Answer:
The correct option is
(e)either (c) or (d) could be correct.
Explanation:
The electric field of a charge radiates out in all directions and the intensity of the electric field strength given by E = F/q₀, diminishes as the lines of force moves further away from the source. The direction of F and E is in the line of potential motion of the source charge in the field.
Equipotential surfaces are locations in the radiated electric that have the same field strength or electric potential. The work done in moving within an equipotential surface is zero and as such since
Work = Force × distance = 0 where distance ≠ 0.
The force acting between two points on an equipotential surface is also zero or the component of the force within an equipotential surface is zero and since there is a force in the electric field, it is acting at right angles to the equipotential surface which could be horizontally to the left or right directions where the equipotential surfaces due to the charge distribution are in the vertical plane.
Therefore it is either horizontally to the left, or horizontally to the right.
Which of the following physical laws or principles can best be used to analyze the collision between the object and the pendulum bob? Which can best be used to analyze the resulting swing? 1. Newton's first law 2. Newton's second law 3. Newton's third law 4. Conservation of mechanical energy 5. Conservation of momentum
Answer:
4,5
Explanation:
The resulting swing converts potential energy to kinetic energy and kinetic energy to potential energy when the swinging stops. This is line with the law of conservation of mechanical energy which deals with inter conversion of energy forms and energy not being able to get lost.
The conservation of momentum is most suited to the collision. This law states that when a collision occurs the initial momentum before and after the collision is the same(without any external force).
Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g = 30~\rm m/s^2. When he releases the ball from chin height without giving it a push, how will the ball's behavior differ from its behavior on Earth? Ignore friction and air resistance. (Select all that apply.)a.It will smash his face.b.It will stop well short of his face.c.It will take less time to return to the point from which it was released.d.Its mass will be greater.e.It will take more time to return to the point from which it was released
Answer:
C
Explanation:
- Let acceleration due to gravity @ massive planet be a = 30 m/s^2
- Let acceleration due to gravity @ earth be g = 30 m/s^2
Solution:
- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:
t = v / a
t = v / 30
- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:
t = v / g
t = v / 9.81
- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C
Two forces act on a 8.50-kg object. One of the forces is 14.0 N. If the object accelerates at 3.50 m/s2 , what is the greatest possible magnitude of the other force
Answer:
The magnitude of the other force is 43.75 N.
Explanation:
Given that,
Mass of the object, m = 8.5 kg
Force 1, [tex]F_1=14\ N[/tex]
Acceleration of the object, [tex]a=3.5\ m/s^2[/tex]
To find,
The greatest possible magnitude of the other force.
Solution,
Let [tex]F_2[/tex] is the magnitude of other force that is acting on the object. For the greatest force, the two forces must be act in opposite direction such that :
[tex]F_2-F_1=ma[/tex]
[tex]F_2=ma+F_1[/tex]
[tex]F_2=14+8.5\times 3.5[/tex]
[tex]F_2=43.75\ N[/tex]
So, the magnitude of the other force is 43.75 N.
What is the effect on the force of gravity between two objects if the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled? Choose one:A. It always increases.B. It always decreases.C. It depends on the specific values of the two quantities.D. It depends on the local value of G.E. It cannot be determined.
Answer:
The correct answer to the question is
B. It always decreases
Explanation:
To solve the question, we note that the foce of gravity is given by
[tex]F_G=\frac{Gm_1m_2}{r^2}[/tex] where
G= Gravitational constant
m₁ = mass of first object
m₂ = mass of second object
r = the distance between both objects
If the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled, we have
[tex]F_{G2} =\frac{Gm_1(2m_2)}{(2r)^2}[/tex] = [tex]\frac{2}{4} \frac{Gm_1m_2}{r^2}[/tex]
Therefore the gravitational force is halved. That is it will always decrease
What do we mean when we say that the sun is in energy balance?
Answer:The amount of energy released by fusion in the Sun's core equals the amount of energy radiated from the Sun's surface into space.
