A wire is stretched between two posts. Another wire is stretched between two posts that are twice as far apart. The tension in the wires is the same, and they have the same mass. A transverse wave travels on the shorter wire with a speed of 249 m/s. What would be the speed of the wave on the longer wire?

Answer:

Explanation:

Given

Speed of ball [tex]u=3.2\ m/s[/tex]

Plane is inclined at an angle [tex]20^{\circ}[/tex]

To win the Game we need to hit the target at [tex]x=2.4\ m[/tex] away

Launch angle of ball [tex]\theta [/tex]

Motion of ball can be considered in two planes i.e. Vertical to the plane and horizontal to the plane

So Net acceleration in vertical plane is [tex]g\sin 20[/tex]

Range of Projectile is given by

[tex]R=\frac{u^2\sin 2\theta }{g}[/tex]

for [tex]R=2.4\ m[/tex]

[tex]2.4=\frac{3.2^2\times sin 2\theta }{g\sin 20}[/tex]

[tex]\sin 2\theta =\frac{2.4\times 9.8\times \sin 20}{3.2^2}[/tex]

[tex]\sin 2\theta =0.7855[/tex]

[tex]2\theta =51.77[/tex]

[tex]\theta =25.88^{\circ}[/tex]

so ball must be launched at an angle of [tex]25.88^{\circ}[/tex]

To answer this, we use projectile motion principles with an equation specifically structured for the problem. Inserting the given values into the equation derived from the horizontal and vertical equations of motion, we can find the required launching angle to hit the target.

To solve this problem, we can apply the concepts found in projectile motion physics. Given the initial speed or velocity (3.2 m/s) and the horizontal distance of the target (2.4 m) we can find the necessary angle (theta) to hit the target. The angle θ can be given by the equation of motion for a projectile which can be derived from the horizontal and vertical equations of motion, θ = atan[(vf² ± sqrt(vf⁴ - g*(g*x² + 2*y*vf²)) / (g*x)] where g = 9.8 m/s² is the acceleration due to gravity, x = 2.4 m is the horizontal distance, y = 0 (height difference), and vf = 3.2 m/s is the final velocity, the speed at which the ball is launched.

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Juanita did more work on the square box

Explanation:

According to the law of conservation of energy, the work done in lifting an object is equal to the increase in gravitational potential energy of the object. This is due to the fact that the work done on the object is converted into potential energy.

The potential energy of an object (GPE) is given by

[tex]GPE=mgh[/tex]

where

m is the mass of the object

g is the acceleration of gravity

h is the height of the object

In this problem, we have two objects:

- The roud box is lifted and its GPE increases by 50 J --> this means that the work done by Juanita on the box is 50 J

- Thr square box is lifted and its GPE increases by 100 J --> this means that the work done by Juanita on the box is 100 J

Therefore, Juanita did more work on the square box.

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The square box received more work by Juanita than the round box based on the increase in gravitational potential energy.

The square box received more work done by Juanita compared to the round box. When the gravitational potential energy increase for the square box is 100 J, which is greater than the 50 J increase in the round box, it indicates that Juanita did more work on the square box.

Answer:

Distance = 26.0m Displacement = 4.0m

Explanation:

Distance specifies only how far an object has traveled while displacement is the distance traveled in a specified direction.

Total distance traveled by the object will be distance travelled through north + distance travelled through south i.e 15.0m + 11.0m = 26.0m

Displacement is gotten by using the Pythagoras theorem. Since the object traveled in the same vertical direction (15.0m through north which is upward i.e positive y direction and 11.0m through south i.e in the negative y direction), the displacement will be 15.0m - 11.0m = 4.0m

The distance traveled is 26.0 m and the magnitude of the displacement vector is 4.0 m

First, we will define the terms distance and displacement

Distance is the total movement of an object without any regard to direction.

Displacement is the difference between the original and final position of a path taken by an object.

Since, the object moves 15.0 m north and then 11.0 m south,

Then,

Distance traveled = 15.0 m + 11.0 m

Distance traveled = 26.0 m

For the magnitude of the displacement,

The object moves 15.0 m north and then 11.0 m south, which is in the opposite (negative) direction

Then,

Magnitude of displacement = 15.0 m - 11.0 m

Magnitude of displacement = 4.0 m

Hence, the distance traveled is 26.0 m and the magnitude of the displacement vector is 4.0 m

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Answer:

after the sun sets or just as it is setting

Explanation:

a crescent moon is thin and reflects less sunlight during the daylight sky so it becomes difficult to spot, but can be spotted when the sun is setting or just sets.

