Final answer:
To determine the height a rocket needs to go above the Earth's surface so that its weight is reduced to 58.8% of its weight at the Earth's surface, we can use the concept of gravitational force.
Explanation:
To determine the height a rocket needs to go above the Earth's surface so that its weight is reduced to 58.8% of its weight at the Earth's surface, we can use the concept of gravitational force. The force of gravity between two objects is given by the equation: F = (G * m1 * m2) / r^2, where G is the universal gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between the objects. We can set up a ratio of the weight of the rocket at a certain height to its weight at the Earth's surface:
Weight at height / Weight at surface = (G * m1 * m2) / (r + h)^2 / (G * m1 * m2) / r^2
By substituting the given values, such as the radius of the Earth and the weight reduction percentage, we can solve for h, which represents the height the rocket needs to reach. The final answer should be in units of kilometers, which can be obtained by converting the radius of the Earth from meters to kilometers.
In outer space, a rod is pushed to the right by a constant force F. Describe the pattern of interatomic distances along the rod. Include a specific comparison of the situations on the left end of the rod, the middle, and the right end. (Hint: Consider how an individual atom in the rod moves.)
Answer:
According to Newtonian Mechanics, when you push the rod with a force F, a work is generated that pushes the individual atoms forward. But all atoms connected to that individual atoms will move along because of the bonding present in solids. At near absolute zero temperatures in space, the molecules of solids gets more confined to their atomic position and the work-done generated by the force F will be enough to push the rod with the initial v forever.
A steel ball is whirled on the end of a chain in a horizontal circle of radius R with a constant period T. If the radius of the circle is then reduced to 0.75R, while the period remains T, what happens to the centripetal acceleration of the ball?
When the radius of the circle is reduced while the period remains constant, the centripetal acceleration of the ball increases.
Explanation:When the radius of the circle is reduced to 0.75R while the period remains T, the centripetal acceleration of the ball increases.
Centripetal acceleration is given by the formula ac = v^2 / r, where v is the tangential velocity and r is the radius of the circle. Since the period T remains constant and the radius is reduced, the velocity of the ball must increase in order to cover the smaller circumference in the same amount of time, resulting in an increase in centripetal acceleration.
The Nichrome wire is replaced by a wire of the same length and diameter, and same mobile electron density but with electron mobility 4 times as large as that of Nichrome. Now what is the electric field inside the wire?
Answer:
The electric field inside the wire will remain the same or constant, while the drift velocity will by a factor of four.
Explanation:
Electron mobility, μ = [tex]\frac{v_d}{E}[/tex]
where
[tex]v_d[/tex] = Drift velocity
E = Electric field
Given that the electric field strength = 1.48 V/m,
Therefore since the electric potential depends on the length of the wire and the attached potential difference, then when the electron mobility is increased 4 times the Electric field E will be the same but the drift velocity will increase four times. That is
4·μ = [tex]\frac{4*v_d}{E}[/tex]
Answer:
Explanation:
Usually, the electron drift velocity in a material is directly proportional to the electric field, which means that the electron mobility is a constant (independent of electric field).
μ × E = Vd
Where,
D = electric field
Vd = drift velocity
μ = electron mobility
μ2 = μ × 4
μ/Vd1 = 4 × μ/Vd2
Vd2 = 4 × Vd1
The electric field is the same but drift velocity increases.
The central atom in a lewis structure tends to be the ____________ atom.
Answer:
The central atom in a lewis structure tends to be the least electronegative atom.
Explanation:
Lewis structures show each atom and its position in the structure of the molecule using its chemical symbol.
In Lewis structure, the central atom tends to be the least electronegative atom. Usually the central atom will be the one that has the most unpaired valence electrons or least electronegative.
Therefore, the best phrase that completes the sentence is "least electronegative".
Answer:
The central atom is also usually the least electronegative.
Explanation:
The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal.
The least electronegative elements in the center of lewis structures because an atom in the central position shares more of its electrons than does a terminal atom. Atoms with higher electronegative are generally more reluctant to share its electrons.
1. At the synaptic terminal, voltage-gated ______________ channels open, thereby stimulating the synaptic vesicles to release their neurotransmitters by exocytosis.
Ion
Explanation:
At the synaptic terminal, voltage-gated ion channels open, thereby stimulating the synaptic vesicles to release the neurotransmitters by exocytosis.
These ion channels are the signaling molecules in neurons. They are the transmembrane proteins that form ion channels. The membrane potential changes the conformation of the channel proteins that regulates their opening and closing. These channels play an important role in neurotransmitter release in presynaptic nerve endings.
For example - Ca²⁺ gated ion channel.
