Answer:
In econ: When supply and Demand are equal
Explanation:
Equilibrium is when something evens out
A car starts its motion from rest and accelerates with an acceleration of 3m/s2 speed reaches to 20.0m/sec. Find time interval during this motion.
Answer:
The time interval is [tex]t = 6.667 \ s[/tex]
Explanation:
From the question we are told that
The acceleration of the car is [tex]a = 3 m/s^2[/tex]
The velocity of the car is [tex]v= 20.0 \ m/s[/tex]
The final velocity of the car can be mathematically represented as
[tex]v =u + at[/tex]
Now since the car start from rest u = 0 So
[tex]v = 0 + at[/tex]
[tex]v = at[/tex]
substituting values
[tex]20 = 3 t[/tex]
[tex]t = \frac{20}{3}[/tex]
[tex]t = 6.667 \ s[/tex]
Puck A, of inertia mm, is attached to one end of a string of length ℓℓ , and the other end of the string is attached to a pivot so that the puck is free to revolve on a smooth horizontal surface. Puck B, of inertia 20m, is attached to one end of a string of length ℓ/4, and the other end of the string is attached to a second pivot so that BB is also free to revolve. In each case, the puck is held as far as possible from the pivot so that the string is taut and then given an initial velocity v perpendicular to the string.
How does the magnitude of the angular momentum of puck A about its pivot compare with that of puck B about its pivot?
Answer:
[tex]L_{B}[/tex] / [tex]L_{A}[/tex] = [tex]\frac{5mvl}{mvl}[/tex]= 5
Explanation:
Find the given attachment
The definitions of angular momentum allow to find the result for the relationship between the two angular moments is:
[tex]\frac{L_B}{L_A} = 5[/tex]
The angular momentum is the vector product of momentum and vector position.
L = r x p
Where the bold letters indicate vectors, r is the position vector and p the moment. In the body that is in a rotation can be written as a function of the moment of inertia and the angular velocity.
L = I w
Linear and angular variables are related.
v = w L
We substitute
L = I v L
The moment of inertia of a disk with respect to its center is:
I = ½ m r²
Let's write the angular momentum for each disk.
Disc A
[tex]L_A[/tex] = ½ [tex]m_A r^2 v_A L_A[/tex]
Disc B
[tex]L_B[/tex] = ½ [tex]m_B r^2 v_B L_B[/tex]
They indicate that the mass of disk A is m and that of disk b is 20m, the length of the string is l for disk A and l / 4 for disk B.
We substitute
[tex]L_A[/tex] = ½ m r² v l
[tex]L_B[/tex] = ½ (20m) r² v ( [tex]\frac{l}{4}[/tex] )
The relationship between the angular moments is
[tex]\frac{L_B}{L_A} = \frac{20m \frac{l}{4} }{m l}\\ \frac{L_B}{L_A} = 5[/tex]
Consequently using the definitions of angular momentum we can find the result for the relationship between the two angular moments is:
[tex]\frac{L_B}{L_A} = 5[/tex]
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a. If the object is not moving relative to the surface it's in contact with. the friction force is static friction. Draw a free-body diagram of the object. The direction of the friction force is such as to oppose sliding of the object relative to the surface.
b. If the object is slicing relative to the surface, then kinetic friction is acting. From Newton's second law, find the normal force n. The friction force is then directed opposite to the motion, and its magnitude is fk= µkn.
c. If the object is rolling along the surface, then rolling friction is acting. From Newton's second law. find the normal force n. The friction force is then directed opposite to the motion. and its magnitude is fr= µrn.
Answer:
a. The free body diagram for this object has been attached. It shows all the forces acting on the body at rest, including the friction force in the opposite direction to sliding of the object (assume it's left to right).
b. Since the object is in contact with the surface, there is a normal force acting on both of them and is equal to the weight exerted by each. This perpendicular force is defined by Newton's second law of motion.
c. The force of friction always acts in a direction opposite to the direction of motion of the body. F = mg ('a' for acceleration is replaced by 'g' gravity because acceleration in this case is just gravity).
Hope that answers the question, have a great day!
Consider two cars, a 700kg Porsche and a 600kg Honda Civic. The Porsche is speeding along at 40 m/s (mph) and the Civic is going half the speed at 20 m/s. If the two cars brake to a stop with the same constant acceleration, lets look at whether the amount of time required to come to a stop or the distance traveled prior to stopping is influenced by their initial velocity.
1. A car traveling 5m/s slams on its brakes, creating an acceleration of -2 m/s^2. How far did the car travel after it applied its brakes?
2. The same car traveling for 10m/s applies the same acceleration of -2 m/s^2. How far did the car travel after it applied its brakes?
Answer:
Explanation:
To find the distance covered by the car after it applied brakes, we use 3rd equation of motion.
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
1.
We have:
Vi = Initial Velocity = 5 m/s
Vf = Final Velocity = 0 m/s (Since, car finally stops)
a = deceleration = - 2 m/s²
s = distance covered by the car = ?
