculated as follows;

[tex]v = 2t^2\\\\ v = 2(2.2)^2\\\\ v = 9.68 \ m/s[/tex]

The acceleration of the particle at 2.2 s is calculated as follows;

[tex]a = \frac{dv}{dt} \\\\ a = 4t\\\\ a = 4(2.2)\\\\ a = 8.8 \ m/s^2[/tex]

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To calculate the pressure in the body we will use the definition of the hydrostatic pressure for which the pressure of a body at a certain distance submerged in a liquid is defined. After calculating this relationship we will apply the equations of the relationship between the volume and the pressure to calculate the volume in state 2,

[tex]P = P_{atm} + \rho gh[/tex]

Here,

[tex]\rho[/tex]= Density of the Fluid (Water)

g = Acceleration due to gravity

h = Height

[tex]P = P_{atm} + 10^3*9.8*8[/tex]

[tex]P = 1.01*10^{5} +10^3*9.8*8[/tex]

[tex]P = 179400Pa[/tex]

Applying the equations of relationship between volume and pressure we have

[tex]P_1V_1 = P_2 V_2[/tex]

[tex]179400*5.7 = 101000*V_2[/tex]

[tex]V_2 = 10.12L[/tex]

Therefore the volume that would his lungs expand if he quickly rose to the surface is 10.12L

Answer:

Ex = -16.51 N/C Ey = 79.14 N/C

Explanation:

As the  electric force is linear, and the electric field, by definition, is just this electric force per unit charge, we can use the superposition principle to get the electric field produced by both charges at any point, as the other charge were not present.

So, we can first the field due to q1.

Due  to q₁ is negative, and located on the y axis, the field due to this charge will be pointing upward,(like the attractive force between q1 and the positive test charge that gives the direction to the field), as follows.

E₁ = k*(3.65 nC) / r₁²

If we choose the upward direction as the positive one (+y), we can find both components of E₁ as follows:

E₁ₓ = 0   E₁y = 8.99*10⁹*3.65*10⁻⁹ / (0.600)²m² = 91.15 N/C (1)

For the field due to q₂, we need first to get the distance along a straight line, between q2 and the origin.

It will be just the pythagorean distance between the points located at the coordinates (1.10 m, 0.800 m) and (0,0), as follows:

r₂² = 1.10²m² + (0.800)²m² = 1.85 m²

The magnitude of the electric field due to  q2 can be found as follows:

E₂ = k*q₂ / r₂² = 8.99*10⁹*(4.2)*10⁹ / 1.85 = 20.41 N/C (2)

Due to q2 is positive, the force on the positive test charge will be repulsive, so E₂ will point away from q2, to the left and downwards.

In order to get the x and y components of E₂, we need to get the projections of E₂ over the x and y axis, as follows:

E₂ₓ = E₂* cosθ, E₂y = E₂*sin θ

the  cosine of  θ, is just, by definition, the opposite  of x/r₂:

⇒ cos θ =- (1.10 m / √1.85 m²) =- (1.10 / 1.36) = -0.809

By the same token, sin θ can be obtained as follows:

sin θ = - (0.800 m / 1.36 m) = -0.588

⇒E₂ₓ = 20.41 N/C * (-0.809) = -16.51 N/C (pointing to the left) (3)

⇒E₂y = 20.41 N/C * (-0.588) = -12.01 N/C (pointing downward) (4)

The total x and y components due to both charges are just the sum of the components of Ex and Ey:

Ex = E₁ₓ + E₂ₓ = 0 + (-16.51 N/C) = -16.51 N/C

From (1) and (4), we can get Ey:

Ey = E₁y + E₂y =  91.15 N/C + (-12.01 N/C) =79.14 N/C

Answer:

Ex = 15.505 N/C

Ey = 79.144 N/C

Explanation:

Particle 1

[tex]E_{1} = \frac{k*q_{1} }{r^2_{1}} \\\\r^2_{1} = 1.1^2+0.8^2\\\\r^2_{1} = 1.85 m^2\\\\Q (angle-with-x-axis) = arctan(\frac{0.8}{1.1}) = 36.03 degree\\\\ E_{1} = \frac{(8.99*10^9)*(4.2*10^(-9)) }{1.85}\\\\E_{1} = 20.4096 N/C[/tex]

