Suppose that we use a heater to boil liquid nitrogen (N2 molecules). 4480 J of heat turns 20 g of liquid nitrogen into gas. Note that the latent heat is equal to the change in enthalpy, and that liquid nitrogen boils at 77 K. The system is kept at a constant pressure of 1 atm. 20) Assuming that you can treat the gas as ideal gas and that the volume of the liquid compute the binding energy of a nitrogen molecule in the liquid. (the binding energy is the difference in internal energy per molecule between the liquid and gas) approximately zero,

a. 9.4 x 10-21 J
b. 3.8 х 1027 J
c. 4.2 x 10-18 J
d. 10-20 J e. 2.1 x 10-19 J

Answer:

[tex]6.67154\times 10^{-9}\ F[/tex]

13.009503 C

Explanation:

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

k = Dielectric constant of air [tex]3\times 10^6\ V/m[/tex]

Side of plate = 0.7 km

A = Area

d = Distance = 650 m

Capacitance is given by

[tex]C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 700^2}{650}\\\Rightarrow C=6.67154\times 10^{-9}\ F[/tex]

The capacitance is [tex]6.67154\times 10^{-9}\ F[/tex]

Electric field is given by

[tex]Q=CV\\\Rightarrow Q=Ckd\\\Rightarrow Q=6.67154\times 10^{-9}\times 3\times 10^6\times 650\\\Rightarrow Q=13.009503\ C[/tex]

The charge on the cloud is 13.009503 C

The capacitance of the cloud-ground system is 6.67 nF, and the cloud can hold up to 13 C of charge before lightning occurs.

To model lightning using a parallel-plate capacitor, assume the ground is one plate and a cloud at 650 m altitude is the other. Let’s calculate the capacitance and the charge the cloud can hold before a lightning strike occurs.

Therefore, the capacitance of the capacitor formed by the cloud and the ground is 6.67 nF, and the cloud can hold up to 13 C of charge before a lightning strike occurs.

The mass that must be added is 0.628 kg

Explanation:

The period of a mass-spring system is given by

[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]

where

m is the mass

k is the spring constant

For the initial mass-spring system in the problem, we have

m = 0.500 kg

T = 1.36 s

Solving for k, we find the spring constant:

[tex]k=(\frac{2\pi}{T})^2 m = (\frac{2\pi}{1.36})^2 (0.500)=10.7 N/m[/tex]

In the second part, we want the period of the same system to be

T = 2.04 s

Therefore, the mass on the spring in this case must be

[tex]m=(\frac{T}{2\pi})^2 k =(\frac{2.04}{2\pi})^2 (10.7)=1.128 kg[/tex]

Therefore, the mass that must be added is

[tex]\Delta m = 1.128 - 0.500 = 0.628 kg[/tex]

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Answer:

Explanation:

Given

No of logs [tex]n=14[/tex]

diameter of log [tex]d=42\ cm[/tex]

Length of log [tex]L=6.4\ m[/tex]

42 % of log volume(V) is above water when no one is on raft

so 58 % of log volume(V) is submerged in the water

Weight of 14 log

[tex]W=14\times \rho _{log}\times V\times g[/tex]

Buoyancy force on 14 logs [tex]F_b=14\times \rho _{water}\times 0.58V\times g[/tex]

as system is in equilibrium so

[tex]W=F_b[/tex]  

[tex]14\times \rho _{log}\times V\times g=14\times \rho _{water}\times 0.58V\times g[/tex]

[tex]\rho _{log}=0.58\rho _{water}[/tex]

[tex]\frac{\rho _{log}}{\rho _{water}}=0.58[/tex]

Specific gravity of log [tex]=0.58[/tex]

Answer:

NORMAL DRIVER: d = 73.3 ft

DRUNK DRIVER: d = 172.3

Explanation:

NORMAL DRIVER:

Distance covered in initial 0.75s = 0.75s *44 =  33ft

USING THE THIRD EQUATION OF MOTION

V^2-U^2 = 2as

0-(44)^2 = 2 (-24) s

s = 1936/48 =40.3 ft

d = 33 + 40.3 = 73.3 ft

DRUNK DRIVER:

Distance covered in initial 3s = 3s *44 =  132 ft

USING THE THIRD EQUATION OF MOTION

V^2-U^2 = 2as

0-(44)^2 = 2 (-24) s

s = 1936/48 =40.3 ft

d = 132 + 40.3 = 172.3 ft

Answer:

n = 1000 pulses

Explanation:

Given that,

The frequency of a pulse waveform, [tex]f=10\ kHz=10^4\ Hz[/tex]

To find,

The number of pulses counted during 100 ms.