Therefore,
s = [(0 m/s)² - (5 m/s)²]/2(- 2 m/s²)
s = 6.25 m
2.
We have:
Vi = Initial Velocity = 10 m/s
Vf = Final Velocity = 0 m/s (Since, car finally stops)
a = deceleration = - 2 m/s²
s = distance covered by the car = ?
Therefore,
s = [(0 m/s)² - (10 m/s)²]/2(- 2 m/s²)
s = 25 m
Hence, the distance traveled by the car is affected by the initial speed in accordance with a direct relationship.
A framed picture hangs from two cords attached to the ceiling.
A picture of a picture frame hanging by two cables at the center of the frame at the same length and angle from the vertical.
Which shows the correct free body diagram of the hanging picture?
A free body diagram with two force vectors, the first pointing downward labeled F Subscript g Baseline, the second pointing upward labeled F Subscript N Baseline.
A free body diagram with three force vectors, the first pointing south labeled F Subscript p Baseline, the second pointing northeast labeled F Subscript T Baseline, and the third pointing northwest labeled F Subscript N.
A free body diagram with three force vectors, the first pointing south labeled F Subscript g Baseline, the second pointing northeast labeled F Subscript T Baseline and the third pointing northwest labeled F Subscript T.
Answer:
Answer is C
Explanation:
I just did it E2020
A free body diagram with three force vectors, the first pointing south labeled F Subscript g Baseline, the second pointing northeast labeled F Subscript T Baseline and the third pointing northwest labeled F Subscript T. Thus option C is correct.
What is free body diagram?Free body diagram is defined as a drawing of an interesting object with all of its surroundings removed and the forces at work on the subject's body clearly depicted.
It is also defined as a diagram that reduces the loads and moments acting on a component or system of components.
Free body diagrams are streamlined depictions of an object and the force vectors operating on it in a problem. Since the diagram will depict this body without its surroundings, it is "free" of its environment.
Thus, a free body diagram with three force vectors, the first pointing south labeled F Subscript g Baseline, the second pointing northeast labeled F Subscript T Baseline and the third pointing northwest labeled F Subscript T. Thus option C is correct.
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floating airboat conditions
Answer:
They are used in icy conditions.
Explanation:
They can also go over grassy plains and mucky swamps.
5.Calculate the entropy changes for the following processes:(a)Melting of one mole of tin at its melting point, 213 ᵒC; ΔHfus = 7.029 kJ/mol(b)Evaporation of one mole of liquid carbon dioxide at its boiling point, 216.6 K. ΔHvap = 15.326 kJ/mol
Answer:
a) ΔS = 14.46 J/k
b) ΔS = 70.76 J/k
Explanation:
The general formula to calculate the entropy change accompanied with a process is:
ΔS = ΔQ/T
where,
ΔS = entropy change for the process
ΔQ = Heat Transfer during the process
T = Absolute Temperature during the process
a)
In this case the heat transfer will be given as:
ΔQ = (ΔHfus)(N)
where,
ΔHfus = Molar Heat of Fusion of Tin = 7.029 KJ/mol
N = No. of moles of tin = 1 mol
Therefore,
ΔQ = (7.029 KJ/mol)(1 mol)
ΔQ = 7.029 KJ = 7029 J
and the absolute temperature is:
T = 213°C +273 = 486 k
using these values in the entropy formula, we get:
ΔS = 7029 J/486 k
ΔS = 14.46 J/k
b)
In this case the heat transfer will be given as:
ΔQ = (ΔHvap)(N)
where,
ΔHvap = Molar Heat of Vaporization of Carbon Dioxide = 15.326 KJ/mol
N = No. of moles of Carbon Dioxide = 1 mol
Therefore,
ΔQ = (15.326 KJ/mol)(1 mol)
ΔQ = 15.326 KJ = 15326 J
and the absolute temperature is:
T = 216.6 k
using these values in the entropy formula, we get:
ΔS = 15326 J/216.6 k
ΔS = 70.76 J/k
A microstate is a state of a physical system described at the finest level of detail. A macrostate is a state of a physical system that is described in terms of the systems overall or average properties at a macroscopic level. A macrostate will generally consist of many different microstates. In defining a macrostate we ignore what is going on at the microscopic (atomic/molecular) level.
Suppose 24 students are enrolled in a class. For simplicity let’s assume that any student who is present is sitting completely still at his or her assigned seat, facing forward (so we don’t have to worry about students being in different locations, having different motions, etc.).
1. Which of the following would constitute a macrostate description of the class attendance and which would constitute a microstate description.
a. A list of the names of each student present today.
b. The number of students in attendance.
Answer:
a. A list of the names of each student present today. (microstate)
b. The number of students in attendance. (macrostate)
Explanation:
You can fins the answer to this question by comparing the situation of the problem with a system of molecules with discrete energy.
