Answer:
The magnitude of the magnetic field is 55μT
Explanation:
Given;
number of turns of the solenoid per length, n = N/L = 1500 turns/m
current in the solenoid, I = 20 mA = 20 x 10⁻³ A
diameter of the solenoid, d = 4 cm = 0.04 m
The magnetic field at a point that is inside the solenoid;
B₁ = μ₀nI
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
B₁ = 4π x 10⁻⁷ x 1500 x 20 x 10⁻³
B₁ = 3.77 x 10⁻⁵ T
Given;
current in the wire, I = 2 A
distance of magnetic field from the wire, r = 1 cm = 0.01 m
The magnetic field at 1.0 cm from the wire;
[tex]B_2 = \frac{\mu_0I}{2\pi r} \\\\B_2 = \frac{4\pi*10^{-7}*2}{2\pi *0.01}\\\\B_2 = 4 *10^{-5} \ T[/tex]
The magnitude of the magnetic field;
[tex]B = \sqrt{B_1^2 +B_2^2} \\\\B = \sqrt{(3.77*10^{-5})^2 + (4*10^{-5})^2} \\\\B = 5.5 *10^{-5} \ T\\\\B = 55 \mu T[/tex]
Therefore, the magnitude of the magnetic field is 55μT
The magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire is [tex]5.5 \times 10^{-5}T[/tex]
Given the following parameters from the question
Number of turns of the solenoid per length, n = N/L = 1500 turns/m current in the solenoid, I = 20 mA = 20 x 10⁻³ A Diameter of the solenoid, d = 4 cm = 0.04 mThe magnetic field at a point that is inside the solenoid is expressed according to the formula;
B₁ = μ₀nIWhere;
μ₀ is the permeability of free space = 4π x 10⁻⁷ m/A
B₁ = 4π x 10⁻⁷ x 1500 x 20 x 10⁻³
B₁ = 3.77 x 10⁻⁵ T
Next is to get the magnetic field strength in the second wire.
Current in the wire, I = 2 A Distance of magnetic field from the wire, r = 1 cm = 0.01 mThe magnetic field at 1.0 cm from the wireSubstitute into the formula:
[tex]B_2=\dfrac{\mu_0 I}{2 \pi r} \\B_2=\frac{4\pi \times 10^{-7}\times 2}{2 \times 3.14\times 0.01} \\B_2 =4.0 \times 10^{-5}T[/tex]
Get the resultant magnetic field:
[tex]B = \sqrt{(0.00003771)^2+(0.00004)^2} \\B =5.5 \times 10^{-7}T[/tex]
Therefore the magnitude of the magnetic field at a point that is inside the solenoid and 1.0 cm from the wire is [tex]5.5 \times 10^{-5}T[/tex]
Learn more on the magnetic field here: https://brainly.com/question/15277459
An inductor is connected to the terminals of a battery that has an emf of 12.0 V and negligible internal resistance. The current is 4.86 mA at 0.700 ms after the connection is completed. After a long time the current is 6.80 mA.
What are
(a) the resistance R of the inductor and
(b) the inductance L of the inductor?
Answer:
a) 1764.71 ohms
b) 1.73 H
Explanation:
From the question, we can identify the following parameters;
Vo =12 V , i = 4.86 mA, t =0.700 ms, io =6.80 mA
(a) Indcued emf V = L di/dt =0
From ohms law Vo = ioR
R = 12/6.80*0.001
R=1764.71 ohms
(b) For LR circuit
i =io (1-e^-t/T)
Time constant T = L/R
4.86 = 6.80 (1-e^-0.7*10^-3/T)
divide both side by 6.8
0.715 = 0.0007/T
L/R = 0.0007/0.715
L/R = 0.000979020979
Substitute R from above
L = 0.000979020979 * 1764.71
L =1.73 H
10. A manufacturer knows from experience that the resistance of resistors she produces is normal with mean µ = 150Ω and the standard deviation σ = 5Ω. What percentage of the resistors will have resistance between 148 Ω and 152 Ω? Between 140 Ω and 160 Ω?
