Answer:
a)
The expression for the electric potential in this problem is:
[tex]V=5x-3x^2y+2yz[/tex]
where
x, y, z are the three spatial coordinates
The relationship between components of the electric field and electric potential is:
[tex]E_x=-\frac{dV}{dx}\\E_y=-\frac{dV}{dy}\\E_z=-\frac{dV}{dz}[/tex]
Therefore, we have to calculate the derivatives of the potential over the three variables.
Doing so, we find:
[tex]E_x=-\frac{d}{dx}(5x-3x^2y+2yz)=-(5-6xy)=6xy-5[/tex]
[tex]E_y=-\frac{d}{dy}(5x-3x^2y+2yz)=-(-3x^2+2z)=3x^2-2z[/tex]
[tex]E_z=-\frac{d}{dz}(5x-3x^2y+2yz)=-(2y)=-2y[/tex]
b)
Here we want to find the magnitude of the electric field at the point P that has coordinates
P (1.00, 0, 22.00) m
First of all, we find the components of the electric field at that point by substituting
x = 1.00
y = 0
z = 22.0
We find:
[tex]E_x=6xy-5=6(1)(0)-5=-5 N/C\\E_y=3x^2-2z=3(1)^2-2(22)=-41 N/C\\E_z=-2y=-2(0)=0[/tex]
Now, the magnitude of the electric field is given by
[tex]E=\sqrt{E_x^2+E_y^2+E_z^2}[/tex]
And by substituting,
[tex]E=\sqrt{(-5)^2+(-41)^2+0}=41.3 N/C[/tex]
The potential difference across the cell membrane is known as
Answer:
the membrane potential
Explanation:
typical values of membrane potential are in the range -40 mv to-70 mv
A falling skydiver has a mass of 101 kg. What is the magnitude of the skydiver's acceleration when the upward force of air resistance has a magnitude that is equal to one-fourth of his weight?
The skydiver has more force acting downward due to gravity than upward from air resistance, and thus he's accelerating downwards. You calculate the acceleration by subtracting the air resistance from the skydiver's weight and dividing by his mass, in this case, yielding an acceleration of 7.36 m/s².
Explanation:The skydiver's situation can be understood by taking into account the forces acting on him, mainly his weight and the air resistance. The weight of the skydiver is determined by multiplying his mass by the gravity constant, 9.81 m/s². This gives a weight of 101 kg × 9.81 m/s² = 991.81 N. The skydiver is said to be experiencing one-fourth that amount in upward force due to air resistance, which would be approximately 247.95 N.
In situations where opposing forces are equal, an object doesn't accelerate and we would expect the skydiver to be in a state of terminal velocity. However, in this case, there's more force acting downward (gravity) than upward (air resistance), so the skydiver is accelerating downwards. To find the amount of that acceleration, we subtract the air resistance from the skydiver's weight, then divide that by his mass. Therefore, (991.81 N - 247.95 N) / 101 kg = 7.36 m/s².
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Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and final kinetic energies.Express your answer in terms of Kinitial and Kfinal.W=
Answer:
[tex]W=K_f-K_i[/tex]
Explanation:
The work done on a particle by external forces is defined as:
[tex]W=\int\limits^{r_f}_{r_i} {F\cdot dr} \,[/tex]
According to Newton's second law [tex]F=ma[/tex]. Thus:
[tex]W=\int\limits^{r_f}_{r_i}{ma\cdot dr} \,\\[/tex]
Acceleration is defined as the derivative of the speed with respect to time:
[tex]W=m\int\limits^{r_f}_{r_i}{\frac{dv}{dt}\cdot dr} \,\\\\W=m\int\limits^{r_f}_{r_i}{dv \cdot \frac{dr}{dt}} \,[/tex]
Speed is defined as the derivative of the position with respect to time:
[tex]W=m\int\limits^{v_f}_{v_i} v \cdot dv \,[/tex]
Kinetic energy is defined as [tex]K=\frac{mv^2}{2}[/tex]:
[tex]W=m\frac{v_f^2}{2}-m\frac{v_i^2}{2}\\W=K_f-K_i[/tex]
A spectrophotometer measures the transmittance or the absorbance. True or False
Answer: FALSE
Explanation: Could you help me with a question?
How does the law of conservation of mass relate to the nitrogen cycle
Explanation:
Nitrogen is a basic component of all life forms.It is found in all proteins and DNA molecules.Nitrogen is available in the earth's atmosphere by 78 % as N2 gas. In nitrogen fixation, bacteria convert nitrogen gas into ammonia or usable form of nitrogen used by plants., When animals eat the plants, they acquire usable nitrogen compounds.During the nitrogen cycle, nitrogen atoms used in the cycle are neither created or destroyed, or changed into other atoms. Thus it can be said that the law of conservation of matter is related to the nitrogen cycle, which states that matter is never created or destroyed in any chemical or physical process.What is the name of the german scientist that proposed the theory of continental drift
Answer:
Alfred Wegener
Explanation:
Alfred Wegener was a German scientist who first discovered the idea of continental drift.
