Answer:
3.6μF
Explanation:
The charge on the capacitor is defined by the formula
q = CV
because the charge will be conserved
q₁ = C₁V₂
q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor and the voltage drop across the two capacitor will be the same
q = q₁ + q₂ = C₁V₂ + C₂V₂
CV = CV₂ + C₂V₂
CV - CV₂ = C₂V₂
C ( V - V₂) = C₂V₂
C ( V/ V₂ - V₂ /V₂) = C₂
C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF
A pitcher claims he can throw a 0.145-kg baseball with as much momentum as a 3.00-g bullet moving with a speed of 1.50 3 103 m/s. (a) What must the baseball’s speed be if the pitcher’s claim is valid? (b) Which has greater kinetic energy, the ball or the bullet?
(a) The baseball's speed must be approximately 6.89 m/s.
(b) The bullet has greater kinetic energy.
Explanation:To determine the baseball's speed, we use the principle of conservation of momentum, which states that the total momentum of an isolated system remains constant. The momentum of the bullet before the pitch must equal the combined momentum of the baseball and the pitcher afterward. By equating the momenta and solving for the baseball's speed, we find it to be approximately 6.89 m/s.
Now, to compare the kinetic energies of the baseball and the bullet, we use the kinetic energy formula, which is proportional to the square of the velocity. Despite the baseball having a larger mass, the bullet's significantly higher velocity results in greater kinetic energy. This is due to the quadratic relationship between velocity and kinetic energy.
In conclusion, the baseball must travel at around 6.89 m/s to match the claimed momentum. However, the bullet still possesses greater kinetic energy due to its higher speed, highlighting the importance of velocity in determining kinetic energy.
a) The speed of the baseball must be [tex]\( {31.0 \, \text{m/s}} \)[/tex] to match the momentum of the bullet.
b) The bullet has significantly greater kinetic energy [tex](\(3375 \, \text{J}\))[/tex] compared to the baseball [tex](\(69.86 \, \text{J}\))[/tex].
To solve the problem, we will use the principles of momentum and kinetic energy.
Part (a): Speed of the Baseball
First, we need to calculate the momentum of the bullet and then find the speed at which the baseball must be thrown to have the same momentum.
1. Momentum of the Bullet:
- Mass of the bullet [tex](\(m_b\))[/tex]: [tex]\(3.00 \, \text{g} = 0.003 \, \text{kg}\)[/tex]
- Speed of the bullet [tex](\(v_b\))[/tex]: [tex]\(1.50 \times 10^3 \, \text{m/s}\)[/tex]
Momentum [tex](\(p\))[/tex] is given by:
[tex]\[ p = m_b \cdot v_b \][/tex]
[tex]\[ p = 0.003 \, \text{kg} \times 1.50 \times 10^3 \, \text{m/s} \][/tex]
[tex]\[ p = 4.50 \, \text{kg} \cdot \text{m/s} \][/tex]
2. Speed of the Baseball:
- Mass of the baseball [tex](\(m_{bb}\)): \(0.145 \, \text{kg}\)[/tex]
- Let the speed of the baseball be [tex]\(v_{bb}\).[/tex]
We want the baseball to have the same momentum as the bullet:
[tex]\[ p = m_{bb} \cdot v_{bb} \][/tex]
[tex]\[ 4.50 \, \text{kg} \cdot \text{m/s} = 0.145 \, \text{kg} \cdot v_{bb} \][/tex]
Solving for [tex]\(v_{bb}\):[/tex]
[tex]\[ v_{bb} = \frac{4.50 \, \text{kg} \cdot \text{m/s}}{0.145 \, \text{kg}} \][/tex]
[tex]\[ v_{bb} = 31.034 \, \text{m/s} \][/tex]
So, the baseball must be thrown with a speed of approximately [tex]\( \boxed{31.0 \, \text{m/s}} \).[/tex]
Part (b): Kinetic Energy Comparison
To compare the kinetic energies of the baseball and the bullet, we use the kinetic energy formula:
[tex]\[KE = \frac{1}{2} m v^2\][/tex]
1. Kinetic Energy of the Bullet:
- Mass [tex](\(m_b\)): \(0.003 \, \text{kg}\)[/tex]
- Speed [tex](\(v_b\)): \(1.50 \times 10^3 \, \text{m/s}\)[/tex]
[tex]\[ KE_b = \frac{1}{2} \cdot 0.003 \, \text{kg} \cdot (1.50 \times 10^3 \, \text{m/s})^2 \][/tex]
[tex]\[ KE_b = \frac{1}{2} \cdot 0.003 \, \text{kg} \cdot 2.25 \times 10^6 \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ KE_b = 0.0015 \cdot 2.25 \times 10^6 \][/tex]
[tex]\[ KE_b = 3375 \, \text{J} \][/tex]
2. Kinetic Energy of the Baseball:
- Mass [tex](\(m_{bb}\)): \(0.145 \, \text{kg}\)[/tex]
- Speed [tex](\(v_{bb}\)): \(31.034 \, \text{m/s}\)[/tex]
[tex]\[ KE_{bb} = \frac{1}{2} \cdot 0.145 \, \text{kg} \cdot (31.034 \, \text{m/s})^2 \][/tex]
[tex]\[ KE_{bb} = \frac{1}{2} \cdot 0.145 \, \text{kg} \cdot 963.1 \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ KE_{bb} = 0.0725 \cdot 963.1 \][/tex]
[tex]\[ KE_{bb} = 69.86 \, \text{J} \][/tex]
Suppose that a meter stick is balanced at its center. A 0.11 kg mass is then positioned at the 17-cm mark. At what cm mark must a 0.38 kg mass be placed to balance the 0.11 kg mass
Answer:
59.55 cm
Explanation:
Note: A meter stick has a length of 100 cm, and it is balanced at 50 cm.
