Answer: C
guide and inspire good conduct
Explanation:
APA are rules of proper source documentation to avert difficulty. This makes you feel that you are not alone. Having to follow detailed citation and formatting guidelines as well may seem like just one more task to add to an already-too-long list of requirements.
Final answer:
Correct answer is C. The five general principles of the APA are meant to guide and inspire good conduct among psychologists, offering a values framework rather than serving as enforceable rules.
Explanation:
The five general principles from the American Psychological Association (APA) are meant to guide and inspire good conduct. These principles serve as ethical guidelines for behaviors and decision-making in psychological practice.
In contrast to enforceable rules, which are mandatory and spell out specific prohibitions and mandates, principles and guidelines such as those provided by the APA are not meant to be enforceable or stand up in court but rather to provide a framework for ethical practice.
They offer a set of values that psychologists are encouraged to aspire to in their professional conduct. These principles cover important areas such as respect for people's rights, integrity, and social responsibility in practice.
To understand the forces between a bar magnet and 1. a stationary charge, 2. a moving charge, and 3. a ferromagnetic object. A bar magnet oriented along the y axis can rotate about an axis parallel to the z axis. Its north pole initially points along j^.
A stationary charge does not experience a force in a magnetic field. A moving charge experiences a force perpendicular to both its velocity and the magnetic field. A ferromagnetic object is attracted to a magnet regardless of its movement.
Explanation:The magnetic force between a bar magnet and different charged materials can be understood in respect to the properties of the magnetic field, and these forces can range from attractive to repulsive depending on the orientation of the objects in the field and the type of materials involved. For a stationary charge, magnetic forces do not act on it because the magnetic force on a charged particle is always perpendicular to the direction of its motion. Hence, a stationary charged particle does not experience a force.
In the case of a moving charge, a magnetic field exerts force on it. This force is perpendicular to both the velocity of the charge and to the magnetic field, and is dependent on the magnitude of the charge, its speed, and the magnetic field. For example, if you place a moving charge in a magnetic field, the charge will experience a force that's oriented at right angles to both its direction of movement and the field itself. Ferromagnetic objects like iron or nickel will experience attraction to the magnet irrespective of their movement. It is the property of the ferromagnetic material to align its domains in the direction of the external magnetic field (from the bar magnet), thus the entire object acts like a small magnet and feels an attraction.
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An early planetary model of the hydrogen atom consisted of a 1.67 X 10-27 kg proton in the nucleus and a 9.11 X 10-31 kg electron in orbit around it at a distance of 5.0 X 10-11 m. In this model, what is the gravitational force between a proton and an electron?
Answer:
The gravitational force between proton and an electron is [tex]4.06 \times 10^{-47}[/tex] N
Explanation:
Given:
Mass of proton [tex]m_{1} = 1.67 \times 10^{-27}[/tex] kg
Mass of electron [tex]m_{2} = 9.11 \times 10^{-31}[/tex] kg
Separation between electron and proton [tex]r = 5 \times 10^{-11}[/tex] m
According to the gravitational law,
[tex]F = \frac{Gm_{1} m_{2} }{r^{2} }[/tex]
Where [tex]G = 6.674 \times 10^{-11}[/tex] = gravitational constant.
[tex]F = \frac{6.674 \times 10^{-11} \times 1.67 \times 10^{-27} \times 9.11 \times 10^{-31} }{(5 \times 10^{-11} )^{2} }[/tex]
[tex]F = 4.06 \times 10^{-47}[/tex] N
Therefore, the gravitational force between proton and an electron is [tex]4.06 \times 10^{-47}[/tex] N
The gravitational force between a proton and an electron is 40 * 10⁻⁴⁸ N
Newton's law of gravitation is given by:
F = Gm₁m₂/r²
Where F is the force, m₁, m₂ are masses, r is the distance between the two masses and G is the gravitational constant = 6.67 * 10⁻¹¹ Nm²/kg²
Given that:
m₁ = 1.67 * 10⁻²⁷ kg, m₂ = 9.11 * 10⁻³¹ kg, r = 5 * 10⁻¹¹ m, hence:
F = 6.67 * 10⁻¹¹ Nm²/kg² * 1.67 * 10⁻²⁷ kg * 9.11 * 10⁻³¹ kg/ (5 * 10⁻¹¹ m)²
F = 40 * 10⁻⁴⁸ N
The gravitational force between a proton and an electron is 40 * 10⁻⁴⁸ N
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A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.010 hour, and the assembly expands from an initial volume of 0.90 L to a final volume of 5.92 L against an external pressure of 1.0 atm. Use the wattage of the lightbulb and the time it is on to calculate ΔE in joules (assume that the cylinder and lightbulb assembly is the system and assume two significant figures). Calculate w and q.