Answer:
Explanation:
Using the formula for finding the Z score
Z = x-µ/σ
x is the sample size
µ is the sample mean
σ is the standard deviation
For percentage of the resistors will have resistance between 148 Ω and 152 Ω, or is calculated as shown
P(148≤x152) = Z(152-150/5) - Z(148-150/5)
P(148≤x152) = Z(0.4)-Z( - 0.4)
P(148≤x152) = 0.6554-0.3446
The Z values are from the normal distribution table.
P(148≤x152) = 0.3108
The percentage of resistor that will have between 148 and 152 ohms is 0.3108×100% = 31.08%
Similarly for resistances between 140 Ω and 160 Ω
P(140≤x160) = Z(160-150/5) - Z(140-150/5)
P(140≤x160) = Z(2.0)-Z( - 2.0)
P(140≤x160) = 0.9775-0.02275
The Z values are from the normal distribution table.
P(140≤x160) = 0.9547
The percentage of resistor that will have between 140 and 160ohms is 0.9547×100% = 95.47%
The percentage of the resistors will have resistance between 148 Ω and 152 Ω is 31.08%
The percentage of the resistors will have resistance between 140 Ω and 160 Ω is 95.47%
Percentage of resistors:To solve for the probability we will use the standard score of the Z score, which is given by:
Z = (x - µ)/σ
where x is the sample size
µ is the sample mean = 150Ω
σ is the standard deviation = 5Ω
The probability of the resistors with resistance between 148 Ω and 152 Ω, will be:
P(148 ≤ 152) = Z((152-150)/5) - Z((148-150)/5)
P(148 ≤ 152) = Z(0.4)-Z( - 0.4)
P(148 ≤ 152) = 0.6554-0.3446
P(148 ≤ 152) = 0.3108
So, the percentage will be:
0.3108×100% = 31.08%
Similarly for resistances between 140 Ω and 160 Ω
P(140 ≤ 160) = Z((160-150)/5) - Z((140-150)/5)
P(140 ≤ 160) = Z(2.0)-Z( - 2.0)
P(140 ≤ 160) = 0.9775-0.02275
P(140 ≤ 160) = 0.9547
The percentage of resistor that will have between 140 and 160ohms is 0.9547×100% = 95.47%
Learn more about standard score:
https://brainly.com/question/25875323?referrer=searchResults
A beach ball filled with air is pushed about 1 m below the surface of a swimming pool and released from rest. Which of the following statements are valid, assuming the size of the ball remains the same?
a) The buoyant force on the ball decreases as the ball approaches the surface of the pool.
b) As the ball rises in the pool, the buoyant force on it increases.
c) The buoyant force on the ball equals its weight and remains constant as the ball rises.
d) The buoyant force on the ball while it is submerged is approximately equal to the weight of the volume of water that could fill the ball.
e) When the ball is released, the buoyant force exceeds the gravitational force, and the ball accelerates upward.
Answer:
e is correct
Explanation:
When a ball is pushed below the surface of a pool, it is submerged when the buoyant force is approximately equal to the water's weight of the volume that could fill the ball. When the ball is released, the buoyant force becomes greater than the gravitational force so that the ball accelerates upward.
What is buoyant force?The buoyant force can be described as the upward force exerted on an object wholly or partially immersed in a fluid and is also called Upthrust. A body submerged partially or fully in a fluid due to the buoyant force appears to lose its weight.
The following factors affect buoyant force the density of the fluid, the volume of the fluid displaced, and the local acceleration due to gravity.
When an object immerses in water, the object experiences a force from the downward direction opposite to the gravitational pull, which causes a decrease in its weight. The difference in this pressure gives the upward force on the object, as buoyancy.
Therefore, options (d), (e) are correct.
Learn more about buoyant force, here:
https://brainly.com/question/21990136
#SPJ2
what is electric field strength
Answer:
Electric field strengh is a measure of the strength of an electric field at a given point in space, equal to the field would induce on a unit electric charge at that point.
