Answer:
The equation that will relate all the given parameters, in other to calculate the potential energy of charge A is:
∆V = ∆U/q, ∆V is potential at charge A position, q is magnitude of charge A, ∆U will be made the subject of the relation, which is the Potential Energy of charge A. The notation "∆" show, the quantities have both in values and final values, in the electric field.(Change in Electric potential and potential energy, due to the effect of the field)
Explanation:
The potential energy of a charged particle (Charge A) in an electric field depends on the magnitude of the charge(Known as stated in the question). However, the potential energy per unit charge has a unique value at any point in the electric field.
Perception is defined as the ____. a.conversion of electromagnetic energy into electrochemical energyb.process of encoding experiencesc.transduction of environmental stimulid.interpretation of sensory information
Answer:
d.interpretation of sensory information
Explanation:
The organizing, recognition, and understanding of sensory information is perception(the Latin perceptio). It is the experience in order to represent and comprehend the knowledge as well as the world.It comprises data collection from sensory organs to brain interpretation
Feng and Isaac are riding on a merry-ground. Feng rides on a horse at the outer rim of the circular platform, twice as far from the center of the circular platform as Isaac, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, what is Feng’s angular speed?
Answer: The question is incomplete or missing details. here is the remaining part of the question ;
1. impossible to determine
2. half of Isaac’s
3. the same as Isaac’s
4. twice Isaac’s
The angular speed of feng will be the same as that of Isaac. Hence the answer is option 3
Explanation:
Since we have been told that both feng and isaac are riding on a merry go round i.e in a circular motion, irrespective of how fast one ride above the other, the angular speed will be constant since they are riding on a merry go round, as such both feng and isaac will maintain equal angular speed, hence the angular speed of feng will be the same as that of Isaac.
When the merry-go-round is rotating at a constant angular speed, what is Feng’s angular speed will be same as Isaac's.
Since, both Feng and Isaac are riding on the same marry-go-round in a circular motion. The speed of the marry-go-round is constant.
Angular speed depends on the frequency of the rotation and frequency of the rotation is irrespective of the position in a circle.
Therefore, when the merry-go-round is rotating at a constant angular speed, what is Feng’s angular speed will be same as Isaac's.
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when electrons are excited to differnt energy levels, the average radii from the nucleus also changes. Rank the following electron energy states according to the average distance of the electron from the nucleus. Rank from largest to smallest distances..
Answer:
The problem can be solved using the equation below
Explanation:
ΔE = E1 − E2
ΔE = 2.179*〖10〗^(-18)*(〖1/〖n_1〗^2 〗-〖1/〖n_2〗^2 〗 )
What is it's speed at this time? A rocket blasts off and moves straight upward from the launch pad with constant acceleration. After 2.7s the rocket is at a height of 86m
Answer: 31.85 m/s
Explanation:
Since we are talking about constant acceleration, we can use the following equation to find the speed [tex]s[/tex] of the rocket:
[tex]s=\frac{d}{t}[/tex]
Where:
[tex]d=86 m[/tex] is the distance the rocket has traveled
[tex]t=2.7 s[/tex] is the time in which the rocket has traveled the distance [tex]d[/tex]
Solving the equation with the given data:
[tex]s=\frac{86 m}{2.7 s}[/tex]
[tex]s=31.85 m/s[/tex] This is the rocket's speed
Tech A says clutch slippage can best be diagnosed by loading the engine while the clutch is released with the transmission in a high gear, such as fourth. Tech B says a slipping clutch can cause friction surfaces warpage or hot spots. Who is correct?a. Technician Ab. Technician Bc. Both Technician A and Technician Bd. Neither Technician A nor Technician B
Answer: c. Both Technician A and Technician B.
Explanation:
In motor enginnering, in case of a clutch slippage. It can best be diagnosed by loading the engine while the clutch is released with the transmission in a high gear, such as fourth gear whilst releasing the clutch pedal slowly.
It's also true that a slipping clutch can cause friction surfaces warpage or hot spots.
This therefore makes both Technicians A and B right in their arguments.
Both Tech A, who suggests diagnosing clutch slippage by loading the engine in high gear, and Tech B, who states that a slipping clutch can lead to friction surface damage, are correct.
