A cylindrical shell of length 220 m and radius 4 cm carries a uniform surface charge density of σ = 14 nC/m^2.
1. What is the total charge on the shell?

Answer:

a)P=462.70 Pa

b)h = 0.047 m of water

Explanation:

Given that

Pressure ,[tex]P=\dfrac{1}{2}\rho V^2[/tex]

[tex]\rho = 1.2\ kg/m^3[/tex]

V= 100 km/h

[tex]V=100\times \dfrac{1000}{3600}\ m/s[/tex]

V=27.77 m/s

The pressure P

[tex]P=\dfrac{1}{2}\rho V^2[/tex]

[tex]P=\dfrac{1}{2}\times 1.2\times 27.77^2\ Pa[/tex]

P=462.70 Pa

We know that density of the water [tex]\rho=1000\ kg/m^3[/tex]

Lets height of the water column = h m

We know that

[tex]_P=\rho _w g h[/tex]

462.70 = 1000 x 9.81 h

[tex]h=\dfrac{462.7}{1000\times 9.81}\ m[/tex]

h = 0.047 m of water

a)P=462.70 Pa

b)h = 0.047 m of water

a)The presuure rise will be P=462.70 Pa

b) The height of the water column h = 0.047 m of water

It is given that

Pressure,

[tex]p= \dfrac{1}{2} \rho v^2[/tex]

Here   [tex]\rho =1.2 \ \dfrac{kg}{m^3}[/tex]

[tex]V=100\ \frac{km}{h} =\dfrac{100\times 1000}{3600} =27.77 \ \dfrac{m}{s}[/tex]

Now to calculate the pressure P

[tex]P=\dfrac{1}{2} \rho v^2[/tex]

[tex]P= \dfrac{1}{2}\times 1.2\times (27.77)^2[/tex]

[tex]P=462.70 \ \frac{N}{m^2}[/tex]

As we know that the density of water

[tex]\rho = 1000\ \frac{kg}{m^3}[/tex]

Lets height of the water column = [tex]h_m[/tex]

As We know that

[tex]P= \rho_w gh[/tex]

[tex]462.70=1000\times 9.81\times h_m[/tex]

[tex]h_m=0.047\ m \ of \ water[/tex]

Thus

a)The presuure rise will be P=462.70 Pa

b) The height of the water column h = 0.047 m of water

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Answer:a) The bullet will miss the monkey because the monkey falls down while the bullet speeds straight forward.

Explanation: The bullet keeps as it aim( the monkey) unless it is redirected by an external force that could redirect it. Hence, the bullet speeds straight forward.

Final answer:

(b) The bullet will hit the monkey because both the monkey and the bullet are subject to gravity's acceleration equally upon being released or fired; both will fall downward at the same rate. Hence, (b) is the correct option.

Explanation:

When a hunter aims horizontally at a monkey in a tree and the monkey drops at the moment the gun is fired, the outcome is determined by Newtonian physics.

According to Newton's laws, the bullet and the monkey are both subject to gravity and will start to fall toward the ground at the same rate, regardless of any horizontal motion. Therefore, the correct answer is:

b) The bullet will strike the monkey because gravity causes both the bullet and the monkey to fall at the same speed.

This scenario illustrates the principle that horizontal and vertical motions are independent of each other. When the gun is fired, the bullet travels forward while also accelerating downward due to gravity.

Since the monkey begins to fall at the same moment the bullet is fired, both the bullet and the monkey undergo the same downward acceleration, meaning they will fall together.

To solve this problem we will apply the concepts related to Coulomb's law for which the Electrostatic Force is defined as,

[tex]F_{initial} = \frac{kq_1q_2}{r^2}[/tex]

Here,

k = Coulomb's constant

[tex]q_{1,2}[/tex] = Charge at each object

r = Distance between them

As the distance is doubled so,

[tex]F_{final} = \frac{kq_1q_2}{( 2r )^2}[/tex]

[tex]F_{final} = \frac{ kq_1q_2}{ 4r^2}[/tex]

[tex]F_{final} = \frac{1}{4} \frac{ kq_1q_2}{r^2}[/tex]

[tex]F_{final} = \frac{1}{4} F_{initial}[/tex]

[tex]\frac{F_{final}}{ F_{initial}} = \frac{1}{4}[/tex]

Therefore the factor is 1/4

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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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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To solve this problem we will apply the concepts related to the linear kinematic movement. We will start by finding the speed of the body from time and the acceleration given.

Through the position equations we will calculate the distance traveled.

Finally, using this same position relationship and considering the previously found speed, we can determine the time to reach your goal.