Without importance of which molecules have a specific energy, but rather, what is the total amount of energy, you can get for different configurations of energy the same amount of the total energy. If different configurations of the energies of the molecules give you the same total energy of the system, you say that the macrostate is the same. In the case of the classroom, it does not matter how are distributed the students in the class, the total number of students is always the same. The macrostate is the same for what ever organization of the students in the class.
If you would interested in the energy of each molecules, you will obtain different configurations. In the case of the classroom. The names of the student will define a microstate because in this case there are many configurations.
a. A list of the names of each student present today. (microstate)
b. The number of students in attendance. (macrostate)
Two tiny conducting spheres are identical and carry charges of - 20.0 µC and +50.0 µC. They are separated by a distance of 2.50 cm. (a) What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive? (b) The spheres are brought into contact and then separated to a distance of 2.50 cm. Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive of repulsive.
Answer:
a
The force experience by the two spheres is [tex]F_1 = 1.44*10^4 N[/tex]
This force is attractive cause the charge are unlike charges
b
The force experienced by the two spheres is [tex]F_2 = 3.24*10^{3} \ N[/tex]
The force is repulsive because the two charges are like charges
Explanation:
From the question we are told that
The charge on the first sphere is [tex]q_1 = -20.0 \mu C = 20 *10^{-6} C[/tex]
The charge on the second sphere is [tex]q_2 = 50 \mu C = 50*10^{-6} C[/tex]
The distance of separation is [tex]d = 2.50 \ cm = \frac{2.50}{100} = 0.025 \ m[/tex]
The electrostatic force experienced by the two spheres is mathematically represented as
[tex]F_1 = \frac{k q_1 q_2}{d^2}[/tex]
Now k is the coulomb’s constant with a value of [tex]k = 9*10^9 N \cdot m^2 /C^2[/tex]
So
[tex]F_1 = \frac{9*10^9 * 20 *10^{-6} * 50*10^{-6}}{0.025}[/tex]
[tex]F_1 = 1.44*10^4 N[/tex]
When the sphere are brought together the charge on each sphere would be the average of the total charge and this can be mathematically evaluated as
[tex]q = \frac{q_1 + q_2 }{2}[/tex]
[tex]q = \frac{(-20 + 50)*10^{-6} }{2}[/tex]
[tex]q = 15 \mu C[/tex]
So when they are seperated the electrostatic force experienced is
[tex]F_2 = \frac{kq^2}{d^2}[/tex]
[tex]F_2 = \frac{ 9*10^9 * (15 *10^{-6})}{0.025}[/tex]
[tex]F_2 = 3.24*10^{3} \ N[/tex]
Jake is a farmer who owns horses and cattle. How often should Jake deworm his animals to keep them healthy?
Explanation:
After every 3 to 4 months
make sure your response is 3-5 sentences.
What are some of the extreme conditions in space that challenge manned space exploration?
Answer: please see below
Explanation:
A manned space exploration is defined as the exploration of individuals --- astronauts in space using a spacecraft as a vehicle and are responsible for operating its controls
The extreme conditions in space that challenge manned space exploration is as follows.
1. extreme loud sound waves cause by the launch of spacecraft which can shatter the spacecraft
2. extreme Temperatures in space ranging from extreme hot temperatures (near the sun) to extreme cold temperatures ( below freezing point out of space.
3.micrometeorite showers responsible for sandblasting can damage spacecraft.
4.Ultra violet Radiation which can alter the control unit of the spacecraft
Because of theses extreme conditions that pose challenges to space explorations, necessary precautions should be taken into consideration to be able to overcome such challenges. These precautions include building the spacecraft and the control unit in such a way that can resist these harmful conditions, also taking in mind safe escape routes for the astronauts in case of failures.
Derive the expression of the mechanical energy for under damped system.
Answer: the expression of the mechanical energy for under damped system is;
x(t)=Ae−γ/2tcos(ωdt+ϕ), where ωd=ω02−γ2/4
γ = damping rate, and
ω0 = the angular frequency of the oscillator without damping.
Explanation:
The physical situation in mechanical energy defined through out the world has three possible results depending on the value of a (which is a constant value), which depends on the value of what is under our radical. This expression can either be positive, negative, or equal to zero which will result in overdamping, underdamping, and critical damping, as the case may be.
γ2 >4ω²0 This is the Over Damped case. Here, the system returns to equilibrium by exponentially decaying towards zero, and the system will not pass that equilibrium position more than once.
γ² < 4ω²0 this is the Under Damped case. Here, the system moves back and forth as it slowly returns to equilibrium and the amplitude of the system decreases over time.
Finally, γ² = 4ω²0
This is the Critically Damped case. Here, the system returns to equilibrium very fast without moving back and forth and without passing the equilibrium position at all.
The 0.100 kg sphere in (Figure 1) is released from rest at the position shown in the sketch, with its center 0.400 m from the center of the 5.00 kg mass. Assume that the only forces on the 0.100 kg sphere are the gravitational forces exerted by the other two spheres and that the 5.00 kg and 10.0 kg spheres are held in place at their initial positions.
