Answer:
d. Potential energy is converted to kinetic energy; the kinetic energy is then converted into the work of bringing the body to a stop.
Explanation:
- At the beginning of the falls, when the person is still at a certain height h, the person has gravitational potential energy:
U = mgh
where m is the mass of the person, g the acceleration due to gravity, h the height above the ground.
- As the person falls down, h decreases, so the potential energy decreases; according to the law of conservation of energy, potential energy is converted into kinetic energy, since the speed of the person increases:
[tex]K=\frac{1}{2}mv^2[/tex]
where v is the speed.
- Just before hitting the ground, all the potential energy has been converted into kinetic energy
- When the person hits the ground, he/she comes to a stop: so work is done by the ground on the person, because the ground applied a force required to stop the person, and the kinetic energy "lost" by the person is equal to the work done by the ground to bring the body to a stop.
What effect does friction have on a roller coaster
Hello There!
Let's first talk about "What Is Friction"
Friction is a force that pulls when two object touch each-other. Friction happens because the molecules on one surface interlock with the molecules on another surface.
Now, let's get back to our original question "What Effect Does Friction Have On A Roller Coaster"
On a roller coaster, friction is a force that opposes motion and significantly slows the cars as they move on the track.
What controls the net direction of molecules?
Molecular diffusion is the process of mass transfer as a result of a concentration difference in a mixture.
This process is related to the thermal movement of the particles and their velocity, which is a function of the temperature, the viscosity of the fluid and the size of the particles. This is how diffusion controls the net direction of the molecular flow from a region of higher concentration to one of lower concentration, as long as there is a difference (or gradient) of concentration.
A coin completes 18 spins in 12 seconds. The centripetal acceleration of the edge of the coin is 2.2 m/s2. The radius of the coin is ____-blank
Answer:
0.025 m
Explanation:
The coin completes 18 revolutions in 12 seconds. The angular velocity of the coin is:
[tex]\omega = \frac{18 rev}{12 s} \cdot (2\pi \frac{rad}{rev})=9.42 rad/s[/tex]
The centripetal acceleration of the edge of the coin is given by
[tex]a=\omega^2 r[/tex]
where we have
a = 2.2 m/s^2 is the acceleration
r is the radius of the coin
Solving for r, we find
[tex]r=\frac{a}{\omega^2}=\frac{2.2 m/s^2}{(9.42 rad/s^2)^2}=0.025 m[/tex]
Answer: 0.025 m or 2.5m
Explanation:
The number of spins given in 12 seconds = 18
So the number of spins in 1 second will be
= 18 / 12 = 1.5rps
The above result is the frequency because it is the number of spins in one second
Let f be frequency
f = 1.5 rps
Using centripetal acceleration given
Ca = 2.2 m/s^2
m is the radius of the given coin
Using the formula
Centripetal acceleration, = m x w^2
Knowing that w = 2πf
2.2 = m x ( 2 x 3.14 x 1.5) ^2
2.2 = m x 88.7364
m = 0.025 m
What is the brightest star in the night sky
the brightest star in the night sky is Sirius A
The brightest star in the night sky is Sirius A.
Sirius A is the brightest star that we see in the night sky. It is a binary system with a white dwarf orbiting a main-sequence star. Despite common misconceptions, Sirius is the actual brightest star, not Polaris.
What causes convection currents to occur?
Answer:
Convection currents are caused by an uneven temperature within something.
For example, within the earth, Convection currents occur when a reservoir of fluid is heated at the bottom, and allowed to cool at the top.. Heat causes the fluid to expand, decreasing its density. If there is cooler material on top, it will be more compact and therefore, will sink to the bottom. The heated material will rise to the top.
Which statement is true?Which statement is true? Gas atoms subjected to electricity emit bright lines. Gas atoms absorb white light and emit a continuous spectrum of light. Gas atoms under high pressure emit dark lines against a continuous spectrum. Gas atoms heated to a high temperature emit white light. Gas atoms subjected to electricity do not emit a spectrum.
Answer:
Gas atoms subjected to electricity emit bright lines.
Explanation:
Gas atoms subject to electricity emit bright lines. This can be seen in fluorescent lamps, for example.
Fluorescent lamps work by ionizing atoms of argon gas and mercury vapor. After receiving an electric current, ionization occurs, at that moment the atoms are accelerated by the potential difference established between the lamp terminals and emit bright light and electromagnetic waves when they return to the natural state.