For time (t) and acceleration (a) we have to,

[tex]t = 8s, a = 165m/s^2[/tex]

The velocity would be,

[tex]u = a*t \\u = 165*8\\u = 1320m/s[/tex]

Now the position is,

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

[tex]h = \frac{1}{2} 165*8^2[/tex]

[tex]h = 5280m[/tex]

Now with the initial speed and position found we will have the time is,

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

[tex]-5280=1320t - \frac{1}{2} 9.8t^2[/tex]

[tex]4.9t^2-1320t-5280=0[/tex]

Solving the polynomian we have,

[tex]t = 273.33s = 4.56minutes[/tex]

Therefore  the rocket will take to hit the ground around to 4.56min

First, we need to determine the velocity of the rocket at the time the rocket is dropped after 8 seconds of powered ascent.
Given the constant acceleration of 165 m/s^2, the velocity (v) at 8 seconds can be found using the formula
v = at
where 'a' is the acceleration and 't' is the time.
Therefore, v = 165 m/s^2 x 8 s = 1320 m/s.

Now that the rocket is dropped, it will initially have the velocity of the rocket at that instant, which is 1320 m/s upward. To find out how long it takes for the rock to reach the highest point, we use the formula
v = u + at
where 'u' is initial velocity, 'v' is final velocity (0 m/s at the highest point), 'a' is the acceleration due to gravity (which is negative since it is in the opposite direction to the initial motion), and 't' is the time.
Solving for time, we get
t = -u/g. With g ≈ 9.81 m/s^2, the time to reach the highest point is
t ≈ -1320 m/s / -9.81 m/s^2 ≈ 134.6 s.

After reaching the highest point, the rocket will start falling back to the ground. Since the rocket starts from rest at the highest point, we can use the formula
s = 0.5gt^2
where 's' is the distance and 't' is time, to calculate the time it takes to fall to the ground.
But since we already have the time to reach the highest point, we can simply double that time to find the total time taken for the round trip, because the time to go up is the same as the time to come down in free fall. So the total time taken for the rocket to hit the ground is
134.6 s up + 134.6 s down = 269.2 s.

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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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

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:

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.

The given speed of Earth in km/sec is converted first to miles/sec and then to miles/hour, resulting in an average speed of Earth around the Sun of approximately 66,661.6 miles/hour.

To solve this, we need to convert kilometers to miles and seconds to hours. First, we should know that 1 kilometer is approximately 0.621371 miles, and 1 hour has 3600 seconds.

Given that, we can first convert Earth's speed from kilometers per second to miles per second by multiplying by the conversion factor:

29.783 km/sec * 0.621371 mile/km = 18.5171 miles/sec.

Next, we convert seconds to hours:

18.5171 miles/sec * 3600 sec/hour = 66,661.6 miles/hour.

So, the average speed of the Earth around the Sun, in miles per hour, to five significant figures is 66,661.6 miles/hour.

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Final answer:

The Earth travels around the Sun at an average speed of 29.783 km/s, which is approximately 66,636.7 miles per hour when converted using the steps of multiplying by 3600 to get km/hr and then by the conversion factor for km to mi.

Explanation:

The student's question is about converting the speed of the Earth's orbit around the Sun from kilometers per second to miles per hour. The given speed is 29.783 km/s. To convert this to miles per hour, we can follow these steps:

Multiply the kilometres per second by 3600, which is the number of seconds in an hour, to get the kilometres per hour.

Convert kilometers per hour to miles per hour by multiplying by the conversion factor (1 kilometer is approximately 0.621371 miles).

Performing these calculations:

29.783 km/s × 3600 s/hr = 107218.8 km/hr

107218.8 km/hr × 0.621371 mi/km = 66636.7 miles per hour

Therefore, Earth travels around the Sun at an average speed of 66,636.7 miles per hour, expressed with five significant figures.

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:

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

F = 124 N

Explanation:

given,

mass, m = 5 Kg

time, t = 4.1 s

displacement = y(t)=(2.80 m/s)t +(0.61 m/s³)t³

velocity

[tex]\dfrac{dy(t)}{dt}=2.80 + 1.83 t^2[/tex]

[tex]v=2.80 + 1.83 t^2[/tex]

again differentiating to get the equation of acceleration

[tex]\dfrac{dv}{dt}= 3.66 t[/tex]

[tex]a= 3.66 t[/tex]

force at time t = 4.10 s

F = m a

F = 5 x 3.66 x 4.1

F = 75 N

the net force when crate is moving upward

F = Mg + Ma

F = 5 x 9.8 + 75

F = 124 N

the magnitude of force is equal to 124 N

Answer:

Yes, there will a force acting on the positive point charge. The options provided are that of negative point charge, rather than a positive point charge stated in the question

Explanation:

Here is the explanation:

When a capacitor is charged, one plate is positively charge and the other is negatively charged. In between the oppositely charged plates, there exist a potential difference. A positive point charge placed in the distance between the two plate will experience an electric force due to the potential difference. The direction of the force will be directed away from the positvely charged plate and towards the negatively charged plate. The reason is due to the law of electrostatic force, which states: like charges repel and unlike charges attract.