Explanation: Energy balance is a used to describe the balance between the amount of energy input and the amount of energy output. When the amount of energy released is Equal to the amount of energy given back to the System we say there is an energy balance.
The Sun is the major source of energy in the Universe, and it produces energy through FUSION OF RADIOACTIVE MATERIALS IN ITS CORE, WHEN THE ENERGY RELEASED BY FUSION THE SUN IS EQUAL TO THE ENERGY RELEASED TO THE OUTER SPACE WE SAY THERE IS ENERGY BALANCE.
If a home uses a large supply of solar panels to generate electricity, but has no battery system, surplus electricity that is produced is usually __________.
Answer:
Passed into the power grid for others to use the electricity
Explanation:
If a home uses a large supply of solar panels to generate electricity, but has no battery system, surplus electricity that is produced is usually passed into the power grid for others to use the electricity, generating a income to the homeowner
When a home using solar panels produces surplus electricity and has no battery system, the excess power is typically fed back into the electricity grid. This process, known as net metering, can often result in a credit from the electricity company.
Explanation:If a home uses a large supply of solar panels to generate electricity, but has no battery system, surplus electricity that is produced is usually fed back into the grid. This feed-in process involves the excess electricity being transmitted to the local electricity network where it can be used by other homes or businesses. In many places, electricity companies provide a credit to the solar power producer for this extra electricity, which can lower the overall utility bill.
This concept is known as net metering which allows homeowners to offset the cost of power drawn from the utility grid by pushing their surplus electricity into the grid. However, it's crucial to note that this process might depend on the policies of the local utility or the regional grid operator.
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The maximum distance at which a highway sign can be read is determined for a sample of young people and a sample of older people. The mean distance is computed for each age group. What's the research hypothesis about the means of the two groups?
Answer:
The population mean are the same
Explanation:
Answer:
The population means are the same.
Explanation:
The hypotheses for a difference in two population means are similar to those for a difference difference two population proportions.
At null point, Ha=0
Let the mean population of the young one be u1
Let the mean population of the old one be u2.
Then, the difference between their mean population distance is given as
Ha=u2-u1
Since, Ha is null point, Ha=0
0=u2-u1
u2=u1
This shows that the mean population distance of the old is equal to the mean population distance of the young.
Therefore their mean population distance is the same
Since it is null alternative then, the population mean are the same.
We must sample the population using
1. Samples must be random to remove or minimize bias.
2. Sample must be representative of the populations in question.
9) Cart 1 has a mass of 4 kg and an initial speed of 4 m/s. It eventually elastically collides with cart 2, whose mass is 6 kg, and which moves at an initial speed of 4 m/s toward cart 1. How fast are the carts moving after their collision? [Enter cart 1's final speed in answer box 1 and cart 2's final speed in answer box 2.]
Answer:
Cart 1 = 4 m/s
Cart 2 = 4 m/s
Explanation:
See attachment
The parallel plates in a capacitor, with a plate area of 8.50 cm2 and an air-filled separation of 3.00 mm, are charged by a 6.00 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 8.00 mm. Neglecting fringing, find (a) the potential difference between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates
Final answer:
The problem about a parallel-plate capacitor requires the application of electromagnetism principles to determine changes in potential difference, initial and final stored energy, and the work required to separate the plates, all based on the capacitor's geometry and the effect of plate separation on capacitance.
Explanation:
To answer the student's question about the parallel-plate capacitor, we need to apply concepts from electromagnetism, specifically the relationships between charge, voltage, capacitance, and energy in capacitors. When the parallel plates of a capacitor are separated, the capacitance changes, but the charge remains the same since it is isolated after being disconnected from the battery. The potential difference between the plates changes as a result of the changing capacitance. The initial energy stored in the capacitor can be calculated using the formula U = 1/2 CV^2, and the final energy stored after increasing the plate separation can be calculated with the same formula but with the new capacitance value. The work done to separate the plates is equivalent to the change in stored energy, which can either be the work done by an external force or the work lost to the system.