Answer:
A. Gas atoms subjected to electricity emit bright lines.
Explanation:
Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. During that time, what are the magnitudes of (a) the angular acceleration and (b) the average angular velocity? (c) What is the instantaneous angular velocity of the disk at the end of the 1.00 s? (d) With the angular acceleration unchanged, through what additional angle (rad) will the disk turn during the next 1.00 s?
With constant angular acceleration [tex]\alpha[/tex], the disk achieves an angular velocity [tex]\omega[/tex] at time [tex]t[/tex] according to
[tex]\omega=\alpha t[/tex]
and angular displacement [tex]\theta[/tex] according to
[tex]\theta=\dfrac12\alpha t^2[/tex]
a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of
[tex]21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}[/tex]
b. Under constant acceleration, the average angular velocity is equivalent to
[tex]\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2[/tex]
where [tex]\omega_f[/tex] and [tex]\omega_i[/tex] are the final and initial angular velocities, respectively. Then
[tex]\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}[/tex]
c. After 1.00 s, the disk has instantaneous angular velocity
[tex]\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}[/tex]
d. During the next 1.00 s, the disk will start moving with the angular velocity [tex]\omega_0[/tex] equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle [tex]\theta[/tex] according to
[tex]\theta=\omega_0t+\dfrac12\alpha t^2[/tex]
which would be equal to
[tex]\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}[/tex]
A submarine can withstand an external pressure of 63 atm before collapsing. If seawater has a density of 1027 kg/m^3 how deep can the sub dive?
Answer:
623.8 m
Explanation:
In 1909 Robert Millikan was the first to find the charge of an electron in his now-famous oil drop experiment. In the experiment tiny oil drops are sprayed into a uniform electric field between a horizontal pair of oppositely charged plates. The drops are observed with a magnifying eyepiece, and the electric field is adjusted so that the upward force q E on some negatively charged oil drops is just sufficient to balance the downward force m g of gravity. Millikan accurately measured the charges on many oil drops and found the values to be whole-number multiples of 1.6 × 10−19 C — the charge of the electron. For this he won the Nobel Prize. If a drop of mass 1.51837 × 10−12 kg remains stationary in an electric field of 1 × 106 N/C, what is the charge on this drop? The acceleration due to gravity is 9.8 m/s 2 . Answer in units of C.
Answer:
[tex]1.49\cdot 10^{-17}C[/tex]
Explanation:
The oil drop remains stationary when the electric force on it and the gravitational force are balanced, so we have:
[tex]F_E = F_G\\qE = mg[/tex]
where
q is the charge of the oil drop
E is the electric field strength
m is the mass of the drop
g is the acceleration due to gravity
here we have
[tex]E=1\cdot 10^6 N/C[/tex]
[tex]m=1.51837\cdot 10^{-12} kg[/tex]
[tex]g=9.8 m/s^2[/tex]
So the charge of the drop is
[tex]q=\frac{mg}{E}=\frac{(1.51837\cdot 10^{-12} kg)(9.8 m/s^2)}{1\cdot 10^6 N/C}=1.49\cdot 10^{-17}C[/tex]
A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if the upper electrode is 17.8 V more positive than the lower electrode. The density of the oil is 885kg/m3. Part A What is the droplet's mass? Express your answer to two significant figures and include the appropriate units. m m = nothing nothing Request Answer Part B What is the droplet's charge? Express your answer to two significant figures and include the appropriate units. q q = nothing nothing Request Answer Part C Does the droplet have a surplus or a deficit of electrons? How many? Does the droplet have a surplus or a deficit of electrons? How many? deficit 9 electrons surplus 9 electrons surplus 16 electrons deficit 7 electrons
A) [tex]2.4\cdot 10^{-16}kg[/tex]
The radius of the oil droplet is half of its diameter:
[tex]r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m[/tex]
Assuming the droplet is spherical, its volume is given by
[tex]V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3[/tex]
The density of the droplet is
[tex]\rho=885 kg/m^3[/tex]
Therefore, the mass of the droplet is equal to the product between volume and density:
[tex]m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg[/tex]
B) [tex]1.5\cdot 10^{-18}C[/tex]
The potential difference across the electrodes is
[tex]V=17.8 V[/tex]
and the distance between the plates is
[tex]d=11 mm=0.011 m[/tex]
So the electric field between the electrodes is
[tex]E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m[/tex]
The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:
[tex]qE=mg[/tex]
So, from this equation, we can find the charge of the droplet:
[tex]q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C[/tex]
C) Surplus of 9 electrons
The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is
[tex]q=-1.5\cdot 10^{-18}C[/tex]
we can think this charge has made of N excess electrons, so the net charge is given by
[tex]q=Ne[/tex]
where
[tex]e=-1.6\cdot 10^{-19}C[/tex] is the charge of each electron
Re-arranging the equation for N, we find:
[tex]N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9[/tex]
so, a surplus of 9 electrons.
