Two loudspeakers are placed next to each other and driven by the same source at 500 Hz. A listener is positioned in front of the two speakers and on the line separating them, thus creating a constructive interference at the listener's ear. What minimum distance would one of the speakers be moved back away from the listener to produce destructive interference at the listener's ear? (The speed of sound = 340 m/s.)

Answer:

F_e = +/- 0.013133 N

Explanation:

Given:

- charge on each sphere q = -56 nC

- separation of spheres r = 3.0 cm

- charge constant k = 8.99*10^9

Find:

- Magnitude of Electric Force F_e on each sphere.

Solution:

The magnitude of electric Force F_e of one sphere on other is:

                                F_e = k*q^2 / r^2

- Plugging the given values :

                                F_e = (8.99*10^9) * (56*10^-19)^2 / (0.03)^2

                                F_e = 0.013133 N

- An equal and opposite force is experienced on the other sphere. Hence, F_e = + / - 0.013133 N

Answer:

2019 light bulbs

Explanation:

So the electrical energy consumed by each 59-W light bulb within 7 hours is the product of the power and time duration

E = Pt = 59 * 7 = 413 Wh or 0.413 kWh

If each light bulb consumes 0.413 kWh, then the total number of light bulbs needed to consume 834 kWh would be

834 / 0.413 = 2019 light bulbs

Answer:

v = 9.824 x 10³ m/s

Explanation:

given,

Linear momentum of photon,λ = 740 n m

for photon,

[tex]p=\dfrac{h}{\lambda}[/tex]

h is the planks constant

P is the momentum

[tex]p=\dfrac{6.626\times 10^{-34}}{740\times 10^{-10}}[/tex]

      p = 8.95 x 10⁻²⁷ kg.m/s

For electron

p = m v

mass of electron = 9.11 x 10⁻³¹ Kg

[tex]v = \dfrac{8.95\times 10^{-27}}{9.11\times 10^{-31}}[/tex]

v = 9.824 x 10³ m/s

hence, the velocity of electron is equal to   v = 9.824 x 10³ m/s

Answer:

4.has gained two electrons

Explanation:

There exist electrovalent bonding the compound MgS . In electrovalent bonding, there is a transfer of electrons from the metal to non-metal.

Magnesium atom has an atomic number 12 and its electron configuration is 2,8,2

Sulfur atom , a non-metal has atomic number of 16 and its electron configuration = 2,8,6

This means that magnesium as a metal needs to loose two electrons from its valence shell to attain its stable structure.Also sulfur requires two more electron to achieve its octet structure.

Hence a transfer of electrons will take place from magnesium atom to sulfur atom, sulfur gaining two electrons.

Final answer:

In MgS, the sulfide ion has gained two electrons, resulting in a charge of -2, represented as S²-. Magnesium loses two electrons to form Mg²+, balancing the electron transfer to create a stable ionic compound.

Explanation:

In the compound MgS, the sulfide ion has gained two electrons to achieve a stable electron configuration. When sulfur (S), which has an atomic number of 16, gains two electrons, it results in an ion with 18 electrons and 16 protons. This gives the ion a charge of -2, since there are two more negative electrons than positive protons. Therefore, the sulfide ion is represented as S²-. The magnesium atom loses its two valence electrons to become a Mg²+ cation, as magnesium is in Group 2A of the periodic table and tends to lose two electrons to achieve a noble gas electron configuration. This electron transfer process is balanced, meaning the number of electrons lost by magnesium is equal to the number of electrons gained by sulfur, forming a stable ionic compound.

Answer:

F_net = - 0.365 N (Down-ward direction)

Explanation:

Given:

- Value of r = 0.75 m

- Charges on x axis are -2*q

- Charge +q on origin

- Charge on - y axis is -2q

- Charge on + y axis is +q

- q = 5.00 * 10^-6 C

Find:

-What is the magnitude of the net force on the ball of mass m that is located on the positive y-axis, because of the other four balls?

Solution:

- Force due to each of the two charges on x axis:

                                  F_x = k*(-2*q)*(+q) / r*^2

                                  r* = sqrt(2)*r

                                  F_x = -k*q^2 / r^2 (Down-wards)

- Force due to +q charge on origin:

                                  F_y = k*(+q)*(+q) / r^2

                                  F_y = + k*q^2 / r^2 (Up-wards)

- Force due to -2*q charge on y-axis:

                                  F_-2y = k*(-2*q)*(+q) / 4*r^2

                                  F_-2y = - k*q^2 / 2*r^2 (Downwards-wards)

- Total net Force on charge +q on + y-axis:

                                  2*F_x*sin(45) + F_y + F_-2y = F_net

                                 -sqrt(2)*k*q^2 / r^2 + k*q^2 / r^2 - k*q^2 / 2*r^2 = F_net

                                 (0.5-sqrt(2))*k*q^2 / r^2 = F_net

                                  F_net =  (0.5-sqrt(2))*(8.99*10^9)*(5*10^-6)^2 / 0.75^2

                                  F_net = - 0.365 N

Answer:

a) [tex]v_{max}=0.4\ m.s^{-1}[/tex]

b) [tex]a_{max}=1.6\ m.s^{-2}[/tex]

c) [tex]v_x=0.32\ m.s^{-1}[/tex]

d) [tex]a_x=0.96\ m.s^{-1}[/tex]

e) [tex]\Delta t=0.232\ s[/tex]

Explanation:

Given:

mass of the object attached to the spring, [tex]m=0.5\ kg[/tex]

spring constant of the given spring, [tex]k=8\ N.m^{-1}[/tex]

amplitude of vibration, [tex]A=0.1\ m[/tex]

a)

Now, maximum velocity is obtained at the maximum Kinetic energy and the maximum kinetic energy is obtained when the whole spring potential energy is transformed.

Max. spring potential energy:

[tex]PE_s=\frac{1}{2} .k.A^2[/tex]

[tex]PE_s=0.5\times 8\times 0.1^2[/tex]

[tex]PE_s=0.04\ J[/tex]

When this whole spring potential is converted into kinetic energy:

[tex]KE_{max}=0.04\ J[/tex]

[tex]\frac{1}{2}.m.v_{max}^2=0.04[/tex]

[tex]0.5\times 0.5\times v_{max}^2=0.04[/tex]

[tex]v_{max}=0.4\ m.s^{-1}[/tex]

b)

Max. Force of spring on the mass:

[tex]F_{max}=k.A[/tex]

[tex]F_{max}=8\times 0.1[/tex]

[tex]F_{max}=0.8\ N[/tex]

Now acceleration:

[tex]a_{max}=\frac{F_{max}}{m}[/tex]

[tex]a_{max}=\frac{0.8}{0.5}[/tex]

[tex]a_{max}=1.6\ m.s^{-2}[/tex]

c)

Kinetic energy when the displacement is, [tex]\Delta x=0.06\ m[/tex]:

[tex]KE_x=PE_s-PE_x[/tex]

[tex]\frac{1}{2} .m.v_x^2=PE_s-\frac{1}{2} .k.\Delta x^2[/tex]

[tex]\frac{1}{2}\times 0.5\times v_x^2=0.04-\frac{1}{2} \times 8\times 0.06^2[/tex]

[tex]v_x=0.32\ m.s^{-1}[/tex]

d)

Spring force on the mass at the given position, [tex]\Delta x=0.06\ m[/tex]:

[tex]F=k.\Delta x[/tex]

[tex]F=8\times 0.06[/tex]

[tex]F=0.48\ N[/tex]

therefore acceleration:

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

[tex]a_x=\frac{0.48}{0.5}[/tex]

[tex]a_x=0.96\ m.s^{-1}[/tex]

e)

Frequency of oscillation:

[tex]\omega=\sqrt{\frac{k}{m} }[/tex]

[tex]\omega=\sqrt{\frac{8}{0.5} }[/tex]

[tex]\omega=4\ rad.s^{-1}[/tex]

So the wave equation is:

[tex]x=A.\sin\ (\omega.t)[/tex]

where x = position of the oscillating mass

put x=0

[tex]0=0.1\times \sin\ (4t)[/tex]

[tex]t=0\ s[/tex]

Now put x=0.08

[tex]0.08=0.1\times \sin\ (4t)[/tex]

[tex]t=0.232\ s[/tex]

So, the time taken in going from point x = 0 cm to x = 8 cm is:

[tex]\Delta t=0.232\ s[/tex]

Final answer:

The problem involves calculating the maximum speed, maximum acceleration, speed and acceleration at a certain distance from equilibrium, and the time interval for an object in simple harmonic motion. Using formulas for SHM including maximum speed (v_max = ωA), maximum acceleration (a_max = ω^2A), and expressions for speed and acceleration at a given position, one can determine these values for the object on the spring.

Explanation:

The student has asked us to calculate various properties of an object undergoing simple harmonic motion (SHM) when attached to a spring with a known force constant and amplitude. To answer this question, one needs to use equations that describe SHM.

Maximum Speed (v_max) Calculation:

The maximum speed (v_max) of an object in SHM occurs when it passes through the equilibrium point and can be calculated using the formula v_max = ωA, where ω is the angular frequency (ω = sqrt(k/m)) and A is the amplitude of the motion.

Maximum Acceleration (a_max) Calculation:

The maximum acceleration (a_max) occurs at the maximum displacement and is given by a_max = ω^2A.

Speed at 6 cm from Equilibrium:

To find the speed at a certain position x, we use the formula v = ω sqrt(A^2 - x^2).

Acceleration at 6 cm from Equilibrium:

Acceleration at any position x is a = -ω^2x.

Time Interval to Move from 0 to 8 cm:

The time interval to move from x = 0 to a certain position x can be found using the formula for time in SHM as a function of position.