A parallel-plate capacitor consists of two large, flat, metal plates held parallel to each other. A positive point charge placed in the gap between the plates and near the center will experience an electric force directed away from the positive plate and towards the negative plate.

A device called a parallel-plate capacitor consists of two large, flat, metal plates held parallel to each other and separated by a small gap. When a positive point charge is placed in the gap between the two plates and near the center of each plate, it experiences an electric force.

The direction of the electric force on the negative charge is from the positive plate towards the negative plate. This means the electric force is directed away from the positive plate and towards the negative plate.

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Answer:

The black body temperature of Earth T_e= 180.4 K

Explanation:

Assuming we have to find Black body temperature of the earth.

[tex]T_e =(\frac{s_o(1-\alpha)}{4\sigma})^{0.25}[/tex]

S0= solar energy striking the earth= 343 Wm^{-2}

\alpaha = 30% = 0.3

\sigma = stephan boltsman constant.= 5.67×10^{-8} Wm^{-2}K^4

[tex]T_e =(\frac{343(1-0.3)}{4\times5.67×10^{-8}})^{0.25}[/tex]

T_e= 180.4 K

Explanation:

Equilibrium position in y direction:

W = Fb (Weight of the block is equal to buoyant force)

m*g = V*p*g

V under water = A*h

hence,

m = A*h*p

Using Newton 2nd Law

[tex]-m*\frac{d^2y}{dt^2} = Fb - W\\\\-m*\frac{d^2y}{dt^2} = p*g*(h+y)*A - A*h*p*g\\\\-A*h*p*\frac{d^2y}{dt^2} = y *p*A*g\\\\\frac{d^2y}{dt^2} + \frac{g}{h} * y =0[/tex]

Hence, T time period

T = 2*pi*sqrt ( h / g )

Answer:

Mass of the wooden Block is 20g.

Explanation:

The buoyant force equation will be used here

Buoyant Force= ρ*g*1/2V Here density used is of water

m*g= ρ*g*1/2V

Simplifying the above equation

2m= ρ*V Eq-1

Also we know from the question that

ρ*V = m + 0.020 Eq-2 ( Density = (Mass+20g)/Volume )

Equating Eq-1 & Eq-2 we get

2m = m+0.020

m = 0.020kg

m = 20g

To solve this problem we will calculate the total volume of inhaled and exhaled water. From the ideal gas equation we will find the total number of moles of water.

An athlete at high performance inhales 4.0L of air at 1atm and 298K.

The inhaled and exhaled air contain 0.5% and 6.2% by volume of water, respectively.

During inhalation, volume of water taken is

[tex]V_i = (4L)(0.5\%)[/tex]

[tex]V_i = 0.02L[/tex]

During exhalation, volume of water expelled is

[tex]V_e = (4L)(6.2\%)[/tex]

[tex]V_e = 0.248L[/tex]

During 40 breathes, total volume of water taken is

[tex]V_{it} = (40L)(0.02L) = 0.8L[/tex]

During 40 breathes, total volume of water expelled out is

[tex]V_{et} = (40L)(0.248L) = 9.92L[/tex]

Therefore resultant volume of water expelled out from the lung is

[tex]\Delta V = 9.92L-0.8L = 9.12[/tex]

From the body through the lung we have that

[tex]n = \frac{PV}{RT}[/tex]

Here,

P = Pressure

R= Gas ideal constant

T= Temperature

V = Volume

Replacing,

[tex]n = \frac{(1atm)(9.12L)}{(8.314J/mol \cdot K)(298K)}[/tex]

[tex]n = 0.373mol/min[/tex]

Therefore the moles of water per minute are expelled from the body through the lungs is 0.373mol/min

The athlete expels 8.89 x 10^-2 moles of water per minute through the lungs.

To calculate the number of moles of water per minute expelled by the athlete through the lungs, we need to determine the amount of water evaporated with each breath and then multiply it by the respiration rate. According to the information provided, an average breath is about 0.5 L, and each breath evaporates 4.0 x 10^-2 g of water. We can convert grams of water to moles by dividing by the molar mass of water (18.02 g/mol). So, the moles of water evaporated with each breath are (4.0 x 10^-2 g)/(18.02 g/mol) = 2.22 x 10^-3 mol/breath.

