Electric charge can accumulate on an airplane in flight. You mayhave observed needle-shaped metal extensions on the wing tips andtail of an airplane. Their purpose is to allow charge to leak offbefore much of it accumulates. The electric field around the needleis much larger than the field around the body of the airplane and,can become large enough to produce dielectric breakdown of the air,discharging the airplane. To model this process, assume that twocharged spherical conductors are connected by a long conductingwire and a charge of 22.0 µC isplaced on the combination. One sphere, representing the body of theairplane, has a radius of 6.00 cm, and the other, representing thetip of the needle, has a radius of 2.00 cm.(a) What is the electric potential of eachsphere?(b) What is the electric field at the surface of each sphere?

Answer:

a) 1.2 m/s

b) 0.350 m

c) 1.2 m/s

d) 0.250 m

Explanation:

a)

       \(v = \lambda * f (1)\)

       \(v = \lambda * f = 6.0 m * 0.2 1/s = 1.2 m/s (3)\)

b)

       ⇒ A = Δd/2 = 0.700m/2 = 0.350 m (4)

c)

d)

The net force on particle q₂, located between particles q₁ and q₃, is approximately 189000 N. The force exerted by particle q₁ on q₂ is positive and equals 252000 N, while the force exerted by particle q₃ on q₂ is negative and equals -63000 N.

To find the net force on particle q₂, we need to calculate the individual forces exerted on q₂ by particles q₁ and q₃ and then determine their sum.

The force between two charged particles can be calculated using Coulomb's law:

F = k * |q₁ * q₂| / r²

Where F is the force between the particles, k is the electrostatic constant (k ≈ 9.0 x \(10^9\) Nm²/C²), q₁ and q₂ are the charges of the particles, and r is the distance between them.

First, let's calculate the force exerted on q₂ by q₁:

F₁₂ = k * |q₁ * q₂| / r₁₂²

F₁₂ = (9.0 x \(10^9\) Nm²/C²) * |(8.0 μC) * (3.5 μC)| / (0.10 m)²

F₁₂ ≈ 252000 N

The force is positive because q₁ and q₂ have opposite charges.

Next, let's calculate the force exerted on q₂ by q₃:

F₂₃ = k * |q₂ * q₃| / r₂₃²

F₂₃ = (9.0 x \(10^9\)Nm²/C²) * |(3.5 μC) * (-2.5 μC)| / (0.15 m)²

F₂₃ ≈ -63000 N

The force is negative because q₂ and q₃ have the same charge.

Finally, we can find the net force on q₂ by summing the individual forces:

Net force = F₁₂ + F₂₃

Net force = 252000 N + (-63000 N)

Net force ≈ 189000 N

The net force on particle q₂ is approximately 189000 N.

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The bug will cross the sticky region once in each cycle of its motion, where a cycle is defined as one complete round trip from the top of the bowl to the bottom and back to the top.

To find the number of cycles the bug goes through, we can use conservation of mechanical energy. At the top of the bowl, the bug has only potential energy, which is converted to kinetic energy as it slides down the bowl. At the bottom of the bowl, all of the potential energy has been converted to kinetic energy, and as the bug slides up the other side of the bowl, the kinetic energy is converted back into potential energy. At the top of the bowl again, the bug has only potential energy, and the cycle repeats.

Because there is no friction (except for the sticky patch), the total mechanical energy of the system is conserved. Therefore, the potential energy at the top of the bowl is equal to the potential energy at the bottom of the bowl, and the kinetic energy at the bottom of the bowl is equal to the kinetic energy at the top of the bowl.

We can set the potential energy at the top of the bowl to zero, and use the conservation of energy to find the potential energy at the bottom of the bowl:

mgh = (1/2)mv^2

where m is the mass of the bug, g is the acceleration due to gravity, h is the depth of the bowl, and v is the speed of the bug at the bottom of the bowl.

Solving for v, we get:

v = sqrt(2gh)

Plugging in the numbers, we get:

v = sqrt(29.810.12) = 0.775 m/s

The time it takes for the bug to slide from the top of the bowl to the bottom and back up to the top is twice the time it takes to slide from the top to the bottom:

t = 2sqrt(2h/g) = 2sqrt(2*0.12/9.81) = 0.774 s

Therefore, the frequency of the bug's motion is:

f = 1/t = 1/0.774 = 1.29 Hz

Since the bug completes one cycle in each oscillation, the bug will cross the sticky region 1.29 times per second, or approximately once every 0.78 seconds.

