A block lies on a plane raised an angle theta from the horizontal. Threeforces act upon the block: F_w_vec, the force of gravity;F_n_vec, the normal force;and F_f_vec, the force of friction. Thecoefficient of friction is large enough to prevent the block fromsliding .Part A)Because the block is not moving, the sum ofthe y components of the forces acting on the block must bezero. Find an expression for the sum of the y componentsof the forces acting on the block, using coordinate system b.Express your answer in terms of someor all of the variables F_n, F_f, F_w, and theta.\sum f_y=Part B)Because the block is not moving, the sum ofthe x components of the forces acting on the block must bezero. Find an expression for the sum of the x componentsof the forces acting on the block, using coordinate system b.Express your answer in terms of someor all of the variables F_n, F_f, F_w, and theta.\sum F_{x}=0Part C)To find the magnitude of the normal force,you must express F_n in terms of F_w since F_f is an unknown. Using the equations youfound in the two previous parts, find an expression for F_n involving F_w and theta but not F_f.F_n =Diagram:In the diagram, a block is placed on an inclined surface with its both x and y components

Answer:

The magnitude of the electric field is [tex]8.6\times10^{2}\ N/C[/tex]

Explanation:

Given that,

Time t = 2.10 s

Speed = 160 m/s

Specific charge =Ratio of charge to mass = 0.100 C/kg

We need to calculate the acceleration

Using equation of motion

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

Put the value into the formula

[tex]a=\dfrac{160-0}{2.10}[/tex]

[tex]a=76.19\ m/s^2[/tex]

We need to calculate the magnitude of the electric field

Using formula of electric field

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

[tex]E=\dfrac{ma}{q}[/tex]

[tex]E=\dfrac{a+g}{\dfrac{q}{m}}[/tex]

Put the value into the formula

[tex]E=\dfrac{76.19+9.8}{0.100}[/tex]

[tex]E=8.6\times10^{2}\ N/C[/tex]

The direction is upward.

Hence, The magnitude of the electric field is [tex]8.6\times10^{2}\ N/C[/tex]

Answer:

E = 8.6 x 10² N/C

Explanation:

given,

initial speed of charge,u = 0 m/s

final speed of charge,v = 160 m/s

time,t = 2.1 s

charge-to-mass ratio = 0.100 C/kg

Electric field of the region = ?

Acceleration of the charge

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

[tex]a = \dfrac{160 - 0}{2.1}[/tex]

a = 76.19 m/s²

specific charge = [tex]\dfrac{q}{m} = 0.1[/tex]

now,

Electric field,

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

charge is moving upwards so,

[tex]E = \dfrac{(a + g)}{\dfrac{q}{m}}[/tex]

[tex]E = \dfrac{(76.19+9.8)}{0.1}[/tex]

E = 860 N/C

electric field , E = 8.6 x 10² N/C

hence, the magnitude of electric field is equal to E = 8.6 x 10² N/C

Answer:

F1 = 1210.65 N

Q = 65.7081 degrees

Explanation:

Sum of Forces in x - direction:

F1 * cos (Q) + F2*sin(30) - F3*(3/5) = Fres*cos (theta)

F1 * cos (Q) + (500)*(0.5) - 450*(3/5) = 570*cos(33)

F1*cos (Q) = 498.0422237 N    .... Eq1

Sum of Forces in y - direction:

F1 * sin (Q) - F2*sin (60) - F3 * (4/5) =  Fres*sin (theta)

F1 * sin (Q) - 500*sin(60) - 450 * (4/5) =  570*sin (33)

F1 * sin (Q) = 1103.45692 N  .... Eq 2

Divide Eq 2 by Eq 1

tan (Q) = 2.21558916

Q = arctan (2.21558916) = 65.7081 degrees

F1 = 1210.65 N

The ideal thermal efficiency for the steam power plant operating on the reheat Rankine cycle is approximately 57.0%.

To determine the thermal efficiency of the steam power plant operating on the ideal reheat Rankine cycle, we will follow the steps in the Problem-Solving Strategies for Thermodynamics:

Identify the Temperatures and Pressures

Boiler Pressure (P1): 17.5 MPa (High-pressure turbine inlet)

Reheater Pressure (P2): 2 MPa (Low-pressure turbine inlet)

Condenser Pressure (P3): 50 kPa

High-Pressure Turbine Inlet Temperature (T1): 550°C = 823 K

Low-Pressure Turbine Inlet Temperature (T2): 300°C = 573 K

Calculate Maximum Efficiency (Carnot Efficiency)

The maximum theoretical efficiency (Carnot efficiency) for a heat engine operating between two temperatures is given by the formula:
[tex]\eta_{max} = 1 - \frac{T_c}{T_h}[/tex]

where:

[tex]T_h[/tex] = Highest temperature (at the boiler, T1) = 823 K

[tex]T_c[/tex] = Lowest temperature (at the condenser) = Convert 50 kPa to temperature

Using steam tables or Mollier diagrams, the saturation temperature corresponding to 50 kPa is approximately 81°C = 354 K.