The continental drift hypothesis says that the large continents drift from one location to another over the broad ocean water bodies with respect to the fixed poles. This was the first step in understanding the interior of the earth and how the tectonic activities take place on earth.
He contributed many pieces of evidence in order to support this hypothesis, but it was initially not accepted as he was not able to explain the main mechanism for the continental motion.
Solar energy heats the surface of the earth including the ground rocks and even roadways as the temperatures of the surfaces increase heat energy is released back
Answer: into the atmosphere
Explanation:
As this energy is released into the atmosphere, bubbles of warm air is formed which is released and it is replaced by cooler air. This process is responsible for many of the weather patterns in our atmosphere, and is known as....
convection
While convection is a form of heat transfer from one place to the other involving the movement of fluids. So in the case narrated above the fluid which serves as a medium is Air.
A block and tackle is used to lift an automobile engine that weighs 1800 N. The person exerts a force of 300 N to lift the engine. How many ropes are supporting the engine
Answer:
1800/300 = 6ropes
Explanation:
The engine weighs 1800N and the person exerts a force of 300N, so for him to lift the engine and exerting a force of 300N all through we divide the weight of the engine by the force exerted to know how many ropes are used. Which makes it 6 thereby each rope uses 300N to lift the engine.
Why does a lone pair of electrons occupy more space around a central atom than a bonding pair of electrons?
Answer:
The lone pair of electrons occupy more space because the electrostatic force becomes weaker.
Explanation:
When there is a bond pair of electrons in the 2 positively charged the atomic nuclei draw the electron density towards them, thereby reducing the bond diameter.
In the case of the lone pair, only 1 nucleus is present, and the enticing electrostatic force becomes weaker and the intensity of the electrons will be increases. Therefore, the lone pair occupies more space than the pair of bonds.
A lone pair of electrons occupies more space around the central atom than bonding pairs due to greater electrostatic repulsions, which are not mitigated by the sharing of electrons with another atom. This concept also applies to multiple bonds, with higher electron density leading to increased space occupancy compared to single bonds and influencing molecular geometry.
The reason why a lone pair of electrons occupies more space around the central atom than a bonding pair of electrons relates to electrostatic repulsions. Lone pairs are not shared with another atom, therefore they tend to occupy a larger region of space due to increased repulsions of their negatively charged electrons. As illustrated in the case of a molecule like ammonia (NH3), the lone pair of electrons on the nitrogen atom occupies more space than the bonding pairs.
Additionally, this concept extends to multiple bonds such as double or triple bonds, which occupy more space around a central atom than a single bond. The higher electron density in multiple bonds leads to greater repulsions, which can influence the bond angles in a molecule, causing deviations from ideal geometry.
The order of space occupied from largest to smallest is as follows: lone pair > triple bond > double bond > single bond. This hierarchy is essential for understanding molecular shape and bond angles, and is exemplified by deviations in expected bond angles due to these repulsions.
If the temperature is held constant during this process and the final pressure is 703 torr , what is the volume of the bulb that was originally filled with gas
Answer:
[tex]2.42liter[/tex]
Explanation:
The question is incomplete; you need some additional data.
I will assume the missing data from a similar question and keep the final pressure of 703 torr given in your question.
An ideal gas at a pressure of 1.20 atm is contained in a bulb of unknown volume. A stopcock is used to connect this bulb with a previously evacuated bulb that has a volume of 0.720 L as shown here. When the stopcock is opened the gas expands into the empty bulb
If the temperature is held constant during this process and the final pressure is 703 torr , what is the volume of the bulb that was originally filled with gas?
SolutionSince the amount of gas and the temperature remain constant, we may use Boyle's law:
[tex]PV=constant\\\\P_1V_1=P_2V_2[/tex]
Form the data:
[tex]P_1=1.20atm\\\\P_2=703torr\\\\V_1=unknown\\\\V_2=V_1+0.720liter[/tex]
1. Convert 703 torr to atm:
[tex]P_2=703torr\times 1atm/760torr=0.925atm[/tex]
2. Sustitute the data in the equation:
[tex]1.2atm\times V_1=0.925atm\times(V_1+0.720liter)[/tex]
3. Solve:
[tex]1.2V_1=0.925V_1+ 0.666liter\\\\0.275V_1=0.666liter\\\\V_1=2.4218liter\approx 2.42liter[/tex]
The answer is rounded to 3 significant figures, according to the data.
Determine the wavelengths for Potassium (violet, λ = 400 nm) and Strontium (red, λ = 700 nm) light emissions. Calculate the frequencies (c = wavelength × frequency) and the energy (E = h × frequency) for each.