Answer:

[tex]{P_2}=146.53\ kPa[/tex]

Explanation:

Volume ,V = 3 L

Initial temperature ,T₁ = 273 K

Initial pressure ,P₁ = 105 kPa

Final temperature ,T₂ = 381 K

Lets take final pressure =P₂

We know that ,If the volume of the gas is constant ,then we can say that

[tex]\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}[/tex]

[tex]{P_2}=P_1\times \dfrac{T_2}{T_1}[/tex]

Now by putting the values in the above equation we get

[tex]{P_2}=105\times \dfrac{381}{273}\ kPa[/tex]

[tex]{P_2}=146.53\ kPa[/tex]

Therefore the final pressure will be 146.53 kPa.

The final pressure of the cylinder after being heated is 75.57 kPa.

To solve this problem, we can use the equation for Charles's Law, which states that at constant pressure, the volume of a gas is directly proportional to its temperature. In this case, the initial volume is 3.0 L and the initial temperature is 273 K. The final temperature is 381 K. Now we can set up a proportion:

(Initial volume) / (Initial temperature) = (Final volume) / (Final temperature)

Plugging in the numbers, we get:

(3.0 L) / (273 K) = (Final volume) / (381 K)

Solving for the final volume gives us:

Final volume = [(3.0 L) / (273 K)] x (381 K) =  4.1732 L

Since the volume remains the same, the pressure is inversely proportional to the new volume. So, if the initial pressure is 105 kPa, the final pressure can be calculated using the following equation:

(Initial pressure) x (Initial volume) = (Final pressure) x (Final volume)

Plugging in the numbers, we get:

(105 kPa) x (3.0 L) = (Final pressure) x (4.1732 L)

Solving for the final pressure gives us:

Final pressure = [(105 kPa) x (3.0 L)] / (4.1732 L) = 75.57 kPa

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Answer:a)1.11×10^-21Nm

b) 1.16×10^-3m

Explanation:see attachment

Answer:

Therefore the magnitude of the net electric filed at the center of the triangle is [tex]=\frac{6K}{A^2}[/tex] N/C

Explanation:

Given,A = side of the triangle. q = 1 C

The center of a triangle is the centroid of the triangle.

The distance between centroid to any vertices of a equilateral triangle is

[tex]=\frac{2}{3}[/tex] of the height of the equilateral triangle

[tex]=\frac{2}{3}\times \frac{\sqrt{3} }{2} \times A[/tex]

[tex]=\frac{1}{\sqrt{3} } A[/tex]

Electric field= [tex]\frac{Kq}{d^2}[/tex]       K=8.99×10⁹ Nm²/C², q=charge  and d = distance

Therefore the magnitude of the net electric filed at the center of the triangle is

=2[tex]\frac{Kq}{d^2}[/tex]         [both charge are at same distance from the centroid]

=[tex]\frac{2k}{(\frac{1}{\sqrt{3} }A)^2 }[/tex]

[tex]=\frac{6K}{A^2}[/tex] N /C      [K=8.99×10⁹ Nm²/C²]

Answer:

0.34m

Explanation:

Final answer:

The wavelength of a radio signal received at a frequency of 880.65 MHz is approximately 0.34 meters, using the formula λ = c / f, where c is the speed of light.

Explanation:

If you receive a call at a frequency of 880.65 MHz, to calculate the wavelength in meters, you can use the formula λ = c / f, where λ is the wavelength in meters, c is the speed of light (approximately 3.00 × 108 m/s), and f is the frequency in hertz (Hz). Since the question provides the frequency in megahertz (MHz), we first convert it to hertz by multiplying by 106.

The frequency is 880.65 MHz, which is equal to 880.65 × 106 Hz. Thus, the wavelength λ = 3.00 × 108 m/s / (880.65 × 106 Hz).

Calculating this gives a wavelength of approximately 0.34 meters.

Answer:

q=3.5*10^-4

Explanation:

concept:

The force acting on both charges is given by the coulomb law:

F=kq1q2/r^2

the centripetal force is given by:

Fc=mv^2/r

The kinetic energy is given by:

KE=1/2mv^2

The tension force:

when the plane is uncharged

T=mv^2/r

T=2(K.E)/r

T=2(50 J)/r

T=100/r

when the plane is charged

T+k*|q|^2/r^2=2(K.E)charged/r

100/r+k*|q|^2/r^2=2(53.5 J)/r

q=√(2r[53.5 J-50 J]/k)                                          √= square root on whole

q=√2(2)(53.5 J-50 J)/8.99*10^9

q=3.5*10^-4

Final answer:

To calculate the new velocity of the boat after being accelerated by the current for 5 seconds, one must use the Pythagorean theorem and arctangent function to find the magnitude and direction of the resultant velocity, resulting in approximately 5.04 m/s at 48.1 degrees from the original motion.

Explanation:

The student's question is about calculating the resultant velocity of a boat being accelerated by a river current perpendicular to its initial direction of motion. With an initial velocity of 3.35 m/s and a river current accelerating the boat at 0.750 m/s2, we need to find the new velocity after 5 seconds.