Answer:
the speed of 0.10 sphere when it is moved to 0.20 kg to the left is : [tex]3.3337*10^{-5} \ m/s[/tex]
Explanation:
Using the expression of the Change in Gravitational Potential Energy:
[tex]U= -(\frac{ Gm_1m_2 }{r_2} - \frac{ Gm_1m_2 }{r_1}) \\ \\ U=Gm_1m_2 (\frac{ 1 }{r_2} - \frac{ 1 }{r_1})[/tex]
So when the sphere exerts just 10 kg mass; the change in the gravitational potential energy is :
[tex]U_1=Gm_1m_2 (\frac{ 1 }{r_2} - \frac{ 1 }{r_1})[/tex]
[tex]U_1=6.67*10^{-11}*0.1 \ kg*10 \ kg(\frac{ 1 }{0.6 \ m} - \frac{ 1 }{0.8 \ m})[/tex]
[tex]U_1 = 2.778*10^{-11} J[/tex]
the change in the gravitational potential energy when the sphere exerts just 5 kg mass is ;
[tex]U_2=Gm_1m_2 (\frac{ 1 }{r_2} - \frac{ 1 }{r_1})[/tex]
[tex]U_2=6.67*10^{-11}*0.1 \ kg*5 \ kg(\frac{ 1 }{0.4 \ m} - \frac{ 1 }{0.2 \ m})[/tex]
[tex]U_2 = -8.335*10^{-11} J[/tex]
The net total change is:
[tex]U_{total } = U_1 +U_2[/tex]
[tex]U_{total} = 2.778*10^{-11} + (-8.335*10^{-11})[/tex]
[tex]U_{total} = -5.557*10^{-11}[/tex]
We all know that for there to be a balance ;loss of gravitational potential energy must be equal to the gain in kinetic energy .
SO;
K.E = [tex]5.557*10^{-11}[/tex]
[tex]\frac{1}{2}mv^2 = 5.557*10^{-11}[/tex]
[tex]v^2 = \frac{2*5.557*10^{-11} \ J}{m_1}[/tex]
[tex]v=\sqrt{ \frac{2*5.557*10^{-11} \ J}{0. 1 \kg}[/tex]
v = [tex]3.3337*10^{-5} \ m/s[/tex]
Thus, the speed of 0.10 sphere when it is moved to 0.20 kg to the left is : [tex]3.3337*10^{-5} \ m/s[/tex]
The speed of 0.10 sphere when it is moved to 0.20 kg to the left is :3.3337 *10^-5 m/s
What is speed ?The speed of any object is defined as the movement of any object with respect to the time.
By using the expression of the Change in Gravitational Potential Energy:
[tex]U=-Gm_1m_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})[/tex][tex]U_2=-Gm_1m_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})[/tex]
So when the sphere exerts just 10 kg mass; the change in the gravitational potential energy is :
[tex]U_1=-Gm_1m_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})[/tex]
[tex]U_1=- 6.67\times 10^{-11}\times 0.1\times 10(\dfrac{1}{0.6}-\dfrac{1}{0.8})[/tex]
[tex]U_1=2.778\times 10^{-11][/tex]
The change in the gravitational potential energy when the sphere exerts just 5 kg mass is ;
[tex]U_2=-Gm_1m_2(\dfrac{1}{r_2}-\dfrac{1}{r_1})[/tex]
[tex]U_2=- 6.67\times 10^{-11}\times 0.1\times 5(\dfrac{1}{0.4}-\dfrac{1}{0.2})[/tex]
[tex]U_2=-8.335\times 10^{-11}\ J[/tex]
The net total change is:
[tex]U_{total}=U_1+U_2[/tex]
[tex]U_{total}=2.778\times 10^{-11}+(-8.335\times 10^{-11})[/tex]
[tex]U_{tot\leq al}=-5.557\times 10^{-11}[/tex]
We all know that for there to be a balance ;loss of gravitational potential energy must be equal to the gain in kinetic energy .
SO;
[tex]\rm KE=5.557\times 10^{-11}[/tex]
[tex]\dfrac{1}{2}mv^2=5.557\times 10^{-11}[/tex]
[tex]v^2=\dfrac{2\times 5.557\times 10^{-11}}{m}[/tex]
[tex]v=\sqrt{\dfrac{2\times 5.557\times 10^{-11}}{m}}[/tex]
[tex]v=3.337\times 10^{-5}\ \frac{m}{s}[/tex]
Thus, the speed of 0.10 sphere when it is moved to 0.20 kg to the left is : [tex]v=3.337\times 10^{-5}\ \frac{m}{s}[/tex]
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Identify three pollutants released into the air when fossil fuels are burned.
Answer: Carbon dioxide (CO2), Carbon monoxide (CO), Methane (CH4)
4
Select the correct answer.
Which of the following is evidence of the past existence of glaciers?