Answer:
For Potassium:
Frequency = 7.5 x 10¹⁴ Hz; E (energy) = 8.83 x 10⁻²¹ J
For Strontium:
Frequency = 4.3 x 0¹⁴ Hz
E (energy) = 2.85 x 10⁻¹⁹ J
Explanation:
Wavelength is represented by λ, and Frequency is represented by ν .
E (energy) = hν = hc/λ, where ν = frequency; c = speed of light = 3 x 10⁸ m/s; 1 s-1 = 1 Hz
h = planck's constant = 6.62 x 10⁻³⁴ J.s; 1 nm = 10⁻⁹ m
1. Potassium λ (wavelength) = 400 nm, Frequency, ν is given by :
ν = c/λ = (3 x 10⁸ m/s) / 400 nm
= (3 x 10⁸ m/s) / 400 x 10⁻⁹ s-1
= 0.0075 x 10¹⁷s-1
= 7.5 x 10¹⁴ s-1
Frequency = 7.5 x 10¹⁴ Hz
E (energy) = hν = (6.62 x 10⁻³⁴ J.s) x (7.5 x 10¹⁴s-1)
E (energy) = 8.83 x 10⁻²¹ J
2. Strontium λ (wavelength) = 700 nm ,Frequency, ν is given by :
ν = c/λ = (3 x 10⁸ m/s) / 700 nm
= (3 x 10⁸ m/s) / 700 x 10⁻⁹s-1
= 0.00428571428 x 10¹⁷s-1
= 4.3 x 10¹⁴ s-1
Frequency = 4.3 x 0¹⁴ Hz
E (energy) = hν = (6.62 x 10⁻³⁴ J.s) x (4.3 x 0¹⁴s-1)
E (energy) = 2.85 x 10⁻¹⁹ J
Explanation:
Below is an attachment containing the solution.
The wave function of a particle in a one-dimensional box of width L is Ψ(x) = Asin(πx/L). If we know the particle must be somewhere in the box, what must be the value of A? Express your answer in terms of L.
Answer: A = square root (2/L)
Explanation: find the attached file for explanation
Final answer:
The wave function of a particle in a one-dimensional box needs to be normalized. The normalization condition requires that the value of A for the wave function Ψ(x) = Asin(πx/L) is found to be A = √(2/L) after solving the normalization integral.
Explanation:
The wave function Ψ(x) of a particle in a one-dimensional box of width L must be normalized so that the total probability of finding the particle within the box is 1. This normalization condition implies that the integral of the square of the absolute value of the wave function over the interval from 0 to L should equal 1.
The normalization integral for the given wave function Ψ(x) = Asin(πx/L) is:
∫ |Asin(πx/L)|² dx = A² ∫ sin²(πx/L) dx = 1
When you solve the integral from 0 to L, the result is:
A² * L/2 = 1
Therefore, solving for A, we get:
A = √(2/L)
So the value of A in terms of L is √(2/L).
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. Which statement is true of the ripples on the pond?
They have a frequency of 2 hertz. (hertz = cycles per second)
Answer:
They have a frequency of 2 hertz
Explanation:
Frequency is the number of occurrences of a repeating event per unit of time and Wavelength is the distance from one crest to another, or from one trough to another.
In the question, it is stated that the leaf does two full up and down bobs, this means that it completes 2 full cycles in one second. Therefore, its frequency is 2/s
where, s⁻¹ is hertz
so, They have a frequency of 2 hertz
Which is true about the spacing of the streamlines in a wire?
The statement " Close spacing represents greater current densities" is true about the spacing of the streamlines in a wire (option F)
Why is this correct?
In a scenario where a conductor is wider on the left and narrower on the right, the electric field lines and current density are depicted by streamlines. With current flowing from the wider end to the narrower end, the total charge and current remain constant. However, the current density fluctuates, being higher at the narrower end.
Hence, when observing streamlines, closer spacing within the wire signifies a higher current density, specifically in the narrower sections compared to the wider ones.
Complete question:
Which is true about the spacing of the streamlines in a wire?
A. Wide spacing represents faster random-motion velocities.
B. Wide spacing represents greater electric field vectors.
C. Wide spacing represents greater current densities.
D. Close spacing represents faster random-motion velocities.
E. Close spacing represents greater electric field vectors.
F. Close spacing represents greater current densities.
A boy takes hold of a rope to pull a wagon (m = 50 kg) on a surface with a static coefficient of friction μS = 0.25. Calculate the force (in newtons) that would need to be applied to the rope to just start the wagon moving.
Answer:
The force that would be applied on the rope just to start moving the wagon is 122 N
Explanation:
Frictional force opposes motion between two surfaces in contact. It is the force that must be applied before a body starts to move. Static friction opposes the motion of two bodies that are in contact but are not moving. The magnitude of static friction to overcome for the body to move can be calculated using equation 1.