If the bonds in the reactants of Figure 7-3 contained 432 kJ of chemical energy and the bonds in the
products contained 478 kJ of chemical energy, what would be the amount of energy change during the
reaction? Would this energy be absorbed or released? Show your work.
Answer:
The energy change would be 46kJThe energy would be absorbedExplanation:
The energy change during a chemical reation, i.e. the reaction energy, is equal to the chemical energy stored in the bonds of the products less the chemical energy stored in the bonds of the reactants.
Hence:
Energy change = 478 kJ - 432kJ = 46kJThe change is positive, this is, the chemical energy of the products is greater than the chemical energy of the reactants.
That corresponds to the second graph, where the level of the energy of the products in the graph is higher than the level of the energy of the reactants. Therefore, the conclusion is that the reaction absorbed energy and it is endothermic.
How much force is required to hold an empty carton of volume 1.5 L beneath the surface in a sink of water? Assume the carton is empty, so its own weight is negligibly small.
Answer:
It will require 14.715 N of force to hold the cartoon beneath the water.
Explanation:
Given the the volume of cartoon is 1.5 liters.
We need to find the force required to hold this cartoon beneath the water.
As we know from the Archimedes principle that the net force is equal to the volume of liquid displaced.
Given volume of the cartoon is 1.5 liters. So, 1.5 liters of water will be displaced.
And we know the density of the water is [tex]1000\ kg/m^3[/tex]. That is [tex]1\ kg/L[/tex]
And [tex]g=9.81\ m/s^2[/tex]
[tex]F_N=\rho Vg\\F_N=1\times 1.5\times 9.81\\F_N=14.715\ N[/tex]
So, it will require 14.715 N of force to hold 1.5 liter volume of cartoon beneath the water.
A(n) ____ line is a dedicated telephone line that can be used for data communications to connect two different locations for continuous point-to-point communications.
Answer:
T- carrier
Explanation:
The T-carriers are frequently used for trunking between switching centers in a telephone network. It makes use if the same twisted pair copper wire that analog trunks employs. One pair for transmitting and the other pair for receiving.
7. Nancy has a mass of 60 kg and sits on the very end of a 3.00 m long plank pivoted in the middle. How much torque must her co-worker provide on the other end of the plank in order to keep Nancy from falling on the ground?
Answer:
Torque = 882Nm
Explanation:
Torque = Mg×distance
But plank's is pivoted ,therefore distance=3/2=1.5m
Mass of Nancy=60jg
Acceleration due to gravity, g=9.8m/s^2
Torque= 60×9.8×1.5
Torque= 882Nm
The drawing shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper right side. The initial speed of the electron is The capacitor is cm long, and its plates are separated by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.
Answer:
The electric field between the plates is 2173 N/C
Explanation:
Given that,
Distance = 0.150 cm
Suppose,The initial speed of the electron is [tex]7.05\times10^6\ m/s[/tex]. The capacitor is 2.00 cm long,
We need to calculate the time
Using formula of time
[tex]t=\dfrac{d}{v}[/tex]
Put the value into the formula
[tex]t=\dfrac{2.00\times10^{-2}}{7.05\times10^{6}}[/tex]
[tex]t=2.8\times10^{-9}\ s[/tex]
We need to calculate the acceleration
Using equation of motion
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
[tex]a=\dfrac{2s}{t^2}[/tex]
Put the value into the formula
[tex]a=\dfrac{2\times0.150\times10^{-2}}{(2.8\times10^{-9})^2}[/tex]
[tex]a=3.82\times10^{14}\ m/s^2[/tex]
We need to calculate the electric field between the plates
Using formula of electric field
[tex]E=\dfrac{F}{q}[/tex]
[tex]E=\dfrac{ma}{q}[/tex]
Put the value into the formula
[tex]E=\dfrac{9.1\times10^{-31}\times3.82\times10^{14}}{1.6\times10^{-19}}[/tex]
[tex]E=2173\ N/C[/tex]
Hence, The electric field between the plates is 2173 N/C
Which of the following physical laws or principles can best be used to analyze the collision between the object and the pendulum bob? Which can best be used to analyze the resulting swing? 1. Newton's first law 2. Newton's second law 3. Newton's third law 4. Conservation of mechanical energy 5. Conservation of momentum
3. Newton's third law
5. Conservation of momentum
Explanation:
Conservation of momentum is mostly used for describing collisions between objects. Here, the type of collision is inelastic collision in which the object when collides with the pendulum bob sticks to it and moves as a combined object. In this process the momentum is conserved.