Answer:

m = \(\frac{k}{g}\) x,

graph of x vs m

Explanation:

For this exercise, the simplest way to determine the mass of the cylinder is to take a spring and hang the mass, measure how much the spring has stretched and calculate the mass, using the translational equilibrium equation

              F_e -W = 0

              k x = m g

              m = \(\frac{k}{g}\) x

We are assuming that you know the constant k of the spring, if it is not known you must carry out a previous step, calibrate the spring, for this a series of known masses are taken and hung by measuring the elongation (x) from the equilibrium position, with these data a graph of x vs m is made to serve as a spring calibration.

  In the latter case, the elongation measured with the cylinder is found on the graph and the corresponding ordinate is the mass

Answer:

Vx = 35.31 [km/h]

Vy = 18.77 [km/h]

Explanation:

In order to solve this problem, we must decompose the velocity component by means of the angle of 28° using the cosine function of the angle.

\(v_{x} = 40*cos(28)\\V_{x} = 35.31 [km/h]\)

In order to find the vertical component, we must use the sine function of the angle.

\(V_{y}=40*sin(28)\\V_{y} = 18.77 [km/h]\)

The maximum angle at which an object will not slip on an incline can be calculated using the coefficient of friction (μ).

When an object is on an inclined plane, there are two main forces acting on it: the gravitational force pulling it downward (mg) and the normal force (N) exerted by the inclined plane perpendicular to its surface. Additionally, there is a frictional force (F) acting parallel to the surface of the incline.

To prevent slipping, the frictional force must be equal to or greater than the force component pulling the object down the incline. This force component is given by the equation F = mg sin(θ), where θ is the angle of inclination.

The maximum frictional force that can be exerted between two surfaces is given by the equation F = μN, where μ is the coefficient of friction.

For an object not to slip, the maximum frictional force (F) must be equal to or greater than the force component pulling the object down the incline (mg sin(θ)). Therefore, we have:

F ≥ mg sin(θ)

Substituting F = μN, we get:

μN ≥ mg sin(θ)

Since N = mg cos(θ) (the normal force is equal to the component of the gravitational force perpendicular to the incline):

μmg cos(θ) ≥ mg sin(θ)

μ cos(θ) ≥ sin(θ)

Now, divide both sides of the equation by cos(θ):

μ ≥ tan(θ)

Taking the inverse tangent (arctan) of both sides, we get:

θ ≤ arctan(μ)

Therefore, the maximum angle at which an object will not slip on an incline is given by θ = arctan(μ).

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Answer:

a). 53.78 m/s

b) 52.38 m/s

c) -75.58 m

Explanation:

See attachment for calculation

In the c part, The negative distance is telling us that the project went below the lunch point.

Answer:

The ground we walk on is flat, not round. Gravity doesn't exist. Objects don't fall down, the flat earth(disk) goes up with a force called dark energy. There is a 45 ft tall ice wall to prevent people from falling off the Earth.

Explanation:

To draw the displacement vector diagram, we start at point A and draw a vector from A to B, representing the athlete's displacement of 120 m South. We then draw a vector from the end of the first vector (B) to the end of the second vector (C), representing the athlete's displacement of 200 m East. Finally, we draw a vector from the end of the second vector (C) to point D, representing the athlete's displacement of 270 m North. The diagram should form a closed triangle.

To find the resultant displacement of the athlete, we can use the Pythagorean theorem and trigonometry. Let's call the displacement from A to B "vector AB," the displacement from B to C "vector BC," and the displacement from C to D "vector CD." The magnitude of the resultant displacement (R) is given by:

R = √(AB² + BC² + CD²)

R = √(120² + 200² + 270²) = 334.4 m (rounded to one decimal place)

To find the direction of the resultant displacement, we can use trigonometry. We can find the angle between the resultant displacement and the North direction using the following formula:

θ = tan⁻¹[(ΣN)/(ΣE)]

Where ΣN is the sum of the Northward components of the displacement vectors (in this case, CD), and ΣE is the sum of the Eastward components of the displacement vectors (in this case, BC).

θ = tan⁻¹[(270 m)/(200 m)] = 53.1° (rounded to one decimal place)

Therefore, the magnitude of the resultant displacement is 334.4 m and the direction of the resultant displacement is 53.1° North of East.

The correct values for heat and work for the piston are;

Q = +42 J, W = -50 J

Option C

We must note that the work that has been done by the piston can be obtained by the formula that we know and that formula states that;

w = PdV

w = work that have been done

P = pressure

dV = change in the volume

As such we would have that the work done is;

w =  1.0 × 10^5 Pa (0.0002 -  0.0007)

w = -50 J

This is the work that has been done but the heat was absorbed hence it is positive.

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The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.

The p-waves travel with a constant velocity of 7 km/s

The time can be calculated by using the formula

t = d / v

where

T1 =  10:05 a.m

d is the distance they take to travel from the epicenter

v is the speed of the p-waves

On average, the speed of p-waves is

v = 7 km/s

d = 5600 km (given)

Substituting the values in the formula;

t = d / v

t = 5600 ÷ 7

t = 800 seconds

Converting into minutes,

t = 800 ÷ 60

t = 13.3

≈ 13 mins

T1 -  13 mins = T2

10:05 - 13 mins = 9.52 am

It means the earthquake occurred prior 13 minutes, that is at 9.52 am.