F = μ x mg .............................. 1
where F is the frictional force;
μ is the coefficient of friction ( μs, in this case, static friction);
m is mass of the object and;
g is the acceleration due to gravity( a constant equal to 9.81 m/[tex]s^{2}[/tex])
from the equation we are provide with;
μs = 0.25
m = 50 kg
g = 9.81 m/[tex]s^{2}[/tex]
F =?
Using equation 1
F = 0.25 x 50 kg x 9.81 m/[tex]s^{2}[/tex]
F = 122.63 N
Therefore a force of 122 N must be applied to the rope just to start the wagon.
Explanation:
Below is an attachment containing the solution.
A hollow conducting sphere with an outer radius of 0.295 m and an inner radius of 0.200 m has a uniform surface charge density of 6.37 10 6 C m2 A charge of 0.370 µC is now introduced into the cavity inside the sphere a What is the new charge density on the outside of the sphere b Calculate the strength of the electric field just outside the sphere
Answer:
a. [tex]6.032\times10^{-6}C/m^2[/tex]
b.[tex]6.816\times10^5N/C[/tex]
Explanation:
#Apply surface charge density, electric field, and Gauss law to solve:
a. Surface charge density is defined as charge per area denoted as [tex]\sigma[/tex]
[tex]\sigma=\frac{Q}{4\pi r_{out}^2}[/tex], and the strength of the electric field outside the sphere [tex]E=\frac{\sigma _{new}}{\epsilon _o}[/tex]
Using Gauss Law, total electric flux out of a closed surface is equal to the total charge enclosed divided by the permittivity.
[tex]\phi=\frac{Q_{enclosed}}{\epsilon_o}\\\\\sigma=\frac{Q}{4\pi r_{out}^2}\\\\\sigma=\frac{0.370\times 10^{-6}}{4\pi \times (0.295m)^2}\\\\=3.383\times10^{-7}C/m^2[/tex] #surface charge outside sphere.
[tex]\sigma_{new}=\sigma_{s}-\sigma\\\\\sigma_{new}=6.37\times10^{-6}C/m^2-3.383\times10^{-7}C/m^2\\\\\sigma_{new}=6.032\times10^{-6}C/m^2[/tex]
Hence, the new charge density on the outside of the sphere is [tex]6.032\times10^{-6}C/m^2[/tex]
b. The strength of the electric field just outside the sphere is calculated as:
From a above, we know the new surface charge to be [tex]6.032\times10^{-6}C/m^2[/tex],
[tex]E=\frac{\sigma _{new}}{\epsilon _o}\\\\=\frac{6.032\times10^{-6}C/m^2}{\epsilon _o}\\\\\epsilon _o=8.85\times10^{-12}C^2/N.m^2\\\\E=\frac{6.032\times10^{-6}C/m^2}{8.85\times10^{-12}C^2/N.m^2}\\\\E=6.816\times10^5N/C[/tex]
Hence, the strength of the electric field just outside the sphere is [tex]6.816\times10^5N/C[/tex]
The lowest-pitch tone to resonate in a pipe of length l that is closed at one end and open at the other end is 200 hz. Which frequencies will not resonate in the same pipe?
Answer:
The frequency 400 hz is not possible .
Explanation:
Given that,
Frequency = 200 hz
Length = l
Suppose, The given frequencies are,
600 Hz, 1000 Hz, 1400 Hz, 1800 Hz and 400 hz.
The possible resonance frequencies are
We need to calculate the fundamental frequency
Using formula of fundamental frequency for pipe
[tex]F=\dfrac{nv}{4L}[/tex]
Where, n = odd number
Put the value of frequency
[tex]200= \dfrac{nv}{4L}[/tex]
We need to calculate the first over tone
Using formula of fundamental frequency
n = 3,
[tex]F=\dfrac{nv}{4L}[/tex]
Put the value into the formula
[tex]F_{2}=3\times\dfrac{v}{4l}[/tex]
[tex]F_{2}=3\times200[/tex]
[tex]F_{2}=600\ Hz[/tex]
We need to calculate the second over tone
Using formula of fundamental frequency
n = 5,
[tex]F=\dfrac{nv}{4L}[/tex]
Put the value into the formula
[tex]F_{3}=5\times200[/tex]
[tex]F_{3}=1000\ Hz[/tex]
We need to calculate the third over tone
Using formula of fundamental frequency
n = 7,
[tex]F=\dfrac{nv}{4L}[/tex]
Put the value into the formula
[tex]F_{4}=7\times200[/tex]
[tex]F_{4}=1400\ Hz[/tex]
We need to calculate the fourth over tone
Using formula of fundamental frequency
n = 9,
[tex]F=\dfrac{nv}{4L}[/tex]
Put the value into the formula
[tex]F_{5}=9\times200[/tex]
[tex]F_{5}=1800\ Hz[/tex]
Hence, The frequency 400 hz is not possible .