Electric field strength is also known as Electric Field Intensity .
Explanation:
Electric Field is also defined as force per charge. The unit will be force unit divided by charge unit. In this case, it will be Newton/Coulomb or N/C.
Please mark me as the brainliest!!!
Thanks!!!
From mechanics, you may recall that when the acceleration of an object is proportional to its coordinate, d2xdt2=−kmx=−ω2x , such motion is called simple harmonic motion, and the coordinate depends on time as x(t)=Acos(ωt+ϕ), where ϕ, the argument of the harmonic function at t=0, is called the phase constant. Find a similar expression for the charge q(t) on the capacitor in this circuit. Do not forget to determine the correct value of ϕ based on the initial conditions described in the problem. Express your answer in terms of q0 , L, and C. Use the cosine function in your answer.
Answer:
q = q₀ sin (wt)
Explanation:
In your statement it is not clear the type of circuit you are referring to, there are two possibilities.
1) The circuit of this problem is a system formed by an Ac voltage source and a capacitor, in this case all the voltage of the source is equal to the voltage at the terminals of the capacitor
ΔV = Δ[tex]V_{C}[/tex]
we assume that the source has a voltage of the form
ΔV = ΔV₀o sin wt
The capacitance of a capacitor is
C = q / ΔV
q = C ΔV sin wt
the current in the circuit is
i = dq / dt
i = c ΔV₀ w cos wt
if we use
cos wt = sin (wt + π / 2)
we make this change by being a resonant oscillation
we substitute
i = w C ΔV₀ sin (wt + π/2)
With this answer we see that the current in capacitor has a phase factor of π/2 with respect to the current
2) Another possible circuit is an LC circuit.
In this case the voltage alternates between the inductor and the capacitor
V_{L} + V_{C} = 0
L di / dt + q / C = 0
the current is
i = dq / dt
they ask us for a solution so that
L d²q / dt² + 1 / C q = 0
d²q / dt² + 1 / LC q = 0
this is a quadratic differential equation with solution of the form
q = A sin (wt + Ф)
to find the constant we derive the proposed solution and enter it into the equation
di / dt = Aw cos (wt + Ф)
d²i / dt²= - A w² sin (wt + Ф)
- A w² + 1 /LC A = 0
w = √ (1 / LC)
To find the phase factor, for this we use the initial conditions for t = 0
in the case of condensate for t = or the charge is zero
0 = A sin Ф
Ф = 0
q = q₀ sin (wt)
At t=0 a 2150kg rocketship in outer space fires the engine which exerts a force=At2, and F(1.25s)=781.25N in the x direction. Find the impulse J during the interval t=2.00s and t=3.5s
Answer:
5.81 X 10^3 Ns
Explanation:
Given that
F = At² and F at t = 1.25 s is 781.25 N ?
A = F/t² at t = 1.25 s => F = 781.25/(1.25)² = 500 N/s²
d(Impulse) = Fdt
Impulse = ∫Fdt =∫At²dt evaluated in the interval 2.00 s ≤ t ≤ 3.50 s
Impulse = At³/3 = (500/3)(t³) = 166.7t³ between t = 2.00 s and t = 3.50 s
Impulse = 166.7[3.5³ - 2³] = 166.7[42.875 - 8] = 166.7[34.875] = 5813.7 Ns
5.81 X 10^3 N.s
what is transmission of heat?
Answer:
Heat transfer is the transmission of heat energy from a body at higher temperature to lower temperature. The three mechanisms of heat transfer are
Conduction ConvectionRadiation.Example of Conduction:
Heating a metal
Example of Convection:
Sea Breeze
Example of Radiation:
Sun
Hope this helps ;) ❤❤❤
Answer:
Transmission of heat is the movement of thermal energy from one thing to another thing of different temperature.
There are three(3) different ways heat can transfer and they are:
a) Conduction (through direct contact).
b) Convection (through fluid movement).
c) Radiation (through electromagnetic waves).
Examples: 1.Heating a saucepan of water using a coalpot.(conduction&convection).