You are participating in the Iditarod, and your sled dogs are pulling you across a frozen lake with a force of 1550 N while a 200 N wind is blowing at you at 150 degrees from your direction of travel. What is the net force (magnitude in newtons and direction in degrees from your direction of travel)? (Give the smallest angle between the direction of travel and the direction of the net force. Assume that you are traveling in the +x-direction and that the wind is blowing from the second quadrant of the xy-plane.)
Answer:
The answer to the question is
The net force (magnitude in newtons and direction in degrees from your direction of travel) is 1726.104 N ∡ 3.315 ° clockwise or 356.685° counterclockwise from the direction of travel.
Explanation:
The net force is given by
The force of pull by the dogs = 1550 N
Direction of force of pull by the dogs = 0° from the x axis
The force of the wind = 200 N
Direction of wind force = 150 ° from the x axis
Therefore x component of the wind force = 200 cos 30 =173.205 N
y component of the wind force = 200 sin 30 = 100 N
Summing the forces in the x and y direction gives
∑Fx = 1550 +173.205 = 1723.205 N
∑Fy = 100 N
Therefore the resultnt force on the Iditarod competitor is given by
√((∑Fx)²+(∑Fy)²) = √((1723.205)²+(100)²) = 1726.104 N and the direction is
[tex]Tan^{-1}\frac{SF_y}{SF_x}[/tex] = [tex]Tan^{-1}\frac{100}{1726.104}[/tex] =3.315 ° in the x, -y plane
The student's question involves calculating the net force acting on a sled during the Iditarod by using vector addition of the force from the sled dogs and the force from the wind, with attention to the angle of the wind force relative to the direction of travel.
Explanation:The student's question involves the combination of forces at different angles and determining the resultant net force and its direction. Using vector addition, the dogs' pulling force (1550 N) is in the +x-direction (the direction of travel), and the wind applies a force of 200 N at an angle of 150 degrees from the direction of travel, implicating that it's blowing from the second quadrant.
To find the net force, we break the 200 N wind force into components along the x (horizontal) and y (vertical) axes. The x-component of the wind force is 200*cos(150°) and the y-component is 200*sin(150°). We then sum the x-components of both forces for the total force in the x-direction, and sum the y-components for the total force in the y-direction.
Next, we calculate the magnitude of the net force by using the Pythagorean theorem on the x and y components, and the direction using the inverse tangent function (arctan) for the angle relative to the positive x-axis, which will give us the smallest angle between the direction of travel and the net force.
You are riding your bicycle directly away from a stationary source of sound and hear a frequency that is 1.0% lower than the emitted frequency. The speed of sound is 343 m/s. What is your speed
Answer:
Explanation:
Given
apparent frequency is 1 % lower than the original frequency
Speed of sound [tex]v=343\ m/s[/tex]
suppose f is the original frequency
apparent frequency [tex]f'=0.99f[/tex]
using Doppler's formula
[tex]f'=f\cdot \frac{v-v_o}{v+v_s} [/tex]
where
[tex]f'[/tex]=apparent frequency
[tex]v[/tex]=velocity of sound in the given media
[tex]v_s[/tex]=velocity of source
[tex]v_o[/tex]=velocity of observer
[tex]0.99f=f(\dfrac{343-v_o}{v+0})[/tex]
[tex]0.99\times 343=343-v_o[/tex]
[tex]v_o=343\times 0.1[/tex]
[tex]v_o=34.3\ m/s[/tex]
Kirchhoff’s loop rule for circuit analysis is an expression of__________?
Answer:
Conservation of energy
Explanation:
Gustav Kirchhoff, a German physicist proposed a rule called Kirchhoff's loop rule also known as Kirchhoff's second rule which states that for any closed loop, the potential difference including the voltage supply is zero. That means energy supplied by a voltage source must absorbed for balance by other components in the loop
This fulfils the law of conservation of energy
Suppose you watch a leaf bobbing up and down as ripples pass it by in a pond. You notice that it does two full up and down bobs each second. What is true of the ripples on the pond?
Answer:
The frequency of the ripples is 2Hz, and their period is 0.5 seconds.
Explanation:
Since the ripples on the pond are making the leaf oscillate up and down at a rate of two times per second, we can calculate the period T and the frequency f of the ripples on the pon[tex]T=\frac{1}{0.5Hz} =[/tex]d.
The frequency, by definition, is the number of waves per unit of time. In this case, we have two waves per second, so the frequency is 2s⁻¹, or 2Hz.
The period is the inverse of the frequency, so
[tex]T=\frac{1}{2Hz} =0.5s[/tex]
Then, the period is equal to 0.5 seconds.
For an object to be classified as a ____, it must meet certain definite criteria: It must be massive enough to pull itself into a spherical shape, it must be able to clear its orbit of most debris, and it must orbit a star.