Answer:
[tex]w = - 508.53[/tex] joules
[tex]q = - 3091.47[/tex] joules
Explanation:
Let us convert the time in hours into seconds
[tex]0.010* 3600\\= 36[/tex]
Change in internal energy
[tex]\delta E = p * \delta t[/tex]
where E is the internal energy in Joules
p is the power in watts
and t is the time in seconds
[tex]\delta E = - 100 * 36\\[/tex]
[tex]\delta E = - 3600[/tex] Joules
Amount of work done by the system
[tex]w = - P * \delta V[/tex]
where P is the pressure and V is the volume
Substituting the given values in above equation, we get -
[tex]w = - 1 * ( 5.92 -0.90)\\[/tex]
[tex]w = -5.02[/tex] liter-atmospheres
Work done in Joules
[tex]- 5.02 * 101.3\\[/tex][tex]= 508.53[/tex]Joules
[tex]q = \delta E - w\\[/tex]
Substituting the given values we get -
[tex]q = - 3600 - (-508.53)\\q = - 3091.47[/tex]
Thus
[tex]w = - 508.53[/tex] joules
[tex]q = - 3091.47[/tex] joules
The change in energy (ΔE) is 1 joule (J), the work (w) is -41 J, and the heat (q) is 42 J.
Explanation:To calculate the change in energy (ΔE) in joules, we can use the formula ΔE = power (P) x time (t). In this case, the power of the lightbulb is 100 W and the time it is on is 0.010 hour. Therefore, ΔE = 100 W x 0.010 hour = 1 joule (J).
To calculate work (w), we can use the equation w = -PΔV, where ΔV is the change in volume of the assembly. Here, ΔV = final volume - initial volume = 5.92 L - 0.90 L = 5.02 L. Given that the external pressure is 1.0 atm, we can convert the volume to liters-atmospheres (L·atm) by multiplying by 0.0821. Therefore, w = -100 W x (5.02 L x 0.0821) = -41 J.
Finally, to calculate heat (q), we can use the first law of thermodynamics, which states that q = ΔE - w. Substituting the values, q = 1 J - (-41 J) = 42 J.
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The rate of change of angular momentum of a particle equals the torque of the net force acting on it is called
Answer:
Explanation:
The rate of change of angular momentum of a particle equals the torque of the net force acting on it is called CONSERVATION OF ANGULAR MOMENTUM
ΣIα = τ
τ(net) = I•α
Where,
α is angular acceleration
τ is Torque
Answer:
Conservation of angular momentum
Explanation:
The rate of change of angular momentum of a particle equals the torque of the net force acting on it is called conservation of angular momentum.
L = ΣF x r = τ
where;
L is the angular momentum of the particle
τ is torque on the particle
r is the distance through which the force act on the particle
ΣF is the net force on the particle
Thus, the rate of change of angular momentum of a particle equals the torque of net force acting on it, (L = τ) and it is called conservation of angular momentum.
When an object is approached electrostatically charged to a neutral body it is possible to charge it by induction, however, if it is touched for a moment the object will be charged by contact, explain this phenomenon, why this happens
Answer:
When the charge body touches the other body, charge pass from it to the other body until get the charge is distributed over the largest possible surface, when separating the two bodies they are charged, but with half the initial charge, this phenomenon is called contact charge
Explanation:
When a charged body approaches another body, it attracts or repels the electrons of the other body, until the net charge is zero, that effect disappears when the body moves away.
When the charge body touches the other body, charge pass from it to the other body until get the charge is distributed over the largest possible surface, when separating the two bodies they are charged, but with half the initial charge, this phenomenon is called contact charge
A particle has a charge of 5.2x10^-19 coulombs and experience a force of 9.5x10^-15 Newton's when it travels through a magnetic field with strength 2.2 x 10^-1 telsa what is the speed of the particle
Answer:
83000 m/s
Explanation:
Using
F = Bqv....................... Equation 1
Where F = Force experienced by the charge, B = magnetic field q = charge of the particle, v = speed of the particle.
make v the subject of the equation
v = F/Bq.................... Equation 2
Given: q = 5.2×10⁻¹⁹ Coulombs, F = 9.5×10⁻¹⁵ Newtons, B = 2.2×10⁻¹ Tesla
Substitute into equation 2
v = 9.5×10⁻¹⁵ /(5.2×10⁻¹⁹×2.2×10⁻¹)
v = 8.3×10⁴ m/s
v = 83000 m/s
At the surface of the moon, the acceleration due to the gravity of the moon is x. At a distance from the center of the moon equal to four times the radius of the moon, the acceleration due to the gravity of the moon is _____.