The subject of this question pertains to automotive technology, specifically the diagnosis of clutch issues in vehicles. When evaluating the statements given by the two technicians:
Tech A is correct in suggesting that clutch slippage can be diagnosed by loading the engine in a high gear while the clutch is released, as this can put enough load on the engine to reveal if the clutch is slipping.
Tech B is correct in saying that a slipping clutch can lead to warpage or the development of hot spots on the friction surfaces.
Therefore, the correct answer is c. Both Technician A and Technician B are correct in their assessments of clutch slippage diagnosis and potential consequences.
A porcelain cup of mass 303 g and specific heat 0.260 cal/g-°C contains 161 cm³ of coffee, which has a specific heat of 1.00 cal/g-°C. If the coffee and cup are initially at 71.0 °C, how much ice at 0.00 °C must be added to lower the temperature to 49.0 °C?
Answer:
[tex] m_i =\frac{1736.702 cal}{129 cal/gr}=13.46 gr[/tex]
So we need to add 13.46 gr of ice in order to reach the final equilibrium temperature of 49 C
Explanation:
For this case we need to use the fact that the sum for all the heats involved in the system are 0, since we assume an equilibrium state.
Data given
[tex]m_p = 303 gr[/tex] mass of the porcelain cup
[tex] cp_p = 0.260 cal/g C[/tex] the specific heat for the porcelain cup
[tex] T_{ip} = T_{ic}= 71 C[/tex] initial temperature for the coffee and the porcelain cup.
[tex] V_{c}= 161 cm^3[/tex] Volume of the coffee.
We can convert this to m^3 and we got 0.000161m^3 and assuming the density fot the coffee equal to the water 1 Kg/m^3 the mass would be:
[tex] m_c = 1 kg/m^3 *0.000161 m^3 = 0.000161 kg=0.161 Kg[/tex]
[tex] Cp_{c} = 1 cal/g C[/tex] Specific heat for the coffee
[tex] m_i =?[/tex] mass of ice required
[tex] T_e= 49C[/tex] equilibrium temperature
[tex]L_f = 80 cal/g [/tex] represent the latent heat of fusin since the ice change the state to liquid.
Solution to the problem
Using this formula:
[tex] \sum_{i=1}^n Q_i = 0[/tex]
We have this:
[tex] m_p cp_p (T_e -T_{ip}) + m_{c} cp_c (T_e -T_{ic}) +m_i L_f + m_i cp_w (T_e -0) =0[/tex]
Now we can replace and we have this:
[tex] 303 gr *(0.260 cal/g C) (49-71)C + 0.161 gr*(1 cal/g C)(49-71)C +m_i [80 cal/gr+(1cal/g C)(49-0)C]=0[/tex]
And now we can solve for [tex] m_i[/tex] and we have:
[tex]-1733.16cal -3.542cal +m_i [129 cal/g]=0[/tex]
[tex]m_i =\frac{1736.702 cal}{129 cal/gr}=13.46 gr[/tex]
So we need to add 13.46 gr of ice in order to reach the final equilibrium temperature of 49 C
Your textbook discusses the cosmic calendar, a model of the history of the universe scaled to a single year. The length of time represented by one month on this cosmic calendar is therefore closest to__________
Answer:
1.15 Billion Years
Explanation:
If the average age of the universe is 13.772 billion years and that equals to a year in cosmic calendar, then the length of time in a month will be 13.772 / 12 = 1.14766666667 ~ 1.15 Billion Years
A particular type of resistor has a tolerance of 3%. Technician A says this indicates that the resistor’s current value can be 3% above or below its stated specification. Technician B says this indicates the resistor’s resistance value can be 3% above or below its stated specification. Who is right?
Answer:
Technician B is correct.
Explanation:
The given value of resistor having 3% tolerance means that the given quantity of the physical parameter can vary by 3% of what is specified. This variation can be either more or less than the quantified value.
When the variation can occur on both sides of the stated value then it is called bilateral tolerance, usually represented as, [tex]\pm3\%[/tex].When the variation is permissible only in one direction then it is called unilateral tolerance, represented by the sign + for the impressibility on the higher side and [tex]-[/tex] sign for the impressibility on the lower side.Calculate the discharge (in cubic meters/second) if a large stream's average velocity is 3 meters/second, its stage is 14 meters, and its width as measured bank-to-bank is 27 meters.