A.
rift valleys
B. fjords
C.
estuaries
D.
oceanic ridges
Answer:
B. fjords
explanation :
Fjords were created by glaciers. In the Earth's last ice age, glaciers covered just about everything. Glaciers move very slowly over time, and can greatly alter the landscape once they have moved through an area. This process is called glaciation.The fjords are one of the glaciers that existed in the past. They are one of the many glacial relief forms that can give us a insight into the size and power of the glaciers.
Answer:
B. fjords
I hope this helps
Find Density of a cylinder with diameter of 5.0 cm and height of 12.0cm and mass of 600.0g
Answer: 2.55 g/cm^3
Explanation:
density is defined as:
Density = mass/volume
Now, the mass of the cylinder is 600g
and the volume of a cylinder is:
V = pi*r^2*h
where r is the radius (half of the diameter), here r = (5/2)cm and h is the height, here 12 cm
So the volume is:
V = 3.14*(2.5cm)^2*12cm = 235.5cm^3
then the density is:
D = 600g/235.5cm^3 = 2.55 g/cm^3
A 0.050 kg bullet strikes a 5.0 kg wooden block with a velocity of 909 m/s and embeds itself in the block which fies off its stand. what was the final velocity of the bullet?
Answer:
The final velocity of the bullet is 9 m/s.
Explanation:
We have,
Mass of a bullet is, m = 0.05 kg
Mass of wooden block is, M = 5 kg
Initial speed of bullet, v = 909 m/s
The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,
[tex]mv=(m+M)V\\\\V=\dfrac{mv}{m+M}\\\\V=\dfrac{0.05\times 909}{0.050+5}\\\\V=9\ m/s[/tex]
So, the final velocity of the bullet is 9 m/s.
Which factor indicates the amount of charge on the source charge?
A. the number of field lines on the test charge
B. the number of field lines on the source charge
C. the direction of lines on the source charge
D. the direction of lines on the test charge
Answer:
B. the number of field lines on the source charge
Explanation:
As we know that electric flux is defined as the number of electric field lines passing through a given area.
So here electric flux due to a point charge "q" is given by
so here we know that flux depends on the magnitude of charge and hence we can say that number of filed lines originating from a point charge will depends on the magnitude of the charge.
The factor indicates the amount of charge on the source charge is the number of field lines on the source charge.
What is electric flux?The electric flux is defined as the number of electric field lines passing through a given area.
The electric flux due to a point charge q is given by the number of filed lines through particular closed area.
We know that flux depends on the magnitude of charge and number of field lines starting from a point charge will depends on the magnitude of the charge.
Thus, the correct option is B.
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A person with normal vision can focus on objects as close as a few centimeters from the eye up to objects infinitely far away. There exist, however, certain conditions under which the range of vision is not so extended. For example, a nearsighted person cannot focus on objects farther than a certain point (the far point), while a farsighted person cannot focus on objects closer than a certain point (the near point). Note that even though the presence of a near point is common to everyone, a farsighted person has a near point that is much farther from the eye than the near point of a person with normal vision.
Both nearsightedness and farsightedness can be corrected with the use of glasses or contact lenses. In this case, the eye converges the light coming from the image formed by the corrective lens rather than from the object itself.If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that the person would need to see an object at infinity clearly?
Answer:
Explanation:
For a person suffering from nearsightedness, far point is 3.5 m . That means , an object beyond 3.5 m will not be clearly visible by the person . Ray of light coming from 3.5 m will be focussed on retina . For an object to be visible from infinity , ray of light coming from infinity should appear to be coming from 3.5 m . In other words , the object placed at infinity should form a virtual image at 3.5 m . This can be done by concave lens.
u = ∝ ; v = 3.5
Using lens formula
[tex]\frac{1}{v} -\frac{1}{u} = \frac{1}{f}[/tex]
[tex]\frac{1}{-3.5} -0 = \frac{1}{f}[/tex]
f = - 3.5 m .
A 6,000-km undersea glass fiber phone line crosses the Atlantic ocean connecting the US and France. (a) How long does it take for a message to transverse this link? (b) How long does it take for a message to travel from US to France by using a satellite link? The satellite is stationed about 22,000 miles above the earth between the US and France. (c) Will two people having a conversation across these two different links notice the travel delays?
Answer:
Explanation:
Message will travel through the fibre at the speed of 3 x 10⁸ m /s which is also the speed of light .
a ) Time for a message to travel from US to France by using glass fibre link
time = distance / speed
= 6000 x 1000 / 3 x 10⁸
= .02 s .
b )
22000 miles = 1.6 x 22000
= 35200 km
Time for a message to travel from US to France by using a satellite link
time = distance / velocity
distance = 2 x √ ( 35200² + 3000² )
= 2 x 35327 km
= .7 x 10⁸ m .
time = .7 x 10⁸ / 3 x 10⁸
= .233 s
c )
Difference of time = .233 - .02
= .213 which is more than 1/6 s for which effect of hearing stays in the brain , which is called persistence of vision . So they will be heard as different sounds.
Which planet is an inner planet, and which one is an
outer planet? Explain your answer. (this is a essay question and you have to type a answer)
Answer:
Inner planets rotate slowly and so they take more time to complete a rotation. Whereas, outer planets rotate faster and so they take less time to complete a rotation.