What are the four types of macromolecules
Hello There!
The 4 types of macromolecules are
NUCLEIC ACIDS
PROTEINS
LIPIDS
CARBOHYDRATES
Answer:
1 2 3 4
Explanation:
What is the shortest distance between two points
Answer: displacement
Explanation:
According to the definition of displacement it is the shortest distance between two points.
Final answer:
The shortest distance between two points is a straight line, which is the displacement in physics. The Pythagorean theorem can be used to calculate this distance in a two-dimensional space. Displacement differs from the total distance traveled as it signifies the most direct path between two points.
Explanation:
The shortest distance between two points is often referred to as a straight line. This concept is not only a geometric truth but also has applications in physics, particularly when discussing displacement and distance traveled.
In a two-dimensional space, such as when navigating a city with a grid layout, the shortest path between two points can be visualized as the hypotenuse of a right triangle.
This forms the basis for utilizing the Pythagorean theorem, which is expressed as a² + b² = c², where a and b are the legs of the triangle and c is the hypotenuse. The theorem helps to quantify the straight-line distance between two points, providing a mathematical model for the physical concept of displacement.
Furthermore, in physics, the term 'displacement' is used to describe this shortest-path scenario between the starting and ending points, which differs from the total distance traveled, which accounts for the actual path taken, regardless of its directness.
proton with an initial speed of 800000 m/s is brought to rest by an electric field.
Part A- Did the proton move into a region of higher potential or lower potential?
Part B - What was the potential difference that stopped the proton?
ΔU = ________V
Part C - What was the initial kinetic energy of the proton, in electron volts?
Ki =_________eV
Best Answer
A) Into a region of higher potential
Explanation:
Let's remind that:
- Like charges repel each other
- Unlike charges attract each other
Here we have a proton, which is a positive charge, which is brought to rest by an electric field. This means that the electric field has slowed down the proton: so, the force exerted by the electric field on the proton was opposite to the direction of motion of the proton. But the lines of an electric field go from points at higher potential to points at lower potential - this means that the proton was actually moving towards a point at higher potential. (for example, it was moving towards another positive charge source of the field, so the potential increases as the proton approaches the source charge).
B) 3,338 V
The initial kinetic energy of the proton is given by:
[tex]K_i = \frac{1}{2}mv^2[/tex]
where
[tex]m=1.67\cdot 10^{-27} kg[/tex] is the proton mass
[tex]v=800,000 m/s=8\cdot 10^5 m/s[/tex] is the initial speed
Substituting,
[tex]K_i = \frac{1}{2}(1.67\cdot 10^{-27}kg)(8\cdot 10^5 m/s)^2=5.34\cdot 10^{-16}J[/tex]
When the proton is brought to rest, all this energy is converted into electric potential energy, given by
[tex]\Delta U = q \Delta V[/tex]
where
[tex]q=1.6\cdot 10^{-19} C[/tex] is the proton charge
[tex]\Delta V[/tex] is the potential difference
Since [tex]\Delta U = K_i[/tex], we can solve to find the potential difference:
[tex]\Delta V=\frac{K_i}{q}=\frac{5.34\cdot 10^{-16} J}{1.6\cdot 10^{-19} C}=3,338 V[/tex]
C) 3,338 eV
We already found the initial kinetic energy of the proton in part B), and it is given by
[tex]K_i =5.34\cdot 10^{-16}J[/tex]
Now we want to convert it into electron volts; keeping in mind the conversion factor between eV and Joules,
[tex]1 eV = 1.6\cdot 10^{-19}J[/tex]
we find:
[tex]K_i = \frac{5.34 \cdot 10^{-16} J}{1.6\cdot 10^{-19} J}=3,338 eV[/tex]
The proton moved into a region of higher potential. The potential difference that brought it to rest and its initial kinetic energy can be calculated using formulas and given values.
