(i) The speed of the journal is 15.7 m/s.(ii) The required wall shear stress per unit length ratio between the prototype and model is 1.
Given,
Speed of journal, n = 3000 rpm
Radius of the journal, R = 50 mm
Density of lubricant oil, ρ = 869 kg/m³
Viscosity of lubricant oil, μ = 2.9 x 10⁻² Pa.s
Clearance of the journal bearing, c = 0.04 mm(i) Speed of journal
For prototype, n₁ = 3000 rpm
Radius of the journal, R₁ = 50 mm
The diameter of the bearing is 100 mm.
Circumferential speed of the journal is given asV = πDN/60
where D is the diameter of the journal
n is the speed of the journal in rpm
.So, the diameter of the journal, D = 100 mm= 0.1 m
Circumferential speed of the journal,
V₁ = π × 0.1 × 3000/60= 15.7 m/s
For model, n₂ = n₁
Radius of the journal, R₂ = R₁/2= 50/2= 25 mm
The diameter of the bearing is 50 mm.
Circumferential speed of the journal is given as
V = πDN/60
where D is the diameter of the journal
n is the speed of the journal in rpm
.So, the diameter of the journal, D = 50 mm= 0.05 m
Circumferential speed of the journal,
V₂ = π × 0.05 × 3000/60= 7.85 m/s
(ii) Wall shear stress per unit length ratio between the prototype and model
The clearance ratio is given ask=c₁/c₂
where
c₁ is the clearance of prototype
c₂ is the clearance of the model
So, the clearance ratio,
k = c₁/c₂= 0.04/0.02= 2
Wall shear stress per unit length is given ask = τ/R
where τ is the shear stress
R is the radius of the journal
For prototype,τ₁ = μV₁/kR₁τ₁ = 2.9 × 10⁻² × 15.7/2 × 0.05τ₁ = 0.225 N/m²
For model,τ₂ = μV₂/R₂τ₂ = 2.9 × 10⁻² × 7.85/25 × 10⁻³τ₂ = 0.225 N/m²
The ratio of wall shear stress per unit length between the prototype and model is given ask₁/k₂= τ₁/τ₂= 1
Therefore, the required wall shear stress per unit length ratio between the prototype and model is 1.
Answer: (i) The speed of the journal is 15.7 m/s.(ii) The required wall shear stress per unit length ratio between the prototype and model is 1.
To know more about speed , visit:
https://brainly.com/question/17661499
#SPJ11
A double-slit experiment is carried out with slit spacing d=0.41 mm. The screen is at a distance of 1.5 m. The bright fringes at the centre of the screen are separated by a distance Δx=1.5 mm. a. Determine the wavelength of the light. b. Determine the spacing of the bright fringes when a source with a wavelength 600 nm is used.
a. The wavelength of light used in the experiment is approximately 407 nm.
b. When a light source with a wavelength of 600 nm is used, the spacing between the bright fringes is approximately 2.19 mm.
a. To determine the wavelength of light in the double-slit experiment, we can use the formula:
λ = (Δx * d) / L
where λ is the wavelength, Δx is the separation between bright fringes, d is the slit spacing, and L is the distance between the screen and the double slits.
Plugging in the given values: Δx = 1.5 mm = 0.0015 m, d = 0.41 mm = 0.00041 m, and L = 1.5 m, we can calculate:
λ = (0.0015 m * 0.00041 m) / 1.5 m ≈ 4.07 × 10^-7 m or 407 nm
Therefore, the wavelength of light used in the experiment is approximately 407 nm.
Part 2:
b. Now, let's calculate the spacing of the bright fringes when a light source with a wavelength of 600 nm is used. We can use the same formula as before:
Δx = (λ * L) / d
Plugging in the new wavelength, λ = 600 nm = 6 × 10^-7 m, and the given values of L = 1.5 m and d = 0.41 mm = 0.00041 m, we can solve for Δx:
Δx = (6 × 10^-7 m * 1.5 m) / 0.00041 m ≈ 0.00219 m or 2.19 mm
Therefore, when a light source with a wavelength of 600 nm is used, the spacing between the bright fringes will be approximately 2.19 mm.
To learn more about double-slit experiment
Click here brainly.com/question/28108126
#SPJ11
A basketball is dropped from a walkway above Bartow Arena. It falls from rest with a constant acceleration of 9.8 m/s
2
until it reaches the floor 2 seconds later. A physics class is modeling the 1-D straight line motion as part of a class project They want construct a simple, but accurate, data table using the following equations: v
i+1
=v
i
+a t and x
i+1
=x
i
+v
i
t which are familiar to you. Their instructor requires them to make the constant time interval Δt between rows in their table sufficiently small so that they have only a 4 percent error in the total distance traveled by the ball during the 2 seconds. What should be the value of Δt in their data table to cover the entire time interval and achieve the desired error?
The value of Δt in the data table to cover the entire time interval and achieve the desired error is 0.1 seconds.
Given that the basketball is dropped from a walkway above Bartow Arena and it falls from rest with a constant acceleration of 9.8 m/s2 until it reaches the floor 2 seconds later.
The equations required to construct the data table are
v i+1 =v i +a t and x i+1 =x i +v i t.
The class project requires that the constant time interval Δt between rows in their table should be sufficiently small so that they have only a 4 percent error in the total distance traveled by the ball during the 2 seconds.
The total distance traveled by the ball in 2 seconds can be calculated using the kinematic equation, x = ½ a t2.
The distance covered is x = 0.5*9.8*22 = 19.6 m.
If Δt is the time interval, then the total number of intervals is 2/Δt.
The distance traveled by the ball for each interval is 0.5 a (Δt)2.
Hence, the total distance covered is given by 0.5 a (Δt)2(2/Δt).
On simplifying, we get the expression,
Δt2 = (0.04*2*19.6/9.8).
Therefore, Δt = 0.1 seconds.
Learn more about kinematic equation:
https://brainly.com/question/28973409
#SPJ11
A hot-air balloon is rising upward with a constant speed of 2.92 m/s. When the balloon is 9.48 m above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?
Therefore, the time it will take for the compass to hit the ground is 1.396 seconds.
The hot-air balloon is rising upward with a constant speed of 2.92 m/s.The height of the balloon from the ground = 9.48 m.
Using the given information, we need to find out the time it will take for the compass to hit the ground.
To find out the time, we will use the kinematic equation for vertically upward motion of an object:
v = u + gtt = (v - u) / gWhere,u = initial velocity of the compass = 0 m/s (as it was dropped from rest)v = final velocity of the compass
g = acceleration due to gravity = 9.81 m/s²t = time taken by the compass to hit the ground
Let's plug the given values in the formula to find the time taken by the compass to hit the ground.t = (v - u) / g ⇒ t = v / g ..... (1)To find the final velocity of the compass, we will use the following formula:
v² = u² + 2ghWhere,h = height of the balloon from the ground = 9.48 m
Let's plug the given values in the formula to find the final velocity of the compass:v² = u² + 2gh⇒ v = √(u² + 2gh)⇒ v = √(0 + 2(9.81)(9.48))⇒ v = √(187.6904)⇒ v = 13.6958 m/s
Substituting the value of v in equation (1), we get:
t = v / g⇒ t = 13.6958 / 9.81⇒ t = 1.396 sTherefore, the time it will take for the compass to hit the ground is 1.396 seconds.