Answer:
Explanation:
Given
average velocity of stream is [tex]v_{avg}=3\ m/s[/tex]
Stage of stream i.e. depth [tex]d=14\ m[/tex]
width of stream [tex]w=27\ m[/tex]
We know that discharge of a stream is given by
[tex]Q=A\cdot v[/tex]
where A=area of cross-section
v=average velocity
[tex]A=14\times 27\ m^2[/tex]
Therefore discharge is given by
[tex]Q=14\times 27\times 3[/tex]
[tex]Q=1134\ m^3/s[/tex]
A 75.3 kg bobsled is pushed along a horizontal surface by two athletes. After the bobsled is pushed a distance of 8.1 m starting from rest, its speed is 7.1 m/s. Find the magnitude of the net force on the bobsled. Answer in units of N
Answer:
233.43 N
Explanation:
Force: This is the product of mass and acceleration of a body.
The formula for force is given as,
F = ma .............. Equation 1
Where F = Net force on the bobsled, m = mass of the bobsled, a = acceleration of the bobsled
We can look for a using the equation of motion
v² = u² + 2as.............. Equation 2
Where V = final velocity, u = initial velocity, a = acceleration, s = distance.
making a the subject of the equation,
a = (v²-u²)/2s................... Equation 3
Given: v = 7.1 m/s, u = 0 m/s ( from rest), s = 8.1 m.
Substitute into equation 3
a = (7.1²-0²)/(2×8.1)
a = 50.41/16.2
a = 3.1 m/s²
Also given: m = 75.3 kg
Substitute into equation 1
F = 75.3×3.1
F = 233.43 N
Hence the net force on the bobsled = 233.43 N
In a transverse wave, the motion of the disturbance is in what direction relative to the wave motion? opposite parallel perpendicular in the same direction
Answer:
[tex]\displaystyle Perpendicular[/tex]
Explanation:
Longitudinal waves are parallel to the direction of the motion of the disturbance, while transverse waves are perpendicular to the direction of the motion of the disturbance.
I am joyous to assist you anytime.
In a transverse wave, the motion of the disturbance is perpendicular to the wave motion.
A transverse wave is a type of wave where movements oscillate along paths at a right angle to the advance of the wave.Examples of this type of wave include seismic waves and electromagnetic waves.The electromagnetic waves can be both radio waves and light waves.In conclusion, in a transverse wave, the motion of the disturbance is perpendicular to the wave motion.
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While driving north at 25 m/s during a rainstorm you notice that the rain makes an angle of 38 degrees with the vertical. While driving back home moments later at the same speed but in the opposite direction, you see that the rain is falling straight down. From these observations, determine the speed of the raindrops relative to the ground. From these observations, determine the angle of the raindrops relative to the ground.
Final answer:
To determine the speed of the raindrops relative to the ground and their angle, we analyze the observations of rain angle in two car speeds using trigonometry and the Pythagoras theorem, concluding with the application of the tangent function and arctan.
Explanation:
When driving north at 25 m/s, rain appears to make an angle of 38 degrees with the vertical due to the combination of the rain's vertical speed and the horizontal speed of the car. However, when driving south at the same speed, the rain appears to fall vertically, indicating the horizontal component of the rain's velocity is equal to the car's speed. To determine the speed of the raindrops relative to the ground and the angle of the raindrops relative to the ground, we can use trigonometry.
Given the observation that the rain appears vertical when driving south at 25 m/s, it implies the horizontal velocity of the rain is 25 m/s (equal but opposite to the car's velocity, thereby canceling it out). From the 38-degree angle observation, we can use tan(38 degrees) = vertical component / horizontal component to find the vertical speed. The answer is derived from tan(38 degrees) = vertical component / 25 m/s, which allows us to calculate the vertical component.
The speed of the raindrops relative to the ground is then found by calculating the resultant of the horizontal and vertical components using Pythagoras theorem, and the angle of the raindrops is found by taking the arctan of the vertical component over the horizontal component.
A wire is stretched between two posts. Another wire is stretched between two posts that are twice as far apart. The tension in the wires is the same, and they have the same mass. A transverse wave travels on the shorter wire with a speed of 249 m/s. What would be the speed of the wave on the longer wire?