Explanation:
The inner planets are the ones that are much closer to the sun and as such they have much smaller orbits around them and take much less time to circle around it. These planets are much smaller, which enables them rotate slowly, so they take more time to complete a rotation..
Whereas, the outer planets differ from the inner planets in that they are much further from the sun and so take much more time for them to be able to make a single circle around the su and also, because they are much bigger, it takes them much more time to spin around their own axis, thus Outer planets rotate faster, so they take less time to complete a rotation
A small particle with positive charge q = +4.25 x 10^-4C and mass m = 5.00 x 10^-5 kg is moving in a region of uniform electric and magnetic fields. The magnetic field is B=4.00 T in the +z-direction. The electric field is also in the +z-direction and has magnitude E = 60.0 N/C. At time t = 0 the particle is on the y-axis at y = +1.00 m and has velocity v = 30.0 m/s in the +x-direction. Neglect gravity.
a) What are the x-, y-, and z-coordinates of the particle at t = 0.0200 s?
b) What is the speed of the particle at t = 0.0200 s?
Answer:
a) r = (0.6 i- 2039 j ^ + 0.102 k⁾ m and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s c) v_{z} = 1.02 10⁻¹m / s
Explanation:
a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration
Fe + Fm = m a
a = (Fe + Fm) / m
the electric force is
Fe = q E k ^
Fe = 4.25 10-4 60 k ^
Fe = 2.55 10-2 k ^
the magnetic force is
Fm = q v x B
Fm = 4.25 10⁻⁴ [tex]\left[\begin{array}{ccc}i&j&k\\30&0&0\\0&0&49\end{array}\right][/tex]
fm = 4.25 10⁻⁴ (-j ^ 30 4)
fm 0 = ^ -5,10 10⁻² j
We look for every component of acceleration
X axis
aₓ = 0
there is no force
Axis y
ay = -5.10 10²/5 10⁻⁵ j ^
ay = -1.02 107 j ^ / s2
z axis
az = 2.55 10⁻² / 5 10⁻⁵ k ^
az = 5.1 10² k ^ m / s²
Having the acceleration in each axis we can encocoar the position using kinematics
X axis
the initial velocity is vo = 30 m / s and an initial position xo = 0
x = vo t + ½ aₓ t₂2
x = 30 0.02 + 0
x = 0.6m
Axis y
acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m
y = I + go t + ½ ay t²
y = 1 + 0 + ½ (-1.02 10⁷) 0.02²
y = 1 - 2.04 10³
y = -2039 m j ^
z axis
acceleration is aza = 5.1 10² m / s², the position and initial speed are zero
z = zo + v₀ t + ½ az t²
z = 0 + 0 + ½ 5.1 10² 0.02²
z = 1.02 10⁻¹ m k ^
therefore the position of the bodies is
r = (0.6 i- 2039 j ^ + 0.102 k⁾ m
b) x axis
since there is no acceleration the speed remains constant
vₓ = 30.0 m / s
Axis y
let's use the equation v = v₀ + [tex]a_{y}[/tex] t
[tex]v_{y}[/tex] = 0 + -1.02 10⁷ 0.02
v_{y} = 2.04 10⁵ m / s
z axis
v_{z} = vo + az t
v_{z} = 0 + 5.1 10² 0.02
v_{z} = 1.02 10⁻¹m / s
A) The coordinates of the particle in x, y and z axis is written as :
( 0.6 i - 2039 j + 0.102 k )m ( i.e. x = 0.6, y = 2039, z = 0.102 )
B) The speed of the particle at t = 0.0200 s
x-axis = 30 m/s y-axis = 2.04 * 10⁵ m/s z-axis = 1.02 * 10⁻¹ m/sA ) Determine the coordinates of the particle in x, y and z axis
applying Newton's second law ; a = ( Fe + Fm ) / m
where : Fe = 2.55 * 10⁻² k
The magnetic force ( Fm ) = qv * B = [tex]\left[\begin{array}{ccc}i&j&k\\30&0&0\\0&0&49\end{array}\right][/tex]
∴ Fm = 0
Acceleration in each axis ( x , y and z )
Ax = 0 because there is no force
Ay = - 1.02 * 10⁷ j/s²
Az = 5.1 * 10² km/s²
i) For the x- axis
x = Vot + ½ * Ax * t²
Vo = 30 m/s
t = 0.02
Ax = 0
∴ x - coordinate of the particle = 0.6m
ii) For the y-axis
y = I + Vo*t + ½ *Ay* t²
y = - 2039 m
iii) For the z-axis
z = zo + v₀* t + ½ * Az* t²
z = 0.102 m .