To know more about velocity visit:
https://brainly.com/question/30559316
#SPJ11
Electrical Engineering (Vol. 95, 2013) studied the power quality of a transformer. Two causes of poor power quality are "sags" and "swells." (A sag is an unusual dip and a swell is an unusual increase in the voltage level of a transformer.) For Turkish transformers built for heavy industry, the mean number of sags per week was 353 and the mean number of swells per week was 184.
Assume the standard deviation of the sag distribution is 30 sags per week and the standard deviation of the swell distribution is 25 swells per week. Also, assume that the number of sags and the number of swells are both normally distributed. Suppose one of the transformers is randomly selected and found to have 410 sags and 130 swells in a week.
a. What is the probability that the number of sags per week is less than
410?
Let, X be the number of sags per week. Then, X ~ N(353, 30) and Y be the number of swells per week. Then, Y ~ N(184, 25).Let Z be the number of sags per week that we are interested in. Then, Z ~ N(410, 30).
To find the probability that the number of sags per week is less than 410, we need to find P(Z < 410).P(Z < 410) = P((Z - 353)/30 < (410 - 353)/30) = P(Z-score < 1.90)Using the Z-table, we find that the probability of Z-score being less than 1.90 is 0.9713.
The problem is asking us to find the probability that the number of sags per week is less than 410. To solve this problem, we need to use the concept of standard normal distribution.The standard normal distribution is a special case of the normal distribution where the mean is 0 and the standard deviation is 1. We can convert any normal distribution into a standard normal distribution using the formula Z = (X - μ)/σ, where X is the variable of interest, μ is the mean, and σ is the standard deviation.If we know the mean and standard deviation of a normal distribution, we can use the Z-score to find the probability of a given value occurring.
The Z-score is the number of standard deviations that a given value is from the mean. The Z-score is calculated using the formula Z = (X - μ)/σ.To find the probability that the number of sags per week is less than 410, we need to convert the distribution of X into a standard normal distribution. We do this by subtracting the mean of X from 410 and dividing by the standard deviation of X.
We get Z-score as 1.90.To find the probability of Z-score being less than 1.90, we use the Z-table. The Z-table provides the probability that a Z-score is less than a given value. The probability of Z-score being less than 1.90 is 0.9713.Therefore, the probability that the number of sags per week is less than 410 is 0.9713.
The probability that the number of sags per week is less than 410 is 0.9713.
To know more about standard deviation :
brainly.com/question/29115611
#SPJ11
Method of modulation used in RADAR metering system. Radar carrier frequencies, Pulse Modulation, and Frequency Modulated Continues-Wave radar. Algorithms and equations for metering distance to objects to be detected from robot
Radar, which is short for radio detection and ranging, is a technology that is used to detect and measure the distance of objects using electromagnetic waves. This technology was initially used in the military to detect enemy aircraft and missiles, but it is now used in various fields such as weather forecasting, aviation, maritime navigation, and traffic control.
There are various methods of modulation used in RADAR metering systems, but two of the most commonly used are pulse modulation and frequency modulated continuous-wave radar. Pulse modulation involves sending out short pulses of energy and then measuring the time it takes for the pulse to be reflected back to the sender. This method is used for measuring the distance of objects that are relatively close to the radar.
On the other hand, frequency modulated continuous-wave radar involves transmitting a continuous wave of energy that is modulated with a varying frequency. The reflected signal is then compared to the original transmitted signal to determine the distance of the object. This method is used for measuring the distance of objects that are further away from the radar.
To know more about radar visit:
https://brainly.com/question/32439909
#SPJ11
A uniform electric field of magnitude 443 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. The electron has moved 3.30 cm. (a) What is the work done by the field on the electron? ] (b) What is the change in potential energy associated with the electron? ] (c) What is the velocity of the electron? magnitude m/s ^2direction A proton is released from rest in a uniform electric field of magnitude 352 N/C. (a) Find the electric force on the proton. magnitude N direction (b) Find the acceleration of the proton. magnitude m/s direction (c) Find the distance it travels in 2.10μs. cmPrevious question
(a) Work done by the field on the electron is given by, W = F × d = qE × d Where, q is the charge on the electron = -1.6 × 10^-19 C;
E is the electric field strength = 443 N/C and d is the displacement of the electron
= 3.30 cm
= 3.30 × 10^-2 m
∴W = qE × d= -1.6 × 10^-19 C × 443 N/C × 3.30 × 10^-2 m
= -2.305 × 10^-19 J
Change in potential energy associated with the electron is given by, ∆U = qV Where V is the potential difference between the final and the initial position of the electron. Initially, the electron is at rest. Hence, the initial kinetic energy of the electron is zero. Therefore, the initial energy of the electron is its potential energy.
U = q
V = -1.6 × 10^-19 C × V
Final kinetic energy of the electron = KE
Final = ½ mv^2 Now, conservation of energy gives,
Initial energy of the electron = Final energy of the electron
∴qV = ½ mv^2
⇒ v = [2qV / m]^0.5
= [2 × -1.6 × 10^-19 C × V / 9.1 × 10^-31 kg]^0.5
Also, V = Ed = 443 N/C × 3.30 × 10^-2 m= 14.619 V
∴v = [2 × -1.6 × 10^-19 C × 14.619 V / 9.1 × 10^-31 kg]^0.5≈ 1.41 × 10^6 m/s
(c) Velocity of the electron is 1.41 × 10^6 m/s, directed in the positive x-direction. A proton is released from rest in a uniform electric field of magnitude 352 N/C.
Distance travelled by the proton in time t is given by,
s = ut + (1/2) at^2= 0 + (1/2) at^2= (1/2) × 3.52 × 10^7 m/s^2 × (2.10 × 10^-6 s)^2= 1.48 × 10^-8 m= 0.148 μm= 0.0148 cm Therefore, the proton travels 0.0148 cm in 2.10 μs.
To know more about electron visit:
https://brainly.com/question/12001116
#SPJ11
small mailbag is released from a helicopter that is descending steadily at 2.77 m/s. (a) After 3.00 s, what is the speed of the mailbag? v= (b) How far is it below the helicopter? d= Your response differs from the correct answer by more than 10%. Double check your calculations. m (c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 2.77 m/s ? v= The response you submitted has the wrong sign. m/s d= The response you submitted has the wrong sign. m
(a) After 3.00 s, the speed of the mailbag is approximately 31.37 m/s.
(b) The mailbag is approximately 48.51 m below the helicopter after 3.00 s.
(c) If the helicopter is rising steadily at 2.77 m/s:
(a) The speed of the mailbag after 3.00 s is approximately -31.37 m/s (downward direction).
(b) The mailbag is approximately +48.51 m above the helicopter after 3.00 s (above the helicopter).
(a) To find the speed of the mailbag after 3.00 s, we need to consider the vertical motion.
The initial velocity of the mailbag (Vi) is the same as the descending speed of the helicopter, but with the opposite sign since the mailbag is released downwards. So, Vi = -2.77 m/s.
Using the equation for the vertical motion:
Vf = Vi + gt
where Vf is the final velocity, g is the acceleration due to gravity (-9.8 m/s²), and t is the time.