Away from the particle at (0,0) due to + charge

Particle 2

[tex]E_{2} = \frac{k*q_{2} }{r^2_{2}} \\\\r^2_{2} = 0.36 m^2\\\\Q (angle-with-x-axis) = arctan(\frac{0.6}{0}) = 90 degree\\\\ E_{2} = \frac{(8.99*10^9)*(3.65*10^(-9)) }{0.36}\\\\E_{2} = 91.149 N/C[/tex]

Towards from the particle at (0,0) due to - charge

Resultant Electric field in y direction

[tex]E_{res,y} = E_{2} -E_{1}*cos(Q)\\E_{res,y} = (91.149) - (20.4096)*sin(36.03)\\\\E_{res,y} = 79.144 N/C[/tex]

Resultant Electric field in x direction

[tex]E_{res,x} = E_{1}*cos(Q)\\E_{res,y} =(20.4096)*cos(36.03)\\\\E_{res,x} = 16.505 N/C[/tex]

The specific kinetic energy of a mass (in kilojoules per kilogram) moving at a velocity of 40 m/s is calculated using the kinetic energy formula, resulting in 0.8 kJ/kg.

The specific kinetic energy of a mass moving with a velocity can be determined using the formula for kinetic energy, K = 0.5*m*v². In this case, where the velocity 'v' is given as 40 m/s, and we want to solve the kinetic energy per kilogram, we can consider the mass 'm' as 1 kg. Hence the specific kinetic energy would be K = 0.5*(1 kg)*(40 m/s)² = 800 J = 0.8 kJ/kg, because 1 kilojoule (kJ) = 1000 joules (J).

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Explanation:

Since the comb has a net charge, it attracts the paper, which has a net charge equal to zero. When the paper touches the comb, an electrical interaction is established between the charge of the comb and the neutral paper, because of this, the paper now has a net charge with the same sign of the comb and they repel.

Answer:

[tex]I_{1}[/tex] = (πD[tex]I_{2}[/tex])/R

Explanation:

If we define the magnitude of the field as B, then we have:

Total magnitude of the field [tex]B_{t}[/tex] = magnitude of the field B_loop + magnitude of the field B_wire. The total magnitude is equivalent to zero. Therefore, the field B_loop has an inward direction while the field B_wire has an outward direction.

B_loop = (μ0)*([tex]I_{2}[/tex])/2*R

B_wire = (μ0)*([tex]I_{1}[/tex])/2*π*D

Thus:

B_loop = B_wire at the center of the loop.

(μ0)*([tex]I_{2}[/tex])/2*R = (μ0)*([tex]I_{1}[/tex])/2*π*D

[tex]I_{1}[/tex] = (πD[tex]I_{2}[/tex])/R

The magnitude of current at the center of the loop is [tex]I_1 = \frac{\pi D I_2}{R}[/tex].

The given parameters;

The magnetic field at the center of the loop is calculated as follows;

[tex]B_o = \frac{\mu_o I_2}{2R}[/tex]

The magnetic field at the distance of the wire is calculated as follows;

[tex]B_o = \frac{\mu_o I_1}{2\pi D}[/tex]

The magnitude of current at the center of the loop is calculated as follows;

[tex]\frac{\mu_o I_2 }{2R} = \frac{\mu_o I_1}{2\pi D} \\\\I_1 = \frac{\pi D I_2}{R}[/tex]

Thus, the magnitude of current at the center of the loop is [tex]I_1 = \frac{\pi D I_2}{R}[/tex].

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Answer:

The conversion factor used here will be 1000 (kg/m^3)/(g/cm^3).

Which is a combination of two conversion factors:

1 kg = 1000 g

1 x 10^6 cm^3 = 1 m^3

Explanation:

We will use unitary method to convert g/cm^3 into kg/m^3. This is shown below:

Since, 1 kilogram is equivalent to 1000 gram and 1 meter is equivalent to 100 centimeter. Therefore:

1 g/cm^3 = (1 g/ cm^3)(1 kg/ 1000 g)(100 cm / 1 m)^3

1 g/cm^3 = 1000 kg/m^3

Hence, the conversion factor that will be multiplied is found to be 1000.