Solution,

The frequency of a pulse waveform is equal to the number of pulses per unit time. It is given by :

[tex]f=\dfrac{n}{t}[/tex]

[tex]n=f\times t[/tex]

[tex]n=10^4\ Hz\times 100\times 10^{-3}\ s[/tex]

n = 1000 pulses

So, there are 1000 pulses counted in a pulse waveform.

Answer:

Distance between both the object will be 404.51 m

Explanation:  

We have given charge on two objects [tex]q_1=q_2=1C[/tex]

Coulomb force between the two objects is given F = 5.5 N

We have to fond the distance between both the object so that force between them is 5.5 N

According to coulomb law force between two charge particle is [tex]F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}[/tex]

So [tex]5.5=\frac{9\times 10^9\times 1\times 1}{r^2}[/tex]

[tex]r^2=16.36\times 10^8[/tex]

r = 404.51 m

So distance between both object will be 404.51 m

Answer:

[tex]F=926.31N[/tex]

Explanation:

To calculate the magnitude of the sum of the vectors we must find their components on the x and y axes, in this case, since they are perpendicular, one vector is on the x axis and the other is on the y axis:

[tex]F=\sqrt{F_x^2+F_y^2}\\F_x=A_x+B_x=A\\F_y=A_y+B_y=B\\F=\sqrt{A^2+B^2}\\F=\sqrt{(655N)^2+(655N)^2}\\F=926.31N[/tex]

To determine the electric power generated by the wind turbine, use the formula P = 0.5 x ρ x A x V^3 x η. Calculate the swept area of the blades using A = π x (d/2)^2. Multiply the power by the number of seconds in a day and convert it to kilowatt-hours to find the energy generated per day. Multiply the energy by the unit price of electricity to determine the revenue generated per day.

To determine the electric power generated by the wind turbine, we need to use the formula P = 0.5 x ρ x A x V^3 x η, where P is the power, ρ is the air density, A is the swept area of the blades, V is the wind speed, and η is the efficiency of the turbine.

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Answer:

E=3.307×10⁻⁴N/C

Explanation:

Given data

Length of the plate side L=24.0 cm =0.24 m

Distance between the plates d= 3.4 mm

Number of electron moves from one plate to others n=1012 electrons

Permittivity of free space ε₀=8.85×10⁻¹²c²/N.m²

Electron charge e=-1.6×10⁻¹⁹C

To find

Electric field between the plates

Solution

E=σ/ε₀

[tex]E=\frac{(Q/A)}{E}\\ E=\frac{Q}{EA}\\ E=\frac{ne}{EA}\\ E=\frac{1012*()1.6*10^{-19} }{8.85*10^{-12}(0.24)(0.24)}\\ E=3.307*10^{-4}N/C[/tex]

Answer:

Answer:

101325 + 10055.25h

//

h = 10.1 m

Explanation:

the pressure at sea level =  1 atm = 101325 Pa

density of sea water = 1025 kg/ m^(3)

pressure due to fluid height = pgh

Absolute pressure = 101325 + 1025*9.81*h

                               = 101325 + 10055.25h

where h= 0 at sea level at increases downwards

//

101325 = 1025* 9.81* h

h = 10.1 m

Explanation:

Answer:

Please see below as the answers are self-explanatory.

Explanation:

1) The resultant force is along the line that joins both charges or both masses (assuming both objects can be represented as points)

2) Both type of forces obey  Newton's 3rd law.

3) Both are proportional to the product of the property that is affected by the force (charges and masses)

4) Both obey an inverse - square law (consequence of our universe being three-dimensional)

1) Main difference, is that while the gravitational force is always attractive, the electrostatic force can be attractive or repulsive, as there are two types of charges, which attract each other being of different type, and repel each other if they are of the same type.