Explanation:Part A: The proton moved into a region of higher potential. This is because the electric field does work on the proton to bring it to a stop, which indicates that the proton moved against the direction of the electric field and hence into a region of higher potential.
Part B: The potential difference that stopped the proton can be calculated using the formula ΔU = ΔK/e, where ΔK is the change in kinetic energy and e is the charge of the proton. Given that the initial speed of the proton is 800000 m/s (which implies a kinetic energy of ½ mv^2), and knowing that the charge of a proton is 1.6 x 10^-19 C, you can solve this equation to find ΔU.
Part C: The initial kinetic energy of the proton can be calculated using the formula K = ½ mv^2. Converting this to electron volts (eV) involves dividing by the charge of an electron (e), which is also 1.6 x 10^-19 C.
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A runner moves 2.88 m/s north. She accelerates at 0.350 m/s^2 at a -52.0 angle. At the point in the motion where she is running directly east, what is Δx?
Answer:
Δx = 11.7 and Δy = 15
The question requires the use of kinematics and vector decomposition to calculate the horizontal displacement of a runner when she changes direction from the north to the east due to acceleration at a given angle.
The student is asking about the projection motion of a runner moving north, who accelerates at an angle. The question focuses on calculating the horizontal displacement (denoted as Δx) when the runner is running directly east. To solve this, one would have to break down the acceleration vector into its northward and eastward components and then use kinematic equations to determine the eastward displacement from the point of initial velocity to the point where the northward velocity component reaches zero and the runner is moving directly east.
At a distance of 0.75 meters from its center, a Van der Graff generator interacts as if it were a point charge, with that charge concentrated at its center. A test charge at that distance experiences an electric field of 4.5 × 10^5 newtons/coulomb. What is the magnitude of charge on this Van der Graff generator?
A. 1.7 × 10^-7 coulombs
B. 2.8 × 10^-7 coulombs
C. 3.0 × 10^-7 coulombs
D. 8.5 × 10^-7 coulombs
Answer: B
i tried putting explanation but its not working
Answer:
B. [tex]2.8 \times 10^{-5} C[/tex]
Explanation:
As we know that the electric field due to Van de graff generator is same as that of a point charge
so it is given by
[tex]E = \frac{kQ}{r^2}[/tex]
here we know that
[tex]E = 4.5 \times 10^5 N/c[/tex]
also we know that
[tex]r = 0.75 m[/tex]
now from above formula we have
[tex]4.5 \times 10^5 = \frac{(9\times 10^9)(Q)}{(0.75)^2}[/tex]
here we will have
[tex]Q = 2.8 \times 10^{-5} C[/tex]
Which of these stars has the coolest surface temperature?A. A starB. F starC. G starD. K star
Answer:
D. K star
Explanation:
Stars are classified into different groups according to their peak wavelength and their surface temperature.
In particular, we have the following group of stars, which correspond to the following surface temperatures:
Group O - Temperature > 25,000 K
Group B - Temperature 11,000 - 25,000 K
Group A - Temperature 7,500 - 11,000 K
Group F - Temperature 6,000 - 7,500 K
Group G - Temperature 5,000 - 6,000 K
Group K - Temperature 3,500 - 5,000 K
Group M - Temperature < 3,500 K
So among the options given, the star with the coolest surface temperature is star in group K.
The coolest surface temperature among the options is a K star. Ait depends on the surface temperature. The correct option is option (D).
Stars are categorized into different spectral classes based on their surface temperature.
The spectral classes are labeled with letters, starting from the hottest to the coolest: O, B, A, F, G, K, and M. So, a K star has a cooler surface temperature compared to an F star, a G star, and an A star.
Therefore, the correct option is option (D) K star has the coolest temperature.
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Why does temperature decrease with increasing altitude in the troposphere
In the troposphere, the temperature generally decreases with altitude. The reason is that the troposphere's gases absorb very little of the incoming solar radiation. Instead, the ground absorbs this radiation and then heats the tropospheric air by conduction and convection.
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A constant voltage is applied across a circuit. If the resistance in the circuit is doubled, what is the effect on the power dissipated by the circuit?
a.The power dissipated is quadrupled
b.The power dissipated is reduced by a factor of 2.
c.The power dissipated is reduced by a factor of 4
d. The power dissipated is doubled.
e. The power dissipated remains constant.