First, calculate the velocity increase caused by the acceleration of the current: velocity increase = acceleration × time, which gives us 0.750 m/s2 × 5 s = 3.75 m/s. This increase is perpendicular to the initial velocity.

Now, we determine the resultant velocity using the Pythagorean theorem since the velocities are perpendicular. The magnitude of the resultant velocity, Vtotal, is given by √(3.35 m/s)2 + (3.75 m/s)2, which equals to approximately 5.04 m/s. To find the direction, we use the arctangent function: θ = tan-1(3.75 m/s / 3.35 m/s), which results in a direction of approximately 48.1 degrees relative to the original direction of motion. Therefore, the boat will be moving with a velocity of approximately 5.04 m/s at a direction of approximately 48.1 degrees from its original direction, due to the current.

Answer:

Explanation:

Given

Man walks 17.5 m straight to west 17.5 m

So position vector is given by

[tex]\vec{r_1}=-17.5\hat{i}[/tex]

Now he walks 22 m North

so position vector is

[tex]r_{21}=22\hat{j}[/tex]

Position of man from initial Position

[tex]\vec{r_{2}}=\vec{r_2}-\vec{r_1}[/tex]

[tex]\vec{r_{2}}=22\hat{j}-(-17.5\hat{i})[/tex]

[tex]\vec{r_{2}}=17.5\hat{i}+22\hat{j}[/tex]

So Magnitude of distance is given by

[tex]|\vec{r_{2}}|=\sqrt{17.5^2+22^2}[/tex]

[tex]|\vec{r_{2}}|=28.11\ m[/tex]  

To determine the distance from the starting point after walking 17.5 m west and 22.0 m north, use the Pythagorean theorem. Calculating this gives a distance of approximately 28.1 meters from the starting point.

Calculating Distance Using Pythagorean Theorem

To find how far you are from your starting point after walking 17.5 m west and then 22.0 m north, we can use the Pythagorean theorem. This is because your path forms a right triangle with the two legs being 17.5 m and 22.0 m.

The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the distance from your starting point) is equal to the sum of the squares of the other two sides:

a² + b² = c²

Substituting these values into the Pythagorean theorem gives:

17.52 + 22.02 = c²

Calculating the squares:

306.25 + 484.00 = c²

Adding them together:

790.25 = c²

Taking the square root of both sides:

c ≈ 28.1 m

Therefore, you are approximately 28.1 meters from your starting point.

Answer: The molar mass of metal (M) is 47.86 g/mol

Explanation:

Let the atomic mass of metal (M) be x

Atomic mass of [tex](MO_2)=(x+32)g/mol[/tex]

To calculate the mass of metal, we use the equation:

[tex]\text{Mass percent of metal}=\frac{\text{Mass of metal}}{\text{Mass of metal oxide}}\times 100[/tex]

Mass percent of metal = 59.93 %

Mass of metal = x g/mol

Mass of metal oxide = (x + 32) g/mol

Putting values in above equation, we get:

[tex]59.93=\frac{x}{(x+32)}\times 100\\\\59.93(x+32)=100x\\\\x=47.86g/mol[/tex]

Hence, the molar mass of metal (M) is 47.86 g/mol

Answer: 1.6 m/s

Explanation:

The relationship between linear speed and angular speed for a rotational motion is

[tex]v=\omega r[/tex]

where

v is the linear speed

[tex]\omega[/tex] is the angular speed

r is the distance from the axis of rotation

For the vinyl record here, we have:

[tex]\omega=90 rev/min[/tex]

Keeping in mind that

[tex]1rev = 2\pi rad\\1 min =60 s[/tex]

We can convert it to rad/s:

[tex]\omega = 90 \cdot \frac{2\pi}{60}=9.4 rad/s[/tex]

The diameter of the disk is 34 cm, so the radius is

[tex]r=\frac{34}{2}=17 cm = 0.17 m[/tex]

Therefore, the linear velocity of  a point on the circumference is

[tex]v=(9.4)(0.17)=1.6 m/s[/tex]

The linear velocity of a point on the circumference of the disk is approximately 1.45 meters per second.

To find the linear velocity of a point on the circumference of the disk, we need to calculate the circumference of the disk and then determine how fast a point on the edge of the disk moves in one minute.