2. Baking a pie in an oven(radiation).
Hope it helps!!Please mark me as the brainliest!!!Thanks!!!!❤❤❤
An insulated beaker with negligible mass contains liquid water with a mass of 0.285 kg and a temperature of 75.2 ∘C How much ice at a temperature of -22.8 ∘C must be dropped into the water so that the final temperature of the system will be 32.0 ∘C ? Take the specific heat of liquid water to be 4190 J/kg⋅K , the specific heat of ice to be 2100 J/kg⋅K , and the heat of fusion for water to be 3.34×105 J/kg
Answer:
Explanation:
We shall apply the theory of
heat lost = heat gained .
heat lost by water = mass x specific heat x temperature diff
= .285 x 4190 x ( 75.2 - 32 ) = 51587.28 J
heat gained by ice to attain temperature of zero
= m x 2100 x 22.8 = 47880 m
heat gained by ice in melting = latent heat x mass
= 334000m
heat gained by water at zero to become warm at 32 degree
= m x 4190 x 32 = 134080 m
Total heat gained = 515960 m
So
515960 m = 51587.28
m = .1 kg
= 100 gm
The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s2. At an instant when the magnitude of the total acceleration is 6.0 m/s2, what is the speed of the particle? Group of answer choices
Answer:
The speed of the particle is 2.86 m/s
Explanation:
Given;
radius of the circular path, r = 2.0 m
tangential acceleration, [tex]a_t[/tex] = 4.4 m/s²
total magnitude of the acceleration, a = 6.0 m/s²
Total acceleration is the vector sum of tangential acceleration and radial acceleration
[tex]a = \sqrt{a_c^2 + a_t^2}\\\\[/tex]
where;
[tex]a_c[/tex] is the radial acceleration
[tex]a = \sqrt{a_c^2 + a_t^2}\\\\a^2 = a_c^2 + a_t^2\\\\a_c^2 = a^2 -a_t^2\\\\a_c = \sqrt{a^2 -a_t^2}\\\\a_c = \sqrt{6.0^2 -4.4^2}\\\\a_c = \sqrt{16.64}\\\\a_c = 4.08 \ m/s^2[/tex]
The radial acceleration relates to speed of particle in the following equations;
[tex]a_c = \frac{v^2}{r}[/tex]
where;
v is the speed of the particle
[tex]v^2 = a_c r\\\\v= \sqrt{a_c r} \\\\v = \sqrt{4.08 *2}\\\\v = 2.86 \ m/s[/tex]
Therefore, the speed of the particle is 2.86 m/s
one arm of a u shaped tube contains water and the other alcohol. if the two fluids meet exactly at the bottom of the U and the alcohol is at a height of 18 cm at what height will water be
Complete Question
One arm of a U-shaped tube (open at both ends) contains water, and the other alcohol.
If the two fluids meet at exactly the bottom of the U, and the alcohol is at a height of 18 cm, at what height will the water be?Assume the density of alcohol is [tex]\rho_a = 790\ kg/m^3[/tex]
Answer:
The height of water is [tex]h_w = 0.142 \ m[/tex]
Explanation:
From the question we are told that
The height of the alcohol is [tex]h_a =18 \ cm = 0.18 \ m[/tex]
The density of the alcohol is [tex]\rho_a = 790\ kg/m^3[/tex]
Generally the pressure on both arm of the tube are equal given that they are both open
i,e [tex]P_a = P_w[/tex]
Where [tex]P_a[/tex] is pressure of alcohol and [tex]P_w[/tex] is pressure of water
So the pressure on the arm of the tube containing the alcohol is mathematically evaluated as
[tex]P_a = g * h * \rho[/tex]
substituting values
[tex]P_a =9.8 * 0.18 * 790[/tex]
[tex]P_a = 1394 \ Pa[/tex]
Generally the pressure on the arm of the tube containing the water is mathematically evaluated as
[tex]P_w = g * h_w * \rho_w[/tex]
where [tex]\rho_w[/tex] is the density of water which has a value [tex]\rho _w = 1000 \ kg/m^3[/tex]
So
[tex]1394 = 9.8 * h_w * 1000[/tex]
=> [tex]h_w = \frac{1394}{9800}[/tex]
=> [tex]h_w = 0.142 \ m[/tex]
A pickup truck starts from rest and maintains a constant acceleration a0. After a time t0, the truck is moving with speed 25 m/s at a distance of 120 m from its starting point. When the truck has travelled a distance of 60 m from its starting point, its speed is v1 m/s.