Answer:
b.The power dissipated is reduced by a factor of 2.
Explanation:
The power dissipated in the circuit is given by
[tex]P=\frac{V^2}{R}[/tex]
where
V is the voltage
R is the resistance
In this problem:
- The voltage V is kept constant
- The resistance is doubled, so R' = 2R
Therefore, the new power dissipated is
[tex]P'=\frac{V^2}{R'}=\frac{V^2}{2R}=\frac{1}{2}\frac{V^2}{R}=\frac{1}{2}P[/tex]
so, the power dissipated is reduced by a factor of 2.
Final answer:
When resistance is doubled and voltage is constant, the power dissipated by the circuit is halved. The power is proportional to the inverse of resistance, so doubling resistance reduces power by a factor of 2.
Explanation:
When the resistance in a circuit is doubled while keeping the voltage constant, the current in the circuit according to Ohm's Law (V = IR) will be halved, because I = V/R. The power dissipated by the circuit can be calculated using the formula P = V2/R. If the resistance is doubled, the new power dissipated becomes Pnew = V2/(2R), which is half the original power. Since the original power is Porig = V2/R, by doubling the resistance, the power is effectively reduced by a factor of 2, not 4.
Thus, the correct answer is: b. The power dissipated is reduced by a factor of 2.
The intensity of sunlight reaching the earth is 1360 w/m2. Assuming all the sunligh is absorbed, what is the radiation pressure force on the earth? Give your answers in (a) newtons (b) as a fraction of the sun's gravitational force on the earth
The radiation pressure force on the Earth due to the Sun's light is approximately 1.09 x 109 Newtons, and it is roughly 3.1 x 10-14 the Sun's gravitational force on Earth.
Explanation:The intensity of sunlight reaching the earth is given as 1360 W/m2. The pressure due to radiation can be calculated using the formula P = E/c, where P is pressure, E is energy and c is the speed of light. From this formula, we find that the radiation pressure force on Earth is roughly 4.53 x 10-6 N/m2.
(a) In Newtons:
The radiation pressure in Newtons will depend on the cross-sectional area exposed to sunlight. For the entire Earth, we would use the cross-sectional area of a circle with radius equals to Earth's radius (6.37 x 106 m). This gives us an approximate radiation pressure force of 1.09 x 109 Newtons.
(b) As a fraction of the Sun's gravitational force:
The gravitational force (Fg) between Earth and the Sun can be calculated using the formula Fg = GMm/r2, with G being the gravitational constant, M the mass of the Sun, m the mass of Earth and r the distance between Earth and the Sun. This gives an approximate Fg of 3.54 x 1022 Newtons. Hence, the radiation force is roughly 3.1 x 10-14 the Sun's gravitational force on Earth.
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The radiation pressure force on the Earth can be calculated using the intensity of sunlight and the surface area of the Earth. The force is approximately 1.09858256 × 10^7 N and is about 1.37 × 10^-38 times the Sun's gravitational force on the Earth.
Explanation:The radiation pressure force on the Earth can be calculated using the formula:
Force = Intensity × Area
Given that the intensity of sunlight reaching the Earth is 1360 W/m2 and assuming all the sunlight is absorbed, we can calculate the force:
Force = 1360 W/m2 × Area
The area of the Earth can be calculated using the formula for the surface area of a sphere:
Area = 4πr2
Where r is the radius of the Earth. Plugging in the value of the radius of the Earth, we can solve for the force:
Area = 4π(6371 km)Area = 4π(6.371 × 10^6 m)Area = 4π(6.371 × 10^6 m)Area = 4(3.14159)(6.371 × 10^6 m)Area = 4(20.0979 × 10^6 m)Area = 80.3916 × 10^6 mArea = 803.916 × 10^4 m2Force = 1360 W/m2 × 803.916 × 10^4 m2Force = 1098.58256 × 10^4 WForce = 1.09858256 × 10^7 WForce = 1.09858256 × 10^7 N(a) The radiation pressure force on the Earth is approximately 1.09858256 × 10^7 N.