Answer:
Gravity at a distance of 4R will be reduced to 1/16 th.
Explanation:
Given:
At the surface of the moon, the acceleration due to the gravity of the moon is x.
We have to find the gravity a t a distance of 4 times from the center of the moon.
Let the radius of the moon be "R".
And
The value of acceleration due to gravity is [tex]g_m[/tex] .
Formula:
⇒ [tex]g_m=\frac{GM}{R^2}[/tex] ...where M is the mass of the moon.
Now
Gravity of the moon at the its surface:
⇒ [tex]g_m=\frac{GM}{R^2}[/tex] ...equation (i)
Gravity of the moon at a distance of [tex]4R[/tex]:
⇒ [tex]g_m_1=\frac{GM}{(4R)^2}[/tex]
⇒ [tex]g_m_1=\frac{GM}{16R^2}[/tex] ...equation (ii)
Dividing equation (i) with (ii) to find the relationship between the two.
⇒ [tex]\frac{g_m_1}{g_m} =\frac{GM}{16R^2}\times \frac{R^2}{GM}[/tex]
⇒ [tex]\frac{g_m_1}{g_m} =\frac{1}{16}[/tex]
⇒ [tex]g_m_1 =g_m(\frac{1}{16})[/tex]
⇒ [tex]g_m_1 =x(\frac{1}{16})[/tex] ...as gm=x at the surface.
So,
We can say that the gravity at a distance of 4R will be reduced to 1/16 th.
Answer: The acceleration due to gravity on the surface of the moon is 1.620 m/s2. 2) The radius of the Earth is 6.38 x 106 m.
A newly proposed device for generating electricity from the sun is a heat engine in which the hot reservoir is created by focusing sunlight on a small spot on one side of the engine. The cold reservoir is ambient air at 20°C. The designer claims that the efficiency will be 50%.
What minimum hot-reservoir temperature, in degrees C, would be required to produce this efficiency?
Answer:
The minimum temperature of hot reservoir is 586K or 313°C
Explanation:
The Carnot cycle is defined as ideal reversible process thermodynamic process which has four successive step. During the expansion and compression of the substance it can done upto desired point and then reversed up.
Here the energy is used from the hot reservoir to do work and deposited into the cold reservoir. As is it reversible the efficiency of the carnot cycle is the theoretical maximum of the heat engine.
The efficiency of carnot cycle is
η = 1 - [tex]\frac{Tc}{Th}[/tex] eq 1
Where Tc is the temperature of cold reservoir = 20° = 20°+273K = 293K
Th is the temperature of hot reservoir.
η is the efficency 50% = 0.5
The minimum temperature of the reservoir is related to the maximum efficiency,
Substituting values in eqn 1
0.05 = 1-[tex]\frac{293}{Th}[/tex]
Th = [tex]\frac{293}{1 - 0.5}[/tex] =586 K
The minimum temperature of hot reservoir is 586K.
ie 586K-273 = 313°C
Final answer:
The minimum hot-reservoir temperature required for a heat engine to have 50% efficiency with a cold reservoir at 20°C is 313.15°C.
Explanation:
To determine the minimum hot-reservoir temperature for a heat engine with 50% efficiency where the cold reservoir is at 20°C, we can use the efficiency formula for a Carnot engine: efficiency (e) = 1 - (Tc/Th), where Tc is the cold reservoir temperature and Th is the hot reservoir temperature, both in kelvins. Rearranging the formula to solve for Th gives: Th = Tc / (1 - e).
First, convert the cold reservoir temperature from Celsius to Kelvin: Tc = 20°C + 273.15 = 293.15 K. Now, plug in the efficiency value: Th = 293.15 K / (1 - 0.50) = 293.15 K / 0.50 = 586.3 K. Convert this back to Celsius to get the minimum hot-reservoir temperature required: Th - 273.15 = 586.3 K - 273.15 = 313.15°C.
Therefore, the minimum hot-reservoir temperature required for the proposed heat engine to reach 50% efficiency is 313.15°C.
What made it possible to construct skyscrapers in the 1800s
Answer:
It was made possible by the new methods of making steel
Explanation:
The Bessemer process was the first inexpensive industrial process for the mass production of steel from molten pig iron prior to the development of the open hearth furnace. The key principle used was the removal of impurities from the iron by process of oxidation with air being blown through the molten iron. The oxidation also enhances the temperature of the iron mass and keeps it molten. So with the presence of this, building of skyscraper was made possible.