From the principle of moment,
Sum of clockwise moment = sum of anti clockwise moment
Taking moment about the center
mg(50-x) = m'(y-50)g.................. Equation 1
Where m = first mass, m' = second mass, x = position of the first mass, y = position of the second mass, g = acceleration
make y the subject of the equation
y = (m(50-x)/m')+50.................... Equation 2
y = (0.11(50-17)/0.38)+50
y = (0.11(33)/0.38)+50
y = 9.55+50
y = 59.55 cm
A massless beam supports two weights as shown.
Find W such that the supporting force at
A is zero (static equilibrium). Force is 471 L' is L/3 and L'' is 9. Answer in nearest whole number.
Taking moments about B
N x L'' = W x L'
Here the moment is = force x perpendicular distance between the axis of rotation and the point of applied force .
Here L' = L/3 and L'' = 9
Thus from figure
471 x 9 = W x [tex]\frac{L}{3}[/tex]
But L'' = [tex]\frac{1}{2}[/tex]( L - [tex]\frac{L}{3}[/tex] ) = [tex]\frac{L}{3}[/tex] = 9
Thus W = 471 N
The value of the weight ( W ) is ; 471 N
Determine the weight ( W) valueFirst step : take moments about B
N * L" = W * L'
Where : L' = L/3, L" = 9
From the figure
471 * 9 = W * [tex]\frac{L}{3}[/tex]
also:
L" = 1/2 ( L - L/3 ) = L/3 = 9
Hence ; W = 471 N
Hence we can conclude that The value of the weight ( W ) is ; 471 N
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Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you know v0 and R, determine the general expressions for the two distinct launch angles θ1 and θ2 that will allow the projectile to hit D. For v0 = 42 m/s and R = 70 m, determine numerical values for θ1 and θ2?
Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.
In an insulated container, liquid water is mixed with ice. What can you conclude about the phases present in the container when equilibrium is established?
a. There would be ice only.
b. There would be liquid water only.
c. There would be both ice and liquid water.
d. There is no way of knowing the phase composition without more information.
Answer:
d. There is no way of knowing the phase composition without more information.
Explanation:
There is no way to ascertain the phases present in the container when equilibrium is established because we are not furnished with enough information.
The most important information needed to have a deeper understanding and provide solution is temperature. Liquid water and ice can both exist at the same temperature. So we cannot conclude based on the information at hand.When its coil rotates at a frequency of 240 Hz, a certain generator has a peak emf of 73 V. (a) What is the peak emf of the generator when its coil rotates at a frequency of 44 Hz? (b) Determine the frequency of the coil's rotation when the peak emf of the generator is 180 V.
Answer: (a) peak emf = 13.38V
(b) frequency = 591.78Hz
Explanation: Please see the attachments below
In a cloud chamber experiment, a proton enters a uniform 0.280 T magnetic field directed perpendicular to its motion. You measure the proton's path on a photograph and find that it follows a circular arc of radius 6.12 cm. How fast was the proton moving?
Answer:
The proton was moving at a speed of 1.64 x 10⁶ m/s
Explanation:
Given;
strength of magnetic field, B = 0.280 T
circular radius, R = 6.12 cm
mass of proton, m = 1.67 x 10⁻²⁷ kg
charge of electron. q = 1.602 x 10⁻¹⁹ C
When the proton follows a circular arc, then magnetic force will be equal to the centripetal force.
Magnetic force, Fm = qvB
Centripetal force, Fr = mv²/R
Fm = Fr
qvB = mv²/R
[tex]qvB = \frac{Mv^2}{R} \\\\\frac{qBR}{M} = \frac{v^2}{v} \\\\v =\frac{qBR}{M} = \frac{1.602*10^{-19} *0.28*0.0612}{1.67*10^{-27}} =1.64 *10^6 \ m/s[/tex]
Therefore, the proton was moving at a speed of 1.64 x 10⁶ m/s
Given Information:
Magnetic field = B = 0.280 T
Radius = r = 6.12 cm = 0.0612 m
Required Information:
Speed of proton = v = ?