Therefore, the earthquake occurred at 9.52 am.

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Answer:

The time taken by the projectile to hit the ground is 6.85 sec.

Explanation:

Given that,

Vertical height of cliff = 230 m

Distance = 300 m

Suppose, determine the time taken by the projectile to hit the ground.

We need to calculate the time

Using second equation of motion

\(s=ut+\dfrac{1}{2}gt^2\)

Where, s = vertical height of cliff

u = initial vertical velocity

g = acceleration due to gravity

Put the value in the equation

\(230=0+\dfrac{1}{2}\times9.8\times t^2\)

\(t=\sqrt{\dfrac{230\times2}{9.8}}\)

\(t=6.85 sec\)

Hence, The time taken by the projectile to hit the ground is 6.85 sec.

Answer:

bc it was a universal explosion and It started the future

Explanation:

FACTS

Answer:

i wouldn't believe so.

Explanation:

because there was no room or air for the sound to move through. this is because of immense heat and the amount of hyperactive neutrons, electrons and protons clouding everywhere. This would mean that even if there was sound it would a. not travel far or b. go in a completely different direction than expected.

w=225000 J  

E=225000 J  

P=9000 W  

P=22500 W

Ew=Fd=E  

P=E/t

The correct answer is C. The level of competition is not very high in most Masters  programs.

Explanation:

In sports, the word "master" is used to define athletes older than 30 and that usually are professional or have trained for many years, although novates are also allowed. This means in most cases in Master programs and teams a high level of competition can be expected due to the experience and extensive training of Master athletes. Indeed, many records in the field of sport belong to Master athletes rather than younger athletes. According to this, the incorrect statement is "The level of competition is not very high in most Masters  programs".

umm actually is 10 points

Answer:

Extensive property

Explanation:

Answer:

Extensive property

Explanation:

Density, mass of a unit volume of a material substance. The formula for density is d = M/V, where d is density, M is mass, and V is volume. Density is commonly expressed in units of grams per cubic centimetre.

Answer:

\(\text{(a)}\\\text{Given that current energy at starting point }h_{1} \text{ is}:\\E_{h1}=mgh+\frac{1}{2}kx^2\\\text{Plug in height and mass: }E_{h1}=10g+250(0.8)^2=10g+160=258J\\\text{Using Conservation of Energy: }E_{h1}=E_{h2} \\\therefore 258J=mg(h_{2}-h_{1})+\frac{1}{2}mv^2\\=258J=10g+v^2\\=160=v^2\\\therefore v=40m/s \text{ at }h_{2}\\\)\(\text{(b)}\\\text{Again, apply conservation of energy: } 258J=mg(h_{3}-h_{2})+\frac{1}{2}mv^2\\\text{Plugging in: }258J=16g+v^2 \\\text{ Simplify: }101.2=v^2\\\therefore v=10.05m/s \text{ at } h_{3}\)

Answer:

violet

Explanation:

Answer:

Warm colours are those with a yellow-ish or black undertone. Imagine how everything looks through your tinted sunglasses or on a beautiful day in autumn when the leaves are falling. The hints of yellow give these colours a sunny warmth, while those with more black mixed in appear darker and can seem more muted.

Answer:

Explanation:

When the elevator is accelerating downwards, the apparent weight of the person is reduced, and when the elevator is accelerating upwards, the apparent weight is increased.

First, we need to determine the actual weight of the person. We can do this by using the formula:

Weight = mass x gravity

where mass is the mass of the person and gravity is the acceleration due to gravity, which is approximately 9.81 m/s^2.

Weight = (598.900 N) / (9.81 m/s^2) = 61.048 kg

Now, when the elevator is not moving, the person is only experiencing the force due to gravity, which is:

Weight = mass x gravity = (61.048 kg) x (9.81 m/s^2) = 598.78 N

Therefore, the scale would read approximately 598.78 Newtons when the elevator is not moving.

Explanation:

Speed is the rate of change of distance with time. It is a scalar quantity with magnitude both no direction.

  Speed  = \(\frac{distance}{time}\)

The unit is m/s or km/hr or mile/hr

Distance covered is simply the length of the path traveled.

 The unit is m or km or miles

Time taken is the duration of an event.

  The unit is seconds or minutes or hour.

In the experiment, place a ringing alarm clock inside a vacuum chamber, and observe that when the chamber is evacuated, the sound from the alarm clock cannot be heard, illustrating that sound does not travel through a vacuum.