Vf = -2.77 m/s + (-9.8 m/s²)(3.00 s)
Vf ≈ -31.37 m/s
The speed of the mailbag after 3.00 s is approximately 31.37 m/s in the downward direction.
(b) To find how far the mailbag is below the helicopter after 3.00 s, we can use the equation for vertical displacement:
d = Vit + (1/2)gt²
where d is the displacement, Vi is the initial velocity, g is the acceleration due to gravity, and t is the time.
d = (-2.77 m/s)(3.00 s) + (1/2)(-9.8 m/s²)(3.00 s)²
d ≈ -48.51 m
The mailbag is approximately 48.51 meters below the helicopter after 3.00 s.
(c) If the helicopter is rising steadily at 2.77 m/s, the signs of the answers in parts (a) and (b) will be opposite.
(a) The speed of the mailbag after 3.00 s is still -31.37 m/s, but now it is in the upward direction.
(b) The mailbag will be approximately +48.51 meters above the helicopter after 3.00 s, meaning it is above the helicopter.
Learn more about the speed at https://brainly.com/question/13943409
#SPJ11
a ball held 0.25m above the ground is tossed straight upward with an initial velocity of 10m/s. Assume that air resistance is negligible.
1. how high will it rise?
2. with what velocity will it impact the ground?
3. plot its acceleration as a function of time.
The height the ball will rise can be determined using the equations of motion. We can use the equation[tex]v^2 = v_0^2[/tex] + 2aΔx, where v is the final velocity, [tex]v_0[/tex]is the initial velocity, a is the acceleration, and Δx is the change in position.
Since the ball is thrown straight upward, its final velocity when it reaches its maximum height will be 0 m/s (taking upward direction as positive). Plugging in the values, we have:
[tex]0 = (10 m/s)^2 + 2(-9.8 m/s^2)[/tex]Δx
Solving for Δx, we find:
Δx =[tex](10 m/s)^2 / (2(-9.8 m/s^2))[/tex]≈ 5.10 m
Therefore, the ball will rise to a height of approximately 5.10 meters.
The velocity with which the ball impacts the ground can be determined using the equation v = v0 + at. Since the ball was thrown straight upward, its initial velocity is 10 m/s (upward direction taken as positive), and the acceleration due to gravity is -9.8 m/s^2 (downward direction taken as negative). We can plug in these values to find the velocity when it impacts the ground:
v = [tex]10 m/s + (-9.8 m/s^2)(t)[/tex]
When the ball impacts the ground, its displacement from the initial position is 0.25 m. Using the equation [tex]x = x_0 + v_0t + (1/2)at^2[/tex], we can solve for time t:
0.25 m = 0 + (10 m/s)t + [tex](1/2)(-9.8 m/s^2)t^2[/tex]
Solving this equation for t, we can find the time it takes for the ball to reach the ground.
Learn more about acceleration here:
https://brainly.com/question/2303856
#SPJ11
The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows three situations in which the current takes different paths througha piece of material. Eac
The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows three situations in which the current takes different paths through a piece of material. Each of the wires is the same material and has the same cross-sectional area.
Explain your answer. The resistance of a material is dependent upon the material's geometry and conductivity, as well as the temperature. The resistance of a material can be calculated using the formula R= V/I, where R is the resistance, V is the voltage, and I is the current. In this drawing, all three wires have the same cross-sectional area and are made of the same material.
This implies that the resistance of each wire is equal, and the current flowing through each wire is inversely proportional to the wire's resistance. The current will take the path with the least resistance, according to Ohm's law.
As a result, the current in wire A is less than that in wire B. Wire B has two separate routes, one of which has a smaller resistance than the other. This leads the current to follow the path of least resistance. Finally, the current in wire C is equal to that in wire B because the resistance of each path is equal. Hence, the magnitude of the current in the wire at the bottom of the drawing is 0.5 A.
To know more about resistance visit-
https://brainly.com/question/29427458
#SPJ11
The position of a 50 g oscillating mass is given by x(t)=(2.2 cm)cos(13t), where t is in seconds. Determine the amplitude: Determine the period: Determine the spring constant: Determine the maximum speed: Determine the total energy: Determine the velocity at t=0.40 s:
Given that the position of an oscillating mass is given by x(t)=(2.2 cm)cos(13t). We need to determine the amplitude, period, spring constant, maximum speed, total energy, and velocity at t = 0.4 s. We will compare the given equation with the standard wave equation i.e., x(t)=A cos(ωt)
Amplitude: The amplitude is the maximum displacement of the oscillating body from its mean position. It is given as Amplitude, A = 2.2 cmPeriod: The period is the time taken by an oscillating body to complete one cycle of oscillation. It is given by the formula T = 2π/ω where ω is the angular frequency of the oscillating body. ω = 2πf where f is the frequency of the oscillating body. T = 2π/ω = 2π/13 Period, T = 0.4846 sSpring constant: The spring constant of an oscillating body is given by the formula, k = 4π²m/T² where m is the mass of the oscillating body. Spring constant, k = 4π² × 0.050 / 0.4846² = 14.52 N/mMaximum speed: The maximum speed of an oscillating body is given by the formula, vmax = Aω Maximum speed, vmax = 2.2 × 13 = 28.6 cm/sTotal energy: The total energy of an oscillating body is given by the formula, E = (1/2)kA² Total energy, E = (1/2) × 14.52 × 2.2² = 17.07 mJ Velocity at t = 0.4 s: Velocity is given as the derivative of displacement with respect to time. v = dx/dt We are given, x(t) = 2.2 cos(13t) Taking the derivative of the above expression, we get v(t) = - 2.2 × 13 sin(13t) At t = 0.4 s,Velocity at t = 0.4 s = - 2.2 × 13 sin (13 × 0.4) Velocity at t = 0.4 s = - 17.8 cm/sTherefore, the amplitude is 2.2 cm, the period is 0.4846 s, the spring constant is 14.52 N/m, the maximum speed is 28.6 cm/s, the total energy is 17.07 mJ, and the velocity at t=0.40 s is -17.8 cm/s.
To learn more about spring oscillations: https://brainly.com/question/29215733
#SPJ11
Location C is 0.030 m from a small sphere that has a charge of 7nC uniformly distributed on its surface. Location D is 0.061 m from the sphere. What is the change in potential along a path from C to D ? V
The sign indicates that the electric potential decreases from C to D. Therefore, the answer is ΔV = -1.05 × 10⁴ V.
The electric potential V at point C due to the point charge Q is given by the equation;V = kQ/r, where r is the distance between the point charge and point C and k is Coulomb's constant whose value is k= 9 × 10⁹ N·m²/C².
We have;
Qc = 7 × 10⁻⁹ Cr
c= 0.030 mV
c = (9 × 10⁹ N·m²/C²) × (7 × 10⁻⁹ C)/(0.030 m)
= 2.10 × 10⁴ V
The electric potential V at point D due to the point charge Q is given by the equation;V = kQ/r, where r is the distance between the point charge and point D. We have;
Qc = 7 × 10⁻⁹
Crd = 0.061 mVd
= (9 × 10⁹ N·m²/C²) × (7 × 10⁻⁹ C)/(0.061 m)
= 1.05 × 10⁴ V
Therefore, the change in potential along a path from C to D is given as;ΔV = Vd - Vc= (1.05 × 10⁴ V) - (2.10 × 10⁴ V)= -1.05 × 10⁴ V
To know more about electric potential visit:-
https://brainly.com/question/28444459
#SPJ11
4. A water reservoir receives water from approximately 100 hectares of land.
a) If the average annual rainfall is 95-cm per year how much water (m3) will the reservoir collect per year?
b) If you need to pump water from the reservoir at 4-gpm through a 8000-ft, 3-inch PVC pipeline to a school that is located at an elevation of 120-ft above the reservoir, what should the rated head of the pump be? Assume the water pressure at the school needs to be 30 psi.