Answer:
a planet
Explanation:
a planet is one which exerts these properties and therefore is the answer
A bugle can be thought of as an open pipe. If a bugle were straightened out, it would be 2.65 mlong.a.) If the speed of sound is 343m/????, find the lowest frequency that is resonant for a bugle (ignoring end corrections)b.) Find the next two resonant frequencies for the bugle.
Answer:
(a). The lowest frequency is 64.7 Hz.
(b). The next two resonant frequencies for the bugle are 129.4 Hz and 194.2 hz.
Explanation:
Given that,
Length = 2.65 m
Speed of sound = 343 m
We need to calculate the wavelength
Using formula of wavelength
[tex]\lambda=2l[/tex]
Put the value into the formula
[tex]\lambda=2\times2.65[/tex]
[tex]\lambda=5.3\ m[/tex]
(a). We need to calculate the lowest frequency
Using formula of frequency
[tex]f_{1}=\dfrac{v}{\lambda_{1}}[/tex]
Put the value into the formula
[tex]f_{1}=\dfrac{343}{5.3}[/tex]
[tex]f_{1}=64.7\ Hz[/tex]
(b). We need to calculate the next two resonant frequencies for the bugle
Using formula of resonant frequency
[tex]f_{2}=\dfrac{v}{\lambda_{2}}[/tex]
[tex]f_{2}=\dfrac{v}{l}[/tex]
Put the value into the formula
[tex]f_{2}=\dfrac{343}{2.65}[/tex]
[tex]f_{2}=129.4\ Hz[/tex]
For third frequency,
[tex]f_{3}=\dfrac{v}{\lambda_{3}}[/tex]
[tex]f_{3}=\dfrac{3v}{2l}[/tex]
Put the value into the formula
[tex]f_{3}=\dfrac{3\times343}{2\times2.65}[/tex]
[tex]f_{3}=194.2\ Hz[/tex]
Hence, (a). The lowest frequency is 64.7 Hz.
(b). The next two resonant frequencies for the bugle are 129.4 Hz and 194.2 hz.
The next two resonant frequencies for the bugle are 64.72Hz and 323.6Hz
The lowest frequency is expressed as:
f = v/2l
[tex]f_0=\frac{343}{2(2.65)} f_0=\frac{343}{5.3}\\f_0= 64.72 Hz[/tex]
Since the bugle is an open pipe the next two resonant frequencies are;
[tex]f_2 =3f_0=3(64.72) = 194.15Hz\\\\f_3 = 5f_0 = 5(64.72) =323.6Hz \\[/tex]
Hence the next two resonant frequencies for the bugle are 64.72Hz and 323.6Hz
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Assume that a planet just like Earth orbits the bright star named Sirius. If this Earth-like planet orbits with a semimajor axis of 1 AU and an orbital period of 7 months, what is the mass of Sirius?
Answer:
5.9x10³⁰ kg = 2.97 solar mass.
Explanation:
The mass of Sirius can be calculated using Kepler's Third Law:
[tex]P^{2} = \frac{4 \pi^{2}}{G (M_{1} + M_{2})} \cdot a^{3}[/tex]
where P: is the period, G: is the gravitational constant = 6.67x10⁻¹¹ m³kg⁻¹s⁻², a: is the size of the orbit, M₁: is the mass of the Earth-like planet = 5.97x10²⁴ kg, and M₂: is the mass of Sirius.
Firts, we need to convert the units:
P = 7 months = 1.84x10⁷ s
a = 1 AU = 1.5x10¹¹ m
Now, from equation (1), we can find the mass of Sirius:
[tex] M_{2} = \frac{4 \pi^{2} a^{3}}{P^{2} G} - M_{1} [/tex]
[tex] M_{2} = \frac{4 \pi^{2} (1.5 \cdot 10^{11} m)^{3}}{(1.84\cdot 10^{7} s)^{2} \cdot 6.67 \cdot 10{-11} m^{3} kg ^{-1} s^{-2}} - 5.97 \cdot 10^{24} kg = 5.9 \cdot 10^{30} kg = 2.97 solar mass [/tex]
Therefore, the mass of Sirius is 5.9x10³⁰ kg = 2.97 solar mass.
I hope it helps you!
Answer: 2.94kg
Explanation: Using Kepler's law of planetary motion which states that, the square of the orbital period of a planet is proportional to the cube of the semi major axis of it's orbit.