Skyscrapers in the 1800s were made possible by the invention of steel girders and elevators, along with the high prices of real estate in city centers.
Explanation:The construction of skyscrapers in the 1800s was made possible by several key advancements. One important development was the invention of steel girders that could support the weight of tall buildings. These girders allowed for the construction of buildings beyond the previous limit of 10 to 12 stories.
Another crucial factor was the invention of elevators, both passenger and freight elevators. The introduction of elevators made it feasible for people and goods to reach higher floors easily, making taller buildings more practical.
Additionally, the price of real estate in city centers played a significant role. As the cost of land increased, developers turned to building upwards to maximize space and value. These advancements in construction technology and the need for efficient use of space contributed to the rise of skyscrapers in the 1800s.
Objects 1 and 2 attract each other with a electrostatic force of 72.0 units. If the distance separating objects 1 and 2 is changed to one- halved the original value (i.e, halved), then the new electrostatic force will be
Un cuerpo de 10N de peso esta apoyado sobre una superficie horizontal. Se le ata una cuerda y se tira de él con una fuerza de 15 N que forma un ángulo de 60° con la horizontal. ¿Cuál es la fuerza normal sobre el cuerpo?
La fuerza normal sobre el cuerpo es de [tex]\(5 \, N\)[/tex].
Explanation:El problema implica un cuerpo de [tex]\(10 \, N\)[/tex] de peso que está siendo sometido a una fuerza de tracción de [tex]\(15 \, N\)[/tex] a un ángulo de [tex]\(60^\circ\)[/tex] con la horizontal. La fuerza normal es la componente perpendicular de la fuerza peso, ya que no hay movimiento vertical. Utilizamos la relación trigonométrica [tex]\(F_{\text{normal}} = F_{\text{peso}} \cdot \cos(\theta)\)[/tex], donde [tex]\(F_{\text{peso}}\)[/tex] es el peso del cuerpo y [tex]\(\theta\)[/tex] es el ángulo entre la fuerza peso y la horizontal.
En este caso, la fuerza peso [tex]\(F_{\text{peso}}\) es \(10 \, N\)[/tex] hacia abajo, y el ángulo [tex]\(\theta\)[/tex] es [tex]\(60^\circ\)[/tex]. Aplicando la fórmula, obtenemos [tex]\(F_{\text{normal}} = 10 \, N \cdot \cos(60^\circ)\)[/tex]. Calculando esto, encontramos [tex]\(F_{\text{normal}} = 10 \, N \cdot 0.5 = 5 \, N\)[/tex].
La fuerza normal es, por lo tanto, [tex]\(5 \, N\)[/tex], lo que significa que la superficie horizontal ejerce una fuerza hacia arriba de [tex]\(5 \, N\)[/tex] para equilibrar la componente vertical de la fuerza aplicada.
Este resultado es consistente con el principio de equilibrio en el plano horizontal. La fuerza normal contrarresta la componente vertical de la fuerza aplicada, manteniendo el cuerpo en equilibrio sin movimiento vertical. Este enfoque, basado en las leyes de la trigonometría y el equilibrio, proporciona una solución clara y precisa para el problema.
A lamp operates at 115 volts with a current of 0.25 ampere. What is the lamp's resistance?
Answer:
460 ohms
Explanation:
Resistance=Voltage/Current
=115/0.25=460
The resistance of the lamp can be calculated using Ohm's law, which states that resistance is equal to voltage divided by current. In this case, the resistance of the lamp is 460 ohms.
Explanation:To find the resistance of a lamp, we can use Ohm's law, which states that resistance is equal to voltage divided by current.
R = V/I
From the given information, the voltage is 115 volts and the current is 0.25 amperes. Plugging these values into the equation, we get:
R = 115 V / 0.25 A = 460 ohms
Therefore, the lamp's resistance is 460 ohms.
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Dolphin echolocation is similar to ultrasound. Reflected sound waves
allow a dolphin to form an image of the object that reflected the waves.
Dolphins can produce sound waves with frequencies ranging from
0.25 kHz to 220 kHz, but only those at the upper end of this spectrum
are used in echolocation. Explain why high-frequency waves work better
than low-frequency waves.
Answer:
Waves with high frequencies have shorter wavelengths that work better than low frequency waves for successful echolocation.
Explanation:
To understand why high-frequency waves work better than low frequency waves for successful echolocation, first we have to understand the relation between frequency and wavelength.