Answer:
Speed of proton = v = 1641.77x10³ m/s
Explanation:
Since the proton is moving in a circular path we can model it in terms of magnetic force and centripetal force.
The magnetic force is given by
F = qvB
The centripetal force is given by
F = mv²/r
equating the both equations yields,
mv²/r = qvB
mv = qBr
v = qBr/m
Where q = 1.6x10⁻¹⁹ C is the of the proton, m = 1.67x10⁻²⁷ kg is the mass of proton, B is the magnetic field and r is the radius o circular arc around which proton is moving.
v = 1.6x10⁻¹⁹*0.280*0.0612/1.67x10⁻²⁷
v = 1641772.45 m/s
v = 1641.77x10³ m/s
Therefore, the proton is moving at the speed of 1641.77x10³ m/s.
A truck moving at 35.0mph collides head on with a car traveling in the opposite direction. The mass of the car is 550 kg, the mass of the truck is 2200 kg. During thecollision both vehicles come to a stop in 0.150s as the front end of both vehicles’crumples. What was the acceleration of the car? Assume that during the collision the acceleration of each vehicle is constant.Also, assume that frictional force are small compared to the forces compared to the contact forces between the vehicles during the colision.
Answer:
Explanation:
Check attachment for solution
Determine the power required for a 1150-kg car to climb a 100-m-long uphill road with a slope of 308 (from horizontal) in 12 s (a) at a constant velocity, (b) from rest to a final velocity of 30 m/s, and (c) from 35 m/s to a final velocity of 5 m/s. Disregard friction, air drag, and rolling resistance.
Explanation:
Below is an attachment containing the solution.
Answer:
the question is incomplete, below is the complete question
"Determine the power required for a 1150-kg car to climb a 100-m-long uphill road with a slope of 30 (from horizontal) in 12 s (a) at a constant velocity, (b) from rest to a final velocity of 30 m/s, and (c) from 35 m/s to a final velocity of 5 m/s. Disregard friction, air drag, and rolling resistance."
a. [tex]47KW[/tex]
b. 90.1KW
c. -10.5KW
Explanation:
Data given
mass m=1150kg,
length,l=100m
angle, =30
time=12s
Note that the power is define as the rate change of the energy, note we consider the potiential and the
Hence
[tex]Power,P=\frac{energy}{time} \\P=\frac{mgh+1/2mv^{2}}{t}[/tex]
to determine the height, we draw the diagram of the hill as shown in the attached diagram
a. at constant speed, the kinetic energy is zero.Hence the power is calculated as
[tex]p=\frac{mgh}{t} \\p=\frac{1150*9.81*100sin30}{12}\\p=47*10^{3}W\\[/tex]
b.
for a change in velocity of 30m/s
we have the power to be
[tex]P=\frac{mgh}{t} +\frac{1/2mv^2}{t}\\ P=47Kw+\frac{ 0.5*1150*30^2}{12}\\ P=47kw +43.1kw\\P=90.1KW[/tex]
c. when i decelerate, we have the power to be
[tex]P=\frac{mgh}{t} +\frac{1/2mv^2}{t}\\ P=47Kw+\frac{ 0.5*1150*\\((5^2)-(35^2))}{12}\\ P=47kw -57.5kw\\P=-10.5KW[/tex]
You are watching people practicing archery when you wonder how fast an arrow is shot from a bow. With a flash of insight you remember your physics and see how you can easily determine what you want to know by a simple measurement. You ask one of the archers to pull back her bow string as far as possible and shoot an arrow horizontally. The arrow strikes the ground 107 feet from the archer making an angle of 3 degrees below the horizontal. What is the initial speed of the arrow?
Answer:
[tex]u_x=55.208\ m.s^{-1}[/tex]
Explanation:
Given:
horizontal distance form the point of shooting where the arrow hits ground, [tex]s=107\ ft[/tex] [tex]=32.614\ m[/tex]
angle below the horizontal form the point of release of arrow where it hits ground, [tex]\theta=3^{\circ}[/tex]
So the height above the ground from where the arrow was shot:
[tex]\tan3^{\circ}=\frac{h}{107}[/tex]
[tex]h=5.6076\ ft=1.71\ m[/tex]
Since the arrow is shot horizontally so the initial vertical component of the velocity is zero ( [tex]u_y=0[/tex] ), we've the final vertical component of the velocity as:[tex]v_y=\sqrt{2g.h}[/tex]
[tex]v_y=\sqrt{2\times 9.8\times 1.71}[/tex]
[tex]v_y=5.789\ m.s^{-1}[/tex]
Using equation of motion:
[tex]v_y=u_y+g.t[/tex]
where:
t = time taken
[tex]5.789=0+9.8\times t[/tex]
[tex]t=0.591\ s[/tex]
Now the horizontal component of speed of the arrow (which remains constant throughout the motion by the Newton's first law of motion):[tex]u_x=\frac{s}{t}[/tex]
[tex]u_x=\frac{32.614}{0.591}[/tex]
[tex]u_x=55.208\ m.s^{-1}[/tex]
A tradesman sharpens a knife by pushing it with a constant force against the rim of a grindstone. The 30-cm-diameter stone is spinning at 200 rpm and has a mass of 28 kg. The coefficient of kinetic friction between the knife and the stone is 0.2. The stone slows steadily to 180 rpm in 10 s of grinding.
a) What is the magnitude of the angular acceleration of the grindstone as it slows down?
b) With what force does the tradesman press the knife against the grindstone?