The best time to find the thinnest crescent moon just after the new moon is in the early evening, just after sunset, when the moon starts to reflect sunlight towards the earth, showing the crescent shape. The appearance of the Moon's surface can vary significantly with its phase and this can be better viewed through binoculars.

To find the thinnest crescent moon just after new moon, the best time to look is usually in the early evening just after sunset. The Moon, moving eastward each day in its 30 days cycle around the Earth, moves roughly 12° in the sky each day. A day or two after the new phase, the thin crescent first appears, as we begin to see a small part of the Moon's illuminated hemisphere reflecting a little sunlight toward us.

Because the Moon is moving eastward away from the Sun, it rises later and later each day. Therefore, after a new moon, the thin crescent will be seen in the west just after sunset.

Keep in mind that the bright crescent increases in size on successive days as the Moon moves farther and farther around the sky away from the direction of the Sun.

Bear in mind that the brighter the Moon is in the night sky, the harder it is to see the faint flashes of meteors. Furthermore, as seen through a good pair of binoculars, the appearance of the Moon's surface changes dramatically with its phase, revealing more topographic details when sunlight streams in from the side, causing topographic features to cast sharp shadows.

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Answer:

a.-6.75 kgm/s

b.[tex]3375 N[/tex]

Explanation:

We are given that

Mass of baseball=0.15 kg

Initial speed=[tex]u=20m/s[/tex]

Final speed=[tex]v=-25m/s[/tex]

a.We know that

Impulse=Change in momentum=[tex]\Delta p=mv-mu=m(v-u)[/tex]

Momentum=[tex]mass\times velocity[/tex]

Using the formula

Impulse=[tex]0.15(-25-20)=-6.75 kgm/s[/tex]

b.Time=[tex]2\times 10^{-3} s[/tex]

Force=[tex]\frac{Impulse}{time}[/tex]

Using the formula

Average force exerted by the bat on the ball=[tex]\frac{-6.75}{2\times 10^{-3}}[/tex] N

Average force exerted by the bat on the ball=[tex]3375N[/tex]

Answer:

What is the coefficient of kinetic friction = 0.432

Explanation:

The detailed steps and derivation with appropriate substitution is as shown in the attached file.

This question involves the concepts of the law of conservation of energy and frictional energy.

a) The speed of the child when he reaches the bottom of the slide is "6 m/s".

b) The coefficient of kinetic friction between the child and slide when the wax paper isn't used is "0.432".

a)

According to the law of conservation of energy in this situation:

Loss in Potential Energy = Gain in Kinetic Energy

[tex]mgh = \frac{1}{2}mv^2\\\\2gh=v^2\\v=\sqrt{2gh}\\\\[/tex]

where,

v = velocity = ?

g = acceleration due to gravity = 9.81 m/s²

h = height lost = 6 ft = 1.83 m

Therefore,

[tex]v=\sqrt{2(9.81\ m/s^2)(1.83\ m)}[/tex]

v = 6 m/s

b)

Now, the velocity becomes half and the friction comes into action. So in this case the law of conservation of energy will be written as:

Loss of Potential Energy = Gain of Kinetic Energy + Frictional Energy

[tex]mgh=\frac{1}{2}m(\frac{v}{2})^2+\mu_kRl\\\\mgh=\frac{1}{2}m(\frac{v}{2})^2+\mu_kmgCos\theta l\\\\gh-\frac{1}{2}(\frac{v}{2})^2+\mu_kgCos\theta l[/tex]

where,

l = length of slide = [tex]\frac{h}{sin\theta}=\frac{1.83\ m}{sin30^o}=3.66\ m[/tex]

[tex]\mu_k[/tex] = coefficient of kinetic friction = ?

Therefore,

[tex](9.81\ m/s^2)(1.83\ m)-\frac{1}{2}(\frac{6\ m/s}{2})^2=\mu_k(9.81\ m/s^2)(Cos30^o)(3.66\ m)\\\\\mu_k=\frac{13.45\ m^2/s^2}{31.09\ m^2/s^2}\\\\\mu_k=0.432[/tex]

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The attached picture explains the law of conservation of energy.

