The displacement of a wave traveling in the positive x-direction is:
y(x,t) = (3.5cm) cos (2.7x−92t), where x is in m and t is in s.
Part A) What is the frequency of this wave?
Part B) What is the wavelength of this wave?
Part C) What is the speed of this wave?

Answer:

B=μ₀I/2r

Explanation:

Produced magnetic field due to an existing electric field through a coil or conductor can be explained by Biot-Savart Law. Formula for this law is:

dB=(μ₀I/4π.r²)dL

Here,

r=Radius of the loop

I and r are constants with respect to length L.

To convert linear displacement L into angular displacement Ф:

dL=r.dФ

So,

dB=(μ₀I/4π.r²)r.dФ

dB=(μ₀I/4π.r)dФ

Integrating both sides over the circle i.e. from 0 radians to 2π radians  (360⁰), while the integration will apply only on dФ as all others are constants.

B=(μ₀I/4πr)(2π-0)

B=(μ₀I/2r)

The magnetic field produced due to current flowing through the coil is [tex]B = \frac{\mu_o I}{2r}[/tex].

The magnetic field produced due to current flowing through a coil given by Biot-Savart Law.

[tex]dB = \frac{\mu_o I}{4\pi r^2} dL[/tex]

where;

[tex]dB = \frac{\mu_o I}{4\pi r^2} .(rd \phi)\\\\dB = \frac{\mu_o I}{4\pi r} \ d \phi\\\\B = \frac{\mu_o I}{4\pi r} [\phi ]^{2\pi} _{0}\\\\B = \frac{\mu_o I}{4\pi r} (2\pi)\\\\B = \frac{\mu_o I}{2 r}[/tex]

Thus, the magnetic field produced due to current flowing through the coil is [tex]B = \frac{\mu_o I}{2r}[/tex].

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The passengers in the car will hear a pitch of approximately 611.03 Hz.

The pitch that the passengers in the car will hear is determined by the Doppler effect. The frequency of a sound wave appears higher when the source is approaching and lower when it is moving away. The formula for calculating the observed frequency is f' = f((v + vd)/(v - vs)), where f' is the observed frequency, f is the source frequency, v is the speed of sound, vd is the velocity of the detector (passengers' car), and vs is the speed of the source (fire engine).

In this case, the frequency of the siren is 400 Hz, the speed of sound is 343 m/s, and the velocity of the detector car is 18 m/s. Since the source is moving south and the detector car is moving north, we take the velocities with opposite signs.

Using the formula f' = f((v + vd)/(v - vs)), we can calculate the observed frequency:

f' = 400((343 + 18)/(343 - (-35)))

f' ≈ 611.03 Hz

Therefore, the passengers in the car will hear a pitch of approximately 611.03 Hz.

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Answer:

1.5T

Explanation:

Since, the amplitude of SHM is A. So, in one time period T the object will travel travel a distance of 4A.

6A-4A= 2A.

Now, this 2A distance must be traveled in T/2 time period.

So, the total time taken to travel a distance of 6A is T+T/2 = 3T/2 = 1.5T

In simple harmonic motion, an object oscillates between a maximum displacement (A) and its equilibrium point, completing a full cycle within a period (T). To move a total distance of 6A, it will need 1.5 cycles, thus taking 1.5T time.

The question you asked is about a simple harmonic oscillator, in this case, an object attached to a spring sliding on a frictionless surface. The motion of this object can be classified as simple harmonic motion (SHM).

In simple harmonic motion, the object oscillates between a maximum displacement or amplitude (A) and its equilibrium point. The time taken for the object to complete one full cycle, i.e., return to its starting point, is called the period (T).

To answer your question, when the object completes one full cycle of movement, it travels a distance of 4A (A to -A and -A to A). Therefore, if the object needs to travel a total distance of 6A, it will need 1.5 cycles. As the period, T, is the time taken for one cycle, the total time taken to travel 6A is 1.5T.

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Answer:

(a) [tex]43.45\frac{km}{h}[/tex]

(b) [tex]12.08\frac{m}{s}[/tex]

Explanation:

(a) 1 mile is equal to 1.609344 kilometers. So, we have:

[tex]27\frac{mi}{h}*\frac{1.609344 km}{1mi}=43.45\frac{km}{h}[/tex]

(b) 1 mile is equal to 1609.34 meters. 1 hour is equal to 3600 seconds. So, we have:

[tex]27\frac{mi}{h}*\frac{1609.34m}{1mi}=4.35*10^{4}\frac{m}{h}\\4.35*10^{4}\frac{m}{h}*\frac{1h}{3600s}=12.08\frac{m}{s}[/tex]

A bicyclist's speed of 27.0 mi/h is equivalent to 43.45 km/h and 12.07 m/s.