(b) To find the force as a fraction of the Sun's gravitational force on the Earth, we need to calculate the gravitational force between the Sun and the Earth. Using Newton's law of gravitation:
Force = G × (mass of the Sun)(mass of the Earth) / (distance between the Sun and the Earth)^2
The mass of the Sun is about 2 × 10^30 kg. The mass of the Earth is about 6 × 10^24 kg. The average distance between the Sun and the Earth is about 1.5 × 10^11m. Plugging these values into the formula, we can calculate the gravitational force:
Force = (6.6743 × 10^-11 N m^2 / kg^2) × (2 × 10^30 kg)(6 × 10^24 kg) / (1.5 × 10^11m)^2Force = (6.6743 × 10^-11 N m^2 / kg^2) × (1.2 × 10^55 kg^2) / (2.25 × 10^22 m^2)Force = 8.00876033 × 10^44 N m^2 / kg^2The radiation pressure force as a fraction of the Sun's gravitational force on the Earth can be calculated as:
Fraction = (Radiation Pressure Force) / (Gravitational Force)
Fraction = (1.09858256 × 10^7 N) / (8.00876033 × 10^44 N)Fraction = 1.37 × 10^-38(b) The radiation pressure force on the Earth is approximately 1.37 × 10^-38 times the Sun's gravitational force on the Earth.
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Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed v at a distance r from the center of the earth. The second satellite travels at a speed that is less than v. At what distance from the center of the earth does the second satellite orbit?
Answer:
The second satellite will orbit at a larger distance
Explanation:
A satellite orbits the Earth due to its gravitational attraction to the Earth, which is equal to the centripetal force, so we can write
[tex]G\frac{Mm}{r^2}=m\frac{v^2}{r}[/tex]
where
G is the gravitational constant
M is the Earth's mass
m is the satellite's mass
r is the distance of the satellite from Earth's center
v is the speed of the satellite
We can rewrite the formula as
[tex]r=\frac{GM}{v^2}[/tex]
so we see that the distance of the satellite from the center of the Earth is inversely proportional to the square of the distance. This means that the second satellite, which travels at a lower speed, will have a larger distance from the centre of the Earth.
A satellite orbiting at a lesser speed than another identical satellite orbits at a greater distance from the center of the earth based on principles of orbital dynamics and Kepler's Second Law.
Explanation:The distance at which the second satellite orbits the earth, with a speed less than v, is greater than r. This is based on principles of orbital dynamics, which show a relationship between orbital speed and the distance from the center of the object being orbited. Looking at the gravitational force that supplies the centripetal acceleration for an orbiting object, we can see that as speed decreases, the gravitational force also decreases, meaning the object must be further from the center of gravity.
Take into consideration Kepler's Second Law, in that the satellite travels an equal area within equal times. If we consider two satellites orbiting, the one with a lesser speed will take a greater time to cover the same area, hence, it will be at a greater distance from the earth's center.
These observations are true for stable, circular orbits. Real world conditions might vary due to additional influences such as atmospheric drag, oblateness of the earth, and gravitational perturbations from the sun and moon.
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A 282 kg bumper car moving 3.50 m/s collides with a 155 kg bumper car moving -1.38 m/s. Afterwards the 282 kg car moves at 1.10 m/s. Find the velocity of the 155 kg car afterwards?
Answer:
4.03 m/s
Explanation:
Initial momentum = final momentum
(282 kg) (3.50 m/s) + (155 kg) (-1.38 m/s) = (282 kg) (1.10 m/s) + (115 kg) v
v = 4.03 m/s
We use the conservation of momentum to find the velocity of the 155 kg bumper car after the collision. Substituting the given values, we solve to get the final velocity as 2.99 m/s. The final velocity of the 155 kg car is 2.99 m/s.
The question is about a collision between two bumper cars and finding the velocity after the collision. We can use the principle of conservation of momentum which states that the total momentum before the collision is equal to the total momentum after the collision.
Let's define the variables:
Mass of first car, m₁ = 282 kgInitial velocity of first car, u₁ = 3.50 m/sMass of second car, m₂ = 155 kgInitial velocity of second car, u₂ = -1.38 m/sFinal velocity of first car, v₁ = 1.10 m/sFinal velocity of second car, v₂ = ?Using the conservation of momentum:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Substituting the given values into the equation:
(282 kg)(3.50 m/s) + (155 kg)(-1.38 m/s) = (282 kg)(1.10 m/s) + (155 kg)v₂
Simplifying:
987 kg·m/s - 213.9 kg·m/s = 310.2 kg·m/s + 155 kg·v₂
773.1 kg·m/s = 310.2 kg·m/s + 155 kg·v₂
Subtracting 310.2 kg·m/s from both sides:
462.9 kg·m/s = 155 kg·v₂
Solving for v₂:
v₂ = 462.9 kg·m/s / 155 kg = 2.99 m/s
Therefore, the final velocity of the 155 kg car after the collision is 2.99 m/s.
