[1] The assembly starts from rest and reaches an angular speed of 150 rev/min under the action of a 20-N force T applied to the staring for t seconds. Determine t. Neglect friction and all masses except those of the four 3-kg spheres, which may be treated as particles.

Answer:

Semimajor axis of the comet's orbit is; α = 1.89 x 10^(13) m

Explanation:

From the question, we have a comet that was seen in April 574 by Chinese astronomers and was seen again in May 1994. That is a period of; T =(April)1994 - (May) 574 = 1420 years plus one month

Converting this to seconds, we have; T = [(1420 x 12) + 1] x 2628002.88 = 4.478 x 10^(10) seconds

Now, the square of the period of any planet is given by the formula;

T² = (4π²/GM)α³

Where;

G is gravitational constant and has a value of 6.67 x 10^(-11) m³/kg.s²

M is mass of sun which has a value of 1.99 x 10^(30) kg

α is the semi major axis of the comets orbit.

Thus, making α the subject, we have;

α = ∛(GMT²/4π²)

So, α = ∛{(6.67 x 10^(-11) x 1.99 x 10^(30) x (4.478 x 10^(10))²}/4π²)

α = 1.89 x 10^(13) m

Answer:

Explanation:

check the picture attached for the explanation. I hope it helps . Thank you

Answer:

Explanation:

a) the period of the motion:

In order to calculate, we have the following formula.

f = 1/T whereas, T = 1/f. So, we need frequency first to calculate period of the motion.

In the question, we have been given amplitude and maximum acceleration. So, we can use the formula of maximum acceleration to calculate the frequency. Here's how:

Maximum acceleration = (2πf)² x A

where, A is amplitude.

a(max) = maximum acceleration

By rearranging the equation and making the frequency our subject and plugging in the values of given quantities then we have:

f = 138.80 hertz

Now, we can calculate, period of the motion by plugging in the value of frequency in the equation of period mentioned above.

Period of the motion = T = 1/f

T= 1/138.80

T = 7.204 x [tex]10^{-3}[/tex] seconds.

b) Maximum Speed of the particle

In order to calculate max. speed of the particle, we have to use following formula:

Max. Speed = 2πf x A

Max. Speed = ( 2 x 3.14 x 138.80 x 7.1x[tex]10^{-3}[/tex])

Max. Speed = 6.19 m/s

c) Total mechanical energy of the oscillator

Total mechanical energy of the oscillator is the sum of kinetic energy and potential energy and for which we have formula to calculate total mechanical energy of the oscillator:

Total Mechanical Energy = T.E = K.E + P.E

T.E =  2m[tex]\pi ^{2}[/tex][tex]f^{2}[/tex][tex]A^{2}[/tex]

we do have values of all the quantities and now by just plugging in the values we will get the total mechanical energy of the oscillator.

T.E = 2 x 0.047 x [tex]3.14^{2}[/tex] x [tex]138.80^{2}[/tex] x [tex]0.0071^{2}[/tex]

T.E = 0.90 J

d) Magnitude of the force on the particle when the particle is at its maximum displacement.

F = -KA here, amplitude is used because it is the maximum displacement from the mean position.

where K = -m ω² x A

ω = 2πf = 2 x 3.14 x 138.80

ω = 871.66

F = -(0.047 x 871.66² x 0.0071²)

F = - 1.8 N

e) when the displacement is at half

F = -Kx A/2

A/2 = 0.0071 / 2

A/2 = 0.0035

F = - m ω² x A/2

F = - (0.047 x 871.66² x 0.0035)

F = - 125 N

Answer:

The magnetic flux through a loop is zero when the B field is perpendicular to the plane of the loop.

Explanation:

Magnetic flux are also known as the magnetic line of force surrounding a bar magnetic in a magnetic field.

It is expressed mathematically as

Φ = B A cos(θ) where

Φ is the magnetic flux

B is the magnetic field strength

A is the area

θ is the angle that the magnetic field make with the plane of the loop

If B is acting perpendicular to the plane of the loop, this means that θ = 90°

Magnetic flux Φ = BA cos90°

Since cos90° = 0

Φ = BA ×0

Φ = 0

This shows that the magnetic flux is zero when the magnetic field strength B is perpendicular to the plane of the loop.