Let the mass of the pendulum be m1 moving with a velocity v1.
Let the mass of the object be m2 moving with a velocity v2.
Since the momentum is conserved during collision, the equation will be
[tex]m1 v1 + m2 v2 = (m1 + m2) v[/tex]
Where, v is the velocity of the combined system.
Conservation of momentum is actually a direct consequence of Newton's third law.
Consider a collision between two objects, object A and object B. When the two objects collide, there is a force on A due to B. However, because of Newton's third law, there is an equal force in the opposite direction, on B due to A
FAB = -FBA
The mechanical energy is not conserved due to the fact that the kinetic energy is not the same before and after the collision.
The conservation of momentum is crucial for analyzing collisions, while the conservation of mechanical energy is ideal for studying swinging motion.
Conservation of momentum can best be used to analyze the collision between the object and the pendulum bob, while Conservation of mechanical energy is best suited to analyze the resulting swing.
The magnetic field about a straight length of current-carrying wire is _________.
Explanation:
The magnetic field about a straight length of current carrying wire is circular in shape. As we know that the electric current passing through the wire is producing magnetic field. One can find out the direction of the magnetic field lines by using right hand rule or a compass. If the conductor is a straight wire then the magnetic field lines will form a concentric circles around the wire. Put you thumb in the direction of the electric current in the wire then the fingers will curl to show the direction of magnetic field which thus form the circular loop.
Answer:
circular in shape.
Explanation:
This magnetic field can be visualized as a pattern of circular field lines surrounding a wire. One way to explore the direction of a magnetic field is with a compass, as shown by a long straight current-carrying wire in.
Because the magnetic field created by the electric current in the wire is changing directions around the wire, it will repel both poles of the magnet by bending away from the wire.
Imagine that you know the mass of a nearby star and you know that there is a planet orbiting around the star with a mass much smaller than the mass of the star. Explain with a sentence or two how, using the Doppler effect technique, you can measure the semi-major axes of a planet orbiting the star.]
Answer:
By measuring the time taken for the stars line of sight velocity to cycle from peak to Peak, and by calculation using newtons version of Kepler's third law
Explanation:
The motion of orbiting planets using planet-hunting techniques can rely on doppler effect. The light from the stars they orbit, as seen from Earth. As the star moves back and forth, the Doppler shift causes a slight change in its apparent colour which can be detected using spectroscopy. The blue shift and the red shift.
The Doppler technique of blue shift and the red shift can be used to estimate the semi major axis of the planets orbit by
Measuring the time it takes for the stars line of sight velocity to cycle from peak to Peak, and using newtons version of Kepler's third law
If a car is taveling with a speed 6 and comes to a curve in a flat road with radius ???? 13.5 m, what is the minumum value the coefficient of friction must be so the car doesn’t slide of the road?
Answer:
The minimum friction coefficient required is 0.3
(friction coefficients have no units)
Explanation:
To find a force we need to know something about a mass, and we haven't been told the mass of the car. Let's just call it 'm' and leave it at that for the moment, because it will cancel out in the end.
The centripetal force is given by F = ma = mv2/r
We have values for the velocity and the radius, so:
Fcent=m×6 × 6/13.5 = 2. 667m N
The frictional force must be equal to or greater than this force in order for the car to successfully make it around the curve without sliding out.
The frictional force will be given by:
Ffrict = μFnorm
Where Fnorm is the normal force, equal to mg.