Answer:

the new volume =93,750 Liters

Explanation:

From Boyles law

P1V1 = P2V2

1.2 X 125000 = 1.6 X V2

V2 = 93,750 Liters

Answer:

Hydrogen peroxides are known to be an active ingredient in hair dyes. Hydrogen peroxide is a damaging chemical although it has been added in diluted amounts in hair dyes. Some hair dye companies have already started looking for alternatives for hydrogen peroxide to use in their products, concerning the damaging effects of the chemical.

Hydrogen peroxide can cause hair loss, dermatitis and scalp burns.

Concerning the harmful chemicals in dyes, many people consider it is best to use henna if you really want a colour change for your hair.

Answer:

1.19×10²² m/s

Explanation:

Kinetic Energy: This can be defined as the the energy of a body due to motion.

The formula for kinetic energy is given as,

Ek = 1/2mv²................... Equation 1

Where Ek = Kinetic energy, m = mass of proton. v = velocity of proton.

Making v the subject of the equation,

v = √(2Ek/m).................. Equation 2.

Given: Ek = 0.995×10⁻⁵ J, m = 1.673×10⁻²⁷ kg.

Substitute into equation 2

v = √(2×0.995×10⁻⁵/1.673×10⁻²⁷ )

v = 1.19×10²² m/s.

Therefore, the velocity of proton = 1.19×10²² m/s

Answer:

d. This statement is false. She and the Space Station share the same orbit and will stay together unless they are pushed apart.

Explanation:

In astronomy, orbit is simply a path of an object around another object in a space. That is, orbit is a path of a body that revolves around a gravitating center of mass. Examples of an orbit is are satellite around a planet, orbit around a center of galaxy, planet around the sun, and among others.

On the other hand, space station refers to a spacecraft that can support a group of human for long time in the orbit. Another names for space stations are orbital space station and orbital station.

Therefore, an astronaut goes on a space walk outside the Space Station shares the same orbit with the space station and they will stay together unless they are pushed apart.

The statement is true that an astronaut would float away during a spacewalk if not tethered. The astronaut and ISS, although influencing each other gravitationally to a small extent, they would continue along their separate paths under the influence of Earth's gravity and microgravity.

This statement is true. When an astronaut steps out for a spacewalk, she and the International Space Station (ISS) are both in orbit around Earth but the astronaut will not stay in place relative to the ISS without a tether holding her to the station.

The reason for this is due to microgravity and the properties of motion in space. The astronaut and the ISS are both in free-fall around the Earth, so in the absence of other forces, they would continue along their separate paths. Without a tether, even a small force (like the push off the astronaut does to get away from the airlock) can cause her to drift away from the station.

While the ISS and the astronaut do influence each other gravitationally, this effect is extremely small compared to the force of Earth's gravity. So, without a tether, the astronaut can float away from the ISS.

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Answer:

A) An interference pattern

Explanation:

the two slit experiment is key to understand the microscopic world. The wave-like properties of light were demonstrated by the famous experiment first performed by Thomas Young in the early nineteenth century. In original experiment, a point source of light illuminates two narrow adjacent slits in a screen, and the image of the light that passes through the slits is observed on a second screen.

Key Points

Answer:

shielded

Explanation:

shielded twisted pair wire is used in environments that have a noticeable amount of electromagnetic interference.

Answer: The answer is attached

Explanation:

Final answer:

In quantum mechanics, the probability of a spin-flip and the potential for achieving a spin flip with certainty are explored through the application of magnetic fields in different directions.

Explanation:

P(a): The probability that the spin-flip would occur can be calculated using quantum mechanics. When applying magnetic fields B along x and y, the probability of finding the spin of the electron oriented in the negative z-direction after is over is determined by the transition probabilities.

P(b): The time to ensure a spin flip with certainty (P = 1) can be calculated by adjusting the duration of the magnetic field application. By manipulating the duration, the spin can be controlled to achieve a certain outcome at a specific time.

Answer:e

Explanation:

Given

Area of parallel plates is A

distance between plates is d

Potential difference between Plates is V

Capacitance is given by

[tex]C=\frac{\epsilon _0A}{d}[/tex]

If separation is doubled then capacitance become half

[tex]C'=\frac{\epsilon _0A}{2d}[/tex]

[tex]C'=\frac{C}{2}[/tex]

Electrical energy stored in the capacitor is given by

[tex]E=\frac{1}{2}CV^2[/tex]

When distance is doubled

[tex]E'=\frac{1}{2}\times \frac{C}{2}\times V^2[/tex]

[tex]E'=\frac{E}{2}[/tex]      

Therefore Energy is halved

If the separation between the plates is doubled, the electrical energy stored in the capacitor will be: d. The electrical energy stored in the capacitor will be quartered.