What causes charges to move in a circuit?
Charges move in a circuit due to the presence of an electrical field created by a voltage difference. The electrical field exerts forces on charged particles, causing them to accelerate and move through the circuit.
Explanation:Circuit charges refer to the movement of electric charge (usually electrons) through an electrical circuit. In a closed circuit, charges flow due to voltage (potential difference), creating an electric current. This flow of charges powers electrical devices and is described by Ohm's law, which relates current, voltage, and resistance in the circuit.
Charges move in a circuit due to the presence of an electrical field created by a voltage difference. The electrical field exerts forces on charged particles, causing them to accelerate and move through the circuit. As charges move, they lose potential energy and gain kinetic energy, traveling from an area of higher potential to an area of lower potential.
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Charges move in a circuit primarily due to the electrical field created by a voltage difference. The electrical field forces the charges, often free electrons, to accelerate, creating an electrical current and eventually reaching a constant 'drift velocity'. The presence and strength of a magnetic field can also affect the flow of charges.
Explanation:In a circuit, charges move due to an electrical field created by a voltage difference, such as a battery. This voltage difference exerts forces on the free electrons, causing them to accelerate and thus creating an electrical current. The exact rate at which these charges flow, meaning the amount of charge per unit of time, is influenced by factors such as the voltage applied, the state and type of the material (conductor or insulator), and the strength of the electrical field.
In the case of a conducting material, which is often the type of material used in circuits, the electrical field forces charge to flow and lose kinetic energy in the process until it reaches a constant velocity, known as the 'drift velocity'. This is essentially an equilibrium state where charges are constantly moving due to the force provided by the electrical field, but their speed or kinetic energy does not increase due to interactions with atoms and free electrons, similar to an object falling and reaching its terminal velocity.
It's also important to acknowledge here the role of a magnetic field, which can also impact the movement of charges in a conductor, causing a change in the direction of the electron flow, and hence, influencing the current within the circuit.
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What can one say about the image produced by a thin lens that produces a positive magnification?
Answer:
The image is virtual and upright
Explanation:
The magnification of a lens can be written as follows:
[tex]M=\frac{y'}{y}=-\frac{q}{p}[/tex]
where
y' is the size of the image
y is the size of the object
q is the location of the image with respect to the lens
p is the location of the object with respect to the lens
In this situation, the magnification is positive. This means that:
- y' (the image) has same sign as y (the object) --> the image is upright (same orientation as the object)
- q has opposite sign to p --> this means that the image is located on the same side as the object, so it is a virtual image
Which term best describes how light waves from a star are affected as the star moves toward earth?
A.Diffraction
B.Redshift
C.Blueshift
D.Cosmological Redshift
Answer:
Answer:C.
Explanation:
Blueshift"If a star is moving towards the earth, its light is shifted to higher frequencies on the color spectrum (towards the green/blue/violet/ultraviolet/x-ray/gamma-ray end of the spectrum). A higher frequency shift is called a "blue shift"
Blueshift is the best term to describe how light waves from a star are affected as the star moves toward earth.
What is Blueshift?Blueshift is the displacement of the spectrum to shorter wavelengths in the light coming from distant celestial objects moving toward the observer.Blueshift is an example of doppler effect.How light waves from a star are affected as the star moves toward earth?Due to blueshift light wave coming from the star shift to longer wavelength to shorter wavelength.As a result we see stars blinking and shining.
So, the correct answer is option C blueshift.
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________ are more likely to be found near rural communities due to the large requirement for space.
Military bases
Farms
Domed stadiums
Coal mines
Complex highway interchanges
Military bases and Farms, Domed stadiums are more likely to be found near rural communities due to the large requirement for space.
What is Rural Communities?
A rural area is an expanse of open ground with few houses or other structures and few inhabitants. The population density in a rural location is very low.
A rural area is an expanse of open ground with few houses or other structures and few inhabitants. The population density in rural areas is quite low. Numerous individuals reside in urban or suburban areas. Their residences and places of business are situated close together.
Most rural communities' main industry is agriculture. On farms or ranches, the majority of people reside or work.
Therefore, Military bases and Farms, Domed stadiums are more likely to be found near rural communities due to the large requirement for space.
