Answer:
B. During condensation, the heavier elements tended to sink nearer the Sun and, being rare, only provided enough material to build the relatively small terrestrial planets.
D. During the collapse of the gaseous nebula, most of the material tended to collect far from the Sun because of the large centrifugal forces, which provided the necessary material to build the large Jovian planets.
Explanation:
The solar nebular hypothesis is the theory of how the earth came to be as proposed and believed by most scientists. It states that the solar system was formed from a nebular which is a large collection of dust. As the condensation of the nebular progressed, the heavier materials which could withstand higher temperatures formed closer to the sun. Because these materials were not so much the planets formed by them were not large in size.
However, the Jovian planets formed beyond the frost line which could sustain icy compounds. The ices that formed these planets were much thus accounting for their larger sizes. Solar winds blew off the lighter hydrogen and helium elements leading to their accumulation far away from the sun.
The aurora borealis is caused by the ____.
A. mesosphere
B. stratosphere
c. thermosphere
D. troposphere
Answer: ionosphere
Explanation: aurora boreal is is caused by the ionosphere which is a part of the thermosphere
A coil has an inductance of 6.00 mH, and the current in it changes from 0.200 A to 1.50 A in a time interval of 0.300 s. Find the magnitude of the average induced emf in the coil during this time interval.
Answer:
0.026 V
Explanation:
Given that,
Inductance of the coil, L = 6 mH
The current changes from 0.2 A to 1.5 A in a time interval of 0.3 s
We need to find the magnitude of the average induced emf in the coil during this time interval. The formula for the induced emf is given by :
[tex]\epsilon=L\dfrac{dI}{dt}\\\\=6\times 10^{-3}\times \dfrac{1.5-0.2}{0.3}\\\\=0.026\ V[/tex]
So, the magnitude of induced emf is 0.026 volts.
For the common-emitter, common-base and emitter-follower amplifier designs,what is the primary benefit of each amplifier?
Answer:
Explanation:
A common emitter amplifier works by inverting. It does have low input impedance. Despite its low input impedance, it poses an otherwise high output impedance.
The common base circuit performs optimally when it acts as a current buffer. It has the ability to take an input current at a low input impedance, and transmit almost the same current to an impedance with a higher output
The primary benefit of emitter follower amplifier is that the transistor is able to provide current and power gain. Although this transistor takes in little current from the input. It still provides an impedance with a low output to a circuit by exercising the output of the follower. This then translates to that the output under load not dropping.
A child wants to throw a rock upward to try to strike a plump apple up in a
tree. If the fastest the child can throw the rock is 7 m/s, what is the highest
the apple can be in order for the child to be able to strike the apple?
Answer:
7m because that is the height that the child can throw.
What is the acceleration due to gravity on a planet with a mass of 4.76x10^23 kg and a radius of 2,635,000 meters? Report your answer to three digits and include the units. (Ex: 7.15 m/s/s).
Answer:
g=4.5727 m/s^2
Explanation:
[tex]We\ know\ that\ , \\F=G\frac{Mm}{d^2} \\As\ F=ma\ and\ a=g\ (g\ is\ the\ acceleration\ caused\ due\ to\ gravity),\\mg=G\frac{Mm}{d^2}\\Canceling\ m\ on\ both\ sides\ we\ get,\\g=G\frac{M}{d^2}\\[/tex]
[tex]We\ are\ given\ that\ ,\\Mass\ of\ the\ planet\ (M) = 4.76*10^{23} kg\\Distance(d) = 2,635,000 m\\G(Universal\ Gravitation\ Constant) = 6.67*10^{-11}\\Hence,\\g=6.67*10^{-11} (\frac{4.76*10^{23}}{2,635,000^2} )\\g= 4.5727 m/s^2[/tex]
An object in motion, will tend to remain in motion until acted upon my an external, unbalanced force is defined to be
Answer:
The first law states that a body at rest will stay at rest until a net external force acts upon it and that a body in motion will remain in motion at a constant velocity until acted on by a net external force. Inertia is the tendency of a body in motion to remain in motion.