To understand why the energy stored in the capacitor is quartered when the separation between the plates is doubled, let's consider the formula for the capacitance of a parallel plate capacitor and the energy stored in a capacitor.

 The capacitance C of a parallel plate capacitor is given by:

[tex]\[ C = \frac{\varepsilon_0 A}{d} \][/tex]

The energy U stored in a capacitor is given by:

[tex]\[ U = \frac{1}{2} C V^2 \][/tex]

[tex]\[ C' = \frac{\varepsilon_0 A}{2d} = \frac{1}{2} \frac{\varepsilon_0 A}{d} = \frac{1}{2} C \][/tex]

[tex]\[ U' = \frac{1}{2} C' V^2 = \frac{1}{2} \left(\frac{1}{2} C\right) V^2 = \frac{1}{4} C V^2 = \frac{1}{4} U \][/tex]

So, when the separation between the plates is doubled, the capacitance is halved, and since the energy is directly proportional to the capacitance, the energy stored in the capacitor is quartered.

Answer:

Based on the data thomson collected in his experiments using cathode rays, the concept of atomicc structure was modified. What were the four things validated by his cathode ray expirement?

Explanation:

Cathode Ray tubes' Experiment of J.J. Thomson, showed that atoms contain  electrons or tiny negative charged subatomic particles.

Based on the Thomson's cathode ray tubes experiment, He validated four things which are as following

1. Cathode rays have mass.

2. Matter contains positive and negative charge.

3. Particles of cathode rays are fundamental to all matter.

4. An atom is divisible.  

Answer:

(a) Total area is 14.5 roods

(b) Total area is 14674.522 square meters

Explanation:

Area occupied by land = 3 acres

1 acre = 40 perches by 4 perches = 160 square perches

3 acres = 3×160 = 480 square perches

Area occupied by livestock = 25 perches by 4 perches = 100 square perches

Total area = 480 + 100 = 580 square perches

1 rood = 4 perches by 1 perch = 4 square perches

580 square perches = 580/4 = 14.5 roods

(b) Total area = 580 square perches

1 perch = 16.5ft = 16.5/3.2808 = 5.03 meters

580 square perches × (5.03 meters/1 perch)^2 = 580 ×25.3009 square meters = 14674.522 square meters

The total area owned by the landowner is 580 perches, which is equivalent to 14.5 roods. When converted to the modern unit of measurement, this totals to approximately 17516 square meters.

To solve this problem, we first need to convert each area to the same unit. In this case, we can convert all areas to perches. According to the old manuscript, the landowner had 3 acres of plowed land. Given that 1 acre is equivalent to 160 perches (40 perches by 4 perches), the plowed land was 480 perches (3 x 160).

The livestock area is 25 perches by 4 perches, which totals to 100 perches. Therefore, the total area owned by the landowner is 580 perches (480 perches + 100 perches).

(a) To convert this to roods, we divide by 40 (since 1 rood is 40 perches by 1 perch) which equals to 14.5 roods

(b) To convert perches to square meters, we need to know that 1 perch is 16.5 ft. and 1 square foot is approximately 0.0929 square meters. So, 1 perch is 325.125 sq ft (16.5 ft * 16.5 ft). Therefore, 1 perch is about 30.2 square meters (325.125 ft * 0.0929), and 580 perches equate to approximately 17516 square meters.

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Answer:

about 10 trillion kilometers

Explanation:

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

t = Seconds in one year = [tex]365.25\times 24\times 60\times 60[/tex]

1 Light year

[tex]1\ ly=ct\\\Rightarrow 1\ ly=3\times 10^8\times 365.25\times 24\times 60\times 60=9.46728\times 10^{15}\ m\\ =9.46728\times 10^{15}\times 10^{-3}\\ =9.46728\times 10^{12}\ km\approx 10\ trillion\ km[/tex]

The answer is a. about 10 trillion kilometers

The frequency of the homozygous recessive genotype in the population is approximately 0.71.