(a) To convert miles per hour (mi/h) to kilometers per hour (km/h), we can use the conversion factor of 1.60934 km/h = 1 mi/h. So, multiplying the given speed of 27.0 mi/h by the conversion factor, we get:

27.0 mi/h * 1.60934 km/h = 43.45 km/h

(b) To convert kilometers per hour (km/h) to meters per second (m/s), we can use the conversion factor of 1 km/h = 0.277778 m/s. So, multiplying the given speed of 43.45 km/h by the conversion factor, we get:

43.45 km/h * 0.277778 m/s = 12.07 m/s

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Answer:

The speed of the proton relative to the laboratory frame is 0.981c

Explanation:

Given that,

Speed of electron v= 0.90c

Speed of proton u= 0.70c

We need to calculate the speed of the proton relative to the laboratory frame

Using formula of speed

[tex]u'=\dfrac{u+v}{1+\dfrac{uv}{c^2}}[/tex]

Where, u = speed of the proton relative to the electron

v = speed of electron relative to the laboratory frame

Put the value into the formula

[tex]u'=\dfrac{0.70+0.90}{1+\dfrac{0.70\times0.90\times c^2}{c^2}}[/tex]

[tex]u'=c\dfrac{0.70+0.90}{1+(0.70\times0.90)}[/tex]

[tex]u'=0.981c[/tex]

Hence, The speed of the proton relative to the laboratory frame is 0.981c

Final answer:

The speed of the proton relative to the laboratory frame, when it moves to the right with a speed of 0.70c relative to an electron moving at 0.90c, is found to be approximately 0.98c using relativistic velocity addition.

Explanation:

To find the speed of the proton relative to the laboratory frame when an electron moves to the right with a speed of 0.90c and a proton moves to the right with a speed of 0.70c relative to the electron, we use the formula for relativistic velocity addition. This formula is given by:

V = (v + u) / (1 + vu/c²),

where V is the velocity of the proton relative to the lab, v is the velocity of the electron relative to the lab (0.90c), u is the velocity of the proton relative to the electron (0.70c), and c is the speed of light. Plugging in these values, we get:

V = (0.90c + 0.70c) / (1 + (0.90*0.70)) = 0.98c.

Therefore, the speed of the proton relative to the laboratory frame is approximately 0.98c, close to the speed of light, demonstrating the relativistic effects when velocities approach the speed of light.

To find the second student's mass, we can use the principle of Pascal's Law which states that when a fluid is at rest in a container, the pressure is transmitted equally in all directions. The height difference between the liquid levels in the right and left pistons can be used to determine the mass.

To find the second student's mass, we can use the principle of Pascal's Law which states that when a fluid is at rest in a container, the pressure is transmitted equally in all directions. In this case, the height difference between the liquid levels in the right and left pistons can be used to determine the mass. The height difference is 40 cm, so the pressure difference is equal to the weight of the second student:

P = ρgh, where P is the pressure difference, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the height difference.

Using the formula, we can solve for the mass of the second student:

m = P/(g*h), where m is the mass of the second student.

Let's plug in the values:

m = (ρgh)/(g*h) = ρ. So, the mass of the second student is equal to the density of the liquid.

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To find the mass of the second student using the difference in liquid levels in a hydraulic system, we can rearrange the pressure difference equation, which takes into account the height difference, gravity, and fluid density.

The subject of the question delves into the physical principles of hydraulics. If the height difference between the liquid levels in the right and left pistons is 40 cm when a second student joins the first, we can apply the principle of hydraulic machines which is essentially the conservation of energy principle.

In hydraulics, pressure applied on one piston gets transmitted to the other without being diminished. Hence, in this context, let's denote the mass of the second student as m2. The difference in heights or pressure difference ΔP can be measured by subtracting fluid devices and can be expressed using the formula ΔP = m2g

In this equation, 'g' is the acceleration due to gravity. Given the height difference Δh as 40cm, we can deduce the density (ρ) of the fluid. Remember, pressure difference ΔP = ρg Δh. Hence, we can obtain the mass of the second student by rearranging the equation as follows: m2 = (ρg Δh) / g. Please note: for this scenario, the value of 'g' needs to be considered as 9.8 m/s² (earth's acceleration) and ρ is the fluid density (e.g., for water, it is 1000 kg/m³).

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Answer:

b)

Explanation:

Assuming that we are talking about average speed during any segment, we can apply the definition of average speed, as follows:

v(avg) = Δx / Δt = (xf-x₀) / (tfi-t₀)

Using this definition for the 4 segments, we have:

1) v(0-100m) = 100 m / 11.20 sec = 8.93 m/s

2) v(100m-200m) = 100 m / (21.32 s - 11.2 s) = 100 m / 10.12 s = 9.88 m/s

3) v(200m -300m) = 100 m / (31.76 s- 21.32s) = 100 m / 10.44 s = 9.58 m/s

4) v(300m-400m) = 100 m / (43.18 s - 31.76 s) = 100 m / 11.42 s = 8.76 m/s

As we can see, the highest speed was reached between the 100m mark and the 200m mark, so the statement b) is the one that results to be true.

Final answer:

The highest speed in Michael Johnson's world-record 400 m sprint was achieved during the segment between the 100 m mark and the 200 m mark with a speed of 9.88 m/s.

Explanation:

To determine which 100 m segment Michael Johnson had the highest speed, we should calculate his speed for each segment separately. Speed is calculated by dividing the distance by the time taken to cover that distance. We are given the total time at each 100 m interval, so we need to calculate the time for each interval separately and then calculate the speed.

Comparing these speeds, we can see that the highest speed was achieved during the segment between the 100 m mark and the 200 m mark.

Answer:

1.58 W

Explanation:

Since the sound spreads uniformly in all directions, it must be in a form of a circle with radius of 12 m. So the area of the circle is

[tex]A = \pi r^2 = \pi 12^2 = 452.389 m^2[/tex]

From the intensity of the sound we can calculate the power at 12 m

[tex]P = AI = 452.389 * 3.5\times10^{-3} = 1.58 W[/tex]

Answer: 171500 times faster.

Explanation: First let's take our Fluid medium to be air. The speed of sound in air is 343m/s or 343000mm/s.