2) It  is possible, artificially, to block the influence of the electrostatic force, shielding a room, for instance, which is not possible for the gravitational force.

Answer: 1. False, an ammeter must be placed in a series connection to experience the same current which it is measuring

2. True, a voltmeter measures voltage (electric potential) in volts

3. False, a voltmeter should have high internal resistance such that it does not significantly alters current in a circuit.

4. True, a meters measure current

5. True, voltmeter would measure electric potential across a resistor

6. False, since they tend to influence the current in a circuit, ideally it should have very low resistance near zero

Explanation:

Answer:

The net force on the 20kg mass m1 is equal to 6.09 x 10^-7 N. This force is due to the sun of the vertical components of the forces F1 and F2 of the masses m2 and m3 respectively on the mass m1.

The law of gravitational force has been applied and the use of the Pythagorean theorem also used.

Explanation:

The full solution can be found in the attachment below.

Thank you for reading this post and I hope it is helpful to you.

The magnitude of the net gravitational force on the masses can be determined using Newton's law when the distance between the two masses is known.

The magnitude of gravitaional force on the two given masses can be determined by applying Newton's law of universal gravitation as shown below;

F = Gm₁m₂/r²

where;

Thus, the magnitude of the net gravitational force on the masses can be determined using Newton's law when the distance between the two masses is known.

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Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

[tex]Q =\frac{KA(\delta T)}{L}[/tex]

[tex]Q =\frac{25.87*1*20}{1}[/tex]

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

Answer: Rate of heat transfer q = 517.4W

Explanation:

Thermal conductivity is a material property that describes its ability to conduct heat. Thermal conductivity is the quantity of heat transmitted through a unit thickness of a material, in a direction normal to a surface of unit area,due to a unit temperature gradient under steady state conditions. It can be shown mathematically using the equation below;

Q/t = kA(∆T)/d

q = kA(∆T)/d

q = Q/t = rate of heat transfer

Q = total amount of heat transfer

t = time

k = thermal conductivity of material

A = Area

d= distance

For the case above, the material used is still air.

k for still air at 20°C and 1bara = 0.02587W/mK

A = 1m^2

d = 1mm = 0.001m

∆T = 20°C = 20K

q = 0.02587×1×20/0.001

q = 517.4W

Therefore, the rate of heat conduction through the still air is 517.4W

Answer:

Stress = 165,521 Pa

Strain = 0.015

Modulus of Elasticity = 11,034,742.72 Pa

Ultimate Tensile Stress = 534,761 Pa

Explanation:

Stress @ Force = 13 N

F = kΔx .... Eq 1

Evaluate k @ F = 16 N and Δx = 0.0035m

k = 16 N / 0.0035 m = 4571.43 N/m

Stress = Force / Area ..... Eq 2

Stress = 13 N *4 / π (0.01²) = 165,521.1408 Pa

Strain @ Force = 13 N

Strain = Δx / original Length = (F / k) / original Length

Hence,

Strain = (13 N / 4571.43 N / m) / 0.19 m = 0.015

Modulus of Elasticity (E)

E = Stress / Strain

E = 165,521.1408 Pa / 0.015 = 11,034,742.72 Pa

Ultimate Tensile Stress

Max Stress that can be achieved by a material

Stress = Fmax/Area

Stress = 42 N *4 / π (0.01²) = 534,761 Pa

Answer:

A) You have to run 73.8 m before you reach the rear of the bus.

B) When you catch the bus, your velocity will be 11.9 m/s.

C) No, an average college student would not reach the bus.

Explanation:

Hi there!

The equations of velocity and position that we will use are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity at time t.