The frequency of the homozygous recessive genotype in the population can be calculated using the Hardy-Weinberg equation. In this case, the frequency of the dominant yellow-flowered plants is given as 50% or 0.5. To calculate the frequency of the homozygous recessive genotype (qq), we need to find the value of q. The equation for the Hardy-Weinberg equilibrium is p² + 2pq + q² = 1.

Given that the frequency of the dominant allele (p) is 0.5, we can substitute the value of p into the equation and solve for q:

0.5² + 2(0.5)(q) + q² = 1

Simplifying the equation gives:

0.25 + q + q² = 1

Combining like terms:

q² + q - 0.75 = 0

Using the quadratic formula to solve for q, we find that q ≈ 0.71 or -1.21. Since the frequency of a trait cannot be negative, we can disregard the negative value. Therefore, the frequency of the homozygous recessive genotype in the population is approximately 0.71.

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A) Initial velocity of ball 1: 9.53 m/s upward

B) Height of the building: 6.48 m

C) Maximum velocity: 11.7 m/s

D) Minimum velocity: 5.83 m/s

Explanation:

A)

The y-position of the 1st ball at time t is given by the equation for free fall motion:

[tex]y_1 = h + v_0 t - \frac{1}{2}gt^2[/tex] (1)

where

h = 21.0 m is the initial height of the ball, the height of the building

[tex]v_0[/tex] is the initial velocity of the ball, upward

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

The y-position of the 2nd ball instead, dropped from the roof 1.19 s later, is given by

[tex]y_2 = h-\frac{1}{2}g(t-1.19)^2[/tex]

where

h = 21.0 m is the initial height of the ball, the height of the building

t' = 1.19 s is the delay in time of the 2nd ball (we can verify that at t = 1.19 s, then [tex]y_2=h[/tex], so the ball is still on the roof

The 2nd ball reaches the ground when [tex]y_2=0[/tex], so:

[tex]0=h-\frac{1}{2}g(t-1.19)^2\\0=(21.0)-4.9(t^2-2.38t+1.42)\\4.9t^2-11.66t-14.04=0[/tex]

Which has two solutions:

t = -0.88 s (negative, we discard it)

t = 3.26 s (this is our solution)

The 1st ball reaches the ground at the same time, so we can substitute t = 3.26 s into eq.(1) and [tex]y_1=0[/tex], so we find the initial velocity:

[tex]0=h+v_0 t -\frac{1}{2}gt^2\\v_0 = \frac{1}{2}gt-\frac{h}{t}=\frac{1}{2}(9.8)(3.26)-\frac{21.0}{3.26}=9.53 m/s[/tex]

B)

In this case, the height of the building h is unknown, while the initial velocity of ball 1 is known:

[tex]v_0 = 8.70 m/s[/tex]

When the two balls reach the ground at the same time, there position is the same, so we can write:

[tex]y_1=y_2\\h+v_0 t - \frac{1}{2}gt^2 = h-\frac{1}{2}g(t-1.19)^2[/tex]

Solving the equation, we find:

[tex]v_0t=1.19gt-\frac{1}{2}g(1.19)^2\\t=\frac{0.5g(1.19)^2}{1.19g-v_0}=2.34 s[/tex]

This is the time at which both balls reache the ground; and substituting into the eq. of ball 2, we find the height of the building:

[tex]0=h-\frac{1}{2}g(t-1.19)^2\\h=0.5g(t-1.19)^2=0.5(9.8)(2.34-1.19)^2=6.48 m[/tex]

C)

If [tex]v_0[/tex] is greater than some value [tex]v_{max}[/tex], then there is no value of h such that the two balls hit the ground at the same time. This situation occurs when the demoninator of the formula found in part b:

[tex]t=\frac{0.5g(1.19)^2}{1.19g-v_0}[/tex]

becomes negative: in that case, the time becomes negative, so no solution is possible.

The denominator becomes negative when

[tex]1.19g-v_0 < 0[/tex]

Therefore when

[tex]v_0>1.19g=(1.19)(9.8)=11.7 m/s[/tex]

So, if the initial velocity of ball 1 is greater than 11.7 m/s, the two balls cannot reach the ground at the same time.