We can equate these two forces, the frictional force and the centripetal force:
Fcent = Ffrict
2.667m=μmg
We can cancel out a factor of m in both sides and rearrange to make μ the subject:
μ = 2.667g
Substituting in the value g=9.8 ms−2,
μ = 2.667/9.8 = 0.27
Approximately = 0.3
Explanation:
Below is an attachment containing the solution.
.A particle moving with a constant acceleration has a velocity of 20 cm/s when its position is x = 10 cm. Its position 7.0 s later is x = –30 cm. What is the acceleration of the particle?
Answer:
[tex]4.08cm/s^2[/tex]
Explanation:
The second equation of a uniformly accelerated motion could be used to solve this problem. This is given by equation (1);
[tex]s=ut+\frac{1}{2}at^2....................(1)[/tex]
where u is the particle's initial velocity, t is the time taken, a is the acceleration and s is the distance travelled.
Given;
u = 20cm/s
t = 7s
a = ?
s = ?
The particle moved from one point [tex]x_1[/tex] to another point [tex]x_2[/tex] along the x-axis, where [tex]x_1=10cm[/tex] and [tex]x_2=-30cm[/tex]. This information could be used to find the distance s covered by the object as follows;
[tex]s=x_1-x_2.................(2)\\s=10-(-30)\\s=10+30\\s=40cm[/tex]
We the make appropriate substitutions into equation (1) and then solve for the acceleration.
[tex]40=(20*7)+\frac{1}{2}*a*7^2\\40=140+\frac{1}{2}*a*49\\40=140+24.5a\\40-140=24.5a\\hence\\24.5a=-100\\a=\frac{-100}{24.5}\\a=-4.08cm/s^2[/tex]
The negative sign is an indication that the particle is decelerating.
Answer:
7.347 cm / s²
Explanation:
Using equation of linear motion
S = ut + 1/2 at²
where total displacement = final displacement - initial displacement
S = - 30 - 10 = - 40 cm
- 40 cm = (20 cm /s × 7 s) + 1/2 a (7²)
- 40 cm = 140 cm + 1/2 49 a
- 40 cm - 140 cm = 1/2 × 49 a
- 180 cm × 2 / 49 s² = a
a = -7.347 cm / s²
It is probably decelerating.
The blades of a ceiling fan start from rest and, after two revolutions, have an angular speed of 0.5 rev/s. The angular acceleration of the blades is constant. What is the angular speed after eight revolutions?
Answer:
[tex]\omega_f = 1 rad/s[/tex]
Explanation:
Given,
After two revolutions, angular speed = 0.5 rev/s
Angular speed after 8 revolution = ?
Using equation of circular motion
[tex]\omega_f^2 = \omega_0^2 + 2\alpha \theta[/tex]
[tex]0.5^2 =0^2 + 2\times \alpha\times 2[/tex]
[tex]\alpha = 0.0625 rad/s^2[/tex]
Now, Angular speed after 8 revolution
[tex]\omega_f^2 = \omega_0^2 + 2\alpha \theta[/tex]
[tex]\omega_f^2 =0^2 + 2\times 0.0625 \times 8[/tex]
[tex]\omega_f = 1 rad/s[/tex]
Hence, the angular speed after 8 revolution is equal to 1 rad/s.
The angular acceleration of the fan blades is 0.25 rev/s². Using this constant angular acceleration, the angular speed of the blades after 8 revolutions is calculated to be 2 rev/s.
Explanation:To solve this problem, we use the equation of motion for rotational systems.
Δω = α * Δθ Where Δθ is the change in angle (in rev), Δω is the change in angular speed (in rev/s), and α is the angular acceleration (in rev/s²). Given that the blades have an angular speed of 0.5 rev/s after 2 revolutions, we can solve for α: α = Δω / Δθ = (0.5 rev/s) / (2 rev) = 0.25 rev/s²
Then, knowing this constant angular acceleration, we can find the angular speed after 8 revolutions: Δω = α * Δθ
Δω = (0.25 rev/s²) * (8 rev) = 2 rev/s
So after 8 revolutions, the angular speed of the ceiling fan is 2 rev/s.