Now substituting:
[tex]\eta_{max} = 1 - \frac{354}{823} \approx 0.570 \text{ or } 57.0\%[/tex]

Answer:

Electric Field is [tex]5.943801*10^6 N/C[/tex]

Explanation:

Electric Field:

It originates from positive charge and ends at negative charge.

General Formula for electric Field:

[tex]E=\frac{kq}{r^2}[/tex]

where:

k is the Coulomb Constant

q is the charge

r is the distance

Given:

q=7.20 mC

r=3.3 meters

k=[tex]8.99*10^9 N.m^2/C^2[/tex]

Find:

Electric Field=?

Solution:

[tex]E=\frac{kq}{r^2}[/tex]

[tex]E=\frac{(8.99*10^9)(7.20*10^{-3})}{3.3^2}\\E=5943801.653 N/C\\E=5.943801653*10^6 N/C[/tex]

Electric Field is [tex]5.943801*10^6 N/C[/tex]

Answer:

Constructive Interference.

Explanation:

Constructive Interference.

Definition:

Two waves meet in such a way their highs(Crests) combine to form a new waves whose magnitude is the sum of magnitude of combining waves.

Since two waves have same wavelength and are in phase so when they combine they well form a way which has the magnitude equal to the sum of the magnitude of both waves.

Reasons why it is Constructive Interference:Waves

Answer:

2. VA

Explanation:

The valency electron or outer electron of a neutral atom of an element determines the group at which an element belong. The electronic configuration of an atom that have valency of 5 can be represented as 2 5 ,  2 8 5 etc.

The electronic configuration 2 5 represent Nitrogen atom while 2 8 5 represent phosphorus atom. The valency 5 depicts the element belongs to group 5A(VA).

This means atoms of valency electron of 3 belong to group IIIA, valency electrons of 4 belongs to group IVA, valency electron of 6 belong to group VIA  and valency electrons of  7 belong to group VIIA.  

Answer:

(A) Distance will be equal to 1.75 km

(B) Displacement will be equal to 1.114 km

Explanation:

We have given circumference of the circular track = 3.5 km

Circumference is given by [tex]2\pi r=3.5[/tex]

r = 0.557 km

(a) It is given that car travels from southernmost point to the northernmost point.

For this car have to travel the distance equal to semi perimeter of the circular track

So distance will be equal to [tex]=\frac{3.5}{2}=1.75km[/tex]

(b) If car go along the diameter of the circular track then it will also go from southernmost point to the northernmost point. and it will be equal to diameter of the track

So displacement will be equal to d = 2×0.557 = 1.114 m

Final answer:

To calculate the centripetal acceleration and average speed of the Sun in its galactic orbit, we utilize relevant formulas and astronomical data. The concept of centripetal acceleration is fundamental in understanding circular motion in celestial bodies like the Sun as it orbits the Milky Way galaxy.

Explanation:

Centripetal acceleration is the acceleration directed toward the center of a circular path. To calculate the centripetal acceleration of the Sun in its galactic orbit, we use the formula: a = v^2 / r, where v is the speed of the Sun and r is the radius of its orbit. Given that the Sun's orbit radius is 3.00 x 10^4 light years and it takes 2.60 x 10^8 years to orbit the Milky Way galaxy, we can also calculate the average speed of the Sun in its galactic orbit. To determine if a nearly inertial frame of reference can be located at the Sun, we need to consider the motion relative to the galactic center.

Answer:

6250 m  

Explanation:

When an object is projected into the air, the distance along the horizontal direction is called the Range.

The Range of a projectile is expressed as;

                                   [tex]R = \frac{u^{2}sin2\alpha }{g}[/tex]

Where,

R is the range of the projectile

α is the angle of inclination with the horizontal

g is the acceleration due to gravity = 9.8 m/s ≈ 10 m/s

Given; α =45° , u = 250 m/s

                                [tex]R = \frac{250^{2}sin2(45)}{10}[/tex]

                                [tex]R = \frac{62500sin90}{10}[/tex]

                                [tex]R = \frac{62500}{10}[/tex]

                                R =  6250 m

The range is 6250 m                              

In a projectile motion, the given object travel 6377.55 m in the horizontal direction.

In a projectile motion, the distance of the object along the horizontal direction is called the Range.  