Now, particle velocity is that which shows the alternating mean velocity of a particle in a medium(say in this case air).

A particle of air exposed to a standard sound pressure of 1 Pascal as a peak velocity amplitude of about 2mm/s.

Recall that the value of speed of sound in Air when converted to mm/s is 343000mm/s.

To get the number of times it's faster than the peak particle Velocity of air. We have to divide it by peak particle Velocity of air which is 2mm/s.

343000/2

=171500 times faster.

I believe this is clear.

Answer:

Explanation:

Given

Race length [tex]L=100\ m[/tex]

Runner reaches maximum velocity in [tex]t_1=2.4\ s[/tex]

[tex]v_{max}=11.3\ m/s[/tex]

distance traveled during this time

[tex]v=u+at[/tex]

[tex]11.3=0+a\times 2.4[/tex]

[tex]a=\frac{11.3}{2.4}=4.708\ m/s^2[/tex]

distance traveled in this time

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

[tex]s=0+0.5\times 4.708\times (2.44)^2[/tex]

[tex]s=14.014\ m[/tex]

Remaining distance will be traveled with [tex]v_{max}[/tex]

remaining distance [tex]d=100-14.014=85.98\ m[/tex]

time taken [tex]t_2=\frac{d}{v_{max}}=\frac{85.98}{11.3}=7.609\ s[/tex]

total time [tex]t=t_1+t_2[/tex]

[tex]t=2.44+7.609=10.049\ s[/tex]        

Answer:

[tex]E=1.8\times 10^{10}\ N.C^{-1}[/tex]

Explanation:

Given:

We know from the Coulomb's law:

[tex]F=\frac{1}{4\pi.\epsilon_0} \times \frac{Q.q}{r^2}[/tex] ...........(1)

where:

[tex]\epsilon_0=[/tex] permittivity of free space

Also we have electric field due to Q at q (which is at a distance of 1 m):

[tex]E=\frac{1}{4\pi.\epsilon_0} \times \frac{Q}{r^2}\ [N.C^{-1}][/tex] ...........(2)

From (1) & (2)

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

[tex]E=\frac{18}{10^{-9}}[/tex]

[tex]E=1.8\times 10^{10}\ N.C^{-1}[/tex]

Answer:

magnitude of the electric field = 1.8 × [tex]10^{10}[/tex] N/C

Explanation:

given data

charge Q =2 C

charge q = 1 nC  = 1 × [tex]10^{-9}[/tex] C

distance r = 1 m

electric force = 18 N

solution

we know that here electric field due to Q is

E = k × [tex]\frac{Q}{r^2}[/tex]   ..............1

here q distance is 1 m

now we  apply here Coulomb’s law and get here electric force

F = k ×[tex]\frac{Q*q}{r^2}[/tex]   ..........2

we know constant k  = 8.988 × [tex]10^{9}[/tex] Nm²/C²

so from above both equation we get

electric filed = [tex]\frac{force}{charge}[/tex]    .................3

put here value

electric filed = [tex]\frac{18}{1*10^{-9}}[/tex] = 1.8 × [tex]10^{10}[/tex] N/C

Positively charged particles always follow the path set by electric field lines. This is because electric field lines are defined by the path that the positive test charges would naturally travel.

The options that best complete the statement are 'always' and 'electric field lines are defined by the path positive test charges travel.' This is because the trajectory of positively charged particle trajectories always follow the direction of electric field lines. The electric field lines are essentially mapped out pathways that positive test charges would naturally follow due to the electric force acting upon them. Hence, a positively charged particle will always move along these field lines, from a region of higher potential to a region of lower potential.

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1) The final  angular speed is 4.85 rad/s

2) The change in kinetic energy is 1746 J

Explanation:

1)

The problem can be solved by applying the law of conservation of angular momentum: in fact, the total angular momentum of the system must be conserved,

[tex]L_1 = L_2[/tex]

The initial angular momentum is given by:

[tex]L_1 = (I_d + I_m) \omega[/tex]

where

[tex]I_d = \frac{1}{2}MR^2[/tex] is the moment of inertia of the cylinder, with

M = 63 kg is its mass

R = 2.5 m is the radius

[tex]I_m = mr^2[/tex] is the moment of inertia of the man, with

m = 97 kg is the mass of the man

r = 2.5 m is the distance of the man fro mthe axis of rotation

[tex]\omega=0.19 rev/s \cdot 2 \pi = 1.19 rad/s[/tex] is the angular speed

The final angular momentum is given by

[tex]L_2 = I_d \omega'[/tex]

where [tex]\omega'[/tex] is the final angular speed, and where the angular momentum of the man is zero because he is at the axis of rotation.

Combining the two equations, we get:

[tex](\frac{1}{2}MR^2 + mr^2) \omega = \frac{1}{2}MR^2 \omega'\\\omega'=(1+2\frac{m}{M})\omega=(1+2\frac{97}{63})(1.19)=4.85 rad/s[/tex]

2)

The initial kinetic energy (which is rotational kinetic energy) of the  system is given by

[tex]K_1 = \frac{1}{2}(I_d + I_m) \omega^2[/tex]

And substituting,

[tex]K_1 = \frac{1}{2}(\frac{1}{2}MR^2+mr^2)\omega^2=\frac{1}{2}(\frac{1}{2}(63)(2.5)^2+(97)(2.5)^2)(1.19)^2=568.6 J[/tex]

The  final kinetic energy is given by

[tex]K_2 = \frac{1}{2}I_d \omega'^2[/tex]

And substituting,

[tex]K_2 = \frac{1}{2}(\frac{1}{2}MR^2)\omega'^2=\frac{1}{2}(\frac{1}{2}(63)(2.5)^2)(4.85)^2=2315 J[/tex]

So, the change in kinetic energy is

[tex]\Delta K = 2315 J - 568.6 J = 1746 J[/tex]

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The angular speed of the merry-go-round changes when the man moves closer to the center due to conservation of angular momentum, resulting in a new angular speed. The change in kinetic energy is calculated by comparing kinetic energy before and after the man's movement.