[tex]\[ v = \frac{C}{t} \][/tex]

First, we calculate the circumference  using the diameter of the disk. The radius is half of the diameter, so:

[tex]\[ r = \frac{diameter}{2} = \frac{34 \text{ cm}}{2} = 17 \text{ cm} \][/tex]

Now, the circumference  is given by:

[tex]\[ C = 2\pi r = 2\pi \times 17 \text{ cm} \][/tex]

Converting centimeters to meters (1 meter = 100 centimeters), we get:

[tex]\[ C = 2\pi \times 0.17 \text{ m} \][/tex]

Next, we know that the record makes 90 rotations in a minute, so the time \( t \) for one rotation is:

[tex]\[ t = \frac{60 \text{ seconds}}{90 \text{ rotations}} \][/tex]

Now we can calculate the linear velocity \( v \):

[tex]\[ v = \frac{C}{t} = \frac{2\pi \times 0.17 \text{ m}}{\frac{60}{90} \text{ seconds}} \] \[ v = \frac{2\pi \times 0.17 \text{ m}}{\frac{2}{3} \text{ seconds}} \] \[ v = \frac{2\pi \times 0.17 \text{ m} \times 3}{2} \] \[ v = \pi \times 0.17 \text{ m} \times 3 \] \[ v = \pi \times 0.51 \text{ m} \] Using the approximation \( \pi \approx 3.14159 \), we get: \[ v \approx 3.14159 \times 0.51 \text{ m/s} \] \[ v \approx 1.594 \text{ m/s} \][/tex]

Rounding to two decimal places, the linear velocity is approximately 1.45 meters per second."

1) Current: 4.5 A

2) Time taken: 4.7 s

Explanation:

1)

The electric current intensity is defined as the rate at which charge flows in a conductor; mathematically:

[tex]I=\frac{q}{t}[/tex]

where

I is the current

q is the amount of charge passing a given point in a time t

For the wire in this problem, we have

q = 9.0 C is the amount of charge

t = 2.0 s is the time interval

Solving for I, we find the current:

[tex]I=\frac{9.0}{2.0}=4.5 A[/tex]

2)

To solve this problem, we can use again the same formula

[tex]I=\frac{q}{t}[/tex]

where

I is the current

q is the amount of charge passing a given point in a time t

In this problem, we have:

I = 3.0 A (current)

q = 14.0 C (charge)

Therefore, the time taken for the charge to move past a particular spot in the wire is

[tex]t=\frac{q}{I}=\frac{14.0}{3.0}=4.7 s[/tex]

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The electric current in the wire in the first scenario is 4.5 A, and it will take 4.7 seconds for 14.0 C of charge to move past a particular spot on the wire with a constant current of 3.0 A in the second scenario.

The amount of electric current in a circuit is defined by the amount of electric charge that passes through it in a given amount of time. This is represented by the formula I = Q / t, where I is the current, Q is the charge and t is the time.

In the first part of your question, we are given that Q = 9.0 C and t = 2.0 s. We can substitute these values into the formula to find the current: I = 9.0 C / 2.0 s = 4.5 A.

In the second part, we are given that I = 3.0 A and Q = 14.0 C. This time, we need to rearrange the formula to find t: t = Q / I = 14.0 C / 3.0 A = 4.7 s.

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Answer:

[tex]P_{abs}=107172.35Pa\\ P_{abs}=107.172kPa[/tex]

Explanation:

Given data

[tex]P_{a}=98kPa\\ p_{oil}=850kg/m^{3}\\ h=110cm=1.10m[/tex]

To find absolute pressure of air in the tank.We must find the pressure athe parallel point from the right tube

So

[tex]P_{abs}=P_{a}+hp_{oil}g\\P_{abs}=98000Pa+(1.10m*850kg/m^{3}*9.81 )\\P_{abs}=107172.35Pa\\ P_{abs}=107.172kPa[/tex]

Answer:

d) Gravity, a normal force, and kinetic friction

Explanation:

Explanation:

a. 1 liter (L) is equal to 61.0237 cubic inches ([tex]in^3[/tex]), So:

[tex]1L*\frac{61.0237in^3}{1L}=61.0237in^3[/tex]

b. 1 J is equal to [tex]9.47817*10^{-4}[/tex] BTU. Thus:

[tex]650J*\frac{9.47817*10^{-4}BTU}{1J}=6.16081*10^{-1}BTU[/tex]

c. 1 kW is equal to 737.56 [tex]\frac{ft\cdot lbf}{s}[/tex]. So:

[tex]0.135kW*\frac{737.56\frac{ft\cdot lbf}{s}}{1kW}=99.57\frac{ft\cdot lbf}{s}[/tex]

d. 1 [tex]\frac{g}{s}[/tex] is equal to 0.1323 [tex]\frac{lb}{min}[/tex]. Therefore:

[tex]378\frac{g}{s}*\frac{0.1323\frac{lb}{min}}{1\frac{g}{s}}=50.01\frac{lb}{min}[/tex]

e. 1 kPa is equal to 0.145[tex]\frac{lbf}{in^2}[/tex]. Thus:

[tex]304kPa*\frac{0.145\frac{lbf}{in^2}}{1kPa}=44.08\frac{lbf}{in^2}[/tex]

f. 1 [tex]\frac{m^3}{h}[/tex] is equal to [tex]9.81*10^{-3}\frac{ft^3}{s}[/tex]. So:

[tex]55\frac{m^3}{h}*\frac{9.81*10^{-3}\frac{ft^3}{s}}{1\frac{m^3}{h}}=5.40*10^{-1}\frac{ft^3}{s}[/tex]

g. 1 [tex]\frac{km}{h}[/tex] is equal to 0.911 [tex]\frac{ft}{s}[/tex]. Therefore:

[tex]50\frac{km}{h}*\frac{0.911\frac{ft}{s}}{1\frac{km}{h}}=45.55\frac{ft}{s}[/tex]

h. 1 N is equal to [tex]1.1*10^{-4}[/tex] ton. Thus:

[tex]8896N*\frac{1.1*10^{-4}ton}{1N}=0.98ton[/tex]

Answer:

[tex]v = 3.81\ m/s[/tex]

Explanation:

given,

Radius of the ball, r₁ = 53 cm

Radius of another ball, r₂ = 26 cm

center to center distance between the balls = 223 cm

time, t = 18.9 s

surface to surface distance between them

S = 223 - (53+26)

S = 144 cm

Speed of the ball = ?

[tex]relative\ speed = \dfrac{distance}{time}[/tex]

[tex]2 v = \dfrac{144}{18.9}[/tex]

both the balls are moving towards each other so, speed doubles.

[tex]v= \dfrac{7.62}{2}[/tex]

[tex]v = 3.81\ m/s[/tex]

Speed of the balls is equal to 3.81 m/s

Answer:

The first possible value of d is 0.85 m

The second possible value of d is 2.55 m

The third possible value of d is 4.25 m

Explanation:

Given that,

Distance =d

Frequency of sound wave= 200 Hz

We need to calculate the wavelength

Using formula of wavelength

[tex]\lambda=\dfrac{v}{f}[/tex]

Put the value into the formula

[tex]\lambda=\dfrac{340}{200}[/tex]

[tex]\lambda=1.7\ m[/tex]

The separation between the speakers in the destructive interference is

[tex]\Delta x= d[/tex]

The equation for destructive interference

[tex]2\pi\times\dfrac{\Delta x}{\lambda}-\Delta\phi_{0}=(m+\dfrac{1}{2})2\pi[/tex]

The loudspeakers are in phase

So, [tex]\Delta\phi_{0}=0[/tex]

The equation for destructive interference is

[tex]2\pi\times\dfrac{d}{\lambda}=(m+\dfrac{1}{2})2\pi[/tex]....(I)

Here, m = 0,1,2,3.....

We need to calculate the first possible value of d

For, m = 0

Put the value in the equation (I)

[tex]2\pi\times\dfrac{d_{1}}{1.7}=(0+\dfrac{1}{2})2\pi[/tex]

[tex]d_{1}=\dfrac{1.7}{2}[/tex]

[tex]d_{1}=0.85\ m[/tex]

We need to calculate the second possible value of d

For, m = 1

Put the value in the equation (I)

[tex]2\pi\times\dfrac{d_{2}}{1.7}=(1+\dfrac{1}{2})2\pi[/tex]

[tex]d_{2}=\dfrac{1.7\times3}{2}[/tex]

[tex]d_{2}=2.55\ m[/tex]

We need to calculate the third possible value of d

For, m = 1

Put the value in the equation (I)

[tex]2\pi\times\dfrac{d_{3}}{1.7}=(2+\dfrac{1}{2})2\pi[/tex]

[tex]d_{3}=\dfrac{1.7\times5}{2}[/tex]

[tex]d_{3}=4.25\ m[/tex]

Hence, The first possible value of d is 0.85 m

The second possible value of d is 2.55 m

The third possible value of d is 4.25 m

Answer:

v_b = 10.628 m /s (13.605 degrees east of south)

Explanation:

Given:

- Velocity of mia v_m = + 6 j m/s

- Velocity of ball wrt mia v_bm = 5.0 m/s     30 degree due east of south

Find:

What are the magnitude and direction of the velocity of the ball relative to the ground? v_b

Solution:

- The relation of velocity in two different frame is given:

                             v_b  - v_m = v_bm

- Components along the direction of v_b,m:

                             v_b*cos(Q) - v_m*cos(30) = 5

                             v_b*cos(Q) = 5 + 6 sqrt(3) / 2

                             v_b*cos(Q) = 5 + 3sqrt(3)  

- Components orthogonal the direction of v_b,m:

                             -v_m*sin(30) = v_b*sin(Q)

                             -6*0.5 = v_b*sin(Q)

                              -3 = v_b*sin(Q)

- Divide two equations:

                               tan(Q) = - 3 / 5 + 3sqrt(3)

                               Q = arctan(- 3 / 5 + 3sqrt(3)

                               Q = -16.395 degrees

                               v_b =  -3 / sin(-16.395)

                               v_b = 10.63 m/s

The magnitude of the soccer ball's velocity relative to the ground is 2.89 m/s, and its direction is approximately 56.31° north of east. This result is found by breaking down vector components and applying vector addition.