The relation between frequency and wavelength is given by
λ = c/f
Where λ is wavelength, c is the speed of light and f is the frequency.
Since the speed of light is constant, the wavelength and frequency are inversely related.
So that means high frequency waves have shorter wavelengths, which is the very reason for the successful echolocation because waves having shorter wavelength are more likely to reach and hit the target and then reflect back to the dolphin to form an image of the object.
Thus, waves with high frequencies have shorter wavelengths that work better than low frequency waves for successful echolocation.
Please Help me
A record player runs at 78 RPMs (revolutions per minute). That means that it spins 78 times (cycles) in a minute. work each one out.
a. Find the frequency of the record player.
b. Find the period of the record player.
Answer:
a. the frequency of the record player
f = 1.3 rev per second = 1.3 Hz
b. the period of the record player.
T = 0.77 seconds
Explanation:
Given;
Speed of record player v = 78 rpm
a) frequency of the record player is the number of revolutions the record player makes per second.
f = 78 rev/minute = 78/60sec = 1.3 rev per second
f = 1.3 Hz
b) period of the record player is the amount of time needed by the record player to complete one revolution.
T = 1/f (or 60s/78rev = 0.77 s)
T = 1/1.3
T = 0.77 seconds
Answer:
(a) The frequency of the record player is 1.3 Hz
(b) The period of the record player is 0.77 s
Explanation:
Given number of revolution per minute = 78 RPM
Part (a) the frequency of the record player
One revolution per minute, 1 RPM = ¹/₆₀ Hz
78 RPM = ?
Thus, 78 RPM = 78 ( ¹/₆₀ HZ) = 1.3 Hz
The frequency of the record player is 1.3 Hz
Part (b) the period of the record player
Period is inverse of frequency
T = 1 / f
T = 1 / 1.3 Hz
T = 0.77 s
The period of the record player is 0.77 s
How do the three opsin molecules respond to different wavelengths of light when their retinal molecular structures are identical?
Answer:
The three opsin molecules respond to different wavelengths of light because of the different structure of the protein bound to the opsin molecules.
Explanation:
Opsin is a protein that is released by the action of light and forms part of the visual pigment rhodopsin. There are two groups of protein termed opsins, type I opsins, which are employed by prokaryotes and by some algae and fungi, and type II opsins which are used by animals.
Though the retinal molecular structures of the opsin molecules are identical, they respond to different wavelengths of light because of the different structure of the protein bound to the opsin molecules.
Scientists have found that the most destructive and deadly tornadoes occur from rotating thunderstorms called
Answer:
Supercells
Explanation:
supercells are rotating thunderstorms that has a well-defined radar circulation called a mesocyclone. They can sometimes produce destructive hail, severe winds, frequent lightning, and flash floods.
You stop for a cappuccino at a coffee shop and notice that the tiny white bubbles of steamed milk remain on the surface of the coffee. These air-filled bubbles stay where they are, rather than descending into the coffee or rising into the air, because they are:
A. more dense than the coffee but less dense than the air above thecoffee.
B. thicker than the coffee but less thick than the air above thecoffee.
C. less dense than the coffee but more dense than the air above thecoffee.
D. lighter than the cup of coffee but heavier than the column ofair above the coffee.
Answer:
C. less dense than the coffee but more dense than the air above thecoffee.
Explanation:
A grinding stone with a mass of 50 kg and a radius of .75 m is rotating with an angular velocity of 30 rev/s. A second, smaller stone with a mass of 20 kg and a radius of .5 m is dropped onto the first without slipping. What is the new angular velocity of the two stones?
Answer:
The new angular velocity of the two stones = 160.064 rad/s
Explanation:
This is a case of conservation of angular momentum.
For initial case:
Mass = 50 kg
Radius = 0.75 m
Angular velocity N = 30 rev/s
We must convert to rad/s w
w = 2¶N = 2 x 3.142 x 30 = 188.52 rad/s
Moment of inertia I = m x r^2
I = 50 x 0.75^2 = 28.125 kgm2
Angular momentum = I x w
= 28.125 x 188.52 = 5302.125 kgm2-rad/s
For second case smaller stone has
m = 20 kg
Radius = 0.5 m
I = m x r^2 = 20 x 0.5^2 = 5 kgm2
Therefore,
Total moment of inertia of new system is
I = 28.125 + 5 = 33.125 kgm2
Final angular momentum = I x Wf
Where Wf = final angular speed of the system.