The Range of a projectile                    

[tex]\bold {R = \dfrac {u^2 sin2\alpha }{g}}[/tex]

Where,  

R - range of the projectile

u - initial speed = 250 m/s  

α - angle of inclination with the horizontal = 45°  

g - gravitational acceleration = 9.8 m/s  

Put the values in the formula,

[tex]\bold {R = \dfrac {(250)^2 sin2(45) }{9.8}}\\\\\bold {R = \dfrac {62500\ sin 90 }{9.8}}\\\\\bold {R =6377.55}[/tex]  Since, sin 90 = 1

Therefore, in a projectile motion, the given object travel 6377.55 m in the horizontal direction.

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To solve this concept we will apply the mathematical equations concerning the calculation of Volume in a sphere and the relation of density as a function of mass and volume, that is

The volume of the neutron star is

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

[tex]V = \frac{4 \pi}{3} (\frac{17*10^{5}cm}{2})^3[/tex]

[tex]V = 25.72^{17}cm^3[/tex]

Now the density of the neutron star is

[tex]\rho = \frac{M}{V}[/tex]

[tex]\rho = \frac{1.989*10^{30}kg(\frac{10^3g}{1kg})}{25.72*10^{17}cm^3}[/tex]

[tex]\rho = 7.733*10^{14}g/cm^3[/tex]

Therefore the density of the neutron star is [tex]\rho = 7.733*10^{14}g/cm^3[/tex]

Answer:

The planet´s orbital period will be one-half Earth´s orbital period.

Explanation:

The planet in orbit, is subject to the attractive force from the sun, which is given by the Newton´s Universal Law of Gravitation.

At the same time, this force, is the same centripetal force, that keeps the planet in orbit (assuming to be circular), so we can put the following equation:

Fg = Fc ⇒ G*mp*ms / r² = mp*ω²*r

As we know to find out the orbital period, as it is the time needed to give a complete revolution around the sun, we can say this:

ω = 2*π / T (rad/sec), so replacing this in the expression above, we get:

Fg = Fc ⇒   G*mp*ms / r² = mp*(2*π/T)²*r

Solving for T²:

T² = (2*π)²*r³ / G*ms (1)

For the planet orbiting the sun in Andromeda, we have:

Ta² = (2*π)*r³ / G*4*ms (2)

As the radius of the orbit (distance to the sun) is the same for both planets, we can simplify it in the expression, so, if we divide both sides in (1) and (2), simplifying common terms, we finally get:

(Te / Ta)² =  4  ⇒ Te / Ta = 2 ⇒ Ta = Te/2

So, The planet's orbital period will be one-half Earth's orbital period.

The planet´s orbital period should be considered as the one-half Earth´s orbital period.

It is subjected to the attractive force from the sun that we called as the

Newton´s Universal Law of Gravitation.

Also, the following equation should be used

Fg = Fc ⇒ G*mp*ms / r² = mp*ω²*r

Now

ω = 2*π / T (rad/sec),

So,

Fg = Fc ⇒   G*mp*ms / r² = mp*(2*π/T)²*r

Now

T² = (2*π)²*r³ / G*ms (1)

And,

Ta² = (2*π)*r³ / G*4*ms (2)

So,

(Te / Ta)² =  4  ⇒ Te / Ta = 2 ⇒ Ta = Te/2

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Answer:

165 Hz, 220 Hz, and 275 Hz belongs to pipe open at both ends

165 Hz, 275 Hz, and 385 Hz belongs to pipe closed at one end

Explanation:

Open ended pips have harmonic frequencies that are multiple of the fundamental frequency

Find the fundamental frequency for each of the samples:

165Hz, 275Hz, 385Hz

(275-165)=110

(385-275)= 110

165 Hz, 220 Hz, and 275 Hz

(220-165)=55

(275-220)=55

F= 55

Note that 165 =3f

220=4f

275=5f

SO these frequencies are multiples of the fundamental frequency

Answer:

we can estimate the _time__.

The three things are;

1. Moon's position in the sky

2. The moon phase

3. The time

Explanation:

In general, knowing any two of the following three things; (the moon's position in the sky, the moon phase, and the time) will allows us to estimate the third. Yes, this statement is true because, position and phase of moon is used to determine the hour of the day and night most especially in the morning. Example of this is the determination of prayer time by Muslims community as it was the major time determinant in the past before the advent of clock or watch.

Also, if will know the time and moon position, we can determine the phase of the moon likewise using time and moon phase to know the moon position.

The different phases of the moon cause changes in the size of the moon. If we know the moon's position in the sky and its phase. we can easily estimate the time.

The moon changes shape every day. This is due to the fact that the celestial body has no light of its own and can only reflect sunlight.