B ) Determine the speed of the particle in all three axis
Vx = 30 m/s
Vy = v₀ + Ay * t
= 30 + - 1.02 * 10⁷ * 0.020
= 2.04 * 10⁵ m/s
Vz = Vo + Az* t
= 1.02 * 10⁻¹
Hence we can conclude that The coordinates of the particle in x, y and z axis is written as : ( 0.6 i - 2039 j + 0.102 k )m ( i.e. x = 0.6, y = 2039, z = 0.102 )The speed of the particle at t = 0.0200 s
x-axis = 30 m/s y-axis = 2.04 * 10⁵ m/s z-axis = 1.02 * 10⁻¹ m/sLearn more : https://brainly.com/question/19915685
Photons with shorter wavelengths have larger
o
A. color
B. amplitude
c. energy
D. speed
SUBMIT
i think photons with shorter wavelengths have larger speed.
I apologise if its incorrect.
Answer:
energy
Explanation:
Why does the gravitational force obey an inverse square law?
Explanation:
the gravity force obeys the inverse square law because according to the newtons law of gravitaion the force of the gravitation is directly proportional to the product of masses ... In this way it depends upon the masses directly.We donot notice any gravitational force because we have very small masses as compared to the plantery universe and the solar system ... due to there greater masses they exert a large force and thus by increasing the distance the force of attraction varies inversely to the force of gravitation...thus gravitional force is a variable quantity ..............varies inversely to the square of distance between the to bodies...
F ∝1/r2
A wire of length L is used in an electric heater. When the potential
difference across the wire is 200 V, the power dissipated in the wire is
1000 W. The same potential difference is applied across a second similar wire of
length 2L. What is the power dissipated in the second wire?
Answer:500 W
Explanation:
Given
When length is L power dissipated is [tex]100\ W[/tex]
Potential difference applied is [tex]200\ V[/tex]
Resistance of wire is directly proportional to length of wire
Suppose initially Resistance is R
When L changes to [tex]2L[/tex]
Resistance changes to [tex]2R[/tex]
Initial Power dissipated is [tex]P_o=\frac{V^2}{R}[/tex]
For 2 R resistance and same voltage, Power dissipated is
[tex]P_1=\frac{V^2}{(2R)}[/tex]
So [tex]P_1=\frac{1}{2}\times \frac{V^2}{R}[/tex]
[tex]P_1=\frac{P_o}{2}[/tex]
So [tex]P_1=\frac{1000}{2}[/tex]
[tex]P_1=500\ W[/tex]
The transfer of energy from one organism to the next in an ecosystem begins with a producer. A producer is an organism that produces its own food. A consumer is an organism in a food chain that obtains energy from producers or other consumers; consumers may be herbivores or carnivores.
Which food chain correctly describes the flow of energy in an ecosystem?
A) grass → grasshopper → fish → human
B) grasshopper → fish → human
C) human → fish → grasshopper
D) human → grasshopper → fish → grass
Answer:
A.)
Explanation:
The answer is food chain A, as plants produce their own food using photosynthesis, therefore making them producers.
A grasshopper then eats this grass, making it a consumer as it is obtaining its energy from the producer. It is then eaten by a fish and a human. This chain accurately describe the flow of energy in an ecosystem, as it goes, producer, consumer, consumer, consumer
Hope this helps a little
Answer:
sorry
Explanation:
patulong
____is the distance traveled during a specific unit of time
What do you mean by peltier coefficient?
Answer:
The Peltier coefficient is a measure of the amount of heat carried by electrons or holes
Explanation:
Answer:
The Peltier coefficient is a measure of the amount of heat carried by electrons or holes.
Explanation:
A metallic circular plate with radius r is fixed to a tabletop. An identical circular plate supported from above by a cable is fixed in place a distance d above the first plate. Assume that dd is much smaller than r. The two plates are attached by wires to a battery that supplies voltage V.
A)What is the tension in the cable? Neglect the weight of the plate.
Express your answer in terms of the variables d, r, V, and constants ϵ0, π.
B)The upper plate is slowly raised to a new height 2d. Determine the work done by the cable by integrating ∫(from d to 2d) F(z)dz, where F(z) is the cable tension when the plates are separated by a distance z.
Express your answer in terms of the variables d, r, V, and constants ϵ0, π.
C)Compute the energy stored in the electric field before the top plate was raised.
Express your answer in terms of the variables d, r, V, and constants ϵ0, π.
D)Compute the energy stored in the electric field after the top plate was raised.
Express your answer in terms of the variables d, r, V, and constants ϵ0, π.
E)Is the work done by the cable equal to the change in the stored electrical energy? If not, why not?
a)The work done in separating the plates is equal to energy change in the plates.
b)The work done in separating the plates is equal to the magnitude of the energy change in the plates. This does not mean that the work done is equal to the change in the energy stored in the plates. The work done on the plates is positive but the plates lose energy. The plates are connected to the battery, so the potential difference across them remains constant as they are separated. Therefore charge is forced off of the plates through the battery, which does work on the battery.
Answer:
the tension in the cable is [tex]\mathbf{F = \frac{\pi E_o v^2r^2}{2d^2}}[/tex]
the work done by the cable is [tex]\mathbf{W= \frac{\pi E_ov^2r^2}{4d}}[/tex]
Explanation:
A)
If we have two circular plate supported by a cable at a fixed distance, then the electric field formed between the two plate of the capacitor can be represented by the equation.