Sirius is the name given to the brightest star in the sky
It could be literally interpreted as;
1/mass of planet = (orbital period in years^2 ÷ semi major axis in AU^3)
Orbital period = 7months is equivalent to 7/12 = 0.5833 years
Semi major axis = 1AU
Therefore, mass of Sirius is given by:
1/Mass = (orbital period^2 ÷ semi major axis^3)
1/Mass = (0.5833^2 ÷ 1^3)
1/Mass = 0.3402 ÷ 1
1/Mass = 0.3402
Therefore,
Mass = 1/0.3402 = 2.94kg
Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both the source and screen are far enough from the slit for Fraunhofer diffraction to apply.
a) If the first diffraction minima are at ±90.0∘, so the central maximum completely fills the screen, what is the width of the slit?
b) For the width of the slit as calculated in part (a), what is the ratio of the intensity at θ=45.0∘ to the intensity at θ=0?
The width of the slit can be found using the formula θ = λ / a and the ratio of the intensity at θ=45.0∘ to θ=0 can be calculated using the formula I(θ) = (I0) (sinc² (πa sin(θ) / λ)).
Explanation:The width of the slit can be calculated using the formula for the angular position of the first minima in single-slit diffraction:
θ = λ / a
where θ is the angular position of the minima, λ is the wavelength of the light, and a is the width of the slit. Rearranging the formula, we can solve for a:
a = λ / θ
Substituting the given values, the width of the slit is:
a = (580 nm) / (90.0∘)
In part (b), the ratio of the intensity at θ=45.0∘ to the intensity at θ=0 can be calculated using the formula for intensity in single-slit diffraction:
I(θ) = (I0) (sinc² (πa sin(θ) / λ))
where I(θ) am the intensity at a certain angle, I0 is the intensity at θ=0, a is the width of the slit, λ is the wavelength, and sinc is the sin function. By substituting the values and calculating the ratio, you can find the answer.
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The width of the slit through which monochromatic light of wavelength 580 nm causes the first diffraction minima at extpm90.0 extdegree is 580 nm. The intensity ratio at heta=45.0 extdegree to heta=0 can be determined by using the formula I(heta) = I_0 * [sinc(eta)]^2.
The question pertains to Fraunhofer diffraction through a single slit and involves finding the width of the slit and the ratio of light intensities at different angles.
Part (a): Width of the slit
According to the single slit diffraction formula, the first minima occur at angles heta where d sin(heta) = ext{m} extlambda, where d is the width of the slit, heta is the angle of the diffraction minima, m is the order of the minima, and extlambda is the wavelength of the light.
Given the first minima at heta = ext{extpm}90.0 ext{extdegree}, we have:
sin(90ext{extdegree}) = ext{1}
ext{m} = ext{1} (since it's the first minima)
d imes 1 = ext{1} imes ext{580 nm}
d = ext{580 nm}
Therefore, the width of the slit, d, is 580 nm.
Part (b): Intensity ratio
The intensity at angle heta in a single-slit diffraction pattern is given by:
I(heta) = I_0 imes ext{[sinc(heta)]}^2
where
heta = rac{ext{extpi} d sin(heta)}{extlambda}
and
ext{sinc}(heta) = rac{ext{sin}(heta)}{heta}
For heta = 45ext extdegree},
ext{beta} can be calculated and the resulting intensity ratio I(45ext{extdegree})/I(0) can be found accordingly.
very long straight wire carries current 15 A. In the middle of the wire a right-angle bend is made. The bend forms an arc of a circle of radius 27 cm, as show. Determine the magnetic field at the center of the arc.
The magnetic field at the center of the circular arc formed by the right-angle bent wire carrying a 15 A current is calculated based on Ampere's law. Since the right-angle bend forms a quarter of a circular loop, the magnetic field strength is a quarter of a whole circular loop. Thus, the magnetic field strength at the center of the turn is approximately 2.21 x 10⁻⁵ Tesla.
Explanation:The magnetic field of a very long straight wire carrying current can be calculated using Ampere's law. The magnetic field strength of a straight wire carrying current is given by B = μI / 2πR, where B is the magnetic field strength, I is the current, R is the distance from the wire, and μ is the permeability of free space, a constant equal to 4π x 10⁻⁷ T.m/A (Tesla square meters per Ampere).
However, in your question, the wire turns and forms an arc, which is part of a circle. The situation is different from a straight wire but similar to a circular loop. For a complete circular loop carrying current, the magnetic field at the center is given by B = μI / 2R. Since we only have a right-angle bend - a quarter of a circular loop, the magnetic field at the center would be a quarter of the magnetic field of a whole circular loop. So, B = μI / 8R = (4π x 10⁻⁷ T.m/A * 15 A) / (8 * 0.27 m) = 2.21 x 10⁻⁵ T.
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The magnetic field at the center of the arc can be determined using Ampere's Law. The formula to calculate the magnetic field created by current in a long straight wire is B = μoI / (2πr), where μo is the permeability of free space, I is the current, and r is the shortest distance to the wire. In this case, the radius of the arc is given as 27 cm, which is equal to 0.27 m. Plugging in the values, we find that the magnetic field at the center of the arc is 0.035 Tesla.