Only the side of the moon facing the sun can reflect this light and seem bright. The opposite side appears black. this is a full moon.

We can only see the black section when it lies between the sun and the earth when a new moon occurs. We witness intermediate phases like a half-moon and crescent in between these two extremes.

In general, we may estimate the third by knowing any two of the following three things

(1) the moon's position in the sky

(2) the moon phase

(3) the time

Yes, this statement is correct since the moon's location and phase are utilized to define the time of day and night, particularly in the morning.

Also, if we know the time and moon position, we can figure out the moon phase by utilizing the time and moon phase to figure out the moon position.

Hence If we know the moon's position in the sky and its phase. we can easily estimate the time.

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Final answer:

Determining if a friction force is required for a car rounding a banked curve depends on the car's speed relative to the curve's ideal speed. At 95 km/h, a frictional force may be needed if this speed is not the ideal speed for the 68 m radius curve banked at 16 degrees.

Explanation:

When a 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees, we need to determine if a friction force is required when the car is traveling at 95 km/h. If the car is traveling at the correct banked curve speed, it could complete the turn without any frictional force. However, if the car travels at a speed higher or lower than this optimal speed, a frictional force will be necessary either to prevent the car from slipping outward or to prevent it from falling inward towards the center of the curve.

To find out whether a friction force is needed, we first need to calculate the ideal speed for this banked turn. This involves calculating the speed at which the components of the normal force provide enough centripetal force for the turn. The ideal speed is reached when no friction force is needed to keep the car on the path, meaning the force of gravity, the normal force, and the centripetal force are in perfect balance.

However, if the car is indeed traveling at 95 km/h, faster or slower than this ideal speed, then either a static frictional force acting upwards along the bank or a static frictional force opposite to the car's direction would be required to maintain its circular path without slipping.

Final answer:

To determine the time it takes for you to reach the open door and jump in, we can use the equations of motion. We know that you are initially 9.0 m away from the door, and the bus is accelerating at 1.0 m/s². The final velocity of the bus is not given, so we can't find the exact time it takes for you to reach the door. However, we can find the maximum time you can wait before starting to run and still catch the bus.

Explanation:

To determine the time it takes for you to reach the open door and jump in, we can use the equations of motion. We know that you are initially 9.0 m away from the door, and the bus is accelerating at 1.0 m/s². The final velocity of the bus is not given, so we can't find the exact time it takes for you to reach the door. However, we can find the maximum time you can wait before starting to run and still catch the bus.

To find the maximum time you can wait, we need to calculate when the distance between you and the door is equal to 0. Since you are moving towards the door at a constant speed of 5.7 m/s and the bus is accelerating away from you, the distance between you and the door will decrease over time. Let's call the time you wait before starting to run as 't'.

The distance traveled by the bus can be calculated using the equation:

S = ut + (1/2)at^2

Where S is the distance traveled, u is the initial velocity which is 0 m/s, a is the acceleration which is 1.0 m/s², and t is the time.

The distance traveled by you can be calculated using the equation:

S = vt

Where S is the distance traveled, v is your constant velocity which is 5.7 m/s, and t is the time.

After the time 't', both the bus and you will be at the same position which is the door. So the total distance traveled by you and the bus will be equal. We can set up the equation:

ut + (1/2)at^2 = vt

Simplifying this equation, we get:

(1/2)at^2 - vt = 0

Since we know that you can wait a maximum of 't' seconds, we need to solve this quadratic equation for 't'.

Using the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac))/2a

Where a = (1/2)a, b = (-v), and c = 0

Substituting the values, we get:

t = (-(-v) ± sqrt((-v)^2 - 4(1/2)a(0)))/2(1/2)a

t = v + sqrt(v^2)/a

Now we can substitute the values of 'v' and 'a' to find the maximum time you can wait before starting to run:

t = 5.7 + sqrt((5.7)^2)/1.0

t = 5.7 + sqrt(32.49)/1.0 ≈ 5.7 + 5.7 = 11.4 seconds

Therefore, the maximum time you can wait before starting to run and still catch the bus is approximately 11.4 seconds.

Answer:

Integrated circuit (IC)

Explanation:

An integrated circuit ( IC ) is a semiconductor which contains multiple electronic components

interconnected to form a complete electronic function. Integrated circuits are the most essential part of all electronic products.

Modern integrated circuits contain as much as billions of circuit components such as transistors , diodes , resistors , and capacitors

onto a single monolithic die .