Final answer:
In a pipe that is closed at one end and open at the other, only odd multiples of the fundamental frequency will resonate. Frequencies that are even multiples of the fundamental 200 Hz, such as 400 Hz, 600 Hz, 800 Hz, will not resonate.
Explanation:
The lowest-pitch tone to resonate in a pipe of length l that is closed at one end and open at the other end is 200 Hz. This pipe supports harmonic frequencies following the sequence fn = n(v/4L), where n = 1, 3, 5..., v is the speed of sound, and L is the length of the pipe. Therefore, frequencies that will not resonate in the same pipe are those that are even multiples of the fundamental frequency, such as 400 Hz, 600 Hz, 800 Hz, and so on.
A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational force exerted on each by the earth.(b)Calculate the magnitude of the acceleration of each object when released.
Answer:
a). Determine the magnitude of the gravitational force exerted on each by the earth.
Rock: [tex]F = 49.06N[/tex]
Pebble: [tex]F = 29.44N[/tex]
(b)Calculate the magnitude of the acceleration of each object when released.
Rock: [tex]a =9.8m/s^{2}[/tex]
Pebble: [tex]a =9.8m/s^{2}[/tex]
Explanation:
The universal law of gravitation is defined as:
[tex]F = G\frac{m1m2}{r^{2}}[/tex] (1)
Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.
Case for the rock [tex]m = 5.0 Kg[/tex]:
m1 will be equal to the mass of the Earth [tex]m1 = 5.972×10^{24} Kg[/tex] and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth [tex]r = 6371000m[/tex].
[tex]F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}[/tex]
[tex]F = 49.06N[/tex]
Newton's second law can be used to know the acceleration.
[tex]F = ma[/tex]
[tex]a =\frac{F}{m}[/tex] (2)
[tex]a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}[/tex]
[tex]a =9.8m/s^{2}[/tex]
Case for the pebble [tex]m = 3.0 Kg[/tex]:
[tex]F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}[/tex]
[tex]F = 29.44N[/tex]
[tex]a =\frac{F}{m}[/tex]
[tex]a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}[/tex]
[tex]a =9.8m/s^{2}[/tex]
(a) The magnitude of the gravitational force on the rock is 49 N, while the magnitude of the gravitational force on the pebble is 2.94 × 10^-3 N. (b) The magnitude of the acceleration for both objects when released is 9.8 m/s^2.
Explanation:(a) Gravitational Force:
The magnitude of the gravitational force exerted on an object near the surface of the Earth can be calculated using the formula:
F = mg
where F is the gravitational force, m is the mass of the object, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
For the rock, m = 5.0 kg:
F = mg = (5.0 kg) * (9.8 m/s^2) = 49 N
For the pebble, m = 3.0 × 10-4 kg:
F = mg = (3.0 × 10-4 kg) * (9.8 m/s^2) = 2.94 × 10-3 N
(b) Acceleration:
The magnitude of the acceleration of an object when released can be calculated using Newton's second law:
F = ma
Rearranging the equation to solve for acceleration, we have:
a = F/m
Substituting the values we calculated in part (a) for the gravitational force exerted on each object, we can determine the magnitude of acceleration.
For the rock, F = 49 N and m = 5.0 kg:
a = (49 N) / (5.0 kg) = 9.8 m/s^2
For the pebble, F = 2.94 × 10-3 N and m = 3.0 × 10-4 kg:
a = (2.94 × 10-3 N) / (3.0 × 10-4 kg) = 9.8 m/s^2
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Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 5.50 m in front of one of the speakers perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible? Express your answer with the appropriate units.
Answer:
The lowest possible frequency of sound is 971.4 Hz.
Explanation:
Given that,
Distance between loudspeakers = 2.00 m
Height = 5.50 m
Sound speed = 340 m/s
We need to calculate the distance
Using Pythagorean theorem
[tex]AC^2=AB^2+BC^2[/tex]
[tex]AC^2=2.00^2+5.50^2[/tex]
[tex]AC=\sqrt{(2.00^2+5.50^2)}[/tex]
[tex]AC=5.85\ m[/tex]
We need to calculate the path difference
Using formula of path difference
[tex]\Delta x=AC-BC[/tex]
Put the value into the formula
[tex]\Delta x=5.85-5.50[/tex]
[tex]\Delta x=0.35\ m[/tex]
We need to calculate the lowest possible frequency of sound
Using formula of frequency
[tex]f=\dfrac{nv}{\Delta x}[/tex]
Put the value into the formula
[tex]f=\dfrac{1\times340}{0.35}[/tex]
[tex]f=971.4\ Hz[/tex]
Hence, The lowest possible frequency of sound is 971.4 Hz.