Next, we can calculate the number of breaths per minute multiplied by the moles of water evaporated per breath to find the moles of water expelled per minute. The respiration rate is given as 40 breaths per minute. Therefore, the moles of water expelled per minute are (2.22 x 10^-3 mol/breath) x 40 breaths/minute = 8.89 x 10^-2 mol/minute.

Answer:

The period of oscillation is 1.33 sec.

Explanation:

Given that,

Mass = 275.0 g

Suppose value of spring constant is 6.2 N/m.

We need to calculate the angular frequency

Using formula of angular frequency

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

Where, m = mass

k = spring constant

Put the value into the formula

[tex]\omega=\sqrt{\dfrac{6.2}{275.0\times10^{-3}}}[/tex]

[tex]\omega=4.74\ rad/s[/tex]

We need to calculate the period of oscillation,

Using formula of time period

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

Put the value into the formula

[tex]T=\dfrac{2\pi}{4.74}[/tex]

[tex]T=1.33\ sec[/tex]

Hence, The period of oscillation is 1.33 sec.

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:

(c)  As 'd' becomes doubled, energy decreases by the factor of 2

Explanation:

Energy stored in a parallel plate capacitor is given by:

[tex]U=\frac{1}{2}CV^2\\\\C=\frac{A\epsilon_{o}}{d}\\\\then\\\\U=\frac{1}{2}\frac{A\epsilon_{o}}{d}V^2--(1)\\\\[/tex]

As capacitor remains connected to the battery so V remains constant. As can be seen from (1) that energy is inversely proportional to the separation between the plates so as 'd' becomes doubled, energy decreases by the factor of 2.

Answer:

(c) It decreases by a factor of 2

Explanation:

Since the capacitor is still connected to the power source, the potential difference remain the same even when the distance is a doubled.

The energy stored in a capacitor can be written as:

E = (1/2)CV^2 .....1

And the capacitance of a capacitor is inversely proportional to the distance between the two plates of the capacitor.

C = kA/d ....2

Therefore, when d doubles, and every other determinant of capacitance remains the same, the capacitance is halved.

Cf = kA/2d = C/2

Cf = C/2

Since the capacitance has been halved and potential difference remains the same, the energy stored would also be halved since the energy stored in the capacitor is directly proportional to the capacitance.

Ef = (1/2)(Cf)V^2

Ef = (1/2)(C/2)V^2 = [(1/2)CV^2]/2

Ef = E/2

Where;

E and Ef are the initial and final energy stored in the capacitor respectively

C and Cf are the initial and final capacitance of the capacitor.

d is the distance between the plates

A is the area of plates

k is the permittivity of dielectrics

Therefore the energy stored in the capacitor is decreased by a factor of 2, when the distance is doubled.

Answer / Explanation

The question in the narrative is incomplete.

Kindly find the complete question below:

If the gap between C and the rigid wall at D is  initially 0.15 mm, determine the support reactions at A and  D when the force is applied. The assembly  is made of A36 steel

Procedure

Recalling the the equation of equilibrium and referencing the free body diagram of the assembly,

Therefore, ∑fₓ  = 0 ,

where, 20 ( 10³) - Fₐ - Fₙ = 0 --------------equation (1)

Now, recalling the compatibility equation, while utilizing the superposition method,

Therefore, δₓ - δfₓ

= 0.15  = 200(10³)(600) ÷ π/4 (0.05²)(200)(10⁹) - [ Fₐ (600) / π/4 (0.05²)(200)(10⁹) + Fₐ (600) π / 4 (0.05²)(200)(10⁹) ]

Solving this further,

We get: Fₐ = 20365.05 N

 Which is equivalent to = 20.4 kN.

Now, substituting the answer (Fₐ) into equation (1)

                Fₙ = 179634.95 N

                        = 180 kN

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:

 U₂ = 400 KJ      

Explanation:

Given that

Initial energy of the tank ,U₁= 800 KJ

Heat loses by fluid ,Q= - 500 KJ

Work done on the fluid ,W= - 100 KJ

Sign -

1.Heat rejected by system - negative

2.Heat gain by system - Positive

3.Work done by system = Positive

4.Work done on the system-Negative

Lets take final internal energy =U₂

We know that

Q= U₂ - U₁ + W

-500 = U₂ - 800 - 100

U₂ = -500 +900 KJ

U₂ = 400 KJ

Therefore the final internal energy = 400 KJ

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.

Learn more about electric field on:

https://brainly.com/question/14557875

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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.

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.

To learn more about the moon's phase refer to the link;

https://brainly.com/question/2285324

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.

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.