A) First, let´s find the time at which you catch the bus. At this time, the position of the bus and yours are the same:

When you catch the bus:

Bus position = your position

The equation of position of the bus with constant velocity is the following:

x = x0 + v · t

You have a constant acceleration, so your position will be given by the equation of position described above:

x = x0 + v0 · t + 1/2 · a · t²

If we place the origin of the frame of reference at the bus stop, your initial position is zero (and the initial position of the bus is 12.0 m). Since you start from rest, your initial velocity is also zero. Then:

Bus position = your position

x0 + v · t = 1/2 · a · t²

12.0 m + 5.00 m/s · t = 1/2 · 0.960 m/s² · t²

0 = 0.48 m/s² · t² - 5.00 m/s · t - 12.0 m

Solving the quadratic equation for t:

t = 12.4 s

You catch the bus in 12.4 s, now, we have to find the distance you run in that time:

x = x0 + v0 · t + 1/2 · a · t²  

Since x0 and v0 = 0

x =  1/2 · a · t²

x = 1/2 · 0.960 m/s² · (12.4 s)²

x = 73.8 m

You have to run 73.8 m before you reach the rear of the bus.

B) Now, let´s use the equation of velocity to find your velocity at t = 12.4 s.

v = v0 + a · t

Since v0 = 0:

v = a · t

v = 0.960 m/s² · 12.4 s

v = 11.9 m/s

When you catch the bus, your velocity will be 11.9 m/s.

C) The record of Usain Bolt is 12.4 m/s. You would have to run almost as fast as Usain Bolt to catch the bus. So the answer is no, an average college student would not reach the bus.

To solve this problem we will convert the values given to international units (meters) and then proceed to calculate the number of wavelengths through the division of the width over the wavelength. This is,

[tex]\lambda = 650 nm =650 * 10^{-9} m[/tex]

[tex]w_1 = 0.16 mm = 0.16 * 10^{-3} m[/tex]

[tex]w_2 = 0.04 mm =0.04 * 10^{-3} m[/tex]

Now the number of wavelengths is the division between the total width over the wavelength therefore

First case,

[tex]\frac{w_1}{\lambda} = \frac{ 0.16 * 10^{-3}}{650 * 10^{-9}} = 4062.5 * 10^6[/tex]

Second case,

[tex]\frac{w_2}{\lambda} = \frac{0.04 * 10^{-3} }{650 * 10^{-9} } = 16250 * 10^6[/tex]

Answer:

D. is always perpendicular to the surface of the conductor

Explanation:

1) Answer is (D) option. Electric field just outside surface of charged conductor is normal to conductor at that point.

It can be explained on the basis of the fact that, Electric field inside conductor under static condition is zero. As a result potential difference between any two points with in conductor is zero. So whole of conductor is equipotential body.

Equipotential surface and Electric field lines always cut at 90 degrees to each other. Conductor being equipotential body, Electric field lines starting or terminating at conductor must be normal to surface. Hence electric field just outside conductor is perpendicular or normal to surface.

Answer:

Average speed of the trip = 52.9 km/h

Distance between initial pairs of cities (start to end) = 70.0 km

Explanation:

Since distance = speed × time

If she drives 30.0 min at 80.0 km/h

Distance covered = (30/60) × 80 = 40.0 km

Again she drives 45.0 min at 40 km/h

Distance covered = (45/60) × 40 = 30.0 km

Again she drives 12.0 min at 100 km/h

Distance covered = (12/60) × 100 = 20.0 km

Total distance covered =  40.0 + 30.0 + 20.0

                                       = 90.0 km

Total time spent = 30.0 + 45.0 + 12.0 +15.0

                           = 102 min

Average speed for the trip = Total distance covered/total time spent

                                             = 90/(102/60)

                                             = 52.9 km/h

Distance between initial cities will be between the start of one city to the end of another

Between the first pairs = 40.0 + 30.0 = 70.0 km

Between the second pairs = 30.0 + 20.0 = 50.0 km

Final answer:

The average speed for the trip, excluding stops, is approximately 62.1 km/h, and the total distance traveled between the initial cities is 90.0 km.

Explanation:

To determine the average speed for the trip, we need to calculate the total distance traveled and divide it by the total time taken, excluding any stops. We will calculate distance as speed multiplied by time for each segment of the trip, convert the minutes into hours, and then sum these distances to find the total distance traveled.