(a). The water reservoir will collect 950,000 m³ of water per year.
(b). The rated head of the pump should be 58.88 m.
(a). To find: How much water (m³) will the reservoir collect per year?
As per data,
Area of land from where water is collected = 100 hectares
Average annual rainfall = 95 cm per year
1 hectare = 10,000 m²
Area of land from where water is collected = 100 hectares
Therefore,
Area = 100 × 10,000
= 1,000,000 m².
Volume of water collected in the reservoir per year = Area × Average annual rainfall
= 1,000,000 m² × 95 cm
= 1,000,000 m² × 0.95 m
= 950,000 m³
Hence, the volume of water is 950,000 m³ of water per year.
(b). To find: The rated head of the pump
As per data,
Flow rate of water = 4 gpm
Length of PVC pipeline = 8000 ft or 2438.4 m
Diameter of PVC pipeline = 3 inch or 0.0762 m
Elevation of school = 120 ft or 36.576 m
Water pressure at the school = 30 psi
Convert flow rate from gallons per minute (gpm) to cubic meter per second (m³/s).
1 US gallon = 0.00378541 m³
Flow rate of water = 4 gpm
= 4 × 0.00378541 m³/s
= 0.01514 m³/s.
Cross-sectional area of 3-inch PVC pipe,
A = π/4 × (0.0762 m)²
= 0.00457 m².
Velocity of water in the PVC pipe,
v = Q/A
Where,
Q = flow rate of water= 0.01514 m³/s
A = cross-sectional area of the pipe= 0.00457 m²
∴ v = 0.01514/0.00457
= 3.31 m/s.
Head loss in PVC pipe,
Hloss = 0.9 × (2438.4/1000) × (v²/2g)
Where,
g = acceleration due to gravity= 9.81 m/s²
∴ Hloss = 0.9 × 2.4384 × (3.31)2/(2 × 9.81)
= 1.62 m.
Total dynamic head (TDH) = Elevation head + friction loss + pressure head
Where,
Elevation head = elevation difference between the reservoir and school
= 120 - 0
= 120 ft
= 36.576 m.
Friction loss = Head loss in PVC pipe = 1.62 m
Pressure head = (30 psi × 0.06895 kg/m³) / g
= 20.685 m
TDH = 36.576 + 1.62 + 20.685
= 58.88 m
Rated head of the pump = TDH = 58.88 m.
Hence, the rated head of the pump is 58.88 m.
To learn more about water reservoir from the given link.
https://brainly.com/question/30829291
#SPJ11
A defibrillator passes 11.24 A of current through the torso of a person for 10 ms. How much charge moves (in Coulombs)? Your answer should be a number with two decimal places, do not include the unit.
The electric current passing through the person’s torso is given as 11.24 A for 10 ms. The amount of electric charge that passes through a person’s torso when a current of 11.24 A is applied for 10 ms can be calculated as follows.
Charge is given byQ = I * twhere,Q = Charge I = Currentt = timeThe value of I = 11.24 A and t = 10 ms = 0.01 sSubstituting these values in the equation givesQ = 11.24 * 0.01Q = 0.1124 C.
Therefore, the amount of charge that passes through a person’s torso when a current of 11.24 A is applied for 10 ms is 0.11 C (rounded to two decimal places).Answer: 0.11
To know more about electric visit:-
https://brainly.com/question/33513737
#SPJ11
A 13.5μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A piece of material having a dielectric constant of 3.65 is placed between the plates, completely filling the space between them. How much energy is stored in the capacitor before the dielectric is inserted? Express your answer in joules. How much energy is stored in the capacitor after the dielectric is inserted? Express your answer in joules. By how much did the energy change during the insertion? Express your answer in joules. Did it increase or decrease? increase decrease
The change in energy during the insertion is 30.146 joules.
Calculate the energy stored in a capacitor, we can use the formula:
E = (1/2) * C * [tex]V^2[/tex]
where E is the energy, C is the capacitance, and V is the potential difference across the capacitor plates.
Before the dielectric is inserted:
C = 13.5 μF = 13.5 ×[tex]10^{(-6)[/tex] F
V = 22.0 V
Using the formula, we can calculate the energy stored in the capacitor before the dielectric is inserted:
E_before = (1/2) * C *[tex]V^2[/tex]
= (1/2) * (13.5 × [tex]10^{(-6))[/tex] * [tex](22.0)^2[/tex]
= 6.417 J (rounded to three decimal places)
The energy stored in the capacitor before the dielectric is inserted is approximately 6.417 joules.
After the dielectric is inserted:
Given:
C' = C * k
where k is the dielectric constant.
k = 3.65
C' = 13.5 μF * 3.65
= 49.275 μF
= 49.275 × [tex]10^{(-6)[/tex] F
Using the formula, we can calculate the energy stored in the capacitor after the dielectric is inserted:
E_after = (1/2) * C' * [tex]V^2[/tex]
= (1/2) * (49.275 × [tex]10^{(-6))[/tex] * [tex](22.0)^2[/tex]
= 36.563 J (rounded to three decimal places)
The energy stored in the capacitor after the dielectric is inserted is approximately 36.563 joules.
Change in energy during the insertion:
The change in energy is given by the difference between the energy stored before and after the dielectric is inserted:
ΔE = E_after - E_before
= 36.563 J - 6.417 J
= 30.146 J (rounded to three decimal places)
The energy increased during the insertion because the dielectric increases the capacitance of the capacitor, which results in more energy being stored.
To know more about energy refer here
https://brainly.com/question/8630757#
#SPJ11
A coil lies flat on a level table top in a region where the magnetic field vector points straight up. The magnetic field suddenly vanishes. When viewed from above, what is the sense of the induced current in this coil as the field fades?
a)Clockwise
b)Counterclockwise
c)There is no induced current
d)Counterclockwise first, and then clockwise as the current fades
When a coil is laid flat on a level table top in a region where the magnetic field vector points straight up, the sense of the induced current in the coil as the field fades is counterclockwise (option b).
When viewed from above, the magnetic field is coming out of the plane of the coil. There is a right-hand rule that can be used to predict the direction of the induced current when a magnetic field changes direction or magnitude.
According to this rule, if you curl the fingers of your right hand in the direction of the field lines, the direction of your thumb points in the direction of the induced current. As the magnetic field decreases, the direction of the induced current is in the opposite direction to that of the magnetic field. Therefore, the direction of the induced current is counterclockwise.
The Faraday's law of electromagnetic induction is the basis of such phenomenon. Faraday's law of induction states that the magnitude of the induced EMF (electromotive force) in a circuit is proportional to the rate of change of the magnetic flux through the circuit.