= 33.125 x Wf = 33.125Wf
Equating the two angular moment, we have,
5302.125 = 33.125Wf
Wf = 5302.125/33.125 = 160.064 rad/s
Answer:
[tex]\dot n = 25.471\,\frac{rev}{s}[/tex]
Explanation:
The situation is described reasonably by the Principle of Angular Conservation:
[tex]\frac{1}{2}\cdot (50\,kg)\cdot (0.75\,m)^{2}\cdot \left(30\,\frac{rev}{s} \right) = \frac{1}{2}\cdot \left[(50\,kg)\cdot (0.75\,m)^{2}+ (20\,kg)\cdot (0.5\,m)^{2} \right] \cdot \dot n[/tex]
The final angular velocity is:
[tex]\dot n = 25.471\,\frac{rev}{s}[/tex]
An object is situated to the left of a lens. A ray of light from the object is close to and parallel to the principal axis of the lens. The ray passes through the lens. Which one of the following statements is true?
The ray passes through a focal point of the lens only if the lens is a converging lens.
The ray passes through the lens without changing direction, no matter whether the lens is converging or diverging.
The ray crosses the principal axis at a distance from the lens equal to twice the focal length, no matter whether the lens is converging or diverging.
The ray passes through a focal point of the lens only if the lens is a diverging lens.
The ray passes through a focal point of the lens, no matter whether the lens is converging or diverging.
Answer:
The ray passes through a focal point of the lens only if the lens is a converging lens.
Explanation:
By the principles of geometric optics, we know that all rays parallel to the principal axis of converging lens, change its direction to a point situated in the axis of the lens. This last point is know as the focal point.
Hence, the only truth choice is:
The ray passes through a focal point of the lens only if the lens is a converging lens.
I attached an image to illustrate this situation
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A simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. The pendulum has a length of 2 m and its period is measured to be 2 s. The value of g obtained in this investigation is most nearly __________.A. 1 m/s²B. 2 m/s²C. 5 m/s²D. 10 m/s²E. 20 m/s²
Answer:
Acceleration due to gravity is 20 [tex]m/sec^2[/tex]
So option (E) will be correct answer
Explanation:
We have given length of the pendulum l = 2 m
Time period of the pendulum T = 2 sec
We have to find acceleration due to gravity g
We know that time period of pendulum is given by
[tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]
[tex]2=2\times 3.14 \sqrt{\frac{2}{g}}[/tex]
[tex]0.3184= \sqrt{\frac{2}{g}}[/tex]
Squaring both side
[tex]0.1014= {\frac{2}{g}}[/tex]
[tex]g=19.71=20m/sec^2[/tex]
So acceleration due to gravity is 20 [tex]m/sec^2[/tex]
So option (E) will be correct answer.
Final answer:
Using the formula for the period of a simple pendulum and given values for length and period, the acceleration due to gravity on the planet is calculated to be approximately 9.87 m/s², which is closest to 10 m/s².
Explanation:
To determine the value of acceleration due to gravity (g) at the surface of a planet using a simple pendulum, we use the formula for the period (T) of a simple pendulum: T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
Rearranging the formula to solve for g gives us: g = (4π²L) / T². Given that the length (L) is 2 m and the period (T) is 2 s, we plug these values into the formula: g = (4π² × 2 m) / (2 s)².
Calculating this gives us: g = (4π² × 2 m) / 4, simplifying further gives g = π² × 2 m. Now substituting π = 3.14159, we find g = (3.14159)² × 2 m ≈ 9.87 m/s², which is closest to choice D: 10 m/s².
In this section we considered a circular parallel-plate capacitor with a changing electric field. Describe the induced magnetic field lines?
Answer:
The study of Maxwell's equations and Ampere's law are used for analysis
Explanation:
Though There's a magnetic field associated with a changing electric field in TEM propagation of an EM wave through space (which is how it propogates, the changing E field begets the M field, the changeing M field begets the E field, leapfrogging each other).
But between capacitor plates, according to Maxwell and Ampere law, the displacement current in Ampere law was exactly to solve cases like that of a capacitor. A magnetic field cannot have discontinuities, unlike the electric field because there are electric charges, but there are no magnetic monopoles, at least as far as we know in the Universe in its current state. We can therefore conclude that there cannot be a magnetic field outside the capacitor and nothing inside
Conventional current is the rate at which positive charge flows in a circuit. In atoms, only the electrons are free to move. What can you conclude from these statements?
A) Electrons must be positively charged.
B) The definition of current must be changed.
C) Electrons do not flow in electric circuits.
D) Charges actually flow opposite the conventional current.