Explanation:
A 5.0 kg hammer strikes a 0.25 kg nail with a force of 10.0 N causing the nail to accelerate at 40.0 m/s^2. What is the acceleration of the hammar?
Answer:
2 m/s².
Explanation:
The following data were obtained from the question:
Mass (m) of hammer = 5 Kg
Mass (m) of nail = 0.25 kg
Force (F) applied = 10 N
Acceleration (a) of nail = 40 m/s².
Acceleration (a) of hammer =?
From Newton's third law which states that to every action, there is an equal but opposite reaction. This implies that the force applied by the hammer on the nail is exactly the force applied by the nail on the hammer. Thus, we can obtain the acceleration of the hammer as follow:
Mass (m) of hammer = 5 Kg
Force (F) applied = 10 N
Acceleration (a) of hammer =?
F = ma
10 = 5 × a
Divide both side by 5
a = 10/5
a = 2 m/s²
Thus, the acceleration of the hammer is 2 m/s².
A uniform solid cylindrical log begins rolling without slipping down a ramp that rises 28.0° above the horizontal. After it has rolled 4.20 m along the ramp, what is the magnitude of the linear acceleration of its center of mass?
Answer:
3.07 m/s
Explanation:
Given:
Angle, [tex]\Theta = 28^{\circ}[/tex]Length = 2.40 mThe acceleration of its center of mass will be:
= [tex]\frac{2}{3}g Sin \Theta[/tex]
By putting the above given values, we get
= [tex]\frac{2}{3}\times 9.8 Sin 28^{\circ}[/tex]
= [tex]\frac{2}{3}\times 9.8\times 0.46[/tex]
= [tex]\frac{9.0167}{3}[/tex]
= [tex]3.07 \ m/s^2[/tex]
Thus the solution above is appropriate.
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Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed on a screen 3.35 m away. Define the width of a bright fringe as the distance between the minima on either side.
(a) What is the width of the central bright fringe?
(b) What is the width of the first bright fringe on either side of the central one?
Answer:
a)[tex]y_{first}=5.3mm[/tex]
b)[tex]y_{second}=10.6-5.3 =5.3 mm[/tex]
Explanation:
a)
The width of the central bright in this diffraction pattern is given by:
[tex]y=\frac{m\lambda D}{a}[/tex] when m is a natural number.
here:
m is 1 (to find the central bright fringe) D is the distance from the slit to the screen a is the slit wide λ is the wavelengthSo we have:
[tex]y_{first}=\frac{633*10^{9}*3.35}{0.0004}[/tex]
[tex]y_{first}=5.3mm[/tex]
b)
Now, if we do m=2 we can find the distance to the second minima.
[tex]y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}[/tex]
[tex]y_{2}=10.6 mm[/tex]
Now we need to subtract these distance, to get the width of the first bright fringe :
[tex]y_{second}=10.6-5.3 =5.3 mm[/tex]
I hope it heps you!
how does the law of conservation energy relate to the system shown in thos model
Answer:
there is no model shown
Explanation:
Therefore it can be thousands of different answers
Steven tries to sneak a look into the freezer to see what kind of ice cream cake he will be eating for his birthday tomorrow. Which of the following best describes what is happening when Steven opens the freezer door right before his dad catches him?
A-Heat energy flows from the kitchen to the freezer
B-Heat energy flows from the freezer to the kitchen
C- Cold energy flows from the kitchen into the freezer
D- Cold energy flows from the freezer into the kitchen
As friction with our atmosphere causes a satellite to spiral inward, closer to Earth, its orbital speed increases. Why?
Answer:
Explanation:
When we go back and look at Kepler's third law. Kepler's third law states that "the square of the period is proportional to the cube of the semi-major axis of the orbit". P²µa³
where a is the semi-major axis of orbit,
P is the period of orbit
The orbit in question is the distance from the center of Earth. If the satellite spirals in as a result of air friction, then it decreases and, so does P.