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Which law states that each planet revolves so that an imaginary line connecting it to the sun sweeps over equal areas in equal time intervals?
Kepler’s Laws are three mathematic laws to describe the movement of the planets around the Sun, but it can be generalized for the movement of any body orbiting a bigger one, for example, The Moon orbiting the Earth.
These laws were formulated by the astronomer Johannes Kepler from observations made by the Danish astronomer Tycho Brahe of the orbit of Mars.
Now, according to the Second Kepler’s Law of Planetary motion:
In equal times, the areas swept by the planet in its orbit around the Sun are equal.
For this to be possible, the speed of the planet must vary. Hence, the planet will move rapidly near the Sun (perihelion) and move slowly when it is away from the Sun (aphelion).
Hydrogen-2 and Hydrogen-3 fuse to form Helium-4 and a neutron. How much energy is released in this nuclear reaction?
Final answer:
In a nuclear fusion reaction, when hydrogen-2 and hydrogen-3 combine to form helium-4 and a neutron, a certain amount of energy is released. The exact amount of energy released can be calculated using the equation E = mc^2, where E is the energy, m is the change in mass, and c is the speed of light.
Explanation:
In a nuclear fusion reaction, when hydrogen-2 (deuterium) and hydrogen-3 (tritium) combine to form helium-4 and a neutron, a certain amount of energy is released. The exact amount of energy released can be calculated using the equation E = mc2, where E is the energy, m is the change in mass, and c is the speed of light.
Based on the given information, we can calculate the change in mass by subtracting the mass of the reactants from the mass of the products. The mass of deuterium (hydrogen-2) is 2 grams, the mass of tritium (hydrogen-3) is 3 grams, the mass of helium-4 is 4 grams, and the mass of a neutron is negligible. Therefore, the change in mass is 2 grams + 3 grams - 4 grams = 1 gram.
Using the equation E = mc2, where c is the speed of light (approximately 3 x 108 m/s), we can calculate the energy released:
E = (1 gram) x (3 x 108 m/s)2 = 9 x 1016 joules
When is the angular momentum of a system constant?
a. When the total kinetic energy is constant.
b. When no net external force acts on the system.
c. When the linear momentum and the energy are constant.
d. When no net torque acts on the system.
e. When the moment of inertia is constant.
The angular momentum of a system constant when no net torque acts on the system. The correct option is d.
When there is no external torque operating on a system in a net way, there will be no change in the system's angular momentum. The principle of the conservation of angular momentum is one of the most fundamental principles in all of physics. The rotational motion of an object or a system of objects can be described using a vector quantity known as the angular momentum of the system.
When there is no net external torque acting on a system, the total angular momentum of the system will not change over time; this property will be known as the system's inertia. This indicates that if an item or system is originally at rest or has a particular angular momentum, it will keep that angular momentum unless an external torque is applied to it and causes it to rotate in the opposite direction.
The other options (a,b,c, and e) do not guarantee continuous angular momentum. Even if some of those conditions could result in particular outcomes for the system, the conservation of angular momentum requires that the system have no net external torque.
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When the core of a star like the Sun uses up its supply of hydrogen for fusion, the core begins to ________.
Answer:
Shrink and heat
Explanation:
Hope my answer has helped you!
Answer:
Shrink and heat up
Explanation:
When the core of a star like the Sun uses up its supply of hydrogen for fusion, the core begins to contract or shrink up which leads to release of energy from the core.
The released energy starts to heat up core until it has gotten to the point where the core is hot enough to able start up the fusion of hydrogen into another element (helium).
The layers on the outside of the sun absorbs the released the energy and starts to enlarge or swell up and the sun develops a very high luminosity which means it start to shine brighter and brighter.
As the outside large swells up, due to the absorption of the released energy, it start to become cool thereby causing a low surface temperature.
It is at the stage that the sun becomes a red giant.
A drunk guy walking on a narrow lane takes 8 steps forward and 6 steps backward and so on.
Each step is 1m long and he requires 1 second to travel. The time that the drunk guy takes to fall in a pit 18m away from the starting point is
a) 18s
b) 126s
c) 78s
d) 62s
Answer: 126 secs
Explanation: he gains 2 steps every time and it takes 9 2 steps to reach 18 meters so what u do is take that 9 and multiply it to 14 (8+6) giving u ur answer of 126 secs.