The correct answer is option b. An Integrated Circuit (IC), or microchip, is a small electronic component consisting of transistors and miniaturized parts etched onto a piece of silicon, pivotal in the development and miniaturization of electronic devices

The small electronic component made up of transistors (tiny switches) and other miniaturized parts is called an Integrated Circuit (IC). An Integrated Circuit, sometimes referred to as a microchip, is an electronic circuit of transistors etched onto a small piece of silicon. This technology allows for complex circuitry to be compacted into a tiny chip, which is crucial for the functionality of modern electronic devices like computers and cell phones. The invention of the IC was pivotal in launching the modern computer revolution because it significantly reduced the size and complexity of electronic devices, replacing bulky vacuum tubes and complicated wiring with a compact, efficient solution.

Integrated Circuits are designed to handle both analog and digital signals, but in the realm of digital electronics, they are essential for managing binary code, the series of ones and zeroes that computers use to process data. This is achieved through the hundreds, thousands, or even millions of transistors that can act as on-off switches within a single IC. The miniaturization and efficiency of ICs have been fundamental in advancing the technological capabilities of electronic devices, making them smaller, faster, and more accessible to the general public.

To solve this problem we will apply the concepts related to the electric field such as the smelting of the Force and the load (In this case the force is equivalent to the weight). Later we will apply the ratio of the total charge as a function of the multiplication of the number of electrons and their individual charge.

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

Here,

m = mass

g = Acceleration due to gravity

Rearranging to find the charge,

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

Replacing,

[tex]q = \frac{(3.37*10^{-9})(9.8)}{11000}[/tex]

[tex]q = 3.002*10^{-12}C[/tex]

Since the field is acting upwards the charge on the drop should be negative to balance it in air. The equation to find the number of electrons then is

[tex]q = ne[/tex]

Here,

n = Number of electrons

e = Charge of each electron

[tex]n = \frac{q}{e}[/tex]

Replacing,

[tex]n = \frac{3.002*10^{-12}}{1.6*10^{-19}}[/tex]

[tex]n = 2.44*10^7[/tex]

Therefore the number of electrons that reside on the drop is [tex]2.44*10^7[/tex]

To determine the number of excess electrons or protons residing on the water drop, we can use the principle of electrostatics. When an electric field is applied to a charged object, the electrostatic force acting on it can be calculated using the equation:

[tex]F = qE[/tex]

Where:

F is the electrostatic force

q is the charge on the object

E is the electric field strength

In this case, the electrostatic force acting on the water drop is balanced by the gravitational force, so we have:

F = mg

Where:

m is the mass of the water drop

g is the acceleration due to gravity

We can equate these two forces and solve for the charge q:

qE = mg

From this equation, we can isolate the charge q:

q = mg / E

Now we can calculate the charge on the water drop:

m = 3.37 × 10^(-9) kg

g = 9.8 m/s^2

E = 11000 N/C

Substituting the values into the equation:

q = (3.37 × 10^(-9) kg * 9.8 m/s^2) / 11000 N/C

Calculating this expression:

q = 3.037 × 10^(-15) C

The elementary charge of an electron or proton is approximately 1.602 × 10^(-19) C. To find the number of excess electrons or protons, we can divide the calculated charge by the elementary charge:

Number of excess electrons or protons = q / elementary charge

Number of excess electrons or protons = (3.037 × 10^(-15) C) / (1.602 × 10^(-19) C)

Calculating this expression:

Number of excess electrons or protons ≈ 1.895 × 10^(4)

Therefore, the water drop has approximately 18,950 excess electrons or protons.

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To develop this problem we will apply the considerations made through the concept of Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. At first the source is moving towards the observer. Than the perceived frequency at first

[tex]F_1 = F \frac{{343}}{(343-V)}[/tex]

Where F is the actual frequency and v is the velocity of the ambulance

Now the source is moving away from the observer.

[tex]F_2 = F\frac{343}{(343+V)}[/tex]

We are also so told the perceived frequency decreases by 11.9%

[tex]F_2 = F_1 - 9.27\% \text{ of } F_1[/tex]

[tex]F_2 = F_1-0.0927F_1[/tex]

[tex]F_2 = 0.9073F_1[/tex]

Equating,

[tex]F\frac{343}{(343+V)}= 0.9073(F\frac{343}{(343-V)})[/tex]

[tex]\frac{1}{(343+V)}= 0.9073\frac{1}{(343-V)}[/tex]

[tex]0.9073(343+V) = 343-V[/tex]

[tex](0.9073)(343)+(0.9073)V = 343-V[/tex]

[tex]V+0.9073V = 343-(0.9073)(343)[/tex]

Solving for V,

[tex]V = 16.67 m/s[/tex]

Answer:

None of the above

Explanation:

When energy is converted from one form to another in a chemical or physical change, none will change. This is due to the law of conservation of energy. It states that the total energy of the system remains constant. It only changes energy from one form of energy to another. So, the correct option is (d) "none of the above".