Note that P = 2πa / V or Vµ1 / a0.5 .
From the formula, we can see that a decrease in the value of a, will lead to an Increase in the value of V.
Which material will heat up the most quickly if placed near a heat source
Answer:
metal
Explanation:
metal heats up the fastest when near a hot source
Developing a question for appointments
Explanation:
how can I help you that?
Is clinical depression associated with eating disorders yes or no
Yes, a symptom of depression is a loss of appetite, and eating disorders are often developed as a result of an unhealthy coping mechanism with life; therefore, people with depression can get eating disorders and vice versa. They go hand in hand.
A force of 35.0 N is required to start a 6.0-kg box movingacross a horizontal concrete floor. (a) What is the coefficientof static friction between the box and the floor?(b) if the 35.0-n force continues, the box accelerates at 0.60 m/s2 . What is the coefficient of kinetic friction.
Answer:
Explanation:
Frictional force acting on the box = 35 N
If μ₁ be the coefficient of static friction and μ₂ be the kinetic friction
μ₁ mg = 35
μ₁ = 35 / 6 x 9.8
= .6
b )
net force acting on box to produce acceleration of .6 m /s²
= 6 x .6 = 3.6 N
net force acting on the box when it is accelerating
= 35 - μ₂ mg
35 - μ₂ x 6 x 9.8 = 3.6
35 - μ₂ x 58.8 = 3.6
μ₂ = .53
Referring to the sketch of a planet around the sun, Area A is three times that of Area B. Compare the times required for the planet to travel from Point 1 to Point 2 and from Point 3 to Point 4 and select the letter of the correct answer.
Answer:
tA is three times tB
Explanation:
tA is three times tB
Referring to the sketch of a planet around the sun, Area A is three times that of Area B. Compare the times required for the planet to travel from Point 1 to Point 2 and from Point 3 to Point 4 and select the letter of the correct answer.
In a liquid metal fast breeder reactor, no neutron moderation is desired and sodium is used as a coolant to minimize fission-neutron thermalization. How many elastic scatters with sodium, on the average, would it take for 2-MeV neutrons to reach an average thermal energy of 0.025 eV?
Answer:
219 scatterings
Explanation:
Given that:
The Coolant used In the liquid metal fast breed reactor = Sodium
The atomic weight (A) of sodium = 23
The initial energy [tex]E_{i}[/tex] = 2 - MeV
The final energy [tex]E_{f}[/tex] = 0.025 eV (thermal energy)
The number of elastic neutron scatterings (n) needed to reach the given average thermal energy can be computed as:
[tex]n = \dfrac{log \bigg(\dfrac{E_f}{E_i} \bigg)}{log \bigg [ \dfrac{A^2+1}{(A+1)^2} \bigg]}[/tex]
[tex]n = \dfrac{log \bigg(\dfrac{0.025}{2 \times 10^6} \bigg)}{log \bigg [ \dfrac{23^2+1}{(23+1)^2} \bigg]}[/tex]
[tex]n = \dfrac{log \bigg(1.25\times 10^{-8} \bigg)}{log \bigg [ 0.92014\bigg]}[/tex]
[tex]n = 218.643[/tex]
n ≅ 219 scatterings
An athlete whirls a 6.41 kg hammer tied to the end of a 1.3 m chain in a simple horizontal circle where you should ignore any vertical deviations. The hammer moves at the rate of 1.64 rev/s. What is the centripetal acceleration of the hammer
The chain makes a circle of radius 1.3 m as it spins, so that 1 revolution of the hammer amounts to it moving a linear distance of 2π (1.3 m) = 2.6π m. Then the hammer's linear speed is
(1.64 rev/s) • (2.6π m/rev) = 4.264π m/s
and so the centripetal acceleration has magnitude
a = v²/r = (4.264π m/s)² / (1.3 m)
a ≈ 140 m/s²
A drop of water dripping from the faucet in your kitchen is electrically neutral. A charged plastic hair comb can attract the drips and drops by the process of __________________.