You're asking about the velocity of the ball relative to the ground when soccer players Mia and Alice are running, and Alice passes the ball to Mia. To solve this, we'll use vector addition. Mia is running due north at 6.00 m/s, and the velocity of the ball relative to Mia is 5.00 m/s at 30° east of south. To find the velocity of the ball relative to the ground, we imagine two vectors: Mia's velocity vector (northward) and the ball's velocity vector relative to Mia. The latter will have an east and a south component due to the 30° angle.

To find the south component of the ball's relative velocity, we use cosine because it's adjacent to the 30° angle:
5.00 m/s * cos(30°) = 4.33 m/s. The east component is found using sine:
5.00 m/s * sin(30°) = 2.50 m/s.

Since Mia is running north, to find the actual velocity of the ball to the south, we subtract Mia's velocity from the south component:
4.33 m/s - 6.00 m/s = -1.67 m/s, where the negative indicates that the ball's actual movement is to the north.

Now, using the Pythagorean theorem for the ball's velocity relative to the ground, we find the magnitude:
(2.50 m/s)² + (-1.67 m/s)² = √(6.25 + 2.7889)
√8.3389 m²/s² = 2.89 m/s.

To find the direction, we use the tangent function, since we have opposite (east component) and adjacent (north component) sides of the right triangle:
an(θ) = 2.50 / 1.67; θ = tan(⁻¹)(2.50 / 1.67);
θ ≈ 56.31° north of east, which is the direction of the ball's velocity relative to the ground.

Answer:

933.804423995 m/s²

Explanation:

[tex]q_1[/tex] = Charge on particle 1 = 67 µC

[tex]q_2[/tex] = Charge on particle 2 = -25 µC

r = Distance between the particles = 47 cm

k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]

m = Mass of particle = 73 g

Electric force is given by

[tex]F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times -25\times 10^{-6}\times 67\times 10^{-6}}{0.47^2}\\\Rightarrow F=-68.1677229516\ N[/tex]

The magnitude of force is 68.1677229516 N

Acceleration is given by

[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{68.1677229516}{0.073}\\\Rightarrow a=933.804423995\ m/s^2[/tex]

The acceleration is 933.804423995 m/s²

Answer:

M_c = 100.8 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two cases shown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

                        M_c = F_a*( 42 / 150 ) *144

                        M_c = 2.5*( 42 / 150 ) *144

                        M_c = 100.8 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

Final answer:

The initial momentum of the bullet before the collision is equal to the momentum of the bullet and block at the instant they reach the maximum height h. Option A

Explanation:

The correct statement of the collision in this scenario is A) The initial momentum of the bullet before the collision is equal to the momentum of the bullet and block at the instant they reach the maximum height h.

According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision, as long as there are no external forces acting on the system. In this case, the bullet and block form a closed system, and therefore, the initial momentum of the bullet is equal to the momentum of the bullet and block at the maximum height.

To understand why this statement is true, we can consider the momentum of the bullet and block system at different points in the motion. Initially, the bullet has momentum in the positive x-direction, and this momentum is transferred to the bullet and block as they rise to the maximum height. Therefore, the statement A is correct.

Answer:

doppler shift's formula for source and receiver moving away from each other:

λ'=λ°√(1+β/1-β)

Explanation:

acceleration of spaceship=α=29.4m/s²

wavelength of sodium lamp=λ°=589nm

as the spaceship is moving away from earth so wavelength of earth should increase w.r.t increasing speed until it vanishes at λ'=700nm

using doppler shift's formula:

λ'=λ°√(1+β/1-β)

putting the values:

700nm=589nm√(1+β/1-β)

after simplifying:

β=0.17

by this we can say that speed at that time is: v=0.17c

to calculate velocity at an acceleration of a=29.4m/s²

we suppose that spaceship started from rest so,

v=v₀+at

where v₀=0

so v=at

as we want to calculate t so:-

t=v/a                                                v=0.17c      ,c=3x10⁸           ,a=29.4m/s²

putting values:

=0.17(3x10⁸m/s)/29.4m/s²

t=1.73x10⁶

Answer:

2.03 Ω

Explanation:

EMF: This can be defined as the potential difference of a cell when it is not delivering any current. The S.I unit of Emf is Volt.

The formula of emf is given as,

E = I(R+r)............................ Equation 1

Where E = Emf, I = current, R = External resistance, r = internal resistance.

Make r the subject of the equation

r = (E-IR)/I........................ Equation 2

Note: From ohm's law, V = IR.

r = (E-V)/I........................ Equation 3

Where V = Terminal voltage

Given: E = 24 V, I = 3.9 A, V = 16.1 V.

Substitute into equation 3

r = (24-16.1)/3.9

r = 7.9/3.9

r = 2.03 Ω

Final answer:

To find the internal resistance of a battery, subtract the terminal voltage from the emf and divide by the current. Given an emf of 24.0 V, a terminal voltage of 16.1 V, and a current of 3.90 A, the internal resistance is about 2.03 Ω.