Answer:
a. 0.21 rad/s2
b. 2.205 N
Explanation:
We convert from rpm to rad/s knowing that each revolution has 2π radians and each minute is 60 seconds
200 rpm = 200 * 2π / 60 = 21 rad/s
180 rpm = 180 * 2π / 60 = 18.85 rad/s
r = d/2 = 30cm / 2 = 15 cm = 0.15 m
a)So if the angular speed decreases steadily (at a constant rate) from 21 rad/s to 18.85 rad/s within 10s then the angular acceleration is
[tex]\alpha = \frac{\Delta \omega}{\Delta t} = \frac{21 - 18.85}{10} = 0.21 rad/s^2[/tex]
b) Assume the grind stone is a solid disk, its moment of inertia is
[tex]I = mR^2/2[/tex]
Where m = 28 kg is the disk mass and R = 0.15 m is the radius of the disk.
[tex] I = 28*0.15^2/2 = 0.315 kgm^2[/tex]
So the friction torque is
[tex]T_f = I\alpha = 0.315*0.21 = 0.06615 Nm[/tex]
The friction force is
[tex]F_f = T_f/R = 0.06615 / 0.15 = 0.441 N[/tex]
Since the friction coefficient is 0.2, we can calculate the normal force that is used to press the knife against the stone
[tex]N = F_f/\mu = 0.441/0.2 = 2.205 N[/tex]
Determine an appropriate size for a square cross-section solid steel shaft to transmit 260 hp at a speed of 550 rev/min if the maximum allowable shear stress is 15 kpsi.
Answer:46.05 mm
Explanation:
Given
[tex]Power=260\ hp\approx 260\times 746=193.96\ KW[/tex]
speed [tex]N=550\ rev/min[/tex]
allowable shear stress [tex](\tau )_{max}=15\ kpsi\approx 103.421\ MPa[/tex]
Power is given by
[tex]P=\frac{2\pi NT}{60}[/tex]
[tex]193.96=\frac{2\pi 550\times T}{60}[/tex]
[tex]T=3367.6\ N-m[/tex]
From Torsion Formula
[tex]\frac{T}{J}=\frac{\tau }{r}-----1[/tex]
where J=Polar section modulus
T=Torque
[tex]\tau [/tex]=shear stress
For square cross section
[tex]r=\frac{a}{2}[/tex]
where a=side of square
[tex]J=\frac{a^4}{6}[/tex]
Substituting the values in equation 1
[tex]\frac{3376.6}{\frac{a^4}{6}}=\frac{103.421\times 10^6}{\frac{a}{2}}[/tex]
[tex]a=0.04605\ m[/tex]
[tex]a=46.05\ mm[/tex]
A square cross-section solid steel shaft of approximately 5 inches on each side can transmit 260 hp at 550 rev/min without exceeding a maximum allowable shear stress of 15 kpsi.
Explanation:The size of the steel shaft can be calculated by using the power-transmitting capacity of a shaft equation, which states:
P = (16πNT) / (60 * 33000), where P is power in hp, N is rotational speed in rev/min, and T is torque in lb-ft.
We also know that the shear stress (τ) is given by the equation: τ = (16T) / (πd^3), where d is the diameter in inches.
To find the correct torque, we start by rearranging the power equation for T: T = (P * 60 * 33000) / (16πN).
Substituting in the given power and rotational speed, we find that the torque is approximately 13400 lb-ft.
We then substitute this value and the allowable shear stress into the shear stress equation, solving for d to get d ≈ 5 inches.
So, a square cross-section solid steel shaft about 5 inches on each side should be able to transmit the given power at the stated speed without exceeding the maximum allowable shear stress.
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A ball is rolling along at speed v without slipping on a horizontal surface when it comes to a hill that rises at a constant angle above the horizontal. In which case will it go higher up the hill: if the hill has enough friction to prevent slipping, or if the hill is perfectly smooth. Justify your answer with a conservation of energy statement
Final answer:
The ball will go higher up the hill if the hill has enough friction to prevent slipping.
Explanation:
The ball will go higher up the hill if the hill has enough friction to prevent slipping. This is because in the case where there is enough friction, the ball can convert some of its kinetic energy to rotational energy, allowing it to roll up the hill. The conservation of energy statement can be used to explain this:
When the ball rolls without slipping, its total mechanical energy is conserved.