Answer: None of the above

Explanation:

The total mass of a system does not change during normal (non-nuclear) chemical reactions or during other processes where energy changes form. The force of gravity is constant and will not change when energy is converted. While energy can be converted between forms or exchanged between species, no additional energy can be created or removed.

To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of [tex] 62 lb / ft ^ 3 [/tex] (Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as

[tex]W = \gamma A * \int_0^15 dy[/tex]

Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing

[tex]W = (62)(14*7)\int^{15}_0 (15-y)dy[/tex]

[tex]W = (14*7*62)\big [15y-\frac{y^2}{2}\big ]^{15}_0[/tex]

[tex]W = (14*7*62)[15(15)-\frac{(15)^2}{2}][/tex]

[tex]W = 683550ft-lbs[/tex]

Therefore the total work in the system is [tex]683550ft-lbs[/tex]

The ratio of the electric force on the proton after the wire segment is shrunk to three times its original length to the force before the segment was shrunk is 3.

The electric force between a point charge and a segment of wire with a distributed charge is given by Coulomb's law.

The formula for the electric force on a point charge q due to a segment of wire with charge Q distributed along its length L is:

[tex]F=\frac{k.q.Q}{L}[/tex]

where:

F is the electric force on the point charge,  

k is Coulomb's constant ( 8.988 × 1 0⁹ Nm²/ C²),

q is the charge of the point charge,  

Q is the charge distributed along the wire segment, and

L is the length of the wire segment.

When the wire segment is shrunk to one-third of its original length, the new length becomes 1/3 L.

The charge distribution remains the same, only the length changes.

So, the new electric force [tex]F_f[/tex] ​ on the proton after the segment is shrunk becomes:

[tex]F_f=\frac{k.q.Q}{\frac{1}{3}L}[/tex]

The original electric force [tex]F_i[/tex]​ on the proton before the segment was shrunk is:

[tex]F_i = \frac{k.q.Q}{L}[/tex]

let's find the ratio [tex]\frac{F_f}{F_i}[/tex] ​:

[tex]\frac{F_f}{F_i}=\frac{\frac{k.q.Q}{\frac{1}{3}L}}{\frac{k.q.Q}{L}}[/tex]

[tex]\frac{F_f}{F_i}=3[/tex]

Hence,  the ratio of the electric force on the proton after the wire segment is shrunk to the force before the segment was shrunk is 3.

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The ratio of the electric force on the proton after the wire segment is shrunk is equal to the ratio of their charges.

The ratio of the electric force on the proton after the wire segment is shrunk to the force before the segment was shrunk can be found using Coulomb's law. Coulomb's law states that the electric force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

In this case, the charges involved are the charge of the wire segment and the charge of the proton. Since the wire segment contains 10 nC of charge, we can consider it as one of the charged objects. The proton is very far from the wire, so we can assume that the distance between them remains the same before and after the wire segment is shrunk. Therefore, the ratio of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk is equal to the ratio of their charges.

Let's assume that the initial force on the proton is Fi and the final force on the proton is Ff. Using the given information, we have:

Fi = k(q1 * q2) / r^2

where k is the electrostatic constant, q1 and q2 are the charges of the wire segment and the proton respectively, and r is the distance between them.

After the wire segment is shrunk to one-third of its original length, the charge of the wire segment remains the same and the distance between the wire segment and the proton also remains the same. Therefore, the ratio Ff/Fi can be calculated as:

Ff/Fi = (q1 * q2) / (q1 * q2) = 1

To solve this problem we will apply the laws of proportion between the resistance / weight ratio that may occur depending on the cases mentioned. The strength-to-weight ratio for each object is,

For the nano tube,

[tex]\frac{S}{W} = \frac{1}{10*10^{-9}} =1*10^{8}[/tex]

For the leg of a flea,

[tex]\frac{S}{W}= \frac{1}{100*10^{-6}}=10000[/tex]

For the elephant,

[tex]\frac{S}{W} = \frac{1}{2}= 0.5[/tex]

From this we can conclude that the resistance / weight ratio of the nano tube is 10 thousand times better than the flea's leg and [tex]10 ^ 8[/tex] times better than the elephant's leg.