Learn more about magnetic field vector here:
https://brainly.com/question/31833405
#SPJ11
A charge of 7pC is distributed uniformly on a spherical surface (radius =2.0 cm ), and a second charge of −5pC is distributed uniformly on a concentric spherical surface (radius −4.0 cm ). Determine the magnitude of the electric field 6 cm from the center of the two surfaces (in N/C).
The magnitude of the electric field 6 cm from the center of the two surfaces is 0.00321 N/C calculated by using the formula for the vector sum of two electric fields.
To determine the magnitude of the electric field 6 cm from the center of the two surfaces, we need to use the formula for the electric field due to two charges:
E = (σ[tex]_{1}[/tex]/2ε[tex]_{0}[/tex]) * (1/[tex]r_1^2[/tex]) - (σ[tex]_{2}[/tex]/2ε[tex]_{0}[/tex]) * (1/[tex]r_2^2[/tex] )
where σ[tex]_{1}[/tex] and σ[tex]_{2}[/tex] are the charges (in C), ε[tex]_{0}[/tex] is the permittivity of free space (in F/m), r[tex]_{1}[/tex] and r[tex]_{2}[/tex] are the distances from the charges (in m), and E is the electric field (in N/C).
First, we need to calculate the electric field due to the first charge. The charge on the first surface is 7pC, so its magnitude is
σ[tex]_{1}[/tex] = 7 × [tex]10^{-6[/tex]C = 0.07 C.
The distance from the first charge to the point where we want to calculate the electric field is r[tex]_{1}[/tex] = 2.0 cm = 0.02 m. Using the formula above, we get:
E[tex]_{1}[/tex] = (0.07/2) × ε[tex]_{0}[/tex] × (1/[tex]0.02^2[/tex]) = 0.00196 N/C
Next, we need to calculate the electric field due to the second charge. The charge on the second surface is -5pC, so its magnitude is
σ[tex]_{2}[/tex] = -5 × [tex]10^{-6[/tex] C = -0.05 C.
The distance from the second charge to the point where we want to calculate the electric field is r[tex]_{2}[/tex] = 4.0 cm = 0.04 m. Using the formula above, we get:
E[tex]_{2}[/tex] = (−0.05/2) × ε[tex]_{0}[/tex] × (1/[tex]0.04^2[/tex]) = -0.00125 N/C
Since the charges are opposite signs and on concentric spherical surfaces, the electric field will be zero at the center of the two spheres and will be directed outward from the center in the radial direction. However, since we want to find the magnitude of the electric field 6 cm from the center, we need to add the two electric fields together. Using the formula for the vector sum of two electric fields, we get:
E = E[tex]_{1}[/tex] + E[tex]_{2}[/tex] = 0.00196 N/C + 0.00125 N/C = 0.00321 N/C
Therefore, the magnitude of the electric field 6 cm from the center of the two surfaces is 0.00321 N/C.
Learn more about magnitude
https://brainly.com/question/30337362
#SPJ11
gauss's law indicates that the flux through a closed surface
Gauss's law indicates that the flux through a closed surface is directly proportional to the charge enclosed by the surface, i.e., the total charge inside the closed surface.
Gauss's law states that the total electric flux via a closed surface enclosing an electric charge of Q is equal to Q/ε0 (where ε0 is the electric constant, also called the vacuum permittivity, which is equal to approximately 8.85×10−12 farad per meter) or φ = Q/ε0.
A closed surface encloses a certain region and acts as a barrier between that region and the rest of the universe. The quantity of electric charge enclosed by the surface is proportional to the electric flux density at each point on the surface and the surface area at that point.
The formula to calculate electric flux (Φ) through an area, A, with an electric field of magnitude E, at an angle θ to the normal to the surface, is
Φ = E × A × cos(θ).
Hence, the electric flux through an area is directly proportional to the electric field strength, the area of the surface, and the cosine of the angle between the electric field and a line perpendicular to the surface.
Learn more about Gauss's law from this link:
https://brainly.com/question/14773637
#SPJ11
Consider a cylindrical resistor of length 1 , radius r and resistivity rho. The resistance of this resistor is 19.3Ω. Then the length of this cylindrical resistor is incmaced hu a fartor of 6 , the radius is increased by a factor of 4 and the resistivity is increased by a factor of 7 . What is the resistance of this new resistor Thes 0/12
The resistance of the new resistor is approximately 2.75 Ω.
The resistance of a cylindrical resistor is given by the formula R = (ρ * L) / (π * r^2), where R is the resistance, ρ is the resistivity, L is the length, and r is the radius.Given that the initial resistance is 19.3 Ω, we can rearrange the formula to solve for the initial length: L = (R * π * r^2) / ρ. Substituting the given values, we have L = (19.3 Ω * π * r^2) / ρ. When the length is increased by a factor of 6, the new length (L') is equal to 6L. Similarly, the new radius (r') is equal to 4r, and the new resistivity (ρ') is equal to 7ρ.Using the same resistance formula for the new resistor, we can calculate the new resistance (R') as follows: R' = (ρ' * L') / (π * (r')^2. Substituting the values, we have R' = ((7ρ) * (6L)) / (π * (4r)^2). Simplifying this expression gives R' = (63 * L) / (8 * π * r^2).By substituting the initial length equation into the new resistance equation, we find R' = (63 * (19.3 Ω * π * r^2) / ρ) / (8 * π * r^2). Simplifying further, we get R' = (19.3 Ω * 63) / (8 * 7) ≈ 2.75 Ω. Therefore, the resistance of the new resistor, after increasing the length by a factor of 6, the radius by a factor of 4, and the resistivity by a factor of 7, is approximately 2.75 Ω.
To learn more about resistance:
https://brainly.com/question/29427458
#SPJ11
Calculate the magnitude of the gravitational force in Newtons between an object of mass (3.736×10∧3)kg and a mass (8.1×10∧7)kg at a distance of (5.786×10∧6) metres. Give your answer with two significant figures. Note: Your answer is assumed to be reduced to the highest power possible. Your Answer: ×10Answer Hide hint for Question 9 Use Newton's Law of Gravity.
The magnitude of the gravitational force between a 3.736×10^3 kg object. An 8.1×10^7 kg object at a distance of 5.786×10^6 meters is approximately 1.1 × 10^10 Newtons.
The magnitude of the gravitational force between two objects can be calculated using Newton's law of universal gravitation:
F = G * (m1 * m2) / r^2
Where F is the gravitational force, G is the gravitational constant (approximately 6.67430 × 10^-11 Nm^2/kg^2), m1 and m2 are the masses of the objects, and r is the distance between their centers.
Plugging in the given values:
m1 = 3.736×10^3 kg
m2 = 8.1×10^7 kg
r = 5.786×10^6 m
F = (6.67430 × 10^-11 Nm^2/kg^2) * ((3.736×10^3 kg) * (8.1×10^7 kg)) / (5.786×10^6 m)^2
Calculating this expression will give us the magnitude of the gravitational force in Newtons. Rounding the final answer to two significant figures, we get:
F ≈ 1.1 × 10^10 N
Therefore, the magnitude of the gravitational force between the two objects is approximately 1.1 × 10^10 Newtons.