Answer: 996m/s
Explanation:
Formula for calculating velocity of wave in a stretched string is
V = √T/M where;
V is the velocity of wave
T is tension
M is the mass per unit length of the wire(m/L)
Since the second wire is twice as far apart as the first, it will be L2 = 2L1
Let V1 and V2 be the speed of the shorter and longer wire respectively
V1 = √T/M1... 1
V2 = √T/M2... 2
Since V1 = 249m/s, M1 = m/L1 M2 = m/L2 = m/2L1
The equations will now become
249 = √T/(m/L1) ... 3
V2 = √T/(m/2L1)... 4
From 3,
249² = TL1/m...5
From 4,
V2²= 2TL1/m... 6
Dividing equation 5 by 6 we have;
249²/V2² = TL1/m×m/2TL1
{249/V2}² = 1/2
249/V2 = (1/2)²
249/V2 = 1/4
V2 = 249×4
V2 = 996m/s
Therefore the speed of the wave on the longer wire is 996m/s
Final answer:
The speed of the wave on the longer wire would be the same as on the shorter wire, which is 249 m/s, because the tension and linear mass density are the same.
Explanation:
The speed of a transverse wave on a string is determined by the tension of the string and the linear mass density of the string, according to the formula v = √{Ft/mu}, where v is the wave speed, Ft is the tension in the string, and \\mu is the linear mass density. Since both wires have the same mass and tension, the linear mass density is the same. Therefore, the speed of the wave on the longer wire would be the same as that on the shorter wire, which is 249 m/s.
Kohlberg believed that_____________.A. moral development progressed through distinct, qualitatively different stages. B. moral development was made up of constant, gradual growth.C. the stages of development were distinct; they did not build on one another.D. children did not necessarily progress through the stages in order.
Answer:
A. moral development progressed through distinct, qualitatively different stages.
Explanation:
Kolhberg’s theory of moral development states that we progress through three levels of moral thinking that build on our cognitive development.
Question: Suppose you see a crescent Moon; how much of the Moon's entire surface (the full globe of the Moon) is in daylight?
Final answer:
When you see a crescent Moon, only a small portion of the Moon's surface is in daylight due to the angle of sunlight. The rest is faintly illuminated by Earthshine.
Explanation:
Crescent Moon: When you see a crescent Moon, only a small portion of the Moon's entire surface is in daylight. The illuminated portion of the Moon's surface is determined by the angle at which sunlight strikes the Moon. The rest of the Moon's surface not in direct sunlight is faintly illuminated by Earthshine, caused by sunlight reflecting off the Earth.
A flywheel with a very low friction bearing takes 1.6 h to stop after the motor power is turned off. The flywheel was originally rotating at 55 rpm. What was the initial rotation rate in radians per second?
Answer: 5.76 rads/s
Explanation:
The initial rotation is 55 rpm
1 rev = 2π radians
55 revs = 55 × 2π/1 = 345.58 radians/min
345.58 rads/min = 345.58rads/60s = 5.76 rads/s
The electric force between objects A and B is F. If the charge of object A were twice as large as it is, but everything else was kept the same, what would be the new electric force between objects A and B?
Answer:
[tex]F'=2F[/tex]
Explanation:
According to Coulomb's law and assuming the objects as point charges, the magnitude of the electric force that each object exerts on the other is defined as:
[tex]F=\frac{kq_Aq_B}{d^2}[/tex]
Here k is thee Coulomb constant, [tex]q_A[/tex] and [tex]q_B[/tex] are the charges of the objects and d is the distance of separation between them. We have [tex]q'_A=2q_A[/tex]:
[tex]F'=\frac{kq'_Aq_B}{d^2}\\F'=\frac{k2q_Aq_B}{d^2}\\F'=2\frac{kq_Aq_B}{d^2}\\F'=2F[/tex]
A small object moves along the xx-axis with acceleration ax(t)ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At tt = 0 the object is at xx = -14.0 mm and has velocity v0xv0x = 8.70 m/sm/s.
What is the xx-coordinate of the object when tt = 10.0 ss?