First segment: 80 km/h  x 0.5 h = 40.0 km
Second segment: 40 km/h x 0.75 h = 30.0 km
Third segment: 100 km/h x 0.2 h = 20.0 km
Total distance = 40.0 km + 30.0 km + 20.0 km = 90.0 km

Now, let's calculate the total travel time in hours, remembering to exclude the time spent not driving (the lunch and gas stop).
Driving time = 30 min + 45 min + 12 min = 87 min = 1.45 h

The average speed is then total distance divided by total driving time.
Average speed = 90.0 km / 1.45 h ≈ 62.1 km/h

To determine the difference between the initial cities along the route, we use the total distance traveled, which we have found to be 90.0 km.

Answer:

Charge of particle 2, [tex]q_2=-7.13\ \mu C[/tex]

Explanation:

Given that,

Charge 1, [tex]q_1=3.11\ \mu C=3.11\times 10^{-6}\ C[/tex]

The distance between charges, r = 0.241 m

Force experienced by particle 1, F = 3.44 N

We need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

[tex]q_2=\dfrac{Fr^2}{kq_1}[/tex]

[tex]q_2=\dfrac{3.44\times (0.241)^2}{9\times 10^9\times 3.11\times 10^{-6}}[/tex]

[tex]q_2=7.13\times 10^{-6}\ C[/tex]

or

[tex]q_2=7.13\ \mu C[/tex]

So, the magnitude of electric charge 2 is [tex]q_2=7.13\ \mu C[/tex]. Since, the force is attractive then the magnitude of charge 2 must be negative.

Final answer:

To determine the magnitude and sign of q2, Coulomb's Law is applied, revealing that q2 has a negative charge of approximately -4.48 μC, due to the attractive force observed.

Explanation:

To find the magnitude and sign of q2, we will use Coulomb's Law, which quantifies the electric force between two charged particles. Coulomb's Law is given by the formula F = k |q1*q2| / r^2, where F is the magnitude of the force between the charges, k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.

Given that q1 = +3.11 μC (or 3.11 x 10^-6 C), r = 0.241 m, and F = 3.44 N, and knowing that the force is attractive (indicating that q2 must have an opposite sign to q1), we can solve for q2 as follows:

F = k |q1*q2| / r^2

3.44 = (8.99 x 10^9) |(3.11 x 10^-6) * q2| / (0.241)^2

Solving for q2, we get q2 ≈ -4.48 x 10^-6 C. The negative sign indicates that q2 is negatively charged, and the magnitude of this charge is 4.48 μC.

Answer:

Work done on the object will be 50 J

Explanation:

We have given force force applied on the object F = 10 N

Object move by a distance of 5 m

So d = 5 m

We have to find the work done on the object by the force

Work done is given by [tex]work=force\times distance[/tex]

So work done [tex]=10\times 5=50J[/tex]

So work done on the object will be 50 J

Answer:

a) In the hand

b) behind the hand

c) in front of the hand

Explanation:

a)

When traveling at a constant speed when an egg is tossed directly upwards it lands back in our hand.  It is so because the instant when the egg is released from the hand the egg has the same velocity as the velocity of the person sitting in the moving car, and hence by the virtue of inertia of motion it land lands back in our hand.

b)

When we toss an egg while the car is accelerating then the egg lands behind the hand because the hand along with the car is accelerating ahead. The instant when the egg is released from the hand the car is accelerating then the car has a lower velocity which the egg acquires in the air, while the  egg is in the air the car further gains the velocity and hence the egg lands behind the hand.

C)

When the speed of the car is decreasing then the egg lands in front of the hand because the moment when the egg was tossed into the air it acquired the instantaneous velocity of the car from that moment and then while it was in the air it had the same velocity as initial while the car further decelerated.

a) 4.95 s when the mass is halved

b) 7.00 s when the amplitude is doubled

c) 4.95 s when spring constant is doubled

d) 9.9 s when mass is doubled

Given:

T₀=7.0 s

The equation for the time period of the object is

[tex]T=2*\pi *\sqrt{\frac{m}{k} }[/tex]        ......................(i)

where m= mass of an object and k= spring constant

m= 1/2

∵T=7.0 s

On substituting value in equation (i)

Thus, T=4.95 s

Time period does not depend on amplitude thus,

T=7.0 s

Spring constant, k=2k

On substituting value in equation (i)

Thus, T=4.95 s

Mass=2m

On substituting value in equation (i)

Thus, T=9.9 s

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The period of a mass-spring oscillating system changes based on variations in mass and spring constant but not amplitude. If the mass is halved or the spring constant is doubled, the period decreases to approximately 4.95 s. If the mass is doubled, the period increases to approximately 9.90 s.