When a 97 kg man on a merry-go-round moves closer to the center, the angular speed of the system changes due to the conservation of angular momentum. The merry-go-round is described as a solid cylinder, and the movement of the man towards the center means his radius of rotation decreases, which leads to an increase in angular velocity. Initially, the man is 2.5 m away from the axis, standing on a merry-go-round that rotates at 0.19 rev/s. Upon reaching the center (0 m from the center), the entire system's angular momentum is conserved but concentrated within a smaller radius.

To calculate the new angular speed, we'll use the concept of moment of inertia (I) and the relation I1 * ω₁ = I2 * ω₂, where ω denotes angular speed. Since the man is no longer contributing to the rotational inertia when he moves to the center, only the merry-go-round's moment of inertia comes into play. Additionally, we'll need to convert the angular speed from rev/s to rad/s using the conversion factor: 1 rev = 2π rad.

The second part concerns the change in kinetic energy (KE) due to the man's movement. The rotational KE of a system is given by (1/2) * I * ω². By finding the KE before and after the man moves to the center, we can determine the change in kinetic energy.

Answer:

95.78125%

18000 m/s

22233.33 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

[tex]v=u+at\\\Rightarrow v=0+20\times 15\times 60\\\Rightarrow v=18000\ m/s[/tex]

The velocity of the rocket at the end of the first 15 minutes is 18000 m/s which is the maximum speed of the rocket in the complete journey.

Distance traveled while speeding up

[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 20\times 900^2\\\Rightarrow s=8100000\ m[/tex]

Distance traveled while slowing down

[tex]s=18000\times 900+\dfrac{1}{2}\times -20\times 900^2\\\Rightarrow s=8100000\ m[/tex]

Distance traveled during constant speed

[tex]384000000-(2\times 8100000)=367800000\ m[/tex]

Fraction

[tex]\dfrac{367800000}{384000000}\times 100=95.78125\ \%[/tex]

Fraction of the total distance is traveled at constant speed is 95.78125%

Time taken at constant speed

[tex]t=\dfrac{367800000}{18000}=20433.33\ s[/tex]

Total time taken is [tex]900+20433.33+900=22233.33\ s[/tex]

The maximum velocity of the spaceship is 18000 m/s. The spaceship travels 95.78% of the total distance at this constant speed. The total time required for the trip is 370 minutes.

To answer the first part of the question, we can use the physics formula for velocity, v = u + at, where u is the initial velocity, a is the acceleration, and t is the time. As the spaceship starts from rest, u = 0. The acceleration a is given as 20 m/s² and the time t for which the spaceship is accelerating is 15min which is equal to 900 seconds. So, the maximum velocity, v = 0 + 20*900 = 18000 m/s.

For the second part, we understand that the spaceship spends an equal amount of time accelerating and decelerating, and the rest of the time it travels at a constant velocity. Therefore, during acceleration and deceleration, it follows a distance d = 1/2 * a * t², with a = 20 m/s² and t = 900 s. We find the distance covered during acceleration and deceleration, d = 1/2 * 20 * 900² = 8100000 m. Given the total distance from Earth to Moon is 384000000 m, the fraction of the distance at a constant speed is 1 - (2*8100000)/384000000 = 0.9578 or 95.78%.

Finally, to find the total time of the trip, we know that the time spent accelerating and decelerating is 15min + 15min = 30min. The distance covered at constant speed is 0.9578 * 384000000m = 367680000m, and the speed is 18000 m/s. So, the time at constant speed is 367680000 / 18000 = 20400s = 340 min. Therefore, the total time of the trip is 30min + 340min = 370 min.

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To solve this problem we will apply the linear motion kinematic equations. In which the speed is defined as the distance traveled in a certain period of time. In turn we must emphasize that 1 light year is equivalent to [tex]9.461*10^{15}m[/tex]. Mathematically the speed is described as,

[tex]v = \frac{x}{t} \rightarrow t = \frac{x}{v}[/tex]

Replacing,

[tex]t = \frac{9.461*10^{15}}{30}[/tex]

[tex]t = 3.15366*10^{14}s (\frac{1 year}{3.154*10^7})[/tex]

[tex]t \approx 10,000,000 years[/tex]

Therefore the correct answer is D.

To solve this problem we will apply the concepts related to energy conservation. Here we will understand that the potential energy accumulated on the object is equal to the work it has. Therefore the relationship that will allow us to calculate the height will be

[tex]W = PE[/tex]

[tex]W = mgh[/tex]

Here,

m = mass

g = Acceleration due to gravity

h = Height

our values are,

[tex]m = 0.345 kg[/tex]

[tex]g = 9.8m/s^2[/tex]

[tex]W = 111J[/tex]

Replacing,

[tex]W =mgh[/tex]

[tex]111 = (0.345)(9.8)h[/tex]

[tex]h = 32.83m[/tex]

Then the height is 32.83m.