As the ball rolls up the hill, its potential energy increases and its kinetic energy decreases.
In the case where there is enough friction, some of the kinetic energy is converted to rotational energy, allowing the ball to reach a higher height on the hill.
Therefore, the ball will go higher up the hill if the hill has enough friction to prevent slipping.
What force (in N) is exerted on the water in an MHD drive utilizing a 25.0 cm diameter tube, if 125 A current is passed across the tube that is perpendicular to a 1.95 T magnetic field? (The relatively small size of this force indicates the need for very large currents and magnetic fields to make practical MHD drives.) N
Final answer:
The force exerted on the water in the MHD drive utilizing a 25.0 cm diameter tube, with a 125 A current and a 1.95 T magnetic field, is 60.94 N.
Explanation:
To find the force exerted on the water in the MHD drive, we can use the formula for the magnetic force:
F = ILB
Where F is the force, I is the current, L is the length of the wire (in this case, the 25.0-cm diameter tube), and B is the magnetic field strength.
Plugging in the given values, we have:
F = (125 A)(0.25 m)(1.95 T)
Solving for F, we get:
F = 60.94 N
Therefore, the force exerted on the water in the MHD drive is 60.94 N.
Final answer:
The force exerted on the water in an MHD drive utilizing a 25.0 cm diameter tube, with a 125 A current and a 1.95 T magnetic field, is approximately 118.13 N.
Explanation:
The force exerted on the water in an MHD drive can be calculated using the formula:
Force (N) = Current (A) * Magnetic Field (T) * Area (m²)
Given that the diameter of the tube is 25.0 cm and the current is 125 A, we can calculate the area as follows:
Area = π * (radius)²
Area = π * (0.125 m)²
Substituting the values into the formula:
Force = 125 A * 1.95 T * (π * (0.125 m)²)
Force ≈ 118.13 N
Therefore, the force exerted on the water in the MHD drive is approximately 118.13 N.
A 15-turn rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are oriented parallel to a uniform magnetic field of strength 0.037 T, and the other two sides are perpendicular to the magnetic field. A)What is the magnitude of the magnetic moment of the loop?B)What torque does the magnetic field exert on the loop?
Answer:
(a) the magnitude of the magnetic moment of the loop is 0.75 Am²
(b) the torque the magnetic field exerts on the loop is 0.028 N.m
Explanation:
Given;
number of turns, N = 15
width of the loop, w = 10 cm = 0.1 m
length of loop, L = 20 cm = 0.2 m
current through the loop, I = 2.5 A
strength of the magnetic field, B = 0.037 T
Area of the loop, A = L x w = 0.2 x 0.1 = 0.02 m²
Part (a) the magnitude of the magnetic moment of the loop
μ = NIA
where;
μ is the magnitude of the magnetic moment of the loop
μ = 15 x 2.5 x 0.02 = 0.75 Am²
Part (b) the torque the magnetic field exerted on the loop
τ = μB
where;
τ is the torque the magnetic field exerts on the loop
τ = μB = 0.75 x 0.037 = 0.028 N.m
Given Information:
Magnetic field = B = 0.037 T
Current = I = 2.5 A
Number of turns = N = 15 turns
Length of rectangular coil = L = 20 cm = 0.20 m
Width of rectangular coil = W = 10 cm = 0.10 m
Required Information:
(a) Magnetic moment = µ = ?
(b) Torque = τ = ?
Answer:
(a) Magnetic moment = 0.75 A.m ²
(b) Torque = 0.0277 N.m
Explanation:
(a) The magnetic moment µ is given by
µ = NIA
Where µ is the magnetic, N is the number of turns, I is the current, A is the area of rectangular loop and is given by
A = W*L
A = 0.10*0.20
A = 0.02 m²
µ = 15*2.5*0.02
µ = 0.75 A.m²
(b) The toque τ exerted on current carrying loop with A area in the presence of a magnetic field B is given by
τ = NIAB
τ = 15*2.5*0.02*0.037
τ = 0.0277 N.m
Alternatively,
τ = µB
τ = 0.75*0.037
τ = 0.0277 N.m
There are two important isotopes of uranium, 235U
and 238U; these isotopes are nearly identical chemically
but have different atomic masses. Only 235U is very
useful in nuclear reactors. Separating the isotopes is called
uranium enrichment (and is often in the news as of this
writing, because of concerns that some countries are
enriching uranium with the goal of making nuclear
weapons.) One of the techniques for enrichment, gas
diffusion, is based on the different molecular speeds of
uranium hexafluoride gas, UF6 . (a) The molar masses of
235U and 238UF6 are 349.0 g/mol and 352.0 g/mol,
respectively. What is the ratio of their typical speeds vrms ?
(b) At what temperature would their typical speeds differ by
1.00 m/s? (c) Do your answers in this problem imply that
this technique may be difficult?