To illustrate that sound requires a medium for its propagation and does not travel through a vacuum, you can perform the following experiment:

Take two identical bells or sound-producing devices and place them in separate chambers, one in a vacuum chamber and the other in a chamber filled with air.

Ensure that both chambers are isolated from external noise sources.

Activate the sound-producing devices simultaneously, creating sound waves in both chambers.

Observe and listen for the sound produced in each chamber.

In the chamber filled with air, you will hear the sound clearly, as air molecules transmit the sound waves, allowing them to propagate and reach our ears.

However, in the vacuum chamber, you will not hear any sound because there is no medium (such as air) for the sound waves to travel through. Without molecules to transmit the vibrations, the sound cannot reach our ears.

This experiment demonstrates that sound requires a medium, such as air or other substances, to propagate and be perceived by our ears. In a vacuum, where there is an absence of molecules, sound cannot travel.

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The sequence "MSKAALTGLLFV" appears to be a protein sequence rather than a DNA sequence. Proteins are formed by linking amino acids together in a specific sequence, and the sequence you provided represents a short section of a protein.

To determine the corresponding DNA sequence, we would need to know the genetic code, which relates the three-letter codons of DNA to specific amino acids. Without that information, we cannot directly derive the DNA sequence from the protein sequence you provided.

Additionally, based solely on the protein sequence "MSKAALTGLLFV," it is challenging to determine the specific function or characteristics of the protein without further information. Proteins have various functions in living organisms, including structural support, enzymatic activity, transport, signaling, and many others.

To understand the function of a protein, it is essential to have more comprehensive information about its structure, cellular location, and any known functional domains or motifs it may contain.

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Answer:

Explanation: Acceleration due to gravity is the acceleration acquired by an object due to the gravitational force. It is a vector quantity, denoted by g and its SI unit is m ... The mass of the earth is about 6E24 kg and the earth radius (Re) is 6371 km which gives G= 9.8m/s/s as the acceleration constant at earth's surface.  I hope this is helpful :)

Acceleration on a body that approached the earth and comes within 6 earth radii of the earth’s surface is 0.2046 m/s²

Acceleration is rate of change of velocity with respect to time.

i.e. a = dv/dt

if an object changes its velocity in short time, we can say that it has grater acceleration.

a= dv/dt =Δv/Δt =  \(\frac{v_{2}- v_{1} }{t_{2}- t_{1}}\)

where    v₂= initial velocity

             v₁= final velocity

             t₂= initial time

             t₁ = final time  

Acceleration due to gravity is nothing but acceleration produced in the body due to gravitational force. According to second newton's second law, F=ma, force is responsible for the acceleration of the body. it is denoted by g and which is equal to 9.8 m/s². Acceleration due to gravity is affected by the height at which a body is from the surface of the earth.

Acceleration due to gravity at height h is given by,

g(h) = GM ÷ (R+h)²

where

g(h) = acceleration due to gravity at height h.

G = universal gravitational constant.

M = mass of the planet(earth)

R= radius of the planet(earth)

h = height of the object from surface.

Given,

G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²

M = 5.97 × 10²⁴ kg

R = 6.3 × 10⁶ m

h=  6R

g(h) = (6.67 × 10⁻¹¹)×(5.97 × 10²⁴) ÷(R+6R)²

g(h) = 3.98×10¹⁴ ÷ (7×6.3 × 10⁶)²

g(h) = 0.2046 m/s²

Hence acceleration due to gravity of body is 0.2046 m/s²

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Answer:

E = 389 MeV

Explanation:

The total energy of particle A, will be equal to the sum of rest mass energy and relative energy of particle A. Therefore,

Total Energy of A = E = Rest Mass Energy + Relative Energy

Using Einstein's Equation: E = mc²

E = m₀c² + mc²

From Einstein's Special Theory of Relativity, we know that:

m = m₀/[√(1-v²/c²)]

Therefore,

E = m₀c² + m₀c²/[√(1-v²/c²)]

E = m₀c²[1 + 1/√(1-v²/c²)]

where,

m₀c² = rest mass energy = 140 MeV

v = relative speed = 0.827 c

Therefore,

E = (140 MeV)[1 + 1/√(1 - (0.827c)²/c²)]

E = (140 MeV)(2.78)

E = 389 MeV

The force of viscosity is F = k r v η. Spherical balls of radius R are falling in a viscous fluid of viscosity η with a velocity v.

What is viscous force?

Viscosity refers to a material's capacity to resist movement between its layers. It is the force that resists relative motion between the layers. It is also known as viscous force.

What is viscous force formula?

In the above equation, the shear force is the viscous force, which is the only internal resistance that causes shear stress to develop along with the fluid layers. As a result, the viscous force formula will be given as. F=Aμdudy.

Hence  F = k r v η is a correct answer.

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