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A policeman investigating an accident measures the skid marks left by a car on the horizontal road. He determines that the distance was 23.74 n. The coefficient of kinetic friction between the tires and the road is μk = 0.29 How fast was the car going when the driver applied the brakes
Final answer:
To find the initial speed of the car, we use the work-energy principle and the equation μk × g × d = ½ × v². By substituting the given coefficient of kinetic friction (0.29) and skid mark distance (23.74 meters), we calculate the initial speed to be approximately 11.68 m/s.
Explanation:
The question involves calculating the initial speed of a car at the moment the driver applied the brakes, which resulted in skid marks on the road. The length of the skid marks and the coefficient of kinetic friction (μk) between the tires and the road are known.
To calculate the initial speed of the car, we use the work-energy principle, which states that the work done by the friction force is equal to the change in kinetic energy of the car:
Work done by friction = Change in kinetic energy
μk × m × g × d = ½ × m × v²
Since mass (m) cancels out and gravitational acceleration (g) is a constant (9.81 m/s²), we can simplify the equation to:
μk × g × d = ½ × v²
Now we can solve for the initial velocity (v):
v = √(2 × μk × g × d)
Plugging in the given values, μk = 0.29 and d = 23.74 meters, we have:
v = √(2 × 0.29 × 9.81 m/s² × 23.74 m)
v ≈ √(2 × 0.29 × 9.81 × 23.74)
v ≈ √(136.4)
v ≈ 11.68 m/s (approximately)
Thus, the initial speed of the car when the brakes were applied was approximately 11.68 meters per second.
An aluminum wire with a diameter of 0.115 mm has a uniform electric field of 0.235 V/m imposed along its entire length. The temperature of the wire is 55.0°C. Assume one free electron per atom. Given that at 20 degrees, rhoo = 2.82x10-8 Ωm and α = 3.9x10-3 /C. Determine:
a) the resistivity of the wire.
b) the current density in the wire.
c) the total current in the wire.
d) the potential different that must exist between the ends of a 2m length of wire if the given electric field is to be produced.
Answer:
Explanation:
a) To get the resistivity ρ at 50 Celsius, given the resitstivity at 20 Celsisus, use:
ρ = ρo(1 + α(T - To))
where To = 20 Celsius
b) Knowing the resistivity at 50 Celsius, and the (uniform) electric field E, you can determine the current density J using:
E = ρJ
(which is actually a density-averaged version of V = IR)
c) Assuming the current is uniform (which is should be in a uniform electric field and constant-diameter wire), the current i can be calculated using:
J = i/A --> i = JA
where A is the cross-sectional area of the wire (given by πr2); make sure to convert the given diameter to a radius, and the radius to base units
d) Since the electric field is given in volts per meter, and you have two meters of length in the wire, you can determine directly from that how many volts difference you need at the ends of the wire to get 0.2 volts per meter.
0.2 = V/d
with d = 2 m. This corresponds to a uniform electric field being related to voltage by V = Ed, where d is distance along the field line.
Explanation:
Below are attachments containing the solution.
If you ride your bike at an average speed of 2 km/h and need to travel a total distance of 20 km, how long will it take you to reach your destination? Show your work.
Answer:
Time taken to reach your destination will be 10hours
Explanation:
Recall the formula for Speed;
speed=Total distance/Total time taken
Speed=2km/h
Total distance=20km
Time taken=x
let x be the unknown time taken
Input each values into the formula;
2=20/x
Making x subject of the equation
x=20/2
x=10
Total time taken =10hours.
Answer:
10 hours
Explanation:
The average speed of a body is given by;
[tex]v_{avg}=\frac{s}{t}..................(1)[/tex]
where s is the total distance travelled and t is the total time spent.
Given;
s = 20km
t = ?
[tex]v_{avg}=2km/h[/tex].
We substitute into equation (1) and then solve for t.
[tex]2=\frac{20}{t}\\2t=20\\t=\frac{20}{2}\\t=10hrs.[/tex]
What is the sum of the kinetic energies of the alpha particle and the new nucleus?