D)

There is also another condition that must be true in order for the two balls to reach the ground at the same time: the time at which ball 1 reaches the ground must be larger than 1.19 s (because ball 2 starts its motion after 1.19 s). This means that the following condition must be true

[tex]t=\frac{0.5g(1.19)^2}{1.19g-v_0}>1.19[/tex]

Solving the equation for [tex]v_0[/tex], we find:

[tex]0.5g(1.19)^2>1.19(1.19g-v_0)\\6.94>13.88-1.19v_0\\1.19v_0>6.94[/tex]

Which gives

[tex]v_0>5.83 m/s[/tex]

Therefore, the minimum speed of ball 1 at the beginning must be 5.83 m/s.

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Answer:

c. about 1/10 as great.

Explanation:

While jumping form a certain height when we bend our knees upon reaching  the ground such that the time taken to come to complete rest is increased by 10 times then the impact force gets reduced to one-tenth of the initial value when we would not do so.

This is in accordance with the Newton's second law of motion which states that the rate of change in velocity is directly proportional to the force applied on the body.

Mathematically:

[tex]F\propto\frac{d}{dt} (p)[/tex]

[tex]\Rightarrow F=\frac{d}{dt} (m.v)[/tex]

since mass is constant

[tex]F=m\frac{d}{dt}v[/tex]

when [tex]dt=10t[/tex]

then,

[tex]F'=m.\frac{v}{10\times t}[/tex]

[tex]F'=\frac{1}{10} \times \frac{m.v}{t}[/tex]

[tex]F'=\frac{F}{10}[/tex] the body will experience the tenth part of the maximum force.

where:

[tex]\frac{d}{dt} =[/tex] represents the rate of change in dependent quantity with respect to time

[tex]p=[/tex] momentum

[tex]m=[/tex] mass of the person jumping

[tex]v=[/tex] velocity of the body while hitting the ground.

Final answer:

The detailed response covers the maximum height reached by the egg, the time it takes to hit the ground, and its velocity just before landing.a)60.0m,b)2.45 seconds,c)-6.0m/s.

Explanation:

a) Maximum Height: The maximum height above the ground reached by the egg can be calculated using the kinematic equation. It will be twice the initial height. In this case, it would be 60.0m.

b) Time to Hit Ground: You can determine the time it takes for the egg to hit the ground by using the kinematic equation for vertical motion. The time would be approximately 2.45 seconds after its release.

c) Velocity Before Landing: The velocity of the egg immediately before hitting the ground would be the same as the initial velocity when it was thrown, but in the opposite direction, which is -6.0m/s.

In a water molecule, the dipole's negative portion is located on the oxygen atom and points towards the hydrogen atoms due to the difference in electronegativity between these elements.

The water molecule, H2O, is a polar molecule, meaning it has a net dipole due to the presence of polar bonds, which result from a significant difference in electronegativities of the atoms involved. In a water molecule, the oxygen atom is more electronegative than hydrogen and therefore pulls the shared electrons toward itself. This creates a charge separation where the oxygen side of the molecule becomes partially negative, and the hydrogen side becomes partially positive. Looking at the provided options, the most accurate statement would be that the negative portion of the dipole is D) pointing from the oxygen through the hydrogen atoms

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Answer:

Explanation:

The electric flux through a Gaussian surface can be calculated using Gauss's law. (a) Calculate electric flux for q1 and q2, (b) Calculate electric flux for q2 and q3, (c) Calculate electric flux for all three charges.

The electric flux through a Gaussian surface can be calculated using Gauss's law. Gauss's law states that the total electric flux through a closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space.

(a) To find the electric flux through the Gaussian surface enclosing only charges q1 and q2, we need to calculate the net charge enclosed by the surface, which is the sum of the two charges. Then, we divide this sum by the permittivity of free space to obtain the electric flux.

(b) Following the same procedure, we can find the electric flux through the Gaussian surface enclosing only charges q2 and q3.

(c) To find the electric flux through the Gaussian surface enclosing all three charges, we calculate the net charge enclosed by the surface, which is the sum of all three charges. Again, we divide this sum by the permittivity of free space to obtain the electric flux.

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Answer:

Gravitational potential energy

Explanation:

Gravitational potential energy is the type of energy an object has due to its position in a gravitational field. Water behind a dam possesses gravitational potential energy due to it being at a higher level than the water on the other side of the dam. When the water falls the gravitational potential energy is converted to kinetic energy, leading to the turning of the turbines to generate electricity.