The period of oscillation for a mass attached to a spring is given by the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant. Keeping this formula in mind, let's evaluate how changes to the system will affect the period of oscillation:

To solve this problem we will apply the concepts related to the conservation of momentum. That is, the final momentum must be the same final momentum. And in each state, the momentum will be the sum of the product between the mass and the velocity of each object, then

[tex]\text{Initial Momentum} = \text{Final Momentum}[/tex]

[tex]m_1u_1 +m_2u_2 = m_1v_1+m_2v_2[/tex]

Here,

[tex]m_{1,2}[/tex]= Mass of each object

[tex]u_{1,2}[/tex]= Initial velocity of each object

[tex]v_{1,2}[/tex]= Final velocity of each object

When they position the final velocities of the bodies it is the same and the car is stationary then,

[tex]m_2u_2 = (m_1+m_2)v_f[/tex]

Rearranging to find the final velocity

[tex]v_f = \frac{m_2u_2}{ (m_1+m_2)}[/tex]

[tex]v_f = \frac{ 12500*7.8}{ 12500+7430}[/tex]

[tex]v_f = 4.8921ft/s[/tex]

The expression for the impulse received by the first car is

[tex]I = m_1 (v-u)[/tex]

[tex]I = \frac{W}{g} (v-u)[/tex]

Replacing,

[tex]I = \frac{12500}{32.2}(4.89-7.8)[/tex]

[tex]I = -1129.65lb\cdot s[/tex]

The negative sign show the opposite direction.

Answer:

[tex]1.152\ \mu m[/tex]

[tex]1.44\ \mu m[/tex]

Explanation:

d = Gap between slits = 0.142 mm

[tex]\lambda[/tex] = Wavelength of light = 576 nm

L = Distance between light and screen = 3.5 m

m = Order = 2

Difference in path length is given by

[tex]\delta=dsin\theta=m\lambda\\\Rightarrow \delta=2\times 576\times 10^{-9}\\\Rightarrow \delta=0.000001152\ m\\\Rightarrow \delta=1.152\times 10^{-6}\ m=1.152\ \mu m[/tex]

The difference in path lengths is [tex]1.152\ \mu m[/tex]

For dark fringe the difference in path length is given by

[tex]\delta=(m+\dfrac{1}{2})\lambda\\\Rightarrow \delta=(2+\dfrac{1}{2})\times 576\times 10^{-9}\\\Rightarrow \delta=0.00000144\ m=1.44\ \mu m[/tex]

The difference in path length is [tex]1.44\ \mu m[/tex]

Answer:

m = 0.255kg

Explanation:

from the formular of a mass - spring system

T = 2π√m/k

making m as the subject of formular

m = T² k/4π²

T =5.91s

k = 0.288 N/m

m = 10.059/39.489

m = 0.255kg

To solve this problem we will start considering the relationship between rms voltage and the maximum voltage. Once the RMS voltage is obtained, we will proceed to find the system current through the given power and the voltage found. Finally by Ohm's law we will find the resistance of the system

The relation of RMS Voltage is,

[tex]V_{rms} = \frac{1}{\sqrt{2}}*V_{max}[/tex]

Note that this conversion is independent from the Frequency.

[tex]V_{rms} = \frac{1}{\sqrt{2}}*(170)[/tex]

[tex]V_{rms}= 120.20V[/tex]

Now the current is,

[tex]I =\frac{P}{V}[/tex]

[tex]I = \frac{(25 W)}{(120.20V)}[/tex]

[tex]I = 0.2079A[/tex]

By Ohm's Law

[tex]V = IR \rightarrow R = \frac{V}{I}[/tex]

Replacing,

[tex]R = \frac{ (120.20 V)}{ ( 0.2079 A)}[/tex]

[tex]R = 578.16 \Omega[/tex]

Therefore the resistance of this lightbulb is[tex]578.16 \Omega[/tex]

The resistance of the lightbulb is 1156.5 Ω. It can be calculated using Ohm's Law, which states that resistance is equal to voltage divided by current.