Answer:

Down

       F1ₓ = 219.6N

       [tex]F1_{y}[/tex]  = 1038.8 N

Top

       F2ₓ = 219.6 N

       [tex]F2_{y}[/tex] = 0  

Explanation:

For this exercise we must make a free body diagram of the ladder, see attached, then use the balance equations on each axis

Transnational Balance

X axis

        F1ₓ -F2ₓ = 0

        F1ₓ = F2ₓ

Y Axis  

         [tex]F1_{y}[/tex] -  [tex]F2_{y}[/tex] - W - W_man = 0           (1)

Rotational balance

The reference system is placed at the bottom of the stairs and we can turn the anti-clockwise direction of rotation as positive

           F2ₓ y - [tex]F2_{y}[/tex] x - W x - W_man x_man = 0

Let us write the data they give, the masses of the ladder (m = 14.0 kg), the mass of man (m_man = 92 kg), the center of mass of the ladder that is 2m from the bottom (the height) and the position of the man which is 3 m high

Let's look with trigonometry for distances

The angle of the stairs is

           cos θ = x / L

           θ = cos⁻¹ x / L

           θ = cos⁻¹ 2 / 5.6

           θ = 69⁰

Height y

          tan 69 = y / x

          y = x tan 69

          y = 2 tan 69

          y = 5.21 m

Distance x

          tan 69 = 2 / x

          x = 2 / tan 69

          x = 0.7677 m

The distance x_man

          x_man = 3 / tan 69

          x_man = 1,152 m

They indicate that between the scalars and the support there is no friction so the vertical force at the top is zero

          [tex]F2_{y}[/tex] = 0

Let's replace in the translational equilibrium equation

         F2ₓ y - [tex]F2_{y}[/tex] x - W x - W_man x_man = 0

         F2ₓ 5.21 -0 - 14.0 9.8 0.7677 - 92.0 9.8 1,152 = 0

         F2ₓ = 1143.97 / 5.21

         F2ₓ = 219.6 N

We use equation 1

         [tex]F1_{y}[/tex] + 0 - W - W_man = 0

        [tex]F1_{y}[/tex] = W + W_man

        [tex]F1_{y}[/tex]  = (m + m_man) g

         [tex]F1_{y}[/tex]  = (14 +92) 9.8

         [tex]F1_{y}[/tex]  = 1038.8 N

We can write the force on each part of the ladder

Down

       F1ₓ = 219.6N

       [tex]F1_{y}[/tex]  = 1038.8 N

Top

       F2ₓ = 219.6 N

       [tex]F2_{y}[/tex] = 0  

Answer:

52.47706 mph

54.5 mph

Explanation:

The average speed is given by

[tex]V_{av}=\dfrac{Distance}{Time}[/tex]

[tex]V_{av}=\dfrac{100}{\dfrac{50}{44}+\dfrac{50}{65}}\\\Rightarrow V_{av}=52.47706\ mph[/tex]

Julie's average speed on the way to Grandmother's house is 52.47706 mph

[tex]V_{av}=\dfrac{44+65}{2}\\\Rightarrow V_{av}=54.5\ mph[/tex]

Average speed on the return trip is 54.5 mph

Answer with Explanation:

We are given that

Mass , m=372 g=[tex]\frac{372}{1000}=0.372 Kg[/tex]

1 kg=1000g

Maximum acceleration, a=[tex]17.6 m/s^2[/tex]

Maximum speed ,v=1.75 m/s

a.We know that

Maximum acceleration, a=[tex]A\omega^2[/tex]

Maximum speed, v=[tex]\omega A[/tex]

[tex]17.6=A\omega^2[/tex]

[tex]1.75=A\omega[/tex]

[tex]\frac{17.6}{1.75}=\frac{A\omega^2}{A\omega}=\omega[/tex]

Angular frequency,[tex]\omega=10.06 rad/s[/tex]

b.Substitute the value of angular frequency

[tex]1.75=A(10.06)[/tex]

[tex]A=\frac{1.75}{10.06}=0.17 m[/tex]

Hence, the amplitude=0.17 m

c.Spring constant,k=[tex]m\omega^2[/tex]

Using the formula

[tex]k=0.372\times (10.06)^2[/tex]

Hence, the spring constant,k=37.6 N/m

Final answer:

The angular frequency is approximately 7.686 rad/s, the amplitude is approximately 0.227 m, and the spring constant is approximately 6.5472 N/m.

Explanation:

To determine the angular frequency, we can use the formula:

ω = √(k/m)

where ω is the angular frequency in radians per second, k is the spring constant in Newtons per meter, and m is the mass in kilograms.

Given that the maximum acceleration is 17.6 m/s^2 and the mass is 372 g (or 0.372 kg), we can calculate the spring constant:

k = m * a

k = 0.372 kg * 17.6 m/s^2 = 6.5472 N/m

Now we can find the angular frequency:

ω = √(6.5472 N/m / 0.372 kg) ≈ 7.686 rad/s

To determine the amplitude, we can use the formula:

A = vmax / ω

where A is the amplitude, vmax is the maximum speed, and ω is the angular frequency.