Answer:

DUNNO

Explanation:

BIT BAD FOR 65% WOULD GET -100%

Answer:

a) v(2) = 2m/s, v(3) = -2m/s

b) speed at t = 2s is 2m/s

speed at t = 3s is 2m/s

c) 0 m/s

Explanation:

We can take the derivative of x(t) to find the equation of velocity

v(t) = x'(t) = 10 - 4t

(a) v(2) = 10 - 4*2 = 10 - 8 = 2 m/s

v(3) = 10 - 4*3 = 10 - 12 = -2 m/s

(b) The speed would be the same as velocity without the direction

speed at t = 2s is 2m/s

speed at t = 3s is 2m/s

(c) The average velocity between t = 2s and t = 3s is distance it travels over period of time

[tex]v_a = \frac{s(3) - s(2)}{\Delta t} = \frac{10*3 - 2*3^2 - (10*2 - 2*2^2)}{3 - 2}[/tex]

[tex]v_a = \frac{12 - 12}{1} = 0/1 = 0 m/s[/tex]

Final answer:

The instantaneous velocity at t = 2 s is 2 m/s and at t = 3 s is -2 m/s. The instantaneous speed at both times is 2 m/s. The average velocity between t = 2 s and t = 3 s is 12 m/s.

Explanation:

(a) To find the instantaneous velocity, we need to find the derivative of the position function x(t) with respect to time. The derivative of x(t) = 10t - 2t² is v(t) = 10 - 4t. Substituting t = 2 and t = 3 into v(t), we get v(2) = 10 - 4(2) = 2 m/s and v(3) = 10 - 4(3) = -2 m/s.

(b) The instantaneous speed is the magnitude of the instantaneous velocity. Since speed is always positive, the speed at t = 2 s and t = 3 s is 2 m/s for both.

(c) The average velocity between t = 2 s and t = 3 s is given by the change in position divided by the change in time. The change in position is x(3) - x(2) = (10(3) - 2(3)²) - (10(2) - 2(2)²) = 12 m, and the change in time is 3 s - 2 s = 1 s. Therefore, the average velocity is 12 m/1 s = 12 m/s.

Answer:985.71 Hz

Explanation:

Given

Speed of ambulance (source) [tex]v_s=108\ kmph\approx 30\ m/s[/tex]

Original Frequency of sound wave [tex]f=900\ Hz[/tex]

speed of sound waves [tex]v=345\ m/s[/tex]

According to Doppler apparent frequency will be different to original frequency when speed of source is moving w.r.t to observer.

[tex]f'=f(\frac{v+v_o}{v-v_s})[/tex]

where f'=apparent frequency

f=original frequency

[tex]v_o[/tex]=observer speed

[tex]f'=900\times (\frac{345+0}{345-30})[/tex]

[tex]f'=900\times (\frac{345}{315})[/tex]

[tex]f'=985.71\ Hz[/tex]        

Answer:

b = 242 m

Explanation:

A = 24200 m²

a = 100 m

b = ?

A seguinte fórmula é aplicada

A = a*b

⇒  b = A / a

⇒  b = (24200 m²) / (100 m)

⇒  b = 242 m

Answer: THE SAME

Explanation: visible light is an electromagnetic wave, which has the properties of both magnet and electrical. The speed of light in air has Generally been estimated to be

299,792 kilometers per second. For for a distant start moving towards the Earth at a speed one-fourth (1/4th) of 299,792kilometers per second compared to a flashing light from the Earth,the speed of the light from the star is expected to be the same.

Answer:

5398.4km/h

Explanation:

IN THIS CASE THE MOMENTUM IS CONSERVED. THE VALUE OF MOMENTUM OF ONE COMBINED ROCKET WILL BE SAME AS OF TWO COMBINED.

Let mass of module be m

then

mass of motor  = 4m  (four times the mass of rocket module)

total mass = m + 4m = 5m

combined velocity = V = 5320kph

Let

absolute (relative to earth)motor velocity after disengagement = v

then

rocket module velocity (relative to earth) after disengagement = v+98  (relative velocity = 98)

momentum conservation equation

combined momentum = module momentum + motor momentum

(m+4m)V = m(v+98) + 4m*v

5mV = 98m+mv + 4mv

5V = 98+v + 4v (m cancels out)

5V - 98 = 5v

((5*5320)-98)/5 = v

v = 5300.4 km/h

velocity of rocket module relative to earth = v +98

                                                                       = 5300.4 + 98

                                                                       = 5398.4km/h