Answer: The total energy created by the Alpha decay.
Explanation: The sum total of the kinetic energy of the alpha particle and the new nucleus is the total energy created by the alpha decay.
Consider the decay of a radioactive nuclide by the spontaneous emission from its nuclei of alpha particles. An alpha particle which is composed of two protons and two neutrons and has a charge of +2. With an appreciable mass and its ejection from the nuclide creates a certain amount of recoil energy in the nucleus. The total energy (Ex) created by alpha decay is therefore the sum of the kinetic energy of the particle, the recoil energy given to the new nucleus, and the total energy of any emitted gamma rays.
The sum of the kinetic energies of an alpha particle and the new nucleus after alpha decay equals the initial energy (Q-value) of the reaction, most of which is carried by the alpha particle due to mass differences.
Explanation:The sum of the kinetic energies of the alpha particle and the new nucleus after an alpha decay can be determined by considering the conservation of energy and momentum. The energy released during the decay, known as the Q-value, is primarily carried away by the alpha particle due to its relatively smaller mass compared to the daughter nucleus.
The energy of the alpha particle can be measured experimentally, which then allows us to determine the energy of the new nucleus. According to conservation of energy, the sum of the kinetic energies of the alpha particle and the new nucleus is equal to the Q-value of the reaction. If the Q-value (Qa) is known to be 4.3 MeV, and assuming the potential energy is zero in the final state, this energy will be distributed between the alpha particle and the new nucleus.
Using the example values provided, if the kinetic energy of the new nucleus (KEnucleus) is calculated to be 23.3 eV, then the remainder of the Q-value minus the kinetic energy of the new nucleus will represent the kinetic energy of the alpha particle. Since the alpha particle carries away most of the kinetic energy, the sum of the kinetic energies will be very close to the initial Q-value.
A piece of copper wire is formed into a single circular loop of radius 9.1 cm. A magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to 0.90 T in a time of 0.66 s. The wire has a resistance per unit length of 2.9 x 10-2 /m. What is the average electrical energy dissipated in the resistance of the wire
Given Information:
Radius of circular loop = r = 9.1 cm = 0.091 m
Change in time = Δt = 0.66 seconds
Change in magnetic field = ΔB = 0.90 T
Resistance of wire per unit length = R = 2.9x10⁻² Ω/m
Number of turns = N = 1
Required Information:
Electrical energy dissipated = E = ?
Answer:
Electrical energy dissipated = 50.09x10⁻³ Joules
Step-by-step explanation:
We know that energy is given by
E = Pt
Where power is given by
P = ξ²/R
Where ξ is the induced EMF in the wire and is given by
ξ = -NΔΦ/Δt
Where ΔΦ is the change in flux and is given by
ΔΦ = ΔBAcosφ
Where φ is the angle between magnetic field and circular loop
A = πr² and R = 2.9x10⁻²*2πr
Substituting the above relations into the energy equation and simplifying yields,
E = [-Nπr²cosφ(ΔB/Δt)²]*t/R
E = [-1*π(0.091)²*cos(0)(0.90/0.66)²*0.66]/2.9x10⁻²*2π*(0.091)
E = 0.050094 Joules
E = 50.09x10⁻³ Joules
Therefore, the average electrical energy dissipated in the circular loop of the wire is 50.09x10⁻³ Joules.
Two large, parallel, conducting plates are 12 cm apart and have charges of equal magnitude and opposite sign on their facing surfaces. An electric force of 15 3.9 10 N − ⋅ acts on an electron if it is placed anywhere between the two plates. (a) Find the electric field magnitude at the position of the electron. (b) What is the potential difference between the plates?