Which of the following statements concerning v1 is true?
a. v1< 12.5m/s
b. v1= 12.5m/s
c. v1 >12.5m/s
Answer:
the correct answer is c v₁> 12.5 m / s
Explanation:
This is a one-dimensional kinematics exercise, let's start by finding the link to get up to speed.
v² = v₀² + 2 a₁ x
as part of rest v₀ = 0
a₁ = v² / 2x
a₁ = 25² / (2 120)
a₁ = 2.6 m / s²
now we can find the velocity for the distance x₂ = 60 m
v₁² = 0 + 2 a1 x₂
v₁ = Ra (2 2,6 60)
v₁ = 17.7 m / s
these the speed at 60 m
we see that the correct answer is c v₁> 12.5 m / s
A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hits the center in 0.455 s. (Neglect any effects due to air resistance.)At what angle relative to the floor was the dart thrown?
Answer:
The angle is [tex]\theta = 15.48^o[/tex]
Explanation:
From the question we are told that
The distance of the dartboard from the dart is [tex]d = 3.66 \ m[/tex]
The time taken is [tex]t = 0.455 \ s[/tex]
The horizontal component of the speed of the dart is mathematically represented as
[tex]u_x = ucos \theta[/tex]
where u is the the velocity at dart is lunched
so
[tex]distance = velocity \ in \ the\ x-direction * time[/tex]
substituting values
[tex]3.66 = ucos \theta * (0.455)[/tex]
=> [tex]ucos \theta = 8.04 \ m/s[/tex]
From projectile kinematics the time taken by the dart can be mathematically represented as
[tex]t = \frac{2usin \theta }{g}[/tex]
=> [tex]usin \theta = \frac{g * t}{2 }[/tex]
[tex]usin \theta = \frac{9.8 * 0.455}{2 }[/tex]
[tex]usin \theta = 2.23[/tex]
=> [tex]tan \theta = \frac{usin\theta }{ucos \theta } = \frac{2.23}{8.04}[/tex]
[tex]\theta = tan^{-1} [0.277][/tex]
[tex]\theta = 15.48^o[/tex]
Why does front side of spoon forms inverted image but the back side form opposite of inverted image?
Answer:
our face is outside the focal length of the concave side of the spoon. We see a virtual inverted image whereas in case of concave mirror we can see a virtual image which is erect.
If your metal car moves over a wide, closed loop of wire embedded in a road surface, is the magnetic field of the Earth within the loop altered? Does this produce a current pulse?
Answer:
Yes it produces current
Explanation:
Because If this enclosed field is somehow changed, then following the law of electromagnetic induction, a pulse of current will be produced in the loop. Dues to a change is produced in the electric field when the iron parts of a car pass over it, momentarily increasing the strength of the field.
7.00 kg piece of solid copper metal at an initial temperature T is placed with 2.00 kg of ice that is initially at -20.0°C. The ice is in an insulated container of negligible mass and no heat is exchanged with the surroundings. After thermal equilibrium is reached, there is 0.90 kg of ice and 1.10 kg of liquid water.
Required:
What was the initial temperature of the piece of copper?
Answer:
122°C
Explanation:
From the data Final temperature is 0 deg C since there is 0.9kg of ice and 1.10kg of liquid water.
That means that 1.10kg of the ice undergoes Heat of Fusion which is 3.34x10^5 J/kg...