Answer:
x = 54.3m (on the +ve x axis)
Explanation:
This is an Initial Value Problem. That means the initial values of certain parameters have been given and that can help solve the problem.
Given that acceleration, a, is:
ax(t) = - 0.032(15.0 - t)
And the initial values are:
x(t = 0) = - 14.0m
v(t = 0) = 8.7m/s
Hence,
a = - 0.032(15 - t)
a = - 0.48 + 0.032t
a = dv/dt = -0.48 + 0.032t
To obtain the velocity, v, integrate the acceleration and apply the initial values of v and t:
v = ∫dv/dt = ∫(-0.48 + 0.032t)
∫dv = ∫(-0.48 + 0.032t)dt
(v - v₀) = -0.48(t - t₀) + 0.032(t²/2 - t₀²/2)
Inputting the initial values t₀ = 0s, v₀ = 8.7m/s:
=> v - 8.7 = -0.48t + 0.032t²/2
v = 8.7 - 0.48t + 0.016t²
To obtain distance, x, integrate the velocity and apply the initial values:
v = dx/dt = 8.7 - 0.48t + 0.016t²
=> ∫dx/dt = ∫(8.7 - 0.48t + 0.016t²)
∫dx= ∫(8.7 - 0.48t + 0.016t²)dt
(x - x₀) = 8.7(t - t₀) - 0.48(t²/2 - t₀²/2) + 0.016(t³/3 - t₀³/3)
Inputting the initial values t₀ = 0s, x₀ = - 14.0m:
(x + 14.0) = 8.7t - 0.48t²/2 + 0.016t³/3
x = 8.7t - 0.48t²/2 + 0.016t³/3 - 14.0
Now that the distance, x, has been obtained, when t = 10s:
x = 8.7*10 - 0.48*10²/2 + 0.016*10³/3 - 14.0
x = 87 - 24 + 5.3 - 14.0
x = 54.3m
Therefore, at time, t = 10s, x = +54.3m. (i.e. 54.3 on the +ve x axis).
Acceleration of a body is the change in velocity with respect to time.
The x-coordinate of the object when value of time is 10 second is 54.3 m.
What is acceleration?Acceleration of a body is the change in velocity with respect to time.
Given information-
A small object moves along the x-axis with acceleration,
[tex]ax(t)=-0.0320(15-t)[/tex]
Initial position and initial velocity of the object is,
[tex]x_0=-14\rm m\\v_0=8.7\rm m/s[/tex]
For the velocity integrate the given equation as,
[tex]ax(t)=-0.0320(15-t)\\\dfrac{dv}{dt} =\int ({-0.48+0.032t} )\, dt\\v-v_0=-0.48t+0.032\times\dfrac{t^2}{2} \\v-8.7=-0.48t+0.016t^2\\v=0.016t^2-0.48t+8.7[/tex]
Integrate it again to find the distance of the object.
[tex]v=0.016t^2-0.48t+8.7\\\int\dfrac{dx}{dt}=\int(0.016t^2-0.48t+8.7)dt\\x-x_0=0.016\times\dfrac{t^3}{3}-0.48\times\dfrac{t^2}{2}+8.7t\\x-14=0.016\times\dfrac{t^3}{3}-0.48\times\dfrac{t^2}{2}+8.7t\\\\x=0.016\times\dfrac{t^3}{3}-0.48\times\dfrac{t^2}{2}+8.7t-14\\[/tex]
Put the value of [tex]t[/tex] as 10 seconds as,
[tex]x=0.016\times\dfrac{10^3}{3}-0.48\times\dfrac{10^2}{2}+8.7\times10-14\\[/tex]
[tex]x=54.3\rm m[/tex]
Hence the x-coordinate of the object when value of time is 10 second is 54.3 m.
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How much charge will have accumulated on the plates of a charging capacitor after a length of time equal to one time constant?