Explanation:The magnetic field at the center of the arc can be determined using Ampere's Law. The formula to calculate the magnetic field created by current in a long straight wire is B = μoI / (2πr), where μo is the permeability of free space, I is the current, and r is the shortest distance to the wire. In this case, the radius of the arc is given as 27 cm, which is equal to 0.27 m.
Plugging in the values, we get:
B = (4π×10-7 T.m/A) × 15 A / (2π×0.27 m) = 0.035 T
Therefore, the magnetic field at the center of the arc is 0.035 Tesla.
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Wire A carries 4 A into a junction, wire B carries 5 A into the same junction, and another wire is connected to the junction. What is the current in this last wire?
Final answer:
According to Kirchhoff's Current Law, the total incoming current to a junction must equal the total outgoing current. Since Wire A and Wire B carry a combined total of 9 A into the junction, the current in the last wire must be 9 A leaving the junction in order to balance the current flow.
Explanation:
The subject of this question is Physics, and it relates to the basic principle of electric current conservation at a junction, also known as Kirchhoff's Current Law. This law states that the total current entering a junction must equal the total current leaving the junction.
Given that Wire A carries 4 A and Wire B carries 5 A into the junction, this means that the total incoming current is 9 A. Therefore, if no other wires are supplying current to the junction, the current in the last wire connected to the junction must be 9 A leaving the junction to satisfy Kirchhoff's Current Law.
A 10-kg bowling ball moving at 4 m/s bounces off a spring at about the same speed that it had before bouncing. What is the change in momentum of the bowling ball
Answer: change in momentum is approximately 0 kgm/s
Explanation: The change in momentum is difference between the momentum before collision and momentum after collision. So since the Velocity after collision is almost the same as that before collision then there will be no change in the momentum since the mass is constant.
Note that momentum is just the product of mass M and Velocity v.
4 The time to failure of an electrical component has a Weibull distribution with parameters λ = 0.056 and a = 2.5. A random collection of 500 components is obtained. Estimate the probability that at least 125 of the 500 components will have failure times larger than 20.
Answer:
the probability that at least 125 of the 500 components will have failure times larger than 20 is 0.7939
Explanation:
see the attached file
What are the steps that produce electricity from this fuel
The process of generating electricity from fuel involves combustion of the fuel to produce heat, using the heat to boil water and produce steam, spinning a turbine with the steam, triggering a generator that produces electricity, and the distribution of that electricity.
Explanation:The process of converting fuel into electricity typically involves several steps. These steps may vary depending on the type of fuel, but let's take coal as an example, a common type of fuel used in power plants.
Combustion: The coal is burned to produce heat. Boiling water: This heat is used to boil water, creating steam. Turbine: The high-pressure steam then spins a turbine. Generator: The spinning turbine triggers a generator which then produces electricity by spinning a coil of wire in a magnetic field. Distribution: The generated electricity is then distributed via power lines to homes and businesses.Learn more about Generating electricity from fuel here:
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In the rough approximation that the density of a planet is uniform throughout its interior, the gravitational field strength (force per unit mass) inside the planet at a distance from the center is , where is the radius of the planet. (For the Earth, at least, this is only a rough approximation, because the outer layers of rock have lower density than the inner core of molten iron). 1. Using the uniform-density approximation, calculate the amount of energy required to move a mass m from the center of the Earth to the surface. 2. For comparison, how much energy would be required to move the mass from the surface of the planet to a very large distance away? 3. Imagine that a small hole is drilled through the center of the Earth from one side to the other. Determine the speed of an object of mass m, dropped into this hole, when it reaches the center of the planet.
Answer:
The answers to the questions are;
1. The amount of energy required to move a mass m from the center of the Earth to the surface is 0.5·m·g·R
2. The amount of energy required to move the mass from the surface of the planet to a very large distance away m·g·R.
3. The speed of an object of mass m, dropped into this hole, when it reaches the center of the planet is 9682.41783 m/s.
Explanation:
We note that the Work done W by the force F on the mass to move a small distance is given by
F×dr
The sum of such work to move the body to a required location is
W =[tex]\int\limits {F} \, dr[/tex]
F = mg' = m[tex]\frac{gr}{R}[/tex]
We integrate from 0 to R (the center to the Earth surface)
Therefore W = [tex]\int\limits^R_0 {m\frac{gr}{R}} \, dr[/tex]
Which gives W = [tex]\frac{mg}{R} [\frac{r^2}{2} ]^R_0 = \frac{1}{2}mgR[/tex]
2. To find the work done we have to integrate from the surface to infinity as follows W = [tex]\int\limits^{inf}_R {m\frac{gr}{R}} \, dr[/tex] = [tex]\frac{mg}{R} [\frac{r^2}{2} ]^{inf}_R = mgR[/tex]
The energy required to move the object to a large distance is equal to twice the energy reqired to move the object to the surface.