What is the significance of electron transport in the photochemical reactions of photosynthesis
Explanation:
Photosynthetic electron transport is helpful in the conversion of solar energy into chemical energy in the process of photosynthesis through transferring electrons sequentially from [tex]H_2O[/tex] through Photosystem II and Photosystem I to NADP+. Cyclically flowing electrons generate ATP molecules, because after passing down the first step of the electron transport chain protons are pumped into the thylakoid lumen, and establishes a gradient in between.However, cyclic electron flow does not involve in the formation of NADPH, nor does it involve in the splitting of water or production of oxygen.A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 34.0° above the horizontal. The car accelerates uniformly to a speed of 2.25 m/s in 10.5 s and then continues at constant speed.
Answer:
11.714 kW
Explanation:
Here is the complete question
A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 34.0∘ above the horizontal. The car accelerates uniformly to a speed of 2.25 m/s in 10.5 s and then continues at constant speed. What power must the winch motor provide when the car is moving at constant speed?
Solution
Since the loaded ore car moves along the mine shaft at an angle of θ = 34° to the horizontal, if F is the force exerted on the cable, then the net force on the laoded ore car is F - mgsinθ = ma where mgsinθ = component of the car's weight along the incline, m = mass of loaded ore car = 950 kg and a = acceleration
F = m(a + gsinθ)
When the car is moving at constant speed, a = 0
So F = m(a + gsinθ) = F = 950(0 + 9.8sin34) = 5206.1 N
Since it continues at a constant speed of v = 2.25 m/s, the power of the winch motor is P = Fv = 5206.1 N × 2.25 m/s = 11713.7 W = 11.714 kW
The force needed to pull a 950 kg ore car up a 34.0° inclined mine shaft is approximately 5518.493 N, involving both the acceleration force and the component of gravitational force parallel to the incline.
The question is about calculating the acceleration and force needed to pull a loaded ore car up a mine shaft inclined at 34.0° above the horizontal.
Step-by-Step Explanation:
Initial Setup: The mass of the ore car (m) is 950 kg, the angle of inclination (θ) is 34.0°, and the final speed (v) of the car is 2.25 m/s after 10.5 seconds (t).Calculating Acceleration: The car starts from rest, so initial velocity (u) = 0 m/s. Using the equation for uniform acceleration, v = u + at, we find a:Therefore, the total force needed to pull the ore car up the inclined shaft is approximately 5518.493 N.
The Flintstones and Rubbles decide to try out the new inclined bowling alley, ``Bedslant Bowling''. Betty's ball and Wilma's ball have the same size, but Wilma's ball is hollow. Fred's ball and Barney's ball are scaled down versions of Betty's ball and Wilma's ball respectively. They all place their bowling balls on the same pitch incline and release them from rest at the same time. Answer choices are (greater than, less than or equal to)
a) The time it takes for Wilma's ball to hit the pins is .... that for Barney's ball to hit.
b) The time it takes for Barney's ball to hit the pins is .... that for Fred's ball to hit.
c) The time it takes for Betty's ball to hit the pins is .... that for Fred's ball to hit.
d) The time it takes for Betty's ball to hit the pins is .... that for Barney's ball to hit.
e) The time it takes for Fred's ball to hit the pins is .... that for Wilma's ball to hit.
f) The time it takes for Wilma's ball to hit the pins is .... that for Betty's ball to hit.
A
Explanation:
fred ball hit the pin the time for all just put a hit in his right knee and do not need to be on bed at the carbon footprint in his first year of practice at all but his teammates were also on bed and you were a good friend to be with you cute guys that he is the most likely and do you have the ability for you I don't want him on you and
The time it takes for Wilma's ball to hit the pins is equal to that for Barney's ball to hit. The time it takes for Barney's ball to hit the pins is greater than that for Fred's ball to hit. The time it takes for Betty's ball to hit the pins is greater than that for Fred's ball to hit.
Explanation:The time it takes for Wilma's ball to hit the pins is equal to that for Barney's ball to hit.
The time it takes for Barney's ball to hit the pins is greater than that for Fred's ball to hit.
The time it takes for Betty's ball to hit the pins is greater than that for Fred's ball to hit.
The time it takes for Betty's ball to hit the pins is equal to that for Barney's ball to hit.
The time it takes for Fred's ball to hit the pins is equal to that for Wilma's ball to hit.
The time it takes for Wilma's ball to hit the pins is greater than that for Betty's ball to hit.
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The front 1.20 m of a 1,550-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a car traveling 24.0 m/s stops uniformly in 1.20 m, how long does the collision last
Answer:
t = 0.1 s
Explanation:
Given:
- The distance crushed (Stopping-Distance) s = 1.20 m
- The mass of the car m = 1,550 kg
- The initial velocity vi = 24.0 m/s
Find:
- How long does the collision last t?