Answer: 63% of the final charge
Explanation:
When a capacitor is connected to a battery/power supply, the capacitor charges following the exponential law:
[tex]Q(t)=Q_0 (1-e^{-\frac{t}{\tau}})[/tex]
where
[tex]Q_0[/tex] is the final charge of the capacitor, which is
[tex]Q_0 =CV_0[/tex]
where C is the capacitance and [tex]V_0[/tex] the potential difference of the battery
t is the time
[tex]\tau[/tex] is the time constant of the circuit
Re-writing the equation,
[tex]Q(t)=CV_0 (1-e^{-\frac{t}{\tau}})[/tex]
After a time equal to one time constant,
[tex]t=\tau[/tex]
Therefore the charge on the capacitor will be
[tex]Q(\tau)=CV_0 (1-e^{-\frac{\tau}{\tau}})=CV_0(1-e^{-1})=0.63CV_0[/tex]
Which means 63% of the final charge.
Final answer:
After one time constant, the charge on a capacitor in an RC circuit will be approximately 63% of its maximum value. The time constant τ is calculated as the product of resistance R and capacitance C. The formula for the charge at any time is Q(t) = Q_{max}(1 - e^{-t/RC}).
Explanation:
When a capacitor charges in an RC circuit, it goes through a process described by an exponential function. After a time period equal to one time constant, represented by the symbol τ and given by the formula τ = RC (where R is the resistance in ohms and C is the capacitance in farads), the charge on a capacitor will reach approximately 63% of its maximum charge. This characteristic time, the time constant, determines how quickly the capacitor charges up to its maximum value, which is limited by the supply voltage.
The charge Q on the capacitor can be determined by the equation Q(t) = Q_{max}(1 - e^{-t/RC}) where Q(t) is the charge at time t, Q_{max} is the maximum charge, R is the resistance, C is the capacitance, and e is the base of the natural logarithm. When t equals the time constant τ (t = RC), the charge on the capacitor is Q(τ) = Q_{max}(1 - e^{-1}) which is approximately 63% of Q_{max}.
Aristotle said that a moving earthly or `mundane' object with nothing pushing or pulling on it will always A slow down and stop. B speed up. C keep moving at the same speed. D follow a circular path.
Answer:
A slow down and stop
Explanation:
When there is no force acting on something it automatically begins to slow down and then stops.Essentially, Aristotle's perspective of motion is that "it requires a force to move an object in an unnatural" way— or, plainly, that "movement involves strength." Indeed, if you propel a book, it keeps moving. Once you stop trying to push, it comes to a stop.
Which of the following would decrease the resistance in a wire?
Increase the thickness of the wire
Increase the mass of the wire
Decrease the thickness of the wire
Decrease the mass of the wire
Increase the thickness of the wire would decrease the resistance in a wire
Explanation:
Thicker wires have a larger cross-section that increases the surface area with which electrons can flow unimpeded. The thicker the wire, therefore, the lower the resistance.
Thin wires have very high resistance the reason the thin tungsten in a bulb glows because it is heated from the high resistance of many electrons trying to pass through a very small cross-section.
Answer:
Increase the thickness of the wire
Explanation:
I just took the test.
008 (part 1 of 3) 10.0 points A 0.338 kg particle has a speed of 3.8 m/s at point A and kinetic energy of 10.1 J at point B. What is its kinetic energy at A? Answer in units of J. 009 (part 2 of 3) 10.0 points What is the particle’s speed at B? Answer in units of m/s. 010 (part 3 of 3) 10.0 points What is the total work done on the particle as it moves from point A to B?
Answer:
1) 2.44 joules
2) 7.73 m/s
3) 7.6 joules
Explanation:
Kinetic energy (K) of a particle is:
[tex] K=\frac{mv^{2}}{2} [/tex] (1)
with m the mass, and v the velocity
1) Because we already now velocity on A (va) and the mass of the object we can calculate its kinetic energy:
[tex]K_{a}=\frac{mv_{a}^{2}}{2}=\frac{(0.338kg)(3.8\frac{m}{s})^{2}}{2}=2.44J [/tex]
2) Because on B we know mass and kinetic energy we should solve (1) for v and use our values to find the velocity on B:
[tex]v_{b}=\sqrt{\frac{2K_{b}}{m}}=\sqrt{\frac{2(10.1J)}{(0.338kg)}}=7.73\frac{m}{s} [/tex]
3) Work-energy theorem states that the change of kinetic energy of an object is equal to the total work done on it, so:
[tex]W=K_b-K_a=10.1J-2.44J= 7.6J [/tex]
What is the net charge on a sphere that has the following? (a) 5.87 106 electrons and 8.11 106 protons C
Answer:
3.6 * 10^-13 C
Explanation:
The net charge of the sphere will be the sum of the total electron charge and total proton charge. Mathematically,
Q = Qe + Qp
TOTAL ELECTRON CHARGE:
An electron has an electronic charge of -1.6022 * 10^-19C.