3 We note that the acceleration due to gravity at the surface is g and reduces to zero at the center of the Earth
v² = u² + 2·g·s
Radius of the Earth = 6371 km
From surface to half radius we have
v₁² = 2×9.81×6371/2×1000 = 62499460.04
v₁ = 7905.66 m/s
From the half the radius of the earth to the Earth center =
v₂² = 7905.66² + 2×9.81/2×6371/2×1000 = 93749215.04
v₂ = 9682.41783 m/s
The speed of an object of mass m, dropped into this hole, when it reaches the center of the planet. is 9682.41783 m/s
1. Energy to Move from Center to Surface: Using the uniform-density approximation, the energy required to move a mass m from Earth's center to the surface is [tex]\(-\frac{GMm}{R}\)[/tex]. 2. Energy to Move from Surface to Infinity: Moving the mass from the surface to infinity requires zero energy. 3. Speed at Earth's Center: The speed of an object dropped through a hole from the surface to the Earth's center is [tex]\(\sqrt{\frac{2GM}{R}}\)[/tex].
1. Energy to Move Mass from Center to Surface:
The gravitational potential energy U is given by:
[tex]\[ U = -\frac{GMm}{r} \][/tex]
where:
- G is the gravitational constant,
- M is the mass of the planet,
- m is the mass being moved,
- r is the distance from the center.
For this scenario, r is the radius of the planet R.
[tex]\[ U_{\text{center to surface}} = -\frac{GMm}{R} \][/tex]
2. Energy to Move Mass from Surface to Infinity:
When moving the mass to a very large distance away [tex](\( \infty \))[/tex], the potential energy becomes zero.
[tex]\[ U_{\text{surface to infinity}} = 0 \][/tex]
3. Speed of Object Dropped Through Earth's Center:[tex]\[ U_{\text{surface}} = K_{\text{center}} \]\[ -\frac{GMm}{R} = \frac{1}{2}mv^2 \][/tex]converted into kinetic energy at the center.
Solve for v:
[tex]\[ v = \sqrt{\frac{2GM}{R}} \][/tex]
Given:
- M is the mass of the Earth,
- m is the mass being moved.
These expressions provide the required energy calculations and the speed of the object dropped through the Earth's center.
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(Problem 9.15) If you are rounding a corner on a bicycle, by twisting the handlebars of your bicycle to the right, which direction does the bicycle wheel precess? Why? Should you lean inward (to the right) or outward (to the left) to counteract this?
Answer:
towards right,
lean inward to avoid tripping
Explanation:
The frictional force becomes centripetal force while turinig and presses on the wheel to maintain the grip of wheel on the road. Otherwise the bicycle will slip. So the press on the wheel is inward or towards the right when turning a corner to right
The rider has to lean inward ( to the right) to avoid tripping. Leaning inward, changes the direction of normal force on the wheels from vertical to slant. This helps to counter the torque produced by the frictional force
A hiker walks 2.00 km north and then 3.00 km east, all in 2.50 hours. Calculate the magnitude and direction of the hiker’s (a) displacement (in km) and (b) average velocity (in km/h) during those 2.50 hours. (c) What was her average
Answer:
Incomplete third question
I think it should be
C. What was her average speed
Explanation:
Check attachment for solution
(a). The displacement is 3.6km.
(b). Average velocity is 2 km/hour
(a). A diagram is attached below which describe situation of question.
In diagram OB represent the displacement.
In right triangle OAB,
[tex](OB)^{2}=(OA)^{2}+(AB)^{2}\\ \\ OB^{2}=2^{2}+3^{2}=4+9=13\\ \\ OB=\sqrt{13}=3.6Km[/tex]
(b). Average velocity is given as,
[tex]v=\frac{Distance}{time}=\frac{2+3}{2.5} =5/2.5=2km/h[/tex]
Learn more about Displacement and Average velocity here:
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What is the minimum diameter mirror on a telescope that would allow you to see details as small as 5.20 km on the moon some 384000 km away? Assume an average wavelength of 550 nm for the light received.
cm
Answer:
4.96cm
Explanation:
distance = 5.20km = 5.2 × 10³ m
Wavelength , λ = 550nm = 5.50 × 10⁻⁷m
distance of moon, L = 384000 km = 3.84 × 10⁸m
formula for resolving power of two objects
d = (1.22 × λ ×L) / D
D = (1.22 × λ ×L) / d
= (1.22 × 5.50 × 10⁻⁷ ×3.84 × 10⁸) / 5.2 × 10³
D = 4.96cm
From the following statements about mechanical waves, identify those that are true for transverse mechanical waves only, those that are true for longitudinal mechanical waves only, and those that are true for both types of waves.A. In longitudinal waves, the particles of the medium move parallel to the direction of the flow of energy.B. Many wave motions in nature are a combination of longitudinal and transverse motion.C. In transverse waves, the particles of the medium move perpendicular to the direction of the flow of energy.D. All of the above
'In transverse waves, the particles of the medium move perpendicular to the direction of the flow of energy' is true for transverse waves only.