[tex]\mathbf{E = \frac{voltage \ \ V}{distance \ \ d}}[/tex]
However; the net electric field i.e the sum of the electric filed produced is represented as:
[tex]\mathbf{E' = \frac{E}{2}} \\ \\ \mathbf{E' = \frac{V}{2d}}[/tex]
So, if we assume that the lower plate and the upper plate possess the charge +q and -q respectively. Then, the tension of the cable which is the same as Force F can be written as:
[tex]\mathbf{F = q* E'}[/tex]
[tex]\mathbf{F = \frac{q*v}{2d}}[/tex] ----- equation (1)
Also ; we know that
[tex]\mathbf{C = \frac{q}{v}= \frac{E_oA}{d}}[/tex]
[tex]\mathbf{\frac{q}{v}= \frac{E_o \pi r^2}{d}} \ \ \ \ \ \mathbf{since \ A = \pi r^2}[/tex]
[tex]\mathbf{{q}= \frac{\pi E_o {v} r^2}{d}}[/tex] ----- equation (2)
Replacing equation 3 into equation (2); we have:
[tex]\mathbf{F = \frac{\pi E_o vr^2}{d}* \frac{v}{2d}}[/tex]
[tex]\mathbf{F = \frac{\pi E_o v^2r^2}{2d^2}}[/tex]
Therefore, the tension in the cable is [tex]\mathbf{F = \frac{\pi E_o v^2r^2}{2d^2}}[/tex]
B)
Assume that the upper plate is displaced by dz in an upward direction ; Then we can express the workdone by the tension as :
[tex]\mathbf{dW = T *dz} \\ \\ \mathbf{dW = F*dz} \\ \\ \mathbf{dW = \frac{\pi E_o v^2r^2}{2z^2}dz }[/tex]
The net workdone to raise the plate from separation d to 2d is:
[tex]\mathbf{W = \int\limits^{2d}_{2zd} {dw} = \frac{\pi E_ov^2r^2}{2} \int\limits^{2d}_d \frac{dz}{z^2} }[/tex]
[tex]\mathbf{W= \frac{\pi E_ov^2r^2}{2} [-\frac{1}{z}]^{2d}_d }[/tex]
[tex]\mathbf{W= - \frac{\pi E_ov^2r^2}{2} [\frac{1}{2d}-\frac{1}{d}]}[/tex]
[tex]\mathbf{W= - \frac{\pi E_ov^2r^2}{2} [\frac{-1}{2d}]}[/tex]
[tex]\mathbf{W= \frac{\pi E_ov^2r^2}{4d}}[/tex]
the work done by the cable is [tex]\mathbf{W= \frac{\pi E_ov^2r^2}{4d}}[/tex]
C) To calculate the energy stored in the Electrical energy Capacitor before the top plate is raised ; we have:
[tex]\mathbf{U_i = \frac{1}{2}Cv^2} \\ \\ \mathbf{U_i = \frac{1}{2}(\frac{E_oA}{d})v^2} \\ \\ \mathbf{U_i = \frac{1}{2}(\frac{E_o \pi r^2}{d})v^2} \\ \\ \mathbf{U_i = \frac{E_o \pi r^2 v^2}{2d}} }[/tex]
D) The energy stored in the plate after the the top plate was raised is as follows:
[tex]\mathbf{U_f = \frac{1}{2}C'v^2} \\ \\ \mathbf{U_f = \frac{1}{2}(\frac{E_oA}{2d})v^2} \\ \\ \mathbf{U_f = \frac{1}{2}(\frac{E_o \pi r^2}{2d})v^2} \\ \\ \mathbf{U_f = \frac{E_o \pi r^2 v^2}{4d}} }[/tex]
E) Yes, work done by the cable equal to the change in the stored electrical energy. The Difference in energy stored before and after the top plate is raised:
[tex]\mathbf{U_i-U_f} = \mathbf{\frac{E_o \pi r^2 v^2}{2d}} }} - \mathbf {\frac{E_o \pi r^2 v^2}{4d}} }}[/tex]
[tex]\mathbf{U_i-U_f}= \mathbf {\frac{E_o \pi r^2 v^2}{4d}} }}[/tex]
Thus;
b)The work done in separating the plates is equal to the magnitude of the energy change in the plates. This does not mean that the work done is equal to the change in the energy stored in the plates.
If you have two spoons of the same
size, one silver and one stainless
steel, there is a quick test to tell
which is which. Hold the end of a
spoon in each hand, then lower them
both into a cup of very hot water.
One spoon will feel hot first. Is that
the silver spoon or the stainless steel
spoon? Explain.
Answer: Silver
Explanation:Silver has ahigh thermal conductivity of 429 watts/meter-k. Stainless steel on the other hand, has low thermal conductivity of 160 watts/meter-k.
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