Solution:
- The stopping distance s is the average velocity v times time t as follows:
s = t*( vf + vi ) / 2
Where,
vf = 0 m/s ( Stopped )
s = t*( vi ) / 2
t = 2*s / vi
t = 2*1.20 / 24
t = 0.1 s
Answer:
t=0.1seconds.
Explanation:
The mass of the car m = 1,550 kg
We know the stopping distance, 1.20m,
we know the final velocity, 0m/s (its stopped),
the starting velocity, 24m/s.
d=t(v2+v1)/2
Rearange to solve for t, and remove the v2^2 as its zero
t=2d/v1
t=2(1.20m)/(24m/s)
t=0.1seconds.
A municipal water supply is provided by a tall water tower. Water from this tower flows to a building. How does the water flow out of a faucet on the ground floor of a building compare with the water flow out of an identical faucet on the second floor of the building
Answer:
THE ANSWER IS: Water flows more rapidly out of the ground-floor faucet.
Explanation:
Water flows more rapidly out of the ground-floor faucet.
Why is the water flow more rapid out of a faucet on the first floor of a building than in an apartment on a higher floor?The first floor of a building has the biggest pressure differential, which is why water flows more quickly out of a faucet there than in an apartment on a higher floor. As we move up the structure, however, the pressure difference reduces.How much water does a water tower hold?An average water tower is usually about 165 feet (50 meters) tall, and its tank can hold about a million gallons of water or more.Why does water flow on the flour easily?Proteins: the higher protein content the higher water absorption.Pentosans: the higher the pentosans content the higher the water absorption.Is water pressure lower on higher floors?In actuality, the idea that a building's age affects water pressure is a fiction. However, it is true that in buildings where the roof tank serves as the supply of water, the water pressure at fixtures is lower in upper floor apartments than in lower level apartments.Learn more about municipal water supply here:
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A Ferris wheel is 30 meters in diameter and boarded from a platform that is 5 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 8 minutes. How many minutes of the ride are spent higher than 22 meters above the ground
Answer:
The ride is above 22m in height for 1.33 minutes.
Explanation:
Let's first find the height required above the boarding platform for the ride to be 22 m above the ground:
Height required = 22 - 5 = 17 m
We can now, using a right angled triangle of height equal to the Ferris wheel radius, calculate the angle from the vertical axis to achieve this height:
Height of triangle = 15 - (17 - 15) = 13 m
Hypotenuse of triangle = radius = 15 m
Angle from the vertical:
Cos( Angle ) = base / hypotenuse = 13 / 15
Angle = 29.92 °
Multiplying this angle by 2 we get the total angle through which the ride is at the required height:
Total Angle = 29.92 * 2 = 59.85 °
To take out the time we can now simply multiply the ratio of this angle /360 by the time taken for one complete revolution:
Time = [tex]\frac{59.85}{360} * 8[/tex]
Time = 1.33 minutes
C2 = 20.0 μF, and C3 = 25.0 μF. If no capacitor can withstand a potential difference of more than 100 V without failure, what are (a) the magnitude of the maximum potential difference that can exist between points A and B and (b) the maximum energy that can be stored in the three-capacitor arrangement?
The maximum potential difference between points A and B is 100 V, and the maximum energy stored in the three-capacitor arrangement is determined by the capacitor with the smallest capacitance.
The maximum potential difference (voltage) between points A and B in a series combination of capacitors is limited by the capacitor with the lowest breakdown voltage. Given C2 = 20.0 μF and C3 = 25.0 μF, and no capacitor can exceed 100 V, the maximum potential difference that can exist between points A and B is 100 V.
The maximum energy that can be stored in the three-capacitor arrangement is determined by the capacitor with the least energy storage capability. In this case, the capacitors are in series, so the total energy stored in the arrangement is the energy stored by the capacitor with the smallest capacitance, which is 20 μF, at the maximum allowed voltage of 100 V.
How does electric force depend on the amount of charge and the distance between charges
Explanation:
The attractive or repulsive forces which act between any two charged species is an electric force.The electric force depends on the distance between the charged species and the amount of charge which can be calculated by the formula given as followsF = k×[tex]\frac{q1q2}{r2}[/tex]
where, K is coulombs constant, which is equal to - 9 x10^9 [tex]Nm^2/C^2.[/tex]
The unit for K is newtons square meters per square coulombs.This is known as Coulomb's Law.When a 70 kg man sits on the stool, by what percent does the length of the legs decrease? Assume, for simplicity, that the stool's legs are vertical and that each bears the same load.
The diameter of one leg of the stool is missing and it's 2cm.
Answer:
(ΔL/L) = 0.00729%
Explanation:
If the Weight of the man is W, the weight will be distributed equally on the 3 legs and so the reactions for each leg will be W/3 or F/3.