Heat lost by copper = Heat gained by ice + Heat of fusion
-> (7.0kg)(390J/kg*C)(0-T) = (2.00kg)(2100J/kg*C)(0 - (-20) + (1.10kg)(3.34x10^5 J/kg)
-> T(2730) = 334001
-> T = 122°C
Three identical capacitors are connected in series to a battery. If a total charge of Q flows from the battery, how much charge does each capacitor carry?
Answer:
Each of the capacitor carries the same charge, Q
Explanation:
When capacitors are connected in series, the battery voltage is divided equally across the capacitors. The total voltage across the three identical capacitors is calculated as follows;
[tex]V_T = V_1 + V_2 + V_3[/tex]
We can also calculate this voltage in terms of capacitance and charge;
[tex]V = \frac{Q}{C} \\\\V_T = V_1 + V_2 +V_3 \\\\(given \ total \ charge \ as \ Q, then \ the \ total \ voltage \ V_T \ can \ be \ written \ as)\\\\V_T = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} \\\\V_T = Q(\frac{1}{C_1 } +\frac{1}{C_2} + \frac{1}{C_3 })\\\\[/tex]
Therefore, each of the capacitor carries the same charge, Q
Do the math: How many seconds would it take an echo sounder’s ping to make the trip from a ship to the Challenger Deep (10,994 meters) and back? Recall that depth in meters = ½ (1500 m/sec × Echo travel time in seconds). Round your answer to two decimal places.
Answer:
14.66secsExplanation:
Given the formula for calculating the depth in metres expressed as
depth in meters = ½ (1500 m/sec × Echo travel time in seconds)
Given depth of the challenger = 10, 994 meters, we will substitute this given value into the formula given to calculate the time take for the echo to travel.
10, 994 = depth in meters = ½ * 1500 m/sec × Echo travel time in seconds
10,994 = 750 * Echo travel time in seconds
Dividing both sides by 750;
Echo travel time in seconds = 10,994 /750
Echo travel time in seconds ≈ 14.66secs (to two decimal places)
Therefore, it would take an echo sounder’s ping 14.66secs to make the trip from a ship to the Challenger Deep and back
A long, current-carrying solenoid with an air core has 1800 turns per meter of length and a radius of 0.0165 m. A coil of 210 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system
Answer:
The mutual inductance is [tex]M = 0.000406 \ H[/tex]
Explanation:
From the question we are told that
The number of turns per unit length is [tex]N = 1800[/tex]
The radius is [tex]r = 0.0165 \ m[/tex]
The number of turns of the solenoid is [tex]N_s = 210 \ turns[/tex]
Generally the mutual inductance of the system is mathematically represented as
[tex]M = \mu_o * N * N_s * A[/tex]
Where A is the cross-sectional area of the system which is mathematically represented as
[tex]A = \pi * r^2[/tex]
substituting values
[tex]A = 3.142 * (0.0165)^2[/tex]
[tex]A = 0.0008554 \ m^2[/tex]
also [tex]\mu_o[/tex] is the permeability of free space with the value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
So
[tex]M = 4\pi * 10^{-7} *1800 * 210 * 0.0008554[/tex]
[tex]M = 0.000406 \ H[/tex]
a cylindrical jar is 10cm long and has a cross sectional area of 36cm. if it is completely filled with a fluid of relative density 0.2, calculate the mass of the fluid in the jar
Answer:
The mass of the fluid is 72 g.
Explanation:
The following data were obtained from the question:
Height (h) = 10 cm
Area of cross section (A) = 36cm²
Relative density = 0.2
Mass =..?
Next, we shall determine the volume of the cylinder. This can be achieved by doing the following:
Volume = Area x Height
Volume = 36 x 10
Volume = 360 cm³
Next, we shall determine the density of the liquid.
This can be obtained as follow:
Relative density = density of substance/density of water.
Relative density = 0.2
Density of water = 1 g/cm³
Density of fluid =...?
Relative density = density of substance/density of water.
0.2 = density of fluid / 1 g/cm³
Cross multiply
Density of fluid = 0.2 x 1 g/cm³
Density of fluid = 0.2 g/cm³
Finally, we shall determine the mass of fluid as follow:
Volume = 360 cm³
Density of fluid = 0.2 g/cm³
Mass of fluid =...?