Answer:

D. depends just on the B field going through the loop

Explanation:

The magnetic flux trough a loop is given by the formula:

[tex]\Phi_B=\vec{B}\cdot \vec{A}[/tex]

[tex]\Phi_B=BAcos\theta[/tex]

Where B is the magnitude of the magnetic field and A is the area of the loop.

It is clear that the flux is maximum when the angle between the direction of B and the direction of the normal vector of A is zero (cos0 = 1).

Furthermore, we can notice that only the magnetic field that crosses the loop contributes to the flux.

Hence, the correct answer is:

D. depends just on the B field going through the loop

hope this helps!!

Answer:

Approximately 4.5 square miles

Explanation:

(surrounding van Nuys and sepulveda stations) and one measuring approximately 2.85 square miles ( surrounding north Hollywood Station)

Final answer:

The 'orange line' referred to in the question likely represents data in a graph related to either the wavelength or frequency of orange light in the visible spectrum or a trend in statistical analysis. The exact answer depends on the specific context of the graph or data set from which the 'orange line' is derived.

Explanation:

The question 'The orange line is measuring the' seems to be related to analyzing data, most likely from a graph or chart that measures certain variables, potentially in the context of physics or mathematics dealing with light spectra or statistical analysis. Given the details provided, including mention of the visible light spectrum and specific wavelengths denoted by numbers, it appears that the 'orange line' may refer to a line on a graph that represents the wavelength or frequency of the color orange within the visible light spectrum.

However, the context provided is ambiguous. For instance, the chapter headings and the Conclusion provided suggests statistical analysis, perhaps in the field of queueing theory and how waiting times are affected by different systems, which implies a more mathematical or business-oriented question. In the field of mathematics, 'orange line' could also be a reference to a trend line or data line in a statistical plot.

Answer with Explanation:

We are given that

Y=(-4,12)

Z=(1,19)

We have to find the ordered pair which represents the vector YZ and magnitude of vector YZ.

Vector YZ=Z-Y=<1,19>-<-4,12>

Vector YZ=<5,7>

Magnitude of vector r

[tex]r=xi+yj[/tex]

[tex]\mid r\mid=\sqrt{x^2+y^2}[/tex]

Using the formula

[tex]\vec{YZ}=\sqrt{5^2+7^2}[/tex]

[tex]\vec{YZ}=\sqrt{25+49}=\sqrt{74} units[/tex]

Answer:

a)9.8

Explanation:

because any object falling from any height is 9.8

(sorry that's the only one i know)

1. The speed of the rock relative to the car when it hits the windshield is [tex]\( {42.00 \, \text{m/s}} \).[/tex]

2. The distance above the point of contact with the windshield from which the rock was dropped is [tex]\( {67.5 \, \text{m}} \).[/tex]

3. The driver would have had [tex]\( {3.71 \, \text{s}} \)[/tex] to react before the rock strikes the windshield.

Let's solve each part of the problem step-by-step.

Part (a): Find the speed of the rock relative to the car when it hits the windshield.

Given:

- Car speed, [tex]\( v_{\text{car}} = 21 \, \text{m/s} \)[/tex]

- Windshield angle with the horizontal,[tex]\( \theta = 60^\circ \)[/tex]

Since the rock is dropped from rest, its initial vertical velocity is [tex]\( 0 \, \text{m/s} \).[/tex]

The rock falls under the influence of gravity, so its vertical velocity [tex]\( v_{\text{vertical}} \)[/tex] at any time t is given by:

[tex]\[ v_{\text{vertical}} = g t \][/tex]

where [tex]\( g = 9.8 \, \text{m/s}^2 \).[/tex]

The horizontal velocity of the rock is [tex]\( 0 \, \text{m/s} \)[/tex] with respect to the ground.

To find the speed of the rock relative to the car when it hits the windshield, we need to consider the components of velocity with respect to the car:

1. The car's horizontal velocity component.

2. The rock's vertical velocity component.

The velocity components with respect to the car:

- Horizontal component relative to the car, [tex]\( v_{\text{horizontal, relative}} = v_{\text{car}} \)[/tex]

- Vertical component, [tex]\( v_{\text{vertical}} = g t \)[/tex]

The speed of the rock relative to the car can be found using the Pythagorean theorem:

[tex]\[ v_{\text{relative}} = \sqrt{(v_{\text{horizontal, relative}})^2 + (v_{\text{vertical}})^2} \][/tex]

We need to find t when the rock strikes the windshield.