Now, Youngs modulus(Y) of douglas fir wood is about 1.3 x 10^(10) N/m^2. Gotten from youngs modulus of common materials.
Now, weight of man is 70kg.
Now diameter of one leg is 2cm.so radius of one leg = 2/2 = 1cm = 1 x 10^(-2)m
Area for one leg is; π( 1 x 10^(-2)m)^2 = 3.14 x 10^(-4)m
Now as stated earlier, the force on one leg is; F/3.
Now F = mg = 70 x 9.81 = 686.7N
So, force on one leg = 686.7/3 = 228. 9N
Now we know youngs modulus(Y) = Stress/Strain.
Stress = F/A while Strain = ΔL/L
Therefore Y = (F/A) / (ΔL/L)
And therefore, (ΔL/L) = F/(AY)
So (ΔL/L) = 228.9/(3.14 x 10^(-4))x(1.3 x 10^(10)) = 7. 29 x 10^(-5)
When expressed in percentage, it becomes 0.00729%
An accurate calculation for the percent decrease in the stool legs' length under a 70 kg man's weight requires knowledge of the materials properties and dimensions of the stool. General concepts of stress, strain, and deformation in materials are discussed, including their relationship to applied forces and material properties as described by Hooke's Law.
Explanation:To determine the percent decrease in the length of the stool legs when a 70 kg man sits on it, we need additional information such as the material properties of the stool legs and any relevant physical dimensions. However, with the given scenario, we can talk generally about stress and strain in materials and how they are calculated in relation to force and deformation. Stress is the force per unit area applied to a material, while strain is the deformation experienced by the material relative to its original length. For an object like a stool leg, the deformation (change in length) due to a person sitting on it could be calculated using Hooke's Law if the deformation remains within the linear elastic range of the material.
Without specific data on the stool's material, cross-sectional area, and elastic modulus, we cannot calculate the exact percent decrease in length of the stool's legs. If such information were provided, we could use the formula σ = F/A, where σ is the stress, F is the force, and A is the cross-sectional area; and the formula ε = δL/L, where ε is the strain, δL is the change in length, and L is the original length. Combined with Hooke's Law (σ = Eε, E being the elastic modulus), we could find the deformation and thus the percent decrease in length.
A block with mass m = 4.90 kg is placed against a spring on a frictionless incline with angle θ = 45° (The block is not attached to the spring). The spring, with spring constant k = 35.0 N/cm, is compressed 16.0 cm and then released.a.) What is the elastic potential energy of the compressed spring?b.) What is the change inn the gravitational potential energy of the block-Earth system as the block moves from the release point to its highest point on the incline?c.) How far along the inline is the highest point from the release point?
Answer:
a) 44.8J
b) 44.8J
c) 127cm
Explanation:
a) The elastic potential energy of a compressed spring is given by the formula:
[tex]U_e=\frac{1}{2} kx^{2}[/tex]
Where U_e is the elastic potential energy, k is the spring constant and x is the distance the spring is compressed. In this case, we have:
[tex]U_e=\frac{1}{2} (35.0N/cm)(16.0cm)^{2} = 4480Ncm[/tex]
To express the result in Joules, we have to use the fact that 1cm=0.01m. Then:
[tex]U_e=4480Ncm=4480N(0.01m)=44.8J[/tex]
In words, the elastic potential energy of the compressed spring is 44.8J.
b) Using the law of conservation of mechanical energy, we have that:
[tex]E_o=E_f\\\\U_{eo}+U_{go}+K_o=U_{ef}+U_{gf}+K_f[/tex]
Taking t=0 the moment in which the block is released, and t=t_f the point of its maximum height, we have that [tex]U_{g0}=0;K_0=0;U_{ef}=0;K_f=0[/tex] because in t=0 the block has no speed and is in tis lowest point; and in t=t_f the block has stopped and isn't in contact with the spring. So, our equation is reduced to:
[tex]U_{gf}=U_{e0}\\\\U_{gf}=44.8J[/tex]
So, the gravitational potential energy of the block in its highest point is 44.8J.
c) Using the gravitational potential energy formula, we have:
[tex]U_g=mgh\\\\\implies h=\frac{U_g}{mg} \\\\h=\frac{44.8J}{(4.90kg)(9.8m/s^{2}) }=0.9m=90cm[/tex]
Using trigonometry, we can compute the distance between the release point and its highest point:
[tex]d=\frac{h}{sin\theta}=\frac{90cm}{sin45\°}=127cm[/tex]
A 2.0 ???????? capacitor and a4.0 ???????? capacitor are connected in parallel across a 300 V potential difference. Calculate the total energy stored in the capacitor?
Answer:
0.27J
Explanation:
[tex]C_eq= C_1 + C_2\\= 2+4\\= 6UF\\U = (1/2) CV^2\\= (1/2)(6 - 6)(300 * 300)\\= 0.27J[/tex]