The resistance of a lightbulb can be found using Ohm's Law, which states that resistance is equal to voltage divided by current. In this case, we need to find the resistance of a lightbulb that uses an average power of 25.0 W when connected to a 60.0-Hz power source with a maximum voltage of 170 V. We can start by finding the current using the formula P = IV, where P is power and I is current. Rearranging the formula gives us I = P/V. Plugging in the given values, we get I = 25.0 W / 170 V = 0.147 A. Now we can use Ohm's Law to find the resistance. R = V/I = 170 V / 0.147 A = 1156.5 Ω.

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Answer:

P = 2i + 5j

Therefore she is 2 blocks east and 5 blocks north.

Resultant P = √(2^2 + 5^2) = √(4+25) = √29 = 5.4 blocks

Angle = taninverse (5/2)

Angle = 68.2°

Explanation:

Given:

Let west be negative and east be positive x axis.

Let north be the positive y axis.

5.00blocks west = -5.00 i

5.00 blocks north = 5.00 j

7.00 blocks east = 7.00i

Addition of the vector form of hee position is;

P = -5i +7i -5j

P = 2i + 5j

Therefore she is 2 blocks east and 5 blocks north.

Resultant P = √(2^2 + 5^2) = √(4+25) = √29 = 5.4 blocks

Angle = taninverse (5/2)

Angle = 68.2°

Answer:

a) [tex]\Delta x =0.32433\ m= 324.33\ mm[/tex]

b) [tex]\delta x=0.087996\ m=87.996\ mm[/tex]

c) [tex]\delta x=0.13227\ m=132.27\ mm[/tex]

Explanation:

Given:

a)

When the object is dropped onto the spring whole of the gravitational potential energy of the mass is converted into the spring potential energy:

[tex]PE_g=PE_s[/tex]

[tex]m.g.h=\frac{1}{2} \times k.\Delta x^2[/tex]

[tex]1.4\times 9.8\times 1.15=0.5\times 300\times \Delta x^2[/tex]

[tex]\Delta x =0.32433\ m= 324.33\ mm[/tex] is the compression in the spring

b)

When there is a constant air resistance force of 0.6 newton then the apparent weight of the body in the medium will be:

[tex]w'=m.g-0.6[/tex]

[tex]w'=1.4\times 9.8-0.6[/tex]

[tex]w'=1.01\ N[/tex]

Now the associated gravitational potential energy is converted into the spring potential energy:

[tex]PE_g'=PE_s[/tex]

[tex]w'.h=\frac{1}{2} \times k.\delta x^2[/tex]

[tex]1.01\times 1.15=0.5\times 300\times \delta x^2[/tex]

[tex]\delta x=0.087996\ m=87.996\ mm[/tex]

c)

On moon, as per given details:

[tex]m.g'.h=\frac{1}{2} \times k.\delta x^2[/tex]

[tex]1.4\times 1.63\times 1.15=0.5\times 300\times \delta x^2[/tex]

[tex]\delta x=0.13227\ m=132.27\ mm[/tex]

Answer:

 t = 40 s

Explanation:

given,

Speed of the P-wave = 7 km/s

distance of the seismic station = 280 Km

time taken by the P-wave = ?

we know,

distance = speed x time

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

[tex]t = \dfrac{280}{7}[/tex]

 t = 40 s

time taken by the P-wave to arrive at the station is equal to 40 s.

Final answer:

The P-wave takes approximately 40 seconds to arrive at the closest seismic station.

Explanation:

The P-waves in the crust travel at a velocity of approximately 7 km/s. To determine the time it takes for the P-wave to arrive at a seismic station, we can use the equation:

Time = Distance / Velocity

In this case, the distance to the epicenter of the earthquake is given as 280 km. Plugging this value into the equation, we get:

Time = 280 km / 7 km/s = 40 seconds

So, it takes approximately 40 seconds for the P-wave to arrive at the closest seismic station.