Explanation:

To determine the internal resistance of a battery, we need to understand a critical relationship between the battery's electromotive force (emf), its terminal voltage, the current flowing through the circuit, and its internal resistance. The formula required for finding the internal resistance is derived from Ohm's Law, and it takes into account the difference between the emf (the voltage the battery would supply in the absence of internal resistance) and the terminal voltage (the actual voltage the battery supplies when connected to a load).

Given are the following parameters:

emf (E) = 24.0 V

Terminal Voltage (V) = 16.1 V

Current (I) = 3.90 A

To calculate the internal resistance (r), we use the formula based on the definition of emf, which is:

E = V + Ir

By rearranging the formula to solve for the internal resistance, we get:

r = (E - V) / I

Substituting the given values:

r = (24.0 V - 16.1 V) / 3.90 A = 7.9 V / 3.90 A ≈ 2.03 Ω

Therefore, the internal resistance of the battery is approximately 2.03 Ω.

Answer:

7. The force is zero

Explanation:

The force is zero when your velocity is parallel to the magnetic field

Answer:

[tex]E=0[/tex] at r < R;

[tex]E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}[/tex] at 2R > r > R;

[tex]E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}[/tex] at r >= 2R

Explanation:

Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,

[tex]\int {E} \, dA=\frac{Q_{enclosed}}{\epsilon}[/tex] (integral over a closed surface)

where,

[tex]E[/tex] = Electric field

[tex]Q_{enclosed}[/tex] = charged enclosed within the closed surface

[tex]\epsilon[/tex] = permittivity of free space

Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .

at r < R,

[tex]Q_{enclosed}[/tex] = 0 and hence [tex]E[/tex] = 0 (since the sphere is conducting, all the charges get repelled towards the surface)

at 2R > r > R,

[tex]Q_{enclosed}[/tex] = Q,

therefore,

[tex]E\times4\pi r^{2}=\frac{Q_{enclosed}}{\epsilon}[/tex]      

(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to [tex]4\pi r^{2}[/tex])

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

at r >= 2R

[tex]Q_{enclosed}[/tex] = 2Q

Hence, by similar calculations, we get,

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

The electric field inside the solid conducting sphere is zero. For r between R and 2R, the electric field can be calculated using Gauss's law.

The electric field inside a solid conducting sphere is zero. Thus, the magnitude of the electric field in the region 0<r<R is zero.

In the region R<r<2R, the charge on the insulating shell induces an equal and opposite charge on the inner surface of the conducting sphere. Therefore, the magnitude of the electric field in this region can be found using Gauss's Law.

The electric field magnitude is given by:

E = Q / (4πε0 r2)

where Q is the charge on the insulating shell, ε0 is the vacuum permittivity, and r is the distance from the center of the sphere.

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Since the Units presented are not in the International System we will proceed to convert them. We know that,

[tex]1 mi/h = 0.447 m/s[/tex]

So the speed in SI would be

[tex]V=95mi/h(\frac{0.447m/s}{1mi/h})[/tex]

[tex]V=42.465 m/s[/tex]

The change in frequency when the wave is reflected is

[tex]f'=f(1+\frac{V}{c})[/tex]

Or we can rearrange the equation as

[tex]f' = f + f\frac{V}{c}[/tex]

f' = Apparent frequency

f = Original Frequency

c = Speed of light

[tex]f'-f = f\frac{V}{c}[/tex]

[tex]\Delta f = f\frac{V}{c}[/tex]

Replacing,

[tex]\Delta f = (10.525*10^9)(\frac{42.465}{3*10^8})[/tex]

[tex]\Delta f =1489.8 Hz[/tex]

Since the waves are reflected, hence the change in frequency at the gun is equal to twice the change in frequency

[tex]\Delta f_T = 2 \Delta f[/tex]

[tex]\Delta f_T = 2(1489.8Hz)[/tex]

[tex]\Delta f_T = 2979.63Hz[/tex]

Therefore the increase in frequency is 2979.63Hz

The increase in frequency of these waves is equal to 2979.6 Hertz.

Given the following data:

Conversion:

1 mi/h = 0.447 m/s.

95.0 mi/h = [tex]95 \times 0.447[/tex] = 42.465 m/s.

To determine the increase in frequency:

Mathematically, the change in frequency of a wave is given by this formula:

[tex]F' = F(1+\frac{V}{c} )\\\\F' = F+F\frac{V}{c}\\\\F' - F=F\frac{V}{c}\\\\\Delta F = F\frac{V}{c}[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]\Delta F = 10.525 \times 10^9 \times \frac{42.465}{3 \times 10^8}\\\\\Delta F = \frac{446.944 \times 10^9}{3 \times 10^8}\\\\\Delta F = 1489.8\;Hertz[/tex]

Since the waves were reflected, the increase in frequency toward the gun is double (twice) the change in frequency. Thus, we have;

[tex]F_2 = 2\Delta F\\\\F_2 = 2\times 1489.8\\\\F_2 = 2979.6\;Hertz[/tex]

Read more on frequency here: brainly.com/question/3841958