'In longitudinal waves, the particles of the medium move parallel to the direction of the flow of energy' is true for longitudinal waves only.
'Many wave motions in nature are a combination of longitudinal and transverse motion' is true for both longitudinal and transverse waves.
Explanation:
Longitudinal waves are those where the direction of propagation of particles are parallel to the medium' particles. While transverse waves propagate perpendicular to the medium' particles.
As wave motions are assumed to be of standing waves which comprises of particles moving parallel as well as perpendicular to the medium, most of the wave motions are composed of longitudinal and transverse motion.
So the option stating the medium' particle moves perpendicular to the direction of the energy flow is true for transverse waves. Similarly, the option stating the medium' particle moves parallel to the direction of flow of energy is true for longitudinal waves only.
And the option stating that wave motions comprises of combination of longitudinal and transverse motion is true for both of them.
Final answer:
The statements A, B, and C are true for both transverse and longitudinal waves.
Explanation:
A. In longitudinal waves, the particles of the medium move parallel to the direction of the flow of energy.
B. Many wave motions in nature are a combination of longitudinal and transverse motion.
C. In transverse waves, the particles of the medium move perpendicular to the direction of the flow of energy.
All of the above statements are true for both transverse and longitudinal waves.
Explanation An is valid for longitudinal waves as it were. In longitudinal waves, for example, sound waves, the particles of the medium vibrate lined up with the bearing of the energy move. This can be pictured as pressure and rarefaction of the particles in the medium.
Only transverse waves are covered by statement C. In cross over waves, for example, light waves, the particles of the medium vibrate opposite to the course of the energy move. The particles can be seen moving side to side or up and down, respectively.
B holds true for both kinds of waves. Many wave movements in nature, for example, water waves and seismic waves, include a mix of longitudinal and cross over movement. Water particles, for instance, move in circular orbits as waves pass by, which is a combination of transverse and longitudinal motion.
A child's train whistle replicates a classic conductor's whistle from the early 1900s. This whistle has two open-open tubes that produce two different frequencies. When you hear these two different frequencies simultaneously, you may have the perception of also hearing a lower note, called a difference tone, that is at the same frequency as the beat frequency between the two notes. The two tubes of the whistle are 14 cm and 12 cm in length.Assuming a sound speed of 350 m/s, what is the frequency of this difference tone?
Answer:
f_beat = 210 Hz
Explanation:
Solution:
- The frequency of the difference tone - beat frequency is given by:
f_beat = |f_1 - f_2|
Where,
f_1 = Frequency of first component
f_2 = Frequency of second component
- The whistle is a tube with openings at both ends. The wavelength (λ) of the fundamental tone in a pipe of length L which is open at both ends is:
λ = 2*L
- For all waves, the relationship between frequency (f) , wavelength (λ), and propagation speed (v) is:
λ*v = c
- So the fundamental frequency of a pipe of length L, open at both ends is:
f = v/(2*L)
- In this case, we are told that v = 350 m/s and the pipes have lengths 12 cm and 14 cm, so the beat (difference) frequency is:
f_beat = | (350 m/s)/(2*0.12 m) - (350 m/s)/(2*0.14 m)|
= 208.33 Hz
- All the input data for this problem were only given to 2 significant digits, so your result should only be reported to 2 significant digits, which in this case would be 210 Hz.
Sebuah kolom udara memiliki panjang 40cm.Jika garpu kala mempunyai frekuensi 320Hz,maka besarnya cepat rambat gelombang bunyi diudara pada saat resonasi pertama adalah....m/s.
Answer:
Velocity of sound is 512 m/s
Explanation:
for first resonance condition we know that
[tex]\frac{\lambda}{4} = L[/tex]
[tex]\frac{\lambda}{4} = 0.40[/tex]
[tex]\lambda = 1.6 m[/tex]
now the frequency is given as
[tex]f = 320 Hz[/tex]
now to find the velocity of sound we have
[tex]v = \lambda f[/tex]
[tex]v = 1.6 (320)[/tex]
[tex]v = 512 m/s[/tex]
The drag is proportional to the square of the speed of the boat, in the form Fd=bv2 where b= 0.5 N⋅ s2/m2. What is the acceleration of the boat just after the rain starts? Take the positive x axis along the direction of motion.