Answer:
In a metal circuit, the charges which are free electrons flow opposite to the flow of conventional current(which is assumed as the flow of positive charges)
Explanation:
Conventional current is defined as the direction that positive charge would flow, which is opposite to the actual flow of electrons in a circuit.
From the statements given, we can conclude that charges actually flow opposite the conventional current. This is because conventional current assumes that positive charge is moving in the direction of the electric field, whereas in actuality, in metal wires, it is the electrons— which have a negative charge—that are moving.
Electrons flow in a direction opposite to the defined conventional current. The designation of the direction of conventional current dates back to Benjamin Franklin's time, and despite later discovery that electrons are the primary charge carriers in circuits, the convention has remained the same.
A spacecraft is separated into two parts by detonating the explosive bolts that hold them together. The masses of the parts are 1200 kg and 1800kg; the magnitude of the impulse on each par from the bolts is 300 N s. With what relative speed do the twe twe parts separate because of the detonation?
Answer:
Explanation:
Given that,
A space ship is separated into two part
Mass of first part
M1 = 1200kg
Mass of second part
M2 = 1800kg
Magnitude of impulse on each part is
I = 300Ns
We want to find the relative velocity at which the two parts separate
Now,
Impulse is give as
I = ft = mv - mu
Since the body is considered to be at rest before detonation
Then, I = mv
So for first part
I = M1•V1
V1 = I / M1
V1 = 300 / 1200
V1 = 0.25m/s
Also, for second part
V2 = —I / M2
V2 = —300 / 1800
V2 = —0.167 m/s
NOTE: the negative sign is due to the fact that the two bodies are moving away from each other.
The relative speed of the two masses because of detonation is
V = V1 — V2
V = 0.25 — (—0.167)
V = 0.25 + 0.167
V = 0.417 m/s
The relative speed of the two masses because of detonation is 0.417 m/s
Answer:
(1) [tex]\frac{5}{12} m/s[/tex]
(2) [tex]-\frac{5}{12} m/s[/tex]
Explanation:
Lets say two parts went left and right, one with 1200Kg, call it A ,went right and one with mass 1800Kg went left. Let's establish right as positive and left as negative for our speed convention.
Both have initial acceleration of 300N, assuming for one second, by newton's second law F=ma, 'A' will have acceleration of 1/4m/s^ which will induce velocity of 0.25m/s as well.
So A has velocity of [tex]\frac{1}{4} m/s[/tex] towards right direction.
and B has velocity of - [tex]\frac{1}{6} m/s[/tex] towards left direction.
Relative Velocities.
and 'B' will have acceleration of [tex]\frac{1}{6} m/s^2[/tex] which will also produce velocity of [tex]\frac{1}{6} m/s[/tex].
[tex]V_{AB} = V_{A} -V_{B}[/tex] = [tex]$\frac{1}{4} --\frac{1}{6}= \frac{5}{12}m/s[/tex] (Read as Velocity A with respect to B) (1)
[tex]V_{BA} =V_{B} -V_{A} = -\frac{1}{6}-\frac{1}{4} =-\frac{5}{12} m/s[/tex] (Read as Velocity B with respect to B). (2).
A cylindrical insulated wire of diameter 5.0 mm is tightly wound 200 times around a cylindrical core to form a solenoid with adjacent coils touching each other. When a 0.10 A current is sent through the wire, what is the magnitude of the magnetic field on the axis of the solenoid near its center
Final answer:
To find the magnetic field inside a solenoid, use the formula B = μ₀ * n * I, where B is the magnetic field, n is the number of turns per unit length, and I is the current through the solenoid.
Explanation:
To find the magnetic field inside the solenoid, we can use the formula B = μ0* n * I, where B is the magnetic field, μ0 is the permeability of free space, n is the number of turns per unit length, and I is the current through the solenoid.
Given that the solenoid has a diameter of 5.0 mm, we can calculate the radius by dividing the diameter by 2, which is 2.5 mm or 0.0025 m. Using the radius, we can find the number of turns per unit length by dividing the total number of turns (200) by the length of the solenoid (unknown in the given information).
Once we know the number of turns per unit length and the current (0.10 A), we can calculate the magnitude of the magnetic field using the formula mentioned earlier.
Final answer:
To calculate the magnitude of the magnetic field on the axis of the solenoid near its center, we can use the formula B = μ0 * n * I.
Explanation:
To calculate the magnitude of the magnetic field on the axis of the solenoid near its center, we can use the formula:
B = μ0 * n * I
Where B is the magnetic field, μ0 is the permeability of free space, n is the number of turns per unit length, and I is the current.