Given that the maximum speed is 1.75 m/s and the angular frequency is 7.686 rad/s, we can calculate the amplitude:

A = 1.75 m/s / 7.686 rad/s ≈ 0.227 m

Therefore, the angular frequency is approximately 7.686 rad/s, the amplitude is approximately 0.227 m, and the spring constant is approximately 6.5472 N/m.

Answer:

r = 0.78 m

Explanation:

If the sound source is emitting the signal evenly in all directions (as from a point source) this means that at any time, the source power is distributed over the surface of a sphere of radius r.

At a distance r of the source, the intensity of the sound is defined as the power per unit area:

I = P/A

As the area is the area of a sphere, we can say the following:

I = P / 4*π*r²

Replacing I and P by the values given, we can solve for r (which is the distance from the listener to the source) as follows:

r² = P / I*4*π ⇒ r = [tex]\sqrt{P/(4*\pi*I)}[/tex] = 0.78 m

The distance between you and the source is 0.78 m.

The given parameters:

The area of the source is calculated as follows;

[tex]A = \frac{P}{I} \\\\ A = \frac{75}{9.8} \\\\ A = 7.65 \ m^2[/tex]

The distance between you and the source is calculated as follows;

[tex]A = 4\pi r^2\\\\ r^2 = \frac{A}{4\pi} \\\\ r = \sqrt{\frac{A}{4\pi}} \\\\ r = \sqrt{\frac{7.65}{4\pi}} \\\\ r = 0.78 \ m[/tex]

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As the gas (air) in the balloon warms, it expands due to thermal expansion, causing its density to decrease. This lower density compared to the surrounding air leads to buoyancy, making the balloon rise.

When the sun warms the gas (air) inside a balloon, the air expands due to an increase in temperature. This process, known as thermal expansion, causes the molecules in the air to move faster and spread out more, occupying a larger volume. As a result, the density of the air inside the balloon decreases because density is defined as mass per unit volume, and while the mass of the air remains constant, its volume increases. In the context of a hot air balloon, or any closed system where air is heated, this decrease in density compared to the cooler surrounding air leads to buoyancy. Buoyancy is the force that makes things float, which in the case of the balloon, causes it to rise since the hot air inside it is less dense than the cooler external environment.

Answer:

(A) position vs time

(B) Force vs position

(C) velocity vs time

Explanation:

Part A

This graph shows that the position of the block increases with time along the x-axis exponentially (that is it increases in unequal amounts in equal time intervals). This is because the velocity of the block is changing with time and as a result the position changes in unequal amounts per time

PartB

The force on the spring increases in a negative direction going from zero to a negative value. This is because the spring is being compressed from configuration 1 to 2. The force of compression on a spring is usually taken to have a negative sign and expansion to have a positive sign. So in this case force becomes increasingly negative with time.

Part C

The velocity of the block decreases from a positive nonzero value (v>0) to zero because the spring resists the motion of the block. As a result the block comes to a stop momentarily. The velocity decreases exponentially because the acceleration of the block is also changing with time since the force of the block is decreasing with time.

Thank you for reading.

Final answer:

The distance from the charge is 30 meters.

Explanation:

The distance from the point charge can be calculated using the equation E = kQ/r^2, where E is the electric field strength, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q is the charge, and r is the distance. In this case, the electric field strength is the same at any point 5.00 mm away from the charge. Therefore, we can set up the equation as follows:

E = kQ/r^2

5.00 x 10^-3 m = (9 x 10^9 Nm^2/C^2)(5 x 10^-9 C)/r^2

By rearranging the equation and solving for r, we can find the distance:

r^2 = (9 x 10^9 Nm^2/C^2)(5 x 10^-9 C)/(5.00 x 10^-3 m)

r^2 = 900 m^2

r = sqrt(900 m^2) = 30 m

Therefore, the charge is located 30 meters away.

Answer:

141.87 kg.

Explanation:

Deduction From Newton's second law of motion.

F = ma....................... Equation 1

Where F = Force acting on the object, m = mass of the object, a = acceleration  of the object.

Making m the subject of the equation,

m = F/a .................. Equation 2

But

a = (v-u)/t............... Equation 3

Where v = final velocity, u = initial velocity, t = time.

Given: v = 107 m/s, u = 0 m/s ( fro rest), t = 2.3 s.

Substituting into equation 3

a = (107-0)/2.3

a = 107/2.3

a = 46.52 m/s².

Also Given, F = 6600 N

Substitute into equation 2

m = 6600/46.52

m = 141.87 kg.

Hence the mass of the object = 141.87 kg.