Answer:

2.121876 mC

Explanation:

[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]

A = Area = [tex]0.037\times 12\ m^2[/tex]

d = Thickness = 0.0225 mm

E = Dielectric strength = [tex]1\times 10^8\ V/m[/tex]

k = Dielectric constant = 5.4

Capacitance is given by

[tex]C=\dfrac{k\epsilon_0A}{d}\\\Rightarrow C=\dfrac{5.4\times 8.85\times 10^{-12}\times 0.037\times 12}{0.0225\times 10^{-3}}\\\Rightarrow C=9.43056\times 10^{-7}\ F[/tex]

Maximum voltage is given by

[tex]V_m=E_md\\\Rightarrow V_m=1\times 10^8\times 0.0225\times 10^{-3}\\\Rightarrow V_m=2250\ V[/tex]

Maximum charge is given by

[tex]Q_m=CV_m\\\Rightarrow Q_m=9.43056\times 10^{-7}\times 2250\\\Rightarrow Q_m=0.002121876\ C=2.121876\ mC[/tex]

The maximum charge that can be stored in this capacitor is 2.121876 mC

The maximum charge that can be stored in this capacitor is 16.016 μC.

To find the maximum charge that can be stored in the capacitor, we need to use the formula Q=CV, where Q is the charge, C is the capacitance, and V is the voltage. The capacitance can be calculated using the formula C = (εrε0A)/d, where εr is the relative permittivity (dielectric constant) of mica, ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

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Final answer:

To compute the ambient temperature, we can use the principle of heat transfer. By applying the formula Q = mcΔT, we can set up two equations to equate the heat lost by the coffee to the heat gained by the surroundings. Solving these equations will give us the ambient temperature.

Explanation:

To compute the ambient temperature, we can use the principle of heat transfer. The heat lost by the coffee is equal to the heat gained by the surroundings. We can use the formula Q = mcΔT, where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

By applying this formula to the coffee, we can calculate the heat lost from t = 0 to t = 10 minutes and from t = 10 minutes to t = 20 minutes, and then equate it to the heat gained by the surroundings.

Let's assume the ambient temperature is T. Using the formula, we can set up the following equations:

By solving these equations, we can find the value of T, which is the ambient temperature.

Explanation:

Given that,

(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :

[tex]V=\dfrac{W}{q}[/tex]

[tex]V=\dfrac{12}{0.0001}[/tex]

[tex]V=12\times 10^4\ Volt[/tex]

(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :

[tex]V=\dfrac{W}{q}[/tex]

[tex]V=\dfrac{24}{0.0002}[/tex]

[tex]V=12\times 10^4\ Volt[/tex]

Therefore, this is the required solution.

Answer:

F_net = 26.512 N

Explanation:

Given:

Q_a = 3.06 * 10^(-4 ) C

Q_b = -5.7 * 10^(-4 ) C

Q_c = 1.08 * 10^(-4 ) C

R_ac = 3 m

R_bc = sqrt (3^2 + 4^2) = 5m

k = 8.99 * 10^9

Coulomb's Law:

F_i = k * Q_i * Q_j / R_ij^2

Compute F_ac and F_bc :

F_ac = k * Q_a * Q_c / R^2_ac

F_ac =  8.99 * 10^9* ( 3.06 * 10^(-4 ))* (1.08 * 10^(-4 )) / 3^2

F_ac = 33.01128 N

F_bc = k * Q_b * Q_c / R^2_bc

F_bc =  8.99 * 10^9* ( 5.7 * 10^(-4 ))* (1.08 * 10^(-4 )) / 5^2  

F_bc = - 22.137 N

Angle a is subtended between F_bc and y axis @ C

cos(a) = 3 / 5

sin (a) = 4 / 5

Compute F_net:

F_net = sqrt (F_x ^2 + F_y ^2)

F_x = sum of forces in x direction:

F_x = F_bc*sin(a) = 22.137*(4/5) = 17.71 N

F_y = sum of forces in y direction:

F_y = - F_bc*cos(a) + F_ac = - 22.137*(3/5) + 33.01128 = 19.72908 N

F_net = sqrt (17.71 ^2 + 19.72908 ^2) = 26.5119 N

Answer: F_net = 26.512 N

Final answer:

The question requires applying Coulomb's law to calculate the net electric force on particle C due to particles A and B, and summing the vector forces to find the resultant.

Explanation:

The question involves finding the net electric force on particle C, which is at the position (0, 3.00 m) in the presence of particles A and B with given charges and positions. The situation can be analyzed using Coulomb's law, which states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.

We would sum the vector forces from both charges on C to find the resultant force. This is a classical physics problem typically addressed in high school or introductory college physics courses.

Answer

given,

F₁ = 15 lb

F₂ = 8 lb

θ₁ = 45°

θ₂ = 25°

Assuming the question's diagram is attached below.

now,

computing the horizontal component of the forces.

F_h = F₁ cos θ₁ - F₂ cos θ₂

F_h = 15 cos 45° - 8 cos 25°

F_h = 3.36 lb

now, vertical component of the forces

F_v = F₁ sin θ₁ + F₂ sin θ₂

F_v = 15 sin 45° + 8 sin 25°

F_v = 13.98 lb

resultant force would be equal to

[tex]F = \sqrt{F_h^2+F_v^2}[/tex]

[tex]F = \sqrt{3.36^2+13.98^2}[/tex]

F = 14.38 lb

the magnitude of resultant force is equal to 14.38 lb

direction of forces

[tex]\theta =tan^{-1}(\dfrac{F_v}{F_h})[/tex]

[tex]\theta =tan^{-1}(\dfrac{13.98}{3.36})[/tex]

   θ = 76.48°

The resultant force will be 14.3844 N making an angle of 76.5° from the horizontal line.