Since the rock strikes the windshield at a right angle, and the windshield makes a 60° angle with the horizontal, the relative speed of the rock [tex]\( v_{\text{relative}} \)[/tex]  must have components that align with the 60° angle.

Thus:

[tex]\[ v_{\text{relative, vertical}} = v_{\text{relative}} \sin(60^\circ) \][/tex]

[tex]\[ v_{\text{relative, horizontal}} = v_{\text{relative}} \cos(60^\circ) \][/tex]

Given:

[tex]\[ v_{\text{relative, vertical}} = v_{\text{vertical}} = g t \][/tex]

[tex]\[ v_{\text{relative, horizontal}} = v_{\text{car}} = 21 \, \text{m/s} \][/tex]

From the component relationship:

[tex]\[ \frac{v_{\text{relative, vertical}}}{v_{\text{relative, horizontal}}} = \tan(60^\circ) = \sqrt{3} \][/tex]

[tex]\[ \frac{g t}{21} = \sqrt{3} \][/tex]

[tex]\[ g t = 21 \sqrt{3} \][/tex]

[tex]\[ t = \frac{21 \sqrt{3}}{9.8} \][/tex]

Calculate t:

[tex]\[ t \approx \frac{21 \cdot 1.732}{9.8} \approx \frac{36.372}{9.8} \approx 3.71 \, \text{s} \][/tex]

Now, we can find the vertical velocity [tex]\( v_{\text{vertical}} \)[/tex]:

[tex]\[ v_{\text{vertical}} = g t = 9.8 \times 3.71 \approx 36.36 \, \text{m/s} \][/tex]

Now, the speed of the rock relative to the car:

[tex]\[ v_{\text{relative}} = \sqrt{(21)^2 + (36.36)^2} \approx \sqrt{441 + 1322.49} \approx \sqrt{1763.49} \approx 42.00 \, \text{m/s} \][/tex]

Part (b): Find the distance above the point of contact with the windshield from which the rock was dropped.

The distance h from which the rock was dropped can be found using the kinematic equation:

[tex]\[ h = \frac{1}{2} g t^2 \][/tex]

Using [tex]\( t \approx 3.71 \, \text{s} \)[/tex]:

[tex]\[ h = \frac{1}{2} \times 9.8 \times (3.71)^2 \approx \frac{1}{2} \times 9.8 \times 13.76 \approx 67.5 \, \text{m} \][/tex]

Part (c): The time available for the driver to react is the time \( t \) it takes for the rock to fall, which we calculated in part (a):

[tex]\[ t \approx 3.71 \, \text{s} \][/tex]

Answer:

[tex]9.05\times 10^{-4} V[/tex]

Explanation:

We are given that

Length of antenna,l=1.45 m

Earth's magnetic field,[tex]B=59\mu T=59\times 10^{-6} T[/tex]

[tex]1\mu=10^{-6}[/tex]

[tex]\theta=65.5^{\circ}[/tex]

Speed,v=90 km/h=[tex]90\times \frac{5}{18}=25m/s[/tex]

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

Magnitude of maximum induced emf in the antenna=[tex]E=Blvcos\theta[/tex]

Substitute the values

[tex]E=59\times 10^{-6}\times 1.45\times 25cos65[/tex]

[tex]E=9.05\times 10^{-4} V[/tex]

Answer:

The order is 2>4>3>1 (TE)

Explanation:

Look up attached file

Answer:

See explanation

Explanation:

Given:-

- The DC power supply, Vo = 600 V

- The resistor, R = 1845 MΩ

- The plate area, A = 58.3 cm^2

- Left plate , ground, V = 0

- The right plate, positive potential.

- The distance between the two plates, D = 0.3 m

- The mass of the charge, m = 0.4 g

- The charge, q = 3*10^-5 C

- The point C = ( 0.25 , 12 )

- The point A = ( 0.05 , 12 )

Find:-

What is the speed, v, of that charge when it reaches point A(0.05,12)?

How long would it take the charge to reach point A?

Solution:-

- The Electric field strength ( E ) between the capacitor plates, can be evaluated by the potential difference ( Vo ) of the Dc power supply.