To know more about Newton's law of universal gravitation,
https://brainly.com/question/9373839
#SPJ11
A turntable A with radius 2.60 meters is built into a stage for use in a theatrical production. It is observed during a rehearsal that a trunk B starts to slide on the turntable 12 s after the turntable begins to rotate. Knowing that the trunk undergoes a constant tangential acceleration of 0.30 m/s ∧
2, determine the coefficient of static friction between the trunk and the turntable.
The coefficient of static friction is found to be 0.30 m/s² divided by the acceleration due to gravity (9.8 m/s²).
To determine the coefficient of static friction between the trunk and the turntable, we can analyze the forces acting on the trunk.
First, let's consider the trunk's tangential acceleration. We know that the trunk undergoes a constant tangential acceleration of 0.30 m/s². This acceleration is caused by the frictional force acting between the trunk and the turntable.
The trunk starts to slide 12 seconds after the turntable begins to rotate. This means that the static friction force has reached its maximum value and is equal to the maximum static friction force (F_max).
We can use the equation of motion to relate the acceleration, time, and displacement:
s = ut + (1/2)at²
In this case, the initial velocity (u) is 0 since the trunk starts from rest. The displacement (s) can be calculated as the distance traveled along the circumference of the turntable, which is 2π times the radius of the turntable.
s = 2πr = 2π(2.60 m)
Using the equation with the given values:
2π(2.60) = 0 + (1/2)(0.30)(12)²
Simplifying the equation:
16.32π = 2.16(12)²
Solving for F_max:
F_max = m * a = m * (0.30 m/s²)
Now, we can determine the normal force (N) acting on the trunk. The normal force is equal to the weight of the trunk since it is in equilibrium:
N = mg
Next, we can calculate the coefficient of static friction (μ_s) using the equation:
μ_s = F_max / N = (m * 0.30 m/s²) / (mg)
Simplifying the equation, the mass (m) cancels out:
μ_s = 0.30 m/s² / g
Finally, substituting the acceleration due to gravity (g = 9.8 m/s²), we can determine the coefficient of static friction (μ_s).
In summary, by analyzing the forces acting on the trunk and using the equation of motion, we can determine the maximum static friction force and the coefficient of static friction between the trunk and the turntable.
Learn more about tangential acceleration here:
https://brainly.com/question/15743294
#SPJ11
Find solutions for your homework
Find solutions for your homework
sciencephysicsphysics questions and answersa ring of charge of radius 1 m is located on the x−y plane with its center at the arigin. how does the magnitude of the electric field due to the ring at =0,0,200>m compare with the magnitude of the electric field at 40,0,100>m ? electric field at 40,0,200>m is equal to the electric field at 40,0,100sm. * electric field at ∠0,0,200>m is twice the
Question: A Ring Of Charge Of Radius 1 M Is Located On The X−Y Plane With Its Center At The Arigin. How Does The Magnitude Of The Electric Field Due To The Ring At =0,0,200>M Compare With The Magnitude Of The Electric Field At 40,0,100≫M ? Electric Field At 40,0,200≫M Is Equal To The Electric Field At 40,0,100sm. * Electric Field At ∠0,0,200≫M Is Twice The
student submitted image, transcription available below
Show transcribed image text
Expert Answer
1st step
All steps
Final answer
Step 1/3
E(Z) = electric field at a distance Z along axis of the ring
Z = distance from the center of the rin...
View the full answer
answer image blur
Step 2/3
Step 3/3
Final answer
Transcribed image text:
A ring of charge of radius 1 m is located on the x−y plane with its center at the arigin. How does the magnitude of the electric field due to the ring at =0,0,200>m compare with the magnitude of the electric field at 40,0,100>m ? Electric field at 40,0,200>m is equal to the electric field at 40,0,100sm. * Electric field at ∠0,0,200>m is twice the electric field at <0,0,100>m. Electric field at 40,0,200×m is half the electric field at <0,0,100×m. Electric field at 40,0,200sm is ane quarter the electric field at 40,0,100sm. Electric field at 40,0,200sm is four times the electric field at <0,0,100sm.
The magnitude of the electric field due to the ring at (0, 0, 200 m) is one-fourth (1/4) of the magnitude of the electric field at (40, 0, 100 m).
To compare the magnitudes of the electric fields due to the ring at two different points, we can use the formula for the electric field due to a ring of charge.
The electric field at a point along the axis of the ring is given by:
E(Z) = (k * Q * z) / (2 * π * ε * [tex](R^2 + z^2)^{(3/2)[/tex])
Where:
E(Z) = electric field at a distance Z along the axis of the ring
k = Coulomb's constant
Q = total charge on the ring
z = distance from the center of the ring along the axis
R = radius of the ring
ε = permittivity of free space
Let's compare the magnitudes of the electric fields at two different points: (0, 0, 200 m) and (40, 0, 100 m).
Electric field at (0, 0, 200 m):
E(200) = (k * Q * 200) / (2 * π * ε * [tex](1^2 + 200^2)^{(3/2)[/tex])
Electric field at (40, 0, 100 m):
E(100) = (k * Q * 100) / (2 * π * ε * [tex](1^2 + 100^2)^{(3/2)[/tex])
To compare the magnitudes, we need to calculate the ratios:
Magnitude ratio = |E(200)| / |E(100)|
Calculating the ratio:
Magnitude ratio = [(k * Q * 200) / (2 * π * ε * [tex](1^2 + 100^2)^{(3/2)[/tex])] / [(k * Q * 100) / (2 * π * ε * (1^2 + 100^2)^(3/2))]
Magnitude ratio = (200 / [tex](1^2 + 100^2)^{(3/2)[/tex]) / (100 / [tex](1^2 + 100^2)^{(3/2)[/tex])
Magnitude ratio = [tex](200 / 200^3) / (100 / 100^3)[/tex]
Magnitude ratio = [tex](1 / 200^2) / (1 / 100^2)[/tex]
Magnitude ratio = [tex](100^2) / (200^2)[/tex]
Magnitude ratio = 1/4
Therefore, the magnitude of the electric field due to the ring at (0, 0, 200 m) is one-fourth (1/4) of the magnitude of the electric field at (40, 0, 100 m).
Learn more about magnitude here:
https://brainly.com/question/30337362
#SPJ11
A 170-loop coil of cross sectional area 7.9 cm 2lies in the plane of the page. An external magnetic field of 0.053 T is directed out of the plane of the page. The external field decreases linearly to 0.023 T in 11 milliseconds. The coil has a resistance of 4.0 Ohms. Determine the magnitude of the average current in the coil.
The magnitude of the average current in the coil can be calculated using Ohm's Law and the change in magnetic field. It is approximately 5.386 Amperes.
The magnitude of the average current in the coil can be determined using Ohm's Law and Faraday's Law of electromagnetic induction. The change in magnetic flux through the coil induces an electromotive force (EMF), which causes a current to flow.
First, we need to calculate the change in magnetic flux (∆Φ). The initial magnetic field strength is 0.053 T, and the final magnetic field strength is 0.023 T. The change in magnetic field (∆B) is given by ∆B = B_final - B_initial = 0.023 T - 0.053 T = -0.03 T.
The change in magnetic flux is then given by ∆Φ = A * ∆B, where A is the cross-sectional area of the coil. Substituting the values, ∆Φ = (7.9 cm^2) * (-0.03 T) = -0.237 T·cm^2.