Given to us

We know that a force is a vector quantity and can be divided into two component a vertical and a horizontal component. As shown below in the image.

The horizontal component is the cosine component while the vertical component is the sine component.

As the forces are pointing in different directions, therefore, the force on the left will be taken as positive while the force on the right is taken as negative.

[tex]F_{H} = F_1 Cos \theta_1 + F_2 Cos \theta_2\\F_{H} = [15\times Cos(45^o)]+[-8\times Cos(25^o)]\\F_H = 3.3561\ N[/tex]

As the forces are pointing in the same direction, therefore, the net force will be in the same direction,

[tex]F_{V} = F_1 Cos \theta_1 + F_2 Cos \theta_2\\F_{V} = [15\times Sin(45^o)]+[8\times Sin(25^o)]\\F_V = 13.9875\ N[/tex]

[tex]F_R = \sqrt{F_H^2+ F_V^2}[/tex]

[tex]F_R = \sqrt{3.3561^2+13.9875^2}\\F_R = 14.3844\ N[/tex]

[tex]Tan(\theta_R) = \dfrac{F_V}{F_H} = \dfrac{13.9875}{3.3561}= 4.167784\\\\(\theta_R) = Tan^{-1}(4.167784)\\(\theta_R) = 76.5^o[/tex]

As both the forces are positive the resultant force will be in the first quadrant, making an angle of 76.5° from the horizontal line.

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To solve this problem we will apply the concepts related to the change in length in proportion to the area and volume. We will define the states of the lengths in their final and initial state and later with the given relationship, we will extrapolate these measures to the area and volume

The initial measures,

[tex]\text{Initial Length} = L[/tex]

[tex]\text{Initial surface Area} = 6L^2[/tex] (Surface of a Cube)

[tex]\text{Initial Volume} = L^3[/tex]

The final measures

[tex]\text{Final Length} = L_f[/tex]

[tex]\text{Final surface area} = 6L_f^2[/tex]

[tex]\text{Final Volume} = L_f^3[/tex]

Given,

[tex]\frac{(SA)_f}{(SA)_i} = 2[/tex]

Now applying the same relation we have that

[tex](\frac{L_f}{L_i})^2 = 2[/tex]

[tex]\frac{L_f}{L_i} = \sqrt{2}[/tex]

The relation with volume would be

[tex]\frac{(Volume)_f}{(Volume)_i} = (\frac{L_f}{L_i})^3[/tex]

[tex]\frac{(Volume)_f}{(Volume)_i} = (\sqrt{2})^3[/tex]

[tex]\frac{(Volume)_f}{(Volume)_i} = (2\sqrt{2})[/tex]

[tex]\frac{(Volume)_f}{(Volume)_i} = 2.83[/tex]

Volume of the cube change by a factor of 2.83

Answer:

Part A:

[tex]E=127500N/C\\E=1.275*10^{5} N/C[/tex]

Part B:

Option B (Towards the South)

Explanation:

Part A:

Magnitude if electric field E:

E=Force/charge

Force=2.04×10−14 N

Charge=1.6×10−19 C

[tex]E=\frac{2.04*10^{-14}}{1.6*10^{-19}} \\E=127500N/C\\E=1.275*10^{5} N/C[/tex]

Part B:

Option B (Towards the South)

As electron is experiencing the force towards south,it means the direction of the electric field is towards the south because direction of field lines is from positive to negative, so proton is moving towards south it means negative charge is in south to which proton is attracted. So electric field is towards South.

Final answer:

The correct option is (b) toward the south. The magnitude of the electric field is 1.275 N/C, and the direction of the electric field is toward the south, as it attracts the positively charged proton in that direction.

Explanation:

To determine the magnitude of the electric field experienced by a proton, we use the formula for electric force: F = Eq, where F is the electric force, E is the electric field strength, and q is the charge of the particle.

Given that the force experienced by the proton is 2.04×10⁻¹⁴ N and the charge of a proton (q) is approximately 1.6×10⁻¹⁹ C, we can rearrange the formula to solve for E: E = F/q. Substituting the values, we obtain E = (2.04×10⁻¹⁴ N) / (1.6×10⁻¹⁹ C) = 1.275 N/C. Therefore, the magnitude of the electric field is 1.275 N/C.

Regarding the direction of the electric field, it is crucial to understand that electric fields exert forces on positive charges away from the source of the field and toward negative charges or areas of lower potential.

Since the proton is positively charged and experiences a force toward the south, the electric field's direction must be toward the south to attract the proton in that direction. So, the correct choice for the direction of the electric field is (b) toward the south.