Answer:
Polarizatión
Explanation:
Polarization is that property of waves that particular to transverse waves. Ir specifies the geometrical orientation of the oscillations. In transverse waves, the direction of the oscillation is often at perpendicular with respect to the direction of motion of the wave itself.
Again, it can be called the property of electromagnetic radiations to which the direction and magnitude of the vibrating electric field are related in a particular way.
Two light bulbs have resistances of 400Ω and 800Ω. The two light bulbs are now connected in parallel across the 120-V line. A. Find the current through each bulb.B. Find the power dissipated in each bulb. C. Find the total power dissipated in both bulbs.
Answer:
a.) O.3A and 0.15A
b.) P1 = 36W, P2 = 18W
c.) 54W
Explanation:
Constant resistance R1 = 400 ohms, R2 = 800 ohms
a.)
The current through each bulb
While in parallel.
I = 120/800/3
= 0.45A
For 400= 800/400+800
= 800/1200 x 0.45
= 0.3A
I-800 = 400 /400+800
= 400/1200*0.45
= 0.15A
B.)
The power dissipated in each bulb
P409 = I²400r
= 0.3A²x400 ohms
= 36 W
P800 = I²800R
= 0.15A²*800 ohms
= 18W
C.)
The total power dissipated
= 36W + 18 W
= 54W
How fast would a(n) 85 kg man need to run in order to have the same kinetic energy as an 8.0 g bullet fired at 410 m/s?
Answer:
V = 3.97 m/s
Explanation:
Mass of a man, M = 85 kg
Mass of a bullet, m = 8 g = 0.008 kg
Speed of bullet, v = 410 m/s
We need to find the speed of a man in order to have the same kinetic energy as that of the bullet. Let the kinetic energy of the bullet is k. So,
[tex]k=\dfrac{1}{2}mv^2\\\\k=\dfrac{1}{2}\times 0.008\times 410^2\\\\k=672.4\ J[/tex]
Since, k = K (K is the kinetic energy of the man and Let V is the speed)
[tex]K=\dfrac{1}{2}MV^2\\\\V=\sqrt{\dfrac{2K}{M}} \\\\V=\sqrt{\dfrac{2\times 672.4}{85}} \\\\V=3.97\ m/s[/tex]
So, the speed of the man is 3.97 m/s.
We have that for the Question "How fast would a(n) 85 kg man need to run in order to have the same kinetic energy as an 8.0 g bullet fired at 410 m/s?" it can be said that the speed is
v=4.0m/s
From the question we are told
How fast would a(n) 85 kg man need to run in order to have the same kinetic energy as an 8.0 g bullet fired at 410 m/s?
Generally the equation for the kinetic energy is mathematically given as
[tex]K.E=1/2*m*v^2\\\\Therefore\\\\K.E=1/2*0.008*(420)^2\\\\K.E=672.4J\\\\[/tex]
Therefore
[tex]v^2=\frac{672.4*2}{85}\\\\v^2=15.821[/tex]
v=4.0m/s
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For the winter, a duck flies 10.0 m/s due south against a gust of wind with a velocity of 2.5 m/s. What is the resultant velocity of the duck?
a. 12.5 m/s south
b. 12.5 m/s north
c. 7.5 m/s south
d. 7.5 m/s north
e. 10.3 m/s south
Answer:
c. 7.5 m/s south
Explanation:
Given the following data;
Velocity of duck = 10m/s due South
Velocity of wind = 2.5m/s
To find the resultant velocity;
Since we know that the duck is flying against a gust of wind, we would have to subtract the velocity of the gust of wind from that of the duck.
This ultimately implies that, the gust of wind (headwind) would decrease the resulting velocity of the duck because it approaches the duck from the front.