Hence, the charge of 5.87 * 10^6 electrons will be:
Qe = - 1.6022 * 10^-19 * 5.87 * 10^6
Qe = - 9.405 * 10^-13 C
TOTAL PROTON CHARGE:
A proton has an electronic charge of 1.6022 * 10^-19. Hence, 8.11 * 10^6 protons will have:
Qp = 1.6022 * 10^-19 * 8.11 * 10^6
Qp = 1.3 * 10^-12 C
=> Q = (-9.405 * 10^-13) + (1.3 * 10^-12)
Q = 3.6 * 10^-13 C
Answer:
3.58*10⁻¹³ C
Explanation:
When we have a net charge on a sphere, this means that there must be a difference in the number of electrons and protons on the sphere, otherwise, the sphere would be electrically neutral.
In this case, we can find this difference as follows:
Np -Ne = 8.11*10⁶ protons - 5.87*10⁶ electrons = 2.24*10⁶ protons.
The total charge carried by all these protons is just the product of the charge of only one proton (which is equal to the elementary charge e) times the number of excess protons, as follows:
Qnet = 2.24*10⁶ protons * 1.6*10⁻19 C/proton = 3.58*10⁻¹³ C
A 10-kg sled carrying a 30-kg child glides on a horizontal, frictionless surface at a speed of 6.0 m/s toward the east. The child jumps off the back of the sled, propelling it forward at 20 m/s. What was the child’s velocity in the horizontal direction relative to the ground at the instant she left the sled?
Answer:
- 1.33m/s
Explanation:
We choose the system to be the child and the sled. The surface is friction less, which means that there are no forces exerted on the system horizontally. This means that the horizontal momentum component of the system is constant and conserved.
So we can use the conservation momentum principle to find the velocity of the child just after he leaves the sled.
This is shown in the attached file.
The child’s velocity in the horizontal direction relative to the ground at the instant she left the sled is 1.33 m/s.
How to calculate the velocity?It should be noted that velocity simply means the directional speed of an object in motion.
In this case, the velocity will be:
= ((30 + 10) × 6) - (10 × 20)/30
= (40 × 6) - (200)/30
= (240 - 200)/30
= 40/30
= 1.33 m/s
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3. If you start at the equator and travel to 100 N latitude, approximately how many kilometers (or miles) north of the equator will you be? Take the circumference of the Earth to be 40,000 kilometers (24,900 miles). Show your calculations.
Answer:
travel = 1111.11 km or ( 691.66 miles ) north of the equator
Explanation:
given data
travel = 10° N latitude
circumference of the Earth = 40,000 kilometers
solution
when we start at equator and travel whole world and finally reach that same point
so we cover 40,000 km and that is circumference of Earth
so that we cover = 360° latitude
and here distance travel in each degree of latitude will be
distance travel in each degree = [tex]\frac{40000}{360}[/tex]
distance travel in each degree = 111.11 km
so here for travel to 10 degrees N
so travel is = 111.11 × 10 =
travel = 1111.11 km or ( 691.66 miles ) north of the equator
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v?
Incomplete question as the car's speed is missing.I have assumed car's speed as 6.0m/s.The complete question is here
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s
Answer:
[tex]F_{B}=-5755N[/tex]
Explanation:
Set up force equation
∑F=ma
∑F=W+FB
[tex]\frac{mv^{2} }{R}=W+F_{B}\\ F_{B}=\frac{mv^{2} }{R}-W\\F_{B}=\frac{(W/g)v^{2} }{R}-W\\F_{B}=\frac{(6000N/9.8m/s^{2} )(6m/s)^{2} }{(15m)}-6000N\\F_{B}=-5755N[/tex]
The minus sign for downward direction
Is this statement true or false? The next generation of nuclear power plants being built in California and South Africa are even safer and more efficient than previous generations of power plants.
Answer:
True
Explanation:
Modern safer and cheaper nuclear reactors can not only meet the range of our long term energy demands, they can also fight global warming.