Answer

given,

torque produced, τ = 22 N.m

Radius of the wheel. r = 10 cm

Moment of inertial = 2 kg.m²

initial angular speed = 0 rad/s

time, t = 5 s

a) we know,

   τ = I α

   22 = 2 x α

    α = 11 rad/s²

  using equation of rotation motion

[tex]\theta = \omega_o t + \dfrac{1}{2}\alpha t^2[/tex]

[tex]\theta =\dfrac{1}{2}\times 11 \times 5^2[/tex]

  θ = 137.5 rad

  θ = 137.5/2π  =  22 revolution.

b) angular velocity of the motor

  [tex]\omega_f = \omega_i + \alpha t[/tex]

  [tex]\omega_f = 0 + 11 x 5[/tex]

  [tex]\omega_f = 55\ rad/s[/tex]

c) acceleration of a point on the rim of the wheel

  radial acceleration

 [tex]a_r = \omega^2 r[/tex]

 [tex]a_r = 55^2\times 0.1[/tex]

 [tex]a_r =302.5 \ m/s^2[/tex]

tangential acceleration of the point on the rim

  [tex]a_t = \alpha r[/tex]

  [tex]a_t = 11\times 0.1[/tex]

  [tex]a_t = 1.1\ m/s^2[/tex]

now, acceleration of the point

[tex]a = \sqrt{a_r^2+a_t^2}[/tex]

[tex]a = \sqrt{302.5^2+1.1^2}[/tex]

[tex]a = 302.5\ m/s^2[/tex]

Answer:

Time, t = 12 minutes

Explanation:

It is given that,

A cyclist rides 16.0 km east, then 8.0 km west, then 8.0 km east, then 32.0 km west, and finally 11.2 km east. Let west direction is negative and east direction is positive. The displacement of the cyclist is :

[tex]d=16-8+8-32+11.2=-4.8\ km[/tex]

d = 4800 m

Let us assumed that the average speed of the cyclist is, v = 24 km/h = 6.66667 m/s

Let t is the time taken by the cyclist to complete the trip. The velocity of an object is given by :

[tex]v=\dfrac{d}{t}[/tex]

[tex]t=\dfrac{d}{v}[/tex]

[tex]t=\dfrac{4800\ m}{6.66667\ m/s}[/tex]

t = 719.99 seconds

t = 720 seconds

or

t = 12 minutes

So, the time taken by the cyclist to complete the trip is 12 minutes. Yes, the time taken by the cyclist to complete the trip is reasonable. Hence, this is the required solution.                                      

Answer:

Here is the complete question.

A charge of 3 μC is at the origin. Sketch the function E x versus x for both positive and negative values of x. (Remember that Ex is negative when E points in the negative x direction.)

The graph is sketched in the attachment.

Explanation:

Since our electric field Eₓ = q/4πε₀x² (electric field for a point charge)

where q = electric charge = 3μC and x = distance from the origin of the charge. 1/4πε₀= 9 × 10⁹ Nm²/C².

Substituting the values into Eₓ  = 3 × 10⁻⁶×9 × 10⁹ /x²= 0.027/x². We plot values of x on the x- axis ranging from -3 to 3 into the equation to give us the graph for Eₓ. Note that the sign of Eₓ changes when x crosses the origin because of the direction of the electric field due to the new position of the charge. The graph is in the attachment below. Note that as x tends to zero, Eₓ tends to infinity and x tends to infinity, Eₓ tends to zero.

Answer: 3 ft

Explanation:

Given:

- the ball is initially at rest before it start moving at constant acceleration.

a = constant

u = initial speed = 0

Using the formula for displacement.

d = ut + 0.5at^2 .....1

Since u = 0, equation 1 becomes;

d = 0.5at^2 .....3

making a the subject of formula

a = 2d/t^2 ....2

Where; a = acceleration, d = distance travelled, t = time taken, u = initial speed

d = 1 ft, t =1 s

Substituting into equation 2 above.

a = (2×1)/1^2 = 2ft/s^2

Since the acceleration is 2ft/s^2, at t= 2 sec the distance it would have covered will be:

Using equation 3;

d = 0.5(2 × 2^2) = 0.5(8) = 4ft

therefore, the distance travelled between t = 1 to t= 2

Is :

d2 = d - d1

Where d = total distance (t=0 to t= 2) =4 ft

d1 = distance covered between (t=0 to t=1) = 1 ft

d2 = 4 - 1 = 3ft

The distance traveled between time 1s and time 2s is 3 ft.

The distance traveled by the ball at time t is calculated as follows;

[tex]s = ut + \frac{1}{2} at^2\\\\s = 0 + \frac{1}{2} at^2\\\\s = \frac{1}{2} at^2\\\\a = \frac{2s}{t^2}[/tex]

when the distance, s = 1 ft and time, t = 1 s

[tex]a = \frac{2(1)}{(1)^2} \\\\a = 2 \ ft/s^2[/tex]

When the time of motion, t = 2s, the distance traveled is calculated as;

[tex]s = \frac{1}{2} (2) (2)^2\\\\s = 4 \ ft[/tex]

The distance traveled between t =1 and t = 2 is calculated as follows;

[tex]s = s_2 - s_1\\\\s = 4 \ ft - 1\ ft\\\\s = 3 \ ft[/tex]

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Answer:

[tex]14m/s^2[/tex]

Explanation:

Initial speed of car=0 Km/h=[tex]0m/s[/tex]

Final speed of car=[tex]100km/h=100\times \frac{5}{18}=\frac{250}{9} m/s[/tex]

By using [tex]1km/h=\frac{5}{18}m/s[/tex]

Time taken by car=2 s

We know that

Acceleration=[tex]a=\frac{v-u}{t}[/tex]

Where v=Final velocity of object

u=Initial velocity of object

t=Time taken by object

Using the formula

Acceleration of car=[tex]\frac{\frac{250}{9}-0}{2}=\frac{250}{9\times 2}=13.9 m/s^2\approx 14m/s^2[/tex]

Hence, the acceleration of car about=[tex]14m/s^2[/tex]

Answer:

the speed of the command module relative to Earth just after the separation = 4943.2 Km/hr

Explanation:

Given:

speed of space vehicle =5000 km/hr

rocket motor speed = 71 km/hr relative to the command module

mass of module = m

mass of motor = 4m

By conservation of linear momentum

Pi = Pf

Pi= initial momentum

Pf=  final momentum

Since, the motion is only in single direction

[tex]MV_i=4mV_{mE}+mV_{cE}[/tex]

Where M is the mass of the space vehicle which equals the sum of motor's mass and the command's mass, Vi its initial velocity, V_mE is velocity of motor relative to Earth, and V_cE is its velocity of the command relative to Earth.