Resultant velocity, /V/ = 10 - 2.5
Resultant velocity, /V/ = 7.5m/s South.
Therefore, the resultant velocity of the duck is 7.5m/s south.
Suppose a rocket is traveling through space and moving at a speed close to the speed of light. Three minutes pass on the rocket as recorded by an observer on Earth.
How many minutes pass on Earth?
less than three minutes because of time dilation
more than three minutes because of time dilation
three minutes because time is absolute
less than or greater than three minutes depending on the speed
Answer:
B
Explanation:
on edge2020
Three minutes pass on the rocket as recorded by an observer on Earth will be more than three minutes because of time dilation. The correct option is B.
What is time dilation?It is a phenomenon a slowing of time according to the theory of relativity when a system is in motion relative to an outside observer and that becomes apparent when the speed of the system approaches that of light is known as time dilatation.
The time dilation results in Three minutes passing on the rocket as recorded by an observer on earth will be more than three minutes.
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In a mercury barometer at atmospheric pressure, the height of the column of mercury in a glass tube is 760 mm. If another mercury barometer is used that has a tube of larger diameter, how high will the column of mercury be in this case?
a. greater than 760 mm
b. Less than 760 mm
c. equal to 760 mm
Answer:
c. equal to 760 mm
Explanation:
We are told that another mercury barometer is used that has a tube of larger diameter. This means a larger area and the weight of the liquid in the tube will have increased since volume = area × height. Also, due to the larger area, the net upward force on the mercury will have also increased by the same amount because force = area × pressure. Therefore, as long as the pressure remains the same, the height of the mercury will also remain the same.
Thus the height of the mercury = 760 mm
5. Salt is often spread on roads during winter to prevent ice from forming. Do you think sugar would
work as well? Explain your answer.
Answer:
No because sugar is more sweet and it would not have the benefits of salt. Think of it like how people used to put salt on meat to preserve it. Would they put sugar on? No they would not because salt is a combination of things that sugar does not have.
Explanation
Consider a one-dimensional crystal (similar to a carbon nanowire) with length 10 um and lattice spacing 0.1 nm.
RequireD:
a. What is the Fermi level assuming one electron per atom?
b. What is the density of states as a function of electron energy?
Answer:
a) Fermi level = 600 electron-volts
b) [tex]\frac{2.04 * 10^{13} }{\sqrt{E} }[/tex]
Explanation:
Given data:
length of one-dimensional crystal = 10 um
Lattice spacing = 0.1 nm
A) Determine the Fermi level assuming one electron per atom
Total length = 10 um
Interatomic separation of a = 0.1 nm
in this case the Atom has one electron therefore the number of electrons = 10^5 and the number of states Ns = gsN = 2 * 10^5 ( attached below is some part of the solution )
hence : Fermi level = 600 electron-volts
B) Determine the density of states as a function of electron energy
attached below is the detailed solution
A 20-kg child is coasting at 3.3 m/s over flat ground in a 4.0-kg wagon. The child drops a 1.0-kg ball out the back of the wagon. What is the final speed of the child and wagon?
Answer:
The final velocity is [tex]u_f = 3.44 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the child is [tex]m_1 = 20 \ kg[/tex]
The initial speed of the child is [tex]u_1 = 3.3 \ m/s[/tex]
The mass of the wagon is [tex]m_w = 4.0 \ kg[/tex]
The initial speed of the wagon is [tex]u_w = 3.3 \ m/s[/tex]
The mass of the ball is [tex]m_2 = 1.0 \ kg[/tex]
The initial speed off the ball is [tex]u_2 = 3.3 \ m/s[/tex]
Generally the initial speed of the system (i.e the child , wagon , ball) is
[tex]u_1 = u_w = u_2 = u =3.3 \ m/s[/tex]
Generally from the law of linear momentum conservation
[tex]p_i = p_f[/tex]
Here [tex]p_i[/tex] is the momentum of the system before the ball is dropped which is mathematically represented as
[tex]p_i = ( m_1 + m_2 + m_3 ) * u[/tex]
=> [tex]p_i = ( 20 + 4 + 1 ) * 3.3[/tex]
=> [tex]p_i = 82.5 \ kg \cdot m/s[/tex]
and
[tex]p_f[/tex] is the momentum of the system after the ball is dropped which is mathematically represented as
[tex]p_f = ( m_1 + m_w ) * u_f[/tex]
=> [tex]p_i = ( 20 + 4 ) * u_f[/tex]
So
[tex]82.5 = 24 * u_f[/tex]
=> [tex]u_f = 3.44 \ m/s[/tex]
PLEASE HELP!!