Density = mass /volume.
0.2 g/cm³ = mass of fluid /360 cm³
Cross multiply
Mass of fluid = 0.2 g/cm³ x 360 cm³
Mass of fluid = 72 g
Therefore, the mass of the fluid in the jar is 72 g.
A bullet with a mass of 20 g and a speed of 960 m/s strikes a block of wood of mass 4.5 kg resting on a horizontal surface. The bullet gets embedded in the block. The speed of the block immediately after the collision is:________.
A) cannot be found because we don't know whether the surface is frictionless.
B) is 0.21 km/s.
C) is 65 m/s.
D) is 9.3 m/s.
E) None of these is correct
Answer:
4.25m/sE. None of the option is correctExplanation:
Using the law of conservation of momentum to solve the problem. According to the law, the sum of momentum of the bodies before collision is equal to the sum of the bodies after collision. The bodies move with the same velocity after collision.
Mathematically.
mu + MU = (m+M)v
m and M are the masses of the bullet and the block respectively
u and U are their respective velocities
v is their common velocity
from the question, the following parameters are given;
m = 20g = 0.02kg
u = 960m/s
M = 4.5kg
U =0m/s (block is at rest)
Substituting this values into the formula above to get v;
0.02(960)+4.5(0) = (0.02+4.5)v
19.2+0 = 4.52v
4.52v = 19.2
Dividing both sides by 4.52
4.52v/4.52 = 19.2/4.52
v = 4.25m/s
Since they have the same velocity after collision, then the speed of the block immediately after the collision is also 4.25m/s
When the reflected path from one surface of a thin film is one full wavelength different in length from the reflected path from the other surface and no phase change occurs, will the result be destructive interference or constructive interference?
Answer:
destructive interference
Explanation:
As we know that , when the phase difference between the path of two wavelength is 180°, then its known as destructive interference . And when the phase difference between the path of two wavelength is 0°, then its known as constructive interference.
In the constructive interference , the resulting amplitude will be maximum while in the destructive interference , the resulting amplitude will be zero(minimum).
Therefore the answer will be destructive interference.
A generator rotates at 95 Hz in a magnetic field of 0.025 T. It has 550 turns and produces an rms voltage of 170 V and an rms current of 60.0 A.
Required:
a. What is the peak current produced?
b. What is the area of each turn of the coil?
Answer:
Peak current= 84.86 A
Area of each turn = 0.029 m^2
Explanation:
The peak value of current can be obtained from Irms= 0.707Io. Where Io is the peak current.
Hence;
Irms= 60.0A
Io= Irms/0.707
Io = 60.0/0.707
Io= 84.86 A
Vrms= 0.707Vo
Vo= Vrms/0.707= 170/0.707 = 240.45 V
From;
V0 = NABω
Where;
Vo= peak voltage
N= number of turns
B= magnetic field
A= area of each coil
ω= angular velocity
But ω= 2πf = 2×π×95= 596.9 rads-1
Substituting values;
A= Vo/NBω
A= 240.45/550×0.025×596.9
A= 0.029 m^2
Consider the uniform electric field E = (8.0ĵ + 2.0 ) ✕ 103 N/C. What is its electric flux (in N · m2/C) through a circular area of radius 9.0 m that lies in the xy-plane? (Enter the magnitude.)
Answer:
5.09 x 10⁵ Nm²/C
Explanation:
The electric flux φ through a planar area is defined as the electric field Ε times the component of the area Α perpendicular to the field. i.e
φ = E A
From the question;
E = (8.0j + 2.0k) ✕ 10³ N/C
r = radius of the circular area = 9.0m
A = area of a circle = π r² [Take π = 3.142]
A = 3.142 x 9² = 254.502m²
Now, since the area lies in the x-y plane, only the z-component of the electric field is responsible for the electric flux through the circular area.