According to Faraday's Law, the induced EMF (∆V) is equal to the rate of change of magnetic flux (∆Φ) divided by the time (∆t): ∆V = -∆Φ/∆t. The negative sign indicates that the induced current opposes the change in magnetic field.
Given that ∆t = 11 milliseconds = 0.011 seconds, we can calculate the induced EMF: ∆V = -(-0.237 T·cm^2) / (0.011 s) = 21.545 V.
Finally, we can calculate the average current (I) using Ohm's Law: I = ∆V / R, where R is the resistance of the coil. Substituting the values, I = (21.545 V) / (4.0 Ω) ≈ 5.386 A.
Therefore, the magnitude of the average current in the coil is approximately 5.386 A.
Learn more about Ohm's Law here
https://brainly.com/question/1247379
#SPJ11
Specific Gas Constant of Unknown Gas
An unknown gas undergoes an isentropic process and expands from 1200 kPa and 1.15 m³/kg to 100 kPa pressure and 5.11 m³/kg. If the specific heat at constant pressure for this gas is 6.19 kJ/kg K, calculate the specific gas constant of this gas in [kJ/kg.K].
The specific gas constant of the gas is 0.347 kJ/kgK The specific gas constant of the gas is 0.347 kJ/kgK which is the answer. The required answer is 0.347 kJ/kgK. Given data: Pressure 1 = 1200 kPa
Volume 1 = 1.15 m³/kg
Pressure 2 = 100 kPa
Volume 2 = 5.11 m³/kg
Specific heat at constant pressure (cp) = 6.19 kJ/kgK
We know that, the gas undergoes an isentropic process which is given by
PV^γ = constant
whereγ is the ratio of specific heats i.e
.γ = cp/cv
As the process is isentropic, the equation can be written as:
P₁V₁^γ = P₂V₂^γ
Also,PV = mRT (where m is mass of gas, R is gas constant, and T is temperature)
For the same gas and the same mass,
P₁V₁/T₁ = P₂V₂/T₂
where,T₁ and T₂ are initial and final temperatures respectively.Rearranging the above equation, we get
T₂ = T₁ * P₂V₂/P₁V₁
This temperature is the final temperature of the gas.We have,
γ = cp/cvγ
= cp/R - 1
We know,cp = 6.19 kJ/kgK
So,γ = 6.19/R - 1
Also,PV = mRT
So,R = PV/mT
For the same mass of gas,P₁V₁/T₁ = P₂V₂/T₂
P₁V₁/T₁ = P₂V₂/(T₁ * P₂V₂/P₁V₁)
P₁V₁² = P₂V₂²T₁
Now, putting the values we get,R = P₁V₁/((γ - 1) * m * T₁) * (P₂V₂/P₁V₁)^((γ - 1)/γ)
Substituting the given values we get,R = 0.347 kJ/kgK
To know more about gas visit:
https://brainly.com/question/31744009
#SPJ11
What is the electric potential due to the nucleus of hydrogen at a distance of 7.50×10
−11
m ? Assume the potential is equal to zero as r→[infinity]. (Express your answer to three significant figures.) V
The electric potential due to the nucleus of hydrogen at a distance of 7.50 × 10^-11 m is approximately 1.92 × 10^7 V.
The electric potential (V) due to the nucleus of hydrogen at a distance (r) can be calculated using the formula:
V = k * (q / r)
where:
k is the Coulomb's constant (9 × 10^9 Nm²/C²)
q is the charge of the nucleus (which is the charge of a proton, +1.6 × 10^-19 C)
r is the distance from the nucleus (7.50 × 10^-11 m)
Plugging in the values:
V = (9 × 10^9 Nm²/C²) * (+1.6 × 10^-19 C) / (7.50 × 10^-11 m)
V ≈ 1.92 × 10^7 V
Therefore, the electric potential due to the nucleus of hydrogen at a distance of 7.50 × 10^-11 m is approximately 1.92 × 10^7 V.
Here you can learn more about electric potential
https://brainly.com/question/28444459#
#SPJ11
Two blocks are connected by a lightweight, flexible cord that passes over a frictionless pulley. If m1 = 2.30 kg and m2 = 9.20 kg, and block 2 is initially at rest 1.40 m above the floor, how long does it take block 2 to reach the floor?
Block 2 takes approximately 0.64 seconds to reach the floor.
To determine the time it takes for block 2 to reach the floor, we can analyze the forces acting on the system. As block 2 descends, block 1 will ascend. The force of gravity acts on both blocks, causing block 2 to accelerate downwards and block 1 to accelerate upwards.
Using Newton's second law of motion, we can write the equations of motion for each block. For block 1, the net force is given by the tension in the cord (T) minus the force of gravity (m1 * g), where g is the acceleration due to gravity. Similarly, for block 2, the net force is the force of gravity acting downwards (m2 * g).
By considering the acceleration of each block, we can relate the masses and acceleration. Since block 2 is initially at rest, its acceleration is constant and equal to the acceleration of block 1. We can express this relationship as m2 * g = (m1 + m2) * a, where a is the common acceleration.
Solving for the acceleration, we find a = (m2 * g) / (m1 + m2). Substituting the given values, we have a = (9.20 kg * 9.8 m/s^2) / (2.30 kg + 9.20 kg) ≈ 7.85 m/s^2.
Next, we can use the kinematic equation s = ut + (1/2)at^2 to determine the time it takes for block 2 to travel the distance of 1.40 m. Here, s is the distance, u is the initial velocity (which is zero for block 2), a is the acceleration, and t is the time.
Rearranging the equation and substituting the known values, we get 1.40 m = (1/2) * 7.85 m/s^2 * t^2. Solving for t, we find t ≈ 0.64 seconds.
Therefore, it takes approximately 0.64 seconds for block 2 to reach the floor.
Learn more about kinematic equation here:
https://brainly.com/question/24458315
#SPJ11
(a) Calculate the electric field between the plates. kV/m (b) Calculate the surface charge density. nC/m
2
(c) Calculate the capacitance. pF (d) Calculate the charge on each plate. PC
The capacitance of the parallel plate capacitor is 7.49 pF and the charge on each plate is 1.12 pC.
Distance between the plates, d = 3mm = 3 × 10⁻³m
Dielectric constant, k = 1.5
Potential difference between the plates, V = 150V
(a) Electric field between the plates:
The electric field between the plates is given by the formula,E = V/d = 150/3 × 10⁻³= 50 × 10⁴ = 5 × 10⁵ V/m
(b) Surface charge density: capacitance = (ε₀kA)/d
Where, ε₀ = permittivity of free space = 8.85 × 10⁻¹² F/m, A = area of the plates
Now, capacitance, C = Q/V
Charge on each plate, Q = CV
Surface charge density, σ = Q/Aσ = C × V/A = ε₀k
V/d= 8.85 × 10⁻¹² × 1.5 × 150/(3 × 10⁻³)= 67.13 × 10⁻⁹= 67.13 nC/m²
(c) The capacitance of the parallel plate capacitor is C = (ε₀kA)/d= 8.85 × 10⁻¹² × 1.5 × A/(3 × 10⁻³)A = (C × d)/(ε₀k) = (2.655 × 10⁻¹¹ × 3 × 10⁻³)/(1.5)= 5.31 × 10⁻¹¹ m²
Capacitance, C = (ε₀kA)/d= 8.85 × 10⁻¹² × 1.5 × 5.31 × 10⁻¹¹/(3 × 10⁻³)= 7.49 pF
(d) Charge on each plate:Charge on each plate, Q = CV= 7.49 × 10⁻¹² × 150= 1.12 × 10⁻⁹ C= 1.12 pC
The electric field between the plates is 5 × 10⁵ V/m.