The first part of the question is missing and it is:
A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 3.00 m/s when it starts to rain. The rain is falling vertically, and it accumulates in the boat at the rate of 10.0 kg/hr.
Answer:
a(o) = 18.03 x 10^(-3) m/s^(-2)
Explanation:
First of all, if we make a momentum balance in the direction of motion (the x-direction), we'll discover that the change of the momentum will be equal to the forces acting on the boat.
Hence, there is only the drag force, which acts against the direction of motion as:
d(m·v)/dt = - k·v² (since f_d = - k·v²)
where k = 0.5 Ns²/m²
Now let's simplify the time derivative on the Left, by applying product rule of differentiation and we obtain:
v·dm/dt + m·dv/dt = - k·v²
where dm/dt = 10kg/hr (this is the change of the mass of the boat)
Furthermore, acceleration is the time derivative of time velocity, Thus;
a = dv/dt = - (k·v² + v·dm/dt) / m
Now, for the moment,when the rain starts; since we know all the values on the right hand side, let's solve for the acceleration ;
a(o) = - (ko·vo² + vo·dm/dt) / mo
= -[ (0.5(3)²) + (3x10/3600)/250
(dt = 3600secs because 1hr = 3600 secs)
So a(o) = 18.03 x 10^(-3) m/s^(-2)
What happens to the bond angle when you add or remove an electron domain
Answer:
when you add the bond angle decreases and when you remove an electron domain the angle increases
Explanation:
Generally, character increase in the hybrid bond, the bond angle increases. Bond angle is affected by the presence or addition of lone pair of electrons at the central atom. Due to this, the bonds are displaced slightly inside resulting in a decrease of bond angle,and when you remove an electron domain the bond angle increases.
After addition of electron, the bond angle decreases and when you remove an electron domain then bond angle increases.
The bond angle decreases due to addition of electron lone pairs, which cause more repulsion on the bond pairs and as a result the bond pairs tend to come closer. However the bond angle decreases when the electro-negativities of ligand atoms are more than that of central atom.The bond angle increases due to remove of electron lone pairs, which decreases repulsion between bond pair and electron lone pair and as a result the bond pairs tend to go slightly away.Learn more:
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A 60.0 kg student runs up a flight of stairs in Rockefeller Hall a height of 4.0 m in a time of 6.0 s. What average mechanical power (in W) is required to do this?
Answer:
400watts
Explanation:
Power is defined as the amount of energy expended in a specific time. It is also defined as the change in work done with respect to time.
Mathematically,
Power = Workdone/time
Work done = Force×distance
Power = Force × Distance/Time
Given force = mass× acceleration due to gravity
Force = 60×10 = 600N
Distance covered = 4.0m
Time taken = 6.0seconds
Power expended = 600×4/6
Power expended = 400Watts
The average mechanical power (in W) required to do this is 400Watts
Answer:
392 W.
Explanation:
Given:
Mass, m = 60 kg
Height, h = 4 m
Time, t = 6 s
Power = energy/time
Energy = mass × acceleration × distance
Power = (60 × 9.8 × 4)/6
= 392 W.
A truck driving on a level highway is acted upon by the following forces: a downward gravitational force of 52 kN (kilonewtons); an upward contact force due to the road of 52 kN; another contact force due to the road of 8.5 kN, directed east; and a drag force due to air resistance of 2.7 kN, directed west. What is the net force acting on the truck
Explanation:
Below is an attachment containing the solution.
The net force acting on the truck can be determined by considering the various forces acting on it and calculating the resultant force.
The net force acting on the truck can be found by adding all the forces together. The gravitational force is balanced by the upward contact force, leaving the forces due to the road and air resistance. When considering the horizontal forces, the net force on the truck is determined by subtracting the force due to air resistance from the force due to the road.
A car is rounding a circular curve of radius r on a banked turn. As the drawing indicates, there are two forces acting on the car, its weight mg and the normal force FN exerted on it by the road. Which force, or force component, provides the centripetal force that keeps the car moving on the circular path?
1. The vertical component, FNcosθ of the normal force.
2. The horizontal component, FNsinθ of the normal force.
3. Both the normal force, FN, and the weight, mg, of the car.
4. The normal force, FN.
5. The weight, mg, of the car.
6. The horizontal component, FNsinθ of the norma.l
Answer:
Check attachment for the body diagram
Explanation:
Centripetal force is a net force that acts on an object to keep it moving along a circular path.
It is given as
F=ma
Where a =v²/r
F=mv²/r
The centripetal force is directed towards the center i.e inward.
Then, analysing the free body diagram, we notice that the horizontal component of the Normal force (FNsinθ) is directed towards the centre. Then, we can say the last option 6 is correct.