Modern techniques provide ways to reduce radioactive waste amount. "A closed fuel cycle may be switched on for new kinds of nuclear plants. Alternatively, the waste is chemically dissuaded to transform the reusable element into fuel. This implies that nuclear waste would not be buried.
Answer:
Truee
Explanation:
A fireworks rocket explodes at a height of 120 m above the ground. An observer on the ground directly under the explosion experiences an average sound intensity of 7.20 10⁻² W/m² for 0.205 s.
(a) What is the total amount of energy transferred away from the explosion by sound?
(b) What is the sound level in decibels heard by the observer?
Answer:
2670.90667586 J
108.573324964 dB
Explanation:
r = Distance = 120 m
A = Area = [tex]4\pi r^2[/tex]
I = Intensity of sound = [tex]7.2\times 10^{-2}\ W/m^2[/tex]
t = Time taken = 0.205 s
[tex]I_0[/tex] = Threshold intensity = [tex]10^{-12}\ W/m^2[/tex]
Power is given by
[tex]P=IA\\\Rightarrow E=7.2\times 10^{-2}\times 4\pi 120^2[/tex]
Energy is given by
[tex]E=Pt\\\Rightarrow E=7.2\times 10^{-2}\times 4\pi 120^2\times 0.205\\\Rightarrow E=2670.90667586\ J[/tex]
The total amount of energy is 2670.90667586 J
Sound intensity level is given by
[tex]\beta=10log\dfrac{I}{I_0}\\\Rightarrow \beta=10log\dfrac{7.2\times 10^{-2}}{10^{-12}}\\\Rightarrow \beta=108.573324964\ dB[/tex]
The sound level is 108.573324964 dB
The total amount of energy transferred away from the explosion by sound is 1.476 x 10^-2 Joules. The sound level in decibels heard by the observer is 97.50 dB.
Explanation:(a) To find the total amount of energy transferred away from the explosion by sound, we can use the formula:
Energy = Intensity x Time
Substituting the given values:
Energy = (7.20 x 10-2 W/m2) x (0.205 s) = 1.476 x 10-2 Joules
(b) The sound level in decibels heard by the observer can be calculated using the formula:
Sound Level (dB) = 10 x log10(Intensity / Threshold Intensity)
Substituting the given values:
Sound Level (dB) = 10 x log10((7.20 x 10-2 W/m2) / (10-12 W/m2)) = 97.50 dB
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When the person is skating from the bottom of the track back up to the top, her potential energy _______ . When the person is skating from the top of the track down to the bottom, his kinetic energy _______ . When the person is skating from the top of the track down to the bottom, her velocity _______ .
Answer:
Explanation:
Potential Energy of an object is a function of elevation from datum thus assuming datum to be ground, the potential energy of the person increases as he moves up to the top.
When a person is skating from the top of the track to the bottom then his potential energy is getting converted into the kinetic energy so his kinetic energy increases as he moves down.
When a person is moving from top to bottom his kinetic energy increases which is a function of mass and velocity. As mass is fixed therefore his velocity increases as he moves down
Final answer:
When the person is skating from the bottom of the track back up to the top, potential energy increases. When the person is skating from the top of the track down to the bottom, kinetic energy increases. When the person is skating from the top of the track down to the bottom, velocity increases.
Explanation:
When the person is skating from the bottom of the track back up to the top, her potential energy increases. When the person is skating from the top of the track down to the bottom, his kinetic energy increases. When the person is skating from the top of the track down to the bottom, her velocity increases.
Several students in the group made comments to the Coach about the back pack. Which student has the correct analysis of the forces and motion of the back pack?
The student who has the correct analysis of the forces and motion of the backpack understands Newton's laws of motion and can apply them to the situation.
Explanation:The student who has the correct analysis of the forces and motion of the backpack would be the one who understands the principles of Newton's laws of motion and can apply them to the situation.
Newton's first law of motion states that an object at rest tends to stay at rest, and an object in motion tends to stay in motion with the same speed and in the same direction, unless acted upon by an external force. This law applies to the backpack on the group's graph, as it initially moves forward and then stops.
Newton's second law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The student who correctly recognizes the forces and acceleration involved in the motion of the backpack would have the correct analysis.