The velocity of motor relative to Earth equals the velocity of motor relative to command plus the velocity of command relative to Earth.

V_mE = V_mc+V_cE

Where V_mc is the velocity of motor relative to command this yields

[tex]5mV_i = 4m(V_{mc}+V_{cE})+mV_{cE}[/tex]

[tex]5V_i = 4V_{mc}+5V_{cE}[/tex]

substituting  the values we get

[tex]V_{cE} = \frac{5V_i-4V_{mc}}{5}[/tex]

[tex]V_{cE} = \frac{5(5000)-4(71)}{5}[/tex]

= 4943.2 Km/hr

the speed of the command module relative to Earth just after the separation = 4943.2 Km/hr

Answer:

The speed of the command module relative to earth just after the separation is 4985.8 km/h

Explanation:

Given that,

Velocity of vehicle = 5000 km/h

Relative velocity = 71 km/h

The mass of the motor is four times the mass of the module.

We need to calculate the velocity of motor

Using formula of relative velocity

[tex]v=v_{2}-v_{1}[/tex]

Put the value into the formula

[tex]71=v_{2}-v_{1}[/tex]

[tex]v_{2}=71+v_{1}[/tex]

We need to calculate the speed of the  command module relative to Earth just after the separation

Using conservation of momentum

[tex]mu=m_{1}v_{1}+m_{2}v_{2}[/tex]

Put the value into the formula

[tex]5m\times5000=4m\times v_{1}+mv_{2}[/tex]

[tex]5\times5000=4\times v_{1}+(71+v_{1})[/tex]

[tex]25000=5v_{1}+71[/tex]

[tex]v_{1}=\dfrac{25000-71}{5}[/tex]

[tex]v_{1}=4985.8\ km/h[/tex]

Hence, The speed of the command module relative to earth just after the separation is 4985.8 km/h

a) The independent variables are x and t

b) The parameters are [tex]k[/tex] (wave number), [tex]A[/tex] (the amplitude), [tex]\omega[/tex] (angular frequency)

c) The phase of this wave is zero

d) The wavelength of the wave is [tex]\lambda=\frac{2\pi}{k}[/tex]

e) The period of the wave is [tex]T=\frac{2\pi}{\omega}[/tex]

f) The speed of propagation of the wave is [tex]v=\frac{\omega}{k}[/tex]

Explanation:

a)

In physics and mathematics:

For the wave in this problem, therefore, we have:

b)

In physics and mathematics, the parameters of a function are the quantities whose value is constant (so, they do not change), and the value of the dependent variable also depends on the values of these parameters.

Therefore in this problem, for the function that represents the y-displacement of the wave, the parameters are all the constant factors in the formula that are not variables. Therefore, they are:

c)

The phase of this wave is zero.

In fact, a general equation for a wave is in the form

[tex]y(x,t)=Asin(kx-\omega t+\phi)[/tex]

where [tex]\phi[/tex] is the phase of the wave, and it represents the initial angular displacement of the wave when x = 0 and t = 0.

However, the equation of the wave in this problem is

[tex]y(x,t)=Asin(kx-\omega t)[/tex]

Therefore, we see that its phase is zero:

[tex]\phi=0[/tex]

d)

The wavelength of a wave is related to the wave number by the following equation

[tex]k=\frac{2\pi}{\lambda}[/tex]

where

k is the wave number

[tex]\lambda[/tex] is the wavelength

For the wave in this problem, we know its wave number, [tex]k[/tex], therefore we can find its wavelength by re-arranging the equation above:

[tex]\lambda=\frac{2\pi}{k}[/tex]

e)

The period of a wave is related to its angular frequency by the following equation

[tex]\omega=\frac{2\pi}{T}[/tex]

where

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

T is the period of the wave

For the wave in this problem, we know its angular frequency [tex]\omega[/tex], therefore we can find its period by re-arranging the equation above:

[tex]T=\frac{2\pi}{\omega}[/tex]

f)

The speed of propagation of a wave is given by the so-called wave equation:

[tex]v=f\lambda[/tex]

where

v is the speed of propagation of the wave

f is the frequency

[tex]\lambda[/tex] is the wavelength

The frequency is related to the period of the wave by

[tex]f=\frac{1}{T}[/tex]

So, we can rewrite the wave equation as

[tex]v=\frac{\lambda}{T}[/tex]

From part d) and e), we found an expression for both the wavelength and the period:

[tex]\lambda=\frac{2\pi}{k}\\T=\frac{2\pi}{\omega}[/tex]

Therefore, we can rewrite the speed of the wave as:

[tex]v=\frac{2\pi/k}{2\pi/\omega}=\frac{\omega}{k}[/tex]

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