Answer:
a) (vᵣₘₛ₁/vᵣₘₛ₂) = 1.00429
where vᵣₘₛ₁ represents the vᵣₘₛ for 235-UF6 and vᵣₘₛ₂ represents the vᵣₘₛ for 238-UF6.
b) T = 767.34 K
c) The answers point to a difficult seperation technique, as the two compounds of the different isotopes have very close rms speeds and to create a difference of only 1 m/s In their rms speeds would require a high temperature of up to 767.34 K.
Explanation:
The vᵣₘₛ for an atom or molecule is given by
vᵣₘₛ = √(3RT/M)
where R = molar gas constant = J/mol.K
T = absolute temperature in Kelvin
M = Molar mass of the molecules.
₁₂
Let the vᵣₘₛ of 235-UF6 be vᵣₘₛ₁
And its molar mass = M₁ = 349.0 g/mol
vᵣₘₛ₁ = √(3RT/M₁)
√(3RT) = vᵣₘₛ₁ × √M₁
For the 238-UF6
Let its vᵣₘₛ be vᵣₘₛ₂
And its molar mass = M₂ = 352.0 g/mol
√(3RT) = vᵣₘₛ₂ × √M₂
Since √(3RT) = √(3RT)
vᵣₘₛ₁ × √M₁ = vᵣₘₛ₂ × √M₂
(vᵣₘₛ₁/vᵣₘₛ₂) = (√M₂/√M₁) = [√(352)/√(349)]
(vᵣₘₛ₁/vᵣₘₛ₂) = 1.00429
b) Recall
vᵣₘₛ₁ = √(3RT/M₁)
vᵣₘₛ₂ = √(3RT/M₂)
(vᵣₘₛ₁ - vᵣₘₛ₂) = 1 m/s
[√(3RT/M₁)] - [√(3RT/M₂)] = 1
R = 8.314 J/mol.K, M₁ = 349.0 g/mol = 0.349 kg/mol, M₂ = 352.0 g/mol = 0.352 kg/mol, T = ?
√T [√(3 × 8.314/0.349) - √(3 × 8.314/0.352) = 1
√T (8.4538 - 8.4177) = 1
√T = 1/0.0361
√T = 27.7
T = 27.7²
T = 767.34 K
c) The answers point to a difficult seperation technique, as the two compounds of the different isotopes have very close rms speeds and to create a difference of only 1 m/s In their rms speeds would require a high temperature of up to 767.34 K.
A farmer uses a tractor to pull a 150 kg bale of hay up a 15∘ incline to the barn at a steady 5.0 km/h. The coefficient of kinetic friction between the bale and the ramp is 0.40.Part A What is the tractor's power output?
Answer:
the tractor's power output = 1318.47 watts
Explanation:
Weight of bale = (mg) = (150 kg * 9.81 m/s²) = 1471.5 newtons
Resolve that weight force into its components at right angles to the ramp and parallel to the slope.
Down the slope we have 1471.5sin(15°) = 380.85 N
At right angles to the ramp we have 1471.5cos(15°) = 1421.35 N
OK, now we need the relationship between the coefficient of friction (μ), the friction force (Ff) and the normal force (Fn) {The normal force is the force at right angles to the friction surface. In this case Fn is equal in magnitude to the component of weight force at right angles to the surface.}
Ff = μ * Fn = (0.40 * 1421.35) = 568.54 N
The bale is not accelerating, so the "pull force" up the incline = component of weight + friction force down
= (380 N + 568.54 N)
= 948.54N
We need the speed in m/s not km/h
5.0 km/h = 5000 m/h
= (5000/3600)
= 1.39 m/s
Power = (948.54 N * 1.39m/s)
= 1318.47 N.m/s
= 1318.47 watts
The power output of the tractor is 1318.47 W
Power:According to the question the weight of bale
mg = 150 × 9.81 = 1471.5 N
Resolving the weight we get:
The normal reaction perpendicular to the ramp is given by:
N = 1471.5cos(15°) = 1421.35 N
The force of friction is given by:
[tex]f = \mu N = 0.40 \times 1421.35\\\\f = 568.54 N[/tex]
The bale is not accelerating, so the force up the incline = component of weight + friction force down
F = mgsin(15°) + f
F = 380 N + 568.54 N
F = 948.54N
Now, the engine speed is:
v = 5.0 km/h = 5000 m/h
v = (5000/3600)
v = 1.39 m/s
The power is defined as the product of force and speed, so:
Power P = Fv
P = 948.54 × 1.39
P = 1318.47 W
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A block with mass 0.50 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 m (Fig. P7.39). When released, the block moves on a horizontal tabletop for 1.00 m before coming to rest. The force constant k is 100 N>m. What is the coefficient of kinetic friction mk between the block and the tabletop?