1.The Law of Conservation of Energy states that?
A. matter cannot be created or destroyed, only stored and transferred.
B. momentum cannot be created or destroyed, only stored and transferred.
C. energy cannot be created or destroyed, only stored and transferred.
2.The total amount of kinetic and potential energy in a system that remains constant is the system's
A .mechanical energy
B. elastic energy
C. chemical energy
D. gravitational energy
3, At the top of a giant hill sits a roller coaster cart, all the energy of the cart is stored as
A. kinetic energy
B. chemical energy
C. elastic energy
D. gravitational potential energy
4. At the bottom of the hill (ground level), the roller coaster cart in the previous problem, all the energy of the cart is stored as
A. kinetic energy
B. chemical energy
C. elastic energy
D. gravitational potential energy
When a pendulum swings when will the potential energy be at a maximum?
A
B
C
D
6. If total mechanical energy = kinetic energy + potential energy, TME = KE + PE, and a roller coaster has 500 J of total mechanical energy. At the very beginning of the track the roller coaster has 500 J of potential energy, how much kinetic energy is there at the beginning of the track?
A 25 kg child is sitting at the top of a 4 m tall slide, what is his potential energy?
It the same child has a velocity of 2 m/s half-way down the slide, what is his kinetic energy?
If the TME energy for the same child on the slide is 100 J, how much potential energy does the child have half-way down the slide?
Answer:
A
A
D
D
when the pendulum is at the lefend and right end.
0 because kinetic energy is only formed when a moving object is moving.
dont know the others sorry
Explanation:
i got about half way for you hope it correct
give me nothing if you got some wrong!
or just give me a message saying that i didn't help
A. If an electron of 16 eV had a head-on collision with a Cs atom at rest, what would be the kinetic energy in eV) of the recoiling Cs atom? Assume an elastic collision. (Take the atom to be the most abundant isotope of the element. B. In a Frank-Hertz experiment, the first excited state of Hg is obtained at an accelerating voltage measured to be 4.87 V. Based on this measured value, determine the wavelength of the ultraviolet radiation you expect to be emitted.
Answer:
A) 1.96 * 10^-8 J
B) 2.55 * 10^-7 m
Explanation:
A) calculate the kinetic energy in eV of the recoiling Cs atom
we have to apply the principle of energy conservation and momentum
Initial kinetic energy of electron = 16 eV = 16 * (1.6 * 10^-19 ) J
Initial kinetic energy of atom = 0
therefore the final kinetic energy after collision ( E )
= [ 16 * ( 1.6 * 10^-19 ) ] + 0
= 1.96 * 10^-8 J
B) Determine the wavelength of the ultraviolet radiation
accelerating Voltage = 4.87 V
K.E = eV = 1.6 * 10^-19 * 4.87
next we will apply Planck's relationship
[tex]\frac{hc}{w} = KE[/tex] --------- ( 1 )
w = wavelength
h = 6.64 * 10^-34 J-S ( Planck's constant )
c = 3 * 10 ^8 m/sec ( speed of light )
KE = 1.6 * 10^-19 * 4.87
substitute given values into equation 1 above
w ( wavelength ) = 2.55 * 10^-7 m