In this case, the solenoid has a diameter of 5.0 mm, so the radius is 2.5 mm. The circumference of the solenoid is 2 π * r, and the length is 200 * this circumference. The number of turns per unit length (n) can be calculated by dividing the number of turns by the length of the solenoid. Finally, plug in the values into the formula to calculate the magnetic field.
Physics - Wave Diagrams Assignment
Wave
Wave
B
CAN SOMEONE HELP ME LOL
Answer:At the top of the page is a transvers wave
C= crest
B= wavelingth
D= trough
A= amplatud
The next wave is a longitudinal wave
What imaginary line marks the southern end of the tropical zone
Answer:
The tropic of Capricorn
Explanation:
This line runs 23.5 degrees south of the equator.
An insoluble solid material that is produced in double replacement reactions is called
Answer:
Precipitate
Explanation:
When a double displacement reaction occurs, the cations and anions switch partners, resulting in the formation of two new ionic compounds AD and CB, one of products is in the solid state and forms an insoluble ionic compound called a precipitate.
A thin uniform rod of mass M and length L is bent at its center so that the two segments are perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through the point where the two segments meet.
Answer:
[tex]\frac{1}{12}ML^2[/tex]
Explanation:
The moments of the whole object is the sum of the moments of the 2 segments of rod at their ends of which length is L/2 and mass M/2:
[tex]I = 2I_{end} = 2\frac{1}{3}\frac{M}{2}\left(\frac{L}{2}\right)^2[/tex]
[tex]I = \frac{1}{3}M\frac{L^2}{4}[/tex]
[tex]I = \frac{1}{12}ML^2[/tex]
Water behind a dam has a certain amount of stored energy that can be released as the water falls over the top of the dam. It may be enough energy to turn a mill wheel or an electricity-generating turbine. Choose the term that best describes the type of energy stored in the water at the top of the dam.
potential energy
kinetic energy
mechanical energy
Answer:
The answer is potential energy
Explanation:
The potential energy is the energy possessed by a body by virtue of it position
For example the water at the top of the dam is being held at a height h above the bottom of the dam
Then the potential energy
PE= weight of the water* the height
PE= m*g*h
5. Suppose a cup of cocoa cooled from 90° C to 60° C after 10 minutes in a room whose temperature was 20° C . Use Newton’s law of cooling [T − Ts = (T0 − Ts )⋅ e−k⋅t ] to answer the following questions: a) How much longer would it take the cocoa to cool to 35° C ? b) Instead of being left to stand in a room, the cup with initial temperature 90° C is placed in a freezer whose temperature is −15° C . How long will it take the cocoa to cool from 90° C to 35° C ?
Answer:
a) t = 1051.6 sec = 17.5 min
b) t = 795.5 sec = 13.25 min
Explanation:
First of all we use the initial data to find out constant 'K'.
T - Ts = (T₀ - Ts) e^(-kt)
Here, we have:
T = Final Temperature = 60° C
Ts = Surrounding Temperature = 20° C
T₀ = Initial Temperature = 90° C
t = time = 10 min = 600 sec
k = constant = ?
Therefore,
60° C - 20° C = (90° C - 20° C).e^(-k600)
40° C/70° C = e^(-k600)
ln (0.57142) = -600k
k = 9.327 x 10⁻⁴ sec⁻¹
a)
Now, for this case we have:
T = Final Temperature = 35° C
Ts = Surrounding Temperature = 20° C
T₀ = Initial Temperature = 60° C
t = time = ?
k = constant = 9.327 x 10⁻⁴ sec⁻¹
Therefore,
35° C - 20° C = (60° C - 20° C).e^(-9.327 x 10⁻⁴ sec⁻¹ x t)
15° C/40° C = e^(-9.327 x 10⁻⁴ sec⁻¹ x t)
ln (15/40) = - 9.327 x 10⁻⁴ sec⁻¹ x t
t = 1051.6 sec = 17.5 min
b)
Now, for this case we have:
T = Final Temperature = 35° C
Ts = Surrounding Temperature = -15° C
T₀ = Initial Temperature = 90° C
t = time = ?
k = constant = 9.327 x 10⁻⁴ sec⁻¹
Therefore,
35° C + 15° C = (90° C + 15° C).e^(-9.327 x 10⁻⁴ sec⁻¹ x t)
50° C/105° C = e^(-9.327 x 10⁻⁴ sec⁻¹ x t)
ln (50/105) = - 9.327 x 10⁻⁴ sec⁻¹ x t
t = 795.5 sec = 13.25 min