Answer:
μk = 0.408
Explanation:
Given:
m=0.50 Kg,
Let compressed distance x = 0.20 m, and
stretched distance after releasing y = 1.00 m
K = 100 N/M
Sol:
Law of conservation of energy
Energy dissipation due to friction = P.E stores in the spring
Ff * y = 1/2 K x ² (Ff = μk Fn) And (Fn = mg) so
μk mgy = 1/2 K x ²
μk = 1/2 K x ² /mgy Putting values
μk = (1/2 ) (100 N/M) (0.20 m)² / (0.50 Kg x 9.8 m/s² x 1 m)
μk = 0.408
0.4
Explanation:(i) Since the mass is forced against the spring, an elastic energy ([tex]E_{E}[/tex]) due to the compression of the spring by the force is produced and is given according to Hooke's law by;
[tex]E_{E}[/tex] = [tex]\frac{1}{2}[/tex] k c² --------------------------------(i)
Where;
k = spring's constant
c = compression caused on the spring.
From the question;
k = 100N/m
c = 0.20m
Substitute these values into equation (i) as follows;
[tex]E_{E}[/tex] = [tex]\frac{1}{2}[/tex] x 100 x 0.20²
[tex]E_{E}[/tex] = 2J
(ii) Now, when the mass is released, it causes the block to move some distance until it stops thereby doing some work within that distance. This means that the elastic energy is converted to workdone. i.e
[tex]E_{E}[/tex] = W ------------(ii)
The work done (W) is given by the product of the net force(F) on the block and the distance covered(s). i.e
W = F x s -----------------(iii)
But, the only force acting on the body as it moves is the frictional force ([tex]F_{R}[/tex]) acting to oppose its motion. i.e
F = [tex]F_{R}[/tex]
Where;
[tex]F_{R}[/tex] = μN [μ = coefficient of kinetic friction, N = mg = normal reaction between the block and the tabletop, m = mass of the block, g = gravity]
[tex]F_{R}[/tex] = μ x mg
Substitute F = [tex]F_{R}[/tex] = μ x mg into equation (iii) as follows;
W = μ x mg x s ----------------(iv)
Now substitute the value of W into equation (ii) as follows;
[tex]E_{E}[/tex] = μ x mg x s ------------------(v)
Where;
[tex]E_{E}[/tex] = 2 J [as calculated above]
m = 0.50 kg
s = distance moved by block = 1.00m
g = 10m/s² [a known constant]
Substitute these values into equation (v) as follows;
2 = μ x 0.50 x 10 x 1
2 = 5μ
μ = 2 / 5
μ = 0.4
Therefore, the coefficient of kinetic friction between the block and the tabletop is 0.4
A 1200-kg car, initially moving at 20 m/s, comes to a stop at a red light over a time of 3 s from the moment the driver hit the brakes. What is the impulse delivered to the car by the static friction force (assumed constant) between the road and the tires
Answer:
Explanation:
Mass of car (M)=1200kg
Initial velocity (u)=20m/s
Stop after time (t)=3sec.
Come to stop implies that the final velocity is zero, v=0m/s
Using newton second law of motion
F=m(v-u)/t
Ft=m(v-u)
Since impulse is Ft
I=Ft
Then, I=Ft=m(v-u)
I=m(v-u)
I=1200(0-20)
I=1200×-20
I=-24,000Ns
The impulse delivered to the car by static friction is -24,000Ns
The impulse delivered to the car by static friction is -24,000Ns
Calculation of the impulse:Since A 1200-kg car, initially moving at 20 m/s, comes to a stop at a red light over a time of 3 s from the moment the driver hit the brakes.
Now here we used second law of motion of newton.
F=m(v-u)/t
Ft=m(v-u)
Since impulse is Ft
So,
I=Ft
Now
, I=Ft=m(v-u)
So,
I=m(v-u)
I=1200(0-20)
I=1200×-20
I=-24,000Ns
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The maximum electric field 9.8 m from a point light source is 3.0 V/m. What are (a) the maximum value of the magnetic field and (b) the average intensity of the light there? (c) What is the power of the source?
Answer:
maximum value of the magnetic field B = 1 ×[tex]10^{-8}[/tex] T
average intensity of the light = 0.011937 W/m²
power of source = 14.40 J
Explanation:
given data
maximum electric field E = 3.0 V/m
distance from a point source r = 9.8 m
solution
first we get here maximum value of the magnetic field
maximum value of the magnetic field = [tex]\frac{E}{C}[/tex] .........1
maximum value of the magnetic field = [tex]\frac{3}{3 \times 10^8}[/tex]
maximum value of the magnetic field B = 1 ×[tex]10^{-8}[/tex] T
and
now we get average intensity of the light that is
average intensity of the light = [tex]\frac{EB}{2\mu _o}[/tex] .........2
average intensity of the light = [tex]\frac{3 \times 1 \times 10^{-8}}{2 \times 4\pi \times 10^{-7}}[/tex]
average intensity of the light = 0.011937 W/m²
and
now we get power of source that is express as
power of source = average intensity × 4×π×r² ..........3
power of source = 0.011937 × 4×π×9.8²
power of source = 14.40 J
The 10-lb block has a speed of 4 ft>s when the force of F = (8t2) lb is applied. Determine the velocity of the block when it moves s = 30 ft. The coefficient of kinetic friction at the surface is ms = 0.2.
v=6ft/sec
To determine the velocity of the block when it moves a distance of 30 ft, we can use the equations of motion and consider the force applied and the friction acting on the block. First, let's find the acceleration of the block using the force applied. Next, we need to consider the frictional force acting on the block. Finally, we can calculate the net force and use the equation of motion to determine the velocity of the block.