Answer:
a) 2.4×10^4N/C
b) 2.9 ×10^3V
Explanation:
Correct ststement: An electric force of 3.9×10^-15N acts on the electron if it is placed anywhere between the two plates.
a) The electric field magnitude is given by E= F/e
Where F = electric force
e= elementary charge carried by a single proton e= 1.6×10^-19C
E= (3.9×10^-15)/(1.6×10^-19)
E= 2.4×10^4NC
b) Ptential difference is given by:
Change in V= E×change in distance
Potential difference= (2.4×10^4)× (0.12)
Potential difgerence= 2.9×10^3V
Would a vibrating proton produce an electromagnetic wave
Yes, a vibrating proton would produce an electromagnetic wave, as accelerating charges emit radiation. This principle is central to many technological and scientific applications, including radio transmissions and the study of galactic structures in astronomy.
Explanation:Accelerating charges such as protons, when they vibrate, indeed produce electromagnetic waves. This effect is due to the fact that a changing electric field generates a magnetic field, and a changing magnetic field, in turn, generates an electric field. As a proton oscillates, it experiences acceleration and therefore can emit radiation. This principle is extensively used in various technologies, like radio transmission, where an alternating current in an antenna accelerates charges and creates electromagnetic waves.
The production and detection of electromagnetic waves are crucial in many fields, including communications and astronomy. Just like an electron, a proton is also a spin 1/2 particle with a magnetic moment and can emit radiation that can be detected, such as the 21-cm line in the hydrogen spectrum, which allows astronomers to map the spiral arms of galaxies.
Considering the mass and charge of particles, a vibrating proton can generate electromagnetic radiation, albeit at different frequencies compared to electrons due to their larger mass. This forms the basis of nuclear magnetic resonance (NMR) utilized in various scientific and medical applications.
Projectile Motion: A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits the ground the horizontal location of the plane will
Answer:
Explanation:
The package had the same velocity as the plane when it was dropped. Newton's 1st Law says that "an object in motion tends to stay in motion, at the same velocity, in a straight line unless acted on by an outside force".
There only outside force acting on the package was its weight -- that force is straight down. The horizontal velocity that the plane gave the package continued (as Newton said it would), so as it fell, horizontally it kept pace with the plane.
If an object oscillates in simple harmonic motion, with the position described by the equation: x(t) = 42.5*cos(21t) What is the angular frequency of oscillation w ?
Answer:
The angular frequency of oscillation w = 21
Explanation:
To solve the question, we note that x(t) is the point in the motion of the object. Therefore to find the angular frequency of oscillation, we find the relationship between the angular velocity and time
The angular frequency, ω is a scalar quantity used to depict the rate of rotation per unit time
When there is a function in simple harmonic motion (SHM) with the following equation then ω is the angular frequency
x(t) = A·cos (2πft) = A·cos(ωt) which is similar to 42.5*cos(21t)
then 21 = angular frequency
What is the speed of a beam of electrons when under the simultaneous influence of E = 1.64×104 V/m B = 4.60×10−3 T Both fields are normal to the beam and to each other and produce no deflection of the electrons. When the electric field is removed, what is the radius of the electron orbit? What is the period of the orbit?
Answer: v = 3.57×10^6 m/s; R = 4.42×10^-3m; T = 7.78×10^-9 s
Explanation:
Magnetic force(B) = 4.60×10^-3 T
Electric force(E) = 1.64×10^4 V/m
Both forces having equal magnitude ;
Magnetic force = electric force
qvB = qE
vB = E
v = (1.64×10^4) ÷ (4.60×10^-3)
v = 3.57×10^6 m/s
2.) Assume no electric field
qvB = ma
Where a = v^2 ÷ r
R = radius
a = acceleration
v = velocity
qvB = m(v^2 ÷ R)
R = (m×v) ÷ (|q|×B)
q=1.6×10^-19C
m = 9.11×10^-31kg
R = (9.11×10^-31 * 3.57×10^6) ÷ (1.6×10^-19 * 4.6×10^-3)
R = 32.5227×10^-25 ÷ 7.36×10^-22
R = 4.42×10^-3m
3.) period(T)
T = (2*pi*R) ÷ v
T = (2* 4.42×10^-3 * 3.142) ÷ (3.57×10^6)
T = (27.775×10^-3) ÷(3.57×10^6)
T = 7.78×10^-9 s