                           E = Vo / D

                           E = 600 / 0.3

                           E = 2,000 V / m

- The electrostatic force (Fe) experienced by the charge placed at point C, can be evaluated:

                           Fe = E*q

                           Fe = (2,000 V / m) * ( 3*10^-5 C)

                           Fe = 0.06 N

- Assuming the gravitational forces ( Weight of the particle ) to be insignificant. The motion of the particle is only in "x" direction under the influence of Electric force (Fe). Apply Newton's equation of motion:

                          Fnet = m*a

Where, a : The acceleration of the object/particle.

- The only unbalanced force acting on the particle is (Fe):

                          Fe = m*a

                          a = Fe / m

                          a = 0.06 / 0.0004

                          a = 150 m/s^2

- The particle has a constant acceleration ( a = 150 m/s^2 ). Now the distance between (s) between two points is:

                         s = C - A

                         s = ( 0.25 , 12 ) - ( 0.05 , 12 )

                         s = 0.2 m

- The particle was placed at point C; hence, velocity vi = 0 m/s. Then the velocity at point A would be vf. The particle accelerates under the influence of electric field. Using third equation of motion, evaluate (vf) at point A:

                        vf^2 = vi^2 + 2*a*s

                        vf^2 = 0 + 2*0.2*150

                        vf = √60

                        vf = 7.746 m/s

- Now, use the first equation of motion to determine the time taken (t) by particle to reach point A:

                       vf - vi = a*t

                       t = ( 7.746 - 0 ) / 150

                       t = 0.0516 s

- The charge placed at point C, the Dc power supply is connected across the capacitor plates. The capacitor starts to charge at a certain rate with respect to time (t). The charge (Q) at time t is given by:

                      [tex]Q = c*Vo*[ 1 - e^(^-^t^/^R^C^)][/tex]

- Where, The constant c : The capacitance of the capacitor.

- The Electric field strength (E) across the plates; hence, the electrostatic force ( Fe ) is also a function of time:

                     [tex]E = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D} \\\\Fe = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D}*q\\\\[/tex]

- Again, apply the Newton's second law of motion and determine the acceleration (a):

                     Fe = m*a

                     a = Fe / m

                     [tex]a = \frac{Vo*q*[ 1 - e^(^-^t^/^R^C^)]}{m*D}[/tex]

- Where the acceleration is rate of change of velocity "dv/dt":

                     [tex]\frac{dv}{dt} = \frac{Vo*q}{m*D} - \frac{Vo*q*[ e^(^-^t^/^R^C^)]}{m*D}\\\\B = \frac{600*3*10^-^5}{0.0004*0.3} = 150, \\\\\frac{dv}{dt} = 150*( 1 - [ e^(^-^t^/^R^C^)])\\\\[/tex]

- Where the capacitance (c) for a parallel plate capacitor can be determined from the following equation:

                      [tex]c = \frac{A*eo}{d}[/tex]

Where, eo = 8.854 * 10^-12  .... permittivity of free space.

                     [tex]K = \frac{1}{RC} = \frac{D}{R*A*eo} = \frac{0.3}{1845*58.3*8.854*10^-^1^2*1000} = 315\\\\[/tex]

- The differential equation turns out ot be:

                     [tex]\frac{dv}{dt} = 150*( 1 - [ e^(^-^K^t^)]) = 150*( 1 - [ e^(^-^3^1^5^t^)]) \\\\[/tex]

- Separate the variables the integrate over the interval :

                    t : ( 0 , t )

                    v : ( 0 , vf )

Therefore,

                   [tex]\int\limits^v_0 {dv} \, = \int\limits^t_0 {150*( 1 - [ e^(^-^3^1^5^t^)])} .dt \\\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^)}{315} )^t_0\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^) - 1}{315} )[/tex]

- The final velocity at point A for the particle is given by the expression derived above. So for t = 0.0516 s, The final velocity would be:

                    [tex]vf = 150*( 0.0516 + \frac{e^(^-^3^1^5^*^0^.^0^5^1^6^) - 1}{315} )\\\\vf = 7.264 m/s[/tex]

- The final velocity of particle while charging the capacitor would be:

                   vf = 7.264 m/s ... slightly less for the fully charged capacitor

Answer:

Horizontal launch

[tex]\vec v = 2.739\cdot i \,\left[\frac{m}{s} \right][/tex]

Vertical launch

[tex]\vec v = 2.739\cdot j \,\left[\frac{m}{s} \right][/tex]

Explanation:

The launch speed is calculated by means of the Principle of Energy Conservation:

[tex]U_{k} = K[/tex]

[tex]\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]

[tex]v = x \cdot \sqrt{\frac{k}{m} }[/tex]

[tex]v = (0.15\,m)\cdot \sqrt{\frac{100\,\frac{N}{m} }{0.3\,kg} }[/tex]

[tex]v \approx 2.739\,\frac{m}{s}[/tex]

The velocities for each scenario are presented herein:

Horizontal launch

[tex]\vec v = 2.739\cdot i \,\left[\frac{m}{s} \right][/tex]

Vertical launch

[tex]\vec v = 2.739\cdot j \,\left[\frac{m}{s} \right][/tex]

Final answer:

The launch velocity of the T-shirt, when the spring with a spring constant of 100 N/m is compressed by 0.15 m, is calculated using conservation of energy. It turns out to be approximately 2.74 m/s for both the horizontal and vertical launches.

Explanation:

To calculate the launch velocity of the T-shirt from the cannon in both horizontal and vertical scenarios, we'll use the conservation of energy principle. This principle states that the potential energy stored in the compressed spring is converted into kinetic energy of the moving T-shirt.

Calculation of Launch Velocity (Horizontal and Vertical)

The energy stored in the compressed spring can be given by the formula:
[tex]E = 1/2 k x^2[/tex]
where E is the energy in joules, k is the spring constant, and x is the compression distance.

For the given spring with k = 100 N/m compressed by 0.15 m, the energy stored is:
[tex]E = 1/2 (100) (0.15)^2 = 1.125 joules[/tex]

Now we set this equal to the kinetic energy of the T-shirt upon release:
[tex]KE = 1/2 m v^2[/tex]
where KE is the kinetic energy, m is the mass of the T-shirt, and v is the velocity.

To find the launch velocity, we solve for v given the mass m = 0.3 kg and energy E = KE:

[tex]v = sqrt((2 * E) / m) = sqrt((2 * 1.125) / 0.3) = sqrt(7.5) = approximately 2.74 m/s[/tex]

This velocity will be the same for both the horizontal and the vertical launch as friction is ignored and the energy is purely transferred into the kinetic energy of the T-shirt.

Final answer:

The direction of the resultant force acting on an electron moving upward on the page through a magnetic field directed into the page is to the left, as per the left-hand rule.

Explanation:

The question deals with the Lorentz force experienced by a charged particle, in this case an electron, moving through a magnetic field. When an electron with initial velocity directed from the bottom edge to the top edge of the page moves through a uniform magnetic field directed into the page, the force acting on it can be determined using the left-hand rule, since electrons have a negative charge. Holding your left thumb up (direction of the electron's velocity) and the fingers into the page (direction of the magnetic field), the resultant force will be directed to the left of the page. Therefore, the correct answer to the direction of the resultant force acting on the electron is c. to the left.

Answer:

(A) [tex]9.14\times 10^{-9}sec[/tex]

(B) [tex]6.20\times 10^{-3}A[/tex]

Explanation:

We have given inductance [tex]L=5.41\mu H=5.41\times 10^{-6}H[/tex]

Resistance [tex]R=0.949kohm=0.949\times 10^3ohm[/tex]

Time constant of RL circuit is equal to [tex]\tau =\frac{L}{R}[/tex]

[tex]\tau =\frac{5.41\times 10^{-6}}{0.949\times 10^3}=5.70\times 10^{-9}sec[/tex]

Battery voltage e = 16 volt

(a) It is given current becomes 79.9% of its final value

Current in RL circuit is given by

[tex]i=i_0(1-e^{\frac{-t}{\tau }})[/tex]

According to question

[tex]0.799i_0=i_0(1-e^{\frac{-t}{\tau }})[/tex]

[tex]e^{\frac{-t}{\tau }}=0.201[/tex]

[tex]{\frac{-t}{\tau }}=ln0.201[/tex]

[tex]{\frac{-t}{5.7\times 10^{-9} }}=-1.6044[/tex]

[tex]t=9.14\times 10^{-9}sec[/tex]

(b) Current at [tex]t=\tau sec[/tex]

[tex]i=i_0(1-e^{\frac{-t}{\tau }})[/tex]

[tex]i=\frac{16}{0.949\times 10^3}(1-e^{\frac{-\tau }{\tau }})[/tex]

[tex]i=6.20\times 10^{-3}A[/tex]