Explanation:To determine the velocity of the block when it moves a distance of 30 ft, we can use the equations of motion and consider the force applied and the friction acting on the block.
First, let's find the acceleration of the block using the force applied. The force applied is given by F = 8t^2 lb, where t is the time in seconds. So, at t = 0, the force is F = 0 lb, and at t = 1, the force is F = 8 lb.
We can now use the equations of motion to find the acceleration. Since the mass of the block is not given, we can assume it to be 10 lb (as mentioned in the question). From the second law of motion, F = ma, where F is the net force, m is the mass, and a is the acceleration. Therefore, 8 = 10a. Solving for a, we get a = 0.8 ft/s^2.
Next, we need to consider the frictional force acting on the block. The coefficient of kinetic friction is given as μs = 0.2. The frictional force, Ff, can be calculated using Ff = μs * N, where N is the normal force. The normal force is equal to the weight of the block, N = mg, where g is the acceleration due to gravity, approximately 32 ft/s^2. So, N = 10 * 32 = 320 lb. Substituting the values, we get Ff = 0.2 * 320 = 64 lb.
Now, we can calculate the net force acting on the block. The net force, Fn, is given as Fn = F - Ff, where F is the force applied. Substituting the values, Fn = 8t^2 - 64 lb.
Finally, we can use the equation of motion, v^2 = u^2 + 2as, to determine the velocity of the block when it moves a distance of 30 ft. Here, v is the final velocity, u is the initial velocity (which is 0 ft/s because the block starts from rest), a is the acceleration, and s is the distance traveled. Substituting the values, v^2 = 0 + 2 * 0.8 * 30. Solving for v, we find
v = 6 ft/s.
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A horizontal board of length 6.1 m and mass 12.8 kg rests on two supports. The first support is at one end of the board. The second support is at a distance of 2.38 m from the other end of the board. What force does this second support exert on the board
Answer:102.84 N
Explanation:
Given
Mass of board [tex]m=12.8\ kg[/tex]
Length of board [tex]l=6.1\ m[/tex]
First support is at one end and second support is at a distance of 2.38 m from the other end
Suppose [tex]R_1[/tex] and [tex]R_2[/tex] are the reactions at the two support
Taking moment about the first end we can write
[tex]mg\times \dfrac{6.1}{2}-R_2(6.1-2.38)=0[/tex]
[tex]R_2=\dfrac{12.8\times 9.8 \times 3.05}{3.72}[/tex]
[tex]R_2=102.84\ N[/tex]
A battery with an emf of 4 V and an internal resistance of 0.7 capital omega is connected to a variable resistance R. Find the current and power delivered by the battery when R is (a) 0, I = 5.714285714 A * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 5.714285714 OK P = 0 W * [1.25 points] 3 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0 OK (b) 5 capital omega, I = 0.701754386 A * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0.701754386 OK P = 2.462296091 W * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 2.462296091 OK (c) 10 capital omega, and I = 0.3738317757 A * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0.3738317757 OK P = 1.397501965 W * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 1.397501965 OK (d) infinite. I = 0 A * [1.25 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 0 OK P = W
Answer:
E = I(R + r)
Making I the subject of the formular by dividing both sides by R + r,
I = E/(R + r)
E = 4V, r = 0.7Ohm, R = 0
I = 4/(0 + 0.7) = 4/0.7
I = 5.174285714A
Explanation:
For a cell of emf E, internal resistance r, connected to an external resistance R, the current flowing through the circuit will be given as:
I = E/(R + r). I is measured in Amperes(A), emf in volts(V), R in Ohms and internal resistance r also in ohms
A block of mass m = 2.5 kg is attached to a spring with spring constant k = 940 N/m. It is initially at rest on an inclined plane that is at an angle of θ = 29° with respect to the horizontal, and the coefficient of kinetic friction between the block and the plane is μk = 0.11. In the initial position, where the spring is compressed by a distance of d = 0.13 m, the mass is at its lowest position and the spring is compressed the maximum amount. Take the initial gravitational energy of the block as zero.
Required:
If the spring pushes the block up the incline, what distance, L in meters, will the block travel before coming to rest?
Final answer:
To find the distance L that the block will travel up the incline before stopping, we apply conservation of energy, accounting for the initial spring potential energy, the gravitational potential energy, and the work done against kinetic friction. By setting up the energy equation and substituting the given values, we can solve for L. Therefore, the value of L is approximately [tex]\( 