Using this in our case, we get:

Density of silicon = (2.33)(1000) kg/m^3

Density of Silicon = 2330 kg/m^3

Answer:

1000

Explanation:

Conversion from 'g' to 'kg': divide by 1000g/kg

Conversion from 'cm^3' to 'm^3': divide by 1000000cm^3/m^3

2.33g/cm^3 = [tex]\frac{2.33*1000000}{100}[/tex]

                    = 2330 kg/m^3

we simply multiply by 1000 to get the units converted to kg/m^3

Answer:

Change in momentum will be -2.61 kgm/sec

Explanation:

We have given mass of the rubber ball m = 0.30 kg

Velocity of the ball before the impact [tex]v_1=4.5m/sec[/tex]

Velocity of ball after impact [tex]v_2=-4.2m/sec[/tex] ( negative sign is due to opposite direction of motion )

Change in momentum is given by [tex]m(v_2-v_1)=0.3\times (-4.2-4.5)=0.3\times =0.3\times -8.7=-2.61kgm/sec[/tex] ( negative sign shows the direction of change in momentum )

Answer:

-0.09 kg m/s

Explanation:

Answer:

[tex]2.537\times 10^{6}\ years[/tex]

Explanation:

Distance to Andromeda galaxy

[tex]2.537\times 10^{6}\ ly=2.537\times 10^{6}\times c\times y[/tex]

Speed of light is

[tex]c=3\times 10^8[/tex]

Time is given by

[tex]t=\dfrac{Distance}{Speed}\\\Rightarrow t=\dfrac{2.537\times 10^{6}\times c\times y}{c}\\\Rightarrow t=2.537\times 10^{6}\ y[/tex]

Hence, the light from Andromeda left [tex]2.537\times 10^{6}\ years[/tex] ago

The light observed from the Andromeda galaxy by the astronomer left Andromeda around 2 million years ago, showcasing the vast distances in space and providing insights into the universe's evolution.

The light from Andromeda galaxy that the astronomer observes left Andromeda approximately 2 million years ago. This is because the distance in light years is the same as the time it takes for the light to reach us.

This phenomenon is due to the vast distances in space. The Andromeda galaxy is the closest large galaxy to the Milky Way, located 2 million light years away from us.

Understanding the age of the light we observe helps us gain insights into the history of distant galaxies and the evolution of the universe.

a) The minimum frequency of the light must be [tex]1.26\cdot 10^{15} Hz[/tex]

b) The maximum velocity of the electrons is [tex]5.93\cdot 10^5 m/s[/tex]

Explanation:

a)

The photoelectric effect is a phenomenon that occurs when electromagnetic radiation hits the surface of a metal causing the release of electrons from the metal's surface.

The equation of the photoelectric effect is:

[tex]hf = \phi +K_{max}[/tex]

where :

[tex]hf[/tex] is the energy of the incoming photons, where

[tex]h[/tex] is the Planck's constant

[tex]f[/tex] is the frequency of the incoming photons

[tex]\phi[/tex] is the work function of the metal, the minimum energy that the photons must have in order to be able to free electrons from the metal

[tex]K_{max}[/tex] is the maximum kinetic energy of the emitted electrons

In order to free electrons, the minimum energy of the photons must be at least  equal to the work function (so that the kinetic energy of the electrons is zero, [tex]K_{max}=0[/tex]. Therefore,

[tex]h f_0 = \phi[/tex]

In this case,

[tex]\phi = 5.22 eV \cdot (1.6\cdot 10^{-19})=8.35\cdot 10^{-19} J[/tex]

Therefore, the minimum frequency of the photons must be

[tex]f_0 = \frac{\phi}{h}=\frac{8.35\cdot 10^{-19}}{6.63\cdot 10^{-34}}=1.26\cdot 10^{15} Hz[/tex]

b)

In this case, the wavelength of the incoming light is

[tex]\lambda = 200 nm = 200 \cdot 10^{-9} m[/tex]

We can find the frequency by using the wave equation:

[tex]f=\frac{c}{\lambda}=\frac{3\cdot 10^8}{200\cdot 10^{-9}}=1.5\cdot 10^{15} Hz[/tex]

Now we can use the equation of the photoelectric effect to find the maximum kinetic energy of the electrons:

[tex]K_{max} = hf-\phi = (6.63\cdot 10^{-34})(1.5\cdot 10^{15})-8.35\cdot 10^{-19}=1.60\cdot 10^{-19} J[/tex]

And therefore, we can find their velocity by using the equation for the kinetic energy:

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

where

[tex]m=9.11\cdot 10^{-31} kg[/tex] is the mass of the electrons

v is their speed

Solving for v,

[tex]v=\sqrt{\frac{2K_{max}}{m}}=\sqrt{\frac{2(1.6\cdot 10^{-19})}{9.11\cdot 10^{-31}}}=5.93\cdot 10^5 m/s[/tex]

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