The surface charge density is 67.13 nC/m².
The capacitance of the parallel plate capacitor is 7.49 pF.
The charge on each plate is 1.12 pC.
Learn more about capacitance with the given link,
https://brainly.com/question/30529897
#SPJ11
A proton (q = 1.60 x 10-19 C, m = 1.67 x 10-27 kg), is released from rest in a 5 N/C uniform electric field. How long (in μs) does it take to move 7 m from its starting point?
μs
The proton will take approximately 11.67 microseconds to move 7 meters from its starting point in a uniform electric field of 5 N/C.
To determine the time it takes for the proton to move a distance of 7 meters, we can use the equations of motion in the presence of a constant force. The force experienced by the proton in the electric field can be calculated using the equation F = qE, where F is the force, q is the charge of the proton, and E is the electric field strength. In this case, the force is given as 5 N and the charge of the proton is 1.60 x [tex]10^{(-19)[/tex] C, so we can rearrange the equation to solve for E: E = F/q. Substituting the values, we find E = 5 N / 1.60 x [tex]10^{(-19)[/tex] C = 3.125 x [tex]10^{19[/tex] N/C.
Next, we can use the equation of motion s = ut + (1/2)[tex]at^2[/tex], where s is the distance, u is the initial velocity (which is zero since the proton is released from rest), a is the acceleration, and t is the time. Since the electric force acts on the proton as a constant force, the acceleration can be calculated using the equation a = F/m, where m is the mass of the proton. Substituting the values, we get a = 3.125 x [tex]10^{19[/tex] N/C / 1.67 x [tex]10^{(-27)[/tex] kg = 1.875 x [tex]10^{46[/tex] [tex]m/s^2[/tex].
Now, we can rearrange the equation of motion to solve for time t: [tex]t = \sqrt(2s/a)[/tex]. Substituting the given distance of 7 meters and the calculated acceleration, we find [tex]t = \sqrt(2 * 7 m / 1.875 * 10^{46} m/s^2)[/tex] ≈ 11.67 x [tex]10^{(-6)[/tex] seconds, which is equivalent to 11.67 microseconds. Therefore, it will take approximately 11.67 microseconds for the proton to move 7 meters from its starting point in the uniform electric field.
Learn more about electric field here:
https://brainly.com/question/30557824
#SPJ11
A long, straight wire lies along the z-axis and carries a 4.20 A current in the +z-direction. Fhd the magnetc feld produced at the point z=200 m,y=200 m,z=0by a 0.600 mm segrent of the wire centered at the origin Enter your answers numerically separated by commas. X incorrect: Try Again: J attempts remaining Part o Eeter your answeis mamerically separated ty commas.
A long, straight wire lies along the z-axis and carries a 4.20 A current in the +z-direction.
The magnetic field produced at the point z = 200 m,
y = 200 m,
z = 0 by a 0.600 mm
To find the magnetic field, we can use the formula,
B=μ₀I2πr
whereB is the magnetic field,
μ₀ is the magnetic constantI is the current in the wire
r is the distance from the wire.
We can divide the wire into small segments, with each segment producing a magnetic field. The magnetic field produced by each segment is perpendicular to the segment and decreases with distance.Now, the wire segment is located at the origin.
So, the distance from the origin to the point where the magnetic field is to be calculated is given by,
√(x²+y²+z²)=√(0²+200²+200²)=282.84
using the above formula
B=μ₀I2πr=4π×10⁻⁷×4.20/(2π×0.600×10⁻³×282.84)=0.00148 TT
the magnetic field produced by the wire segment at the point z = 200 m, y = 200 m, z = 0 is 0.00148 T.
To know more about current visit:
https://brainly.com/question/31686728
#SPJ11
Suppose a car is traveling at +18.3 m/s, and the driver sees a traffic light turn red. After 0.249 s has elapsed (the reaction time), the driver applies the brakes, and the car decelerates at 7.00 m/s
2
. What is the stopping distance of the car, as measured from the point where the driver first notices the red light? Number Units
Given the velocity of the car, the driver’s reaction time, and the car’s deceleration, we can calculate the stopping distance of the car. The stopping distance is 25.6 m.
Given dataInitial velocity of the car, u = +18.3 m/s
Time elapsed before the driver applies brake, t = 0.249 s
Deceleration of the car, a = -7.00 m/s²
We know that acceleration is defined as:
a = (v - u) / t
Where,v = Final velocity of the car u = Initial velocity of the car t = Time elapsed after applying brake
We can rearrange the above formula to find the final velocity of the car:
v = u + at
On substituting the given values, we get:
v = 18.3 + (-7.00) × 0.249= 16.214 m/s
Now, we know that the stopping distance can be given as:
s = (v² - u²) / 2as
We can rearrange the above formula to find the stopping distance of the car, we get:
s = v² / 2a - u² / 2a
On substituting the given values, we get:
s = (16.214)² / [2 (-7.00)] - (18.3)² / [2 (-7.00)]= 25.6 m
Therefore, the stopping distance of the car is 25.6 m
Given the velocity of the car, the driver’s reaction time, and the car’s deceleration, we can calculate the stopping distance of the car. The stopping distance is 25.6 m.
To know more about Deceleration visit:
brainly.com/question/13802847
#SPJ11
An object is thrown downwards with an initial velocity of 3.0 m/s. a. Find the final velocity and the displacement of the object after 2.0 s. (4 marks) b. If the building is 30 meters tall, would the object hit the ground before or after a time of 2.0 seconds? (1 mark)
The final velocity of the object after 2.0 seconds can be determined using the equations of motion. The displacement of the object can also be calculated. Additionally, based on the given initial velocity and the height of the building, it can be concluded whether the object will hit the ground before or after 2.0 seconds.
a. To find the final velocity of the object after 2.0 seconds, we can use the equation of motion: final velocity = initial velocity + (acceleration * time). In this case, since the object is thrown downwards, the acceleration due to gravity is considered positive. So, the final velocity would be 3.0 m/s + ([tex]9.8 m/s^2 * 2.0 s[/tex]) = 21.6 m/s.
To calculate the displacement, we can use the equation: displacement = (initial velocity * time) + (0.5 * acceleration * [tex]time^2[/tex]). Plugging in the given values, the displacement would be (3.0 m/s * 2.0 s) + (0.5 * 9.8 [tex]m/s^2[/tex] * [tex](2.0 s)^2[/tex]) = 19.6 m.
b. If the building is 30 meters tall, we compare the displacement of the object after 2.0 seconds (which we calculated as 19.6 m) with the height of the building. Since the displacement is less than the height of the building, the object would not hit the ground within 2.0 seconds. It would take more time for the object to reach the ground, given its downward motion and the given initial velocity.
Learn more about equations of motion here:
https://brainly.com/question/29278163
#SPJ11