Therefore;
φ = (2.0) x 10³ x 254.502
φ = 5.09 x 10⁵ Nm²/C
The electric flux is 5.09 x 10⁵ Nm²/C
At which temperature do the lattice and conduction electron contributions to the specific heat of Copper become equal.
Answer:
At 3.86K
Explanation:
The following data are obtained from a straight line graph of C/T plotted against T2, where C is the measured heat capacity and T is the temperature:
gradient = 0.0469 mJ mol−1 K−4 vertical intercept = 0.7 mJ mol−1 K−2
Since the graph of C/T against T2 is a straight line, the are related by the straight line equation: C /T =γ+AT². Multiplying by T, we get C =γT +AT³ The electronic contribution is linear in T, so it would be given by the first term: Ce =γT. The lattice (phonon) contribution is proportional to T³, so it would be the second term: Cph =AT³. When they become equal, we can solve these 2 equations for T. This gives: T = √γ A .
We can find γ and A from the graph. Returning to the straight line equation C /T =γ+AT². we can see that γ would be the vertical intercept, and A would be the gradient. These 2 values are given. Substituting, we f ind: T =
√0.7/ 0.0469 = 3.86K.
A cylindrical shell of radius 7.00 cm and length 2.59 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 20.1 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C.
A) Use approximate relationships to find the net charge on the shell.
B) Use approximate relationships to find the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell.
If theta is 30 degrees and there is no friction, what would be the block's acceleration down the incline, in meters per second squared?
Answer:
a= 4.9m/s²
Explanation:
Using Fnet= mgsintheta = ma
But a= gsintheta
a= 9.8xsin 30
= 4.9m/s²
what help in reversing direction of current of current
Answer:
To reverse the direction of an electric current, we simply reverse the voltage either automatically with the help of some switching circuitry or manually by changing the voltage source terminals connection.
Explanation:
For electric current to flow, there must be a potential difference, usually referred to as the voltage. The electric current flow is analogous to the flow of water under the action of a pump, through a series of pipe connections. The voltage is similar to the driving action of the pump, and current flows the same way water flows. The resistance due to drag on the pipe wall is equivalent to electric resistance. For current to flow in the reverse direction, the voltage or rather, the potential difference is changed, causing the current to flow in the opposite direction. This can be done by switching the terminals of the voltage source, or by automatic means. The automatic switching can be done with a transistor based circuitry.
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string
Answer:
The time interval is [tex]t = 5.48 *10^{-3} \ s[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]l = 3.00 \ m[/tex]
The mass of the string is [tex]m = 5.00 \ g = 5.0 *10^{-3}\ kg[/tex]
The tension on the string is [tex]T = 500 \ N[/tex]
The velocity of the pulse is mathematically represented as
[tex]v = \sqrt{ \frac{T}{\mu } }[/tex]
Where [tex]\mu[/tex] is the linear density which is mathematically evaluated as
[tex]\mu = \frac{m}{l}[/tex]
substituting values
[tex]\mu = \frac{5.0 *10^{-3}}{3}[/tex]
[tex]\mu = 1.67 *10^{-3} \ kg /m[/tex]
Thus
[tex]v = \sqrt{\frac{500}{1.67 *10^{-3}} }[/tex]
[tex]v = 547.7 m/s[/tex]
The time taken is evaluated as
[tex]t = \frac{d}{v}[/tex]
substituting values
[tex]t = \frac{3}{547.7}[/tex]
[tex]t = 5.48 *10^{-3} \ s[/tex]
A person standing 180m from the foot of a high building claps hi
hand and hears the echo 0.03minutes later. What is the speed
sound in air at that temperature?
A) 331m/s
B) 240m/s C) 200m/s D) 300m/s
Answer:
C) 200 m/s
Explanation:
The sound travels a total distance of 360 m in 0.03 minutes.
v = (360 m) / (0.03 min × 60 s/min)
v = 200 m/s
The principles of magnetism apply everywhere on earth. What does this tell us about God and His character?
Answer:
God is omnipresent.
Explanation:
This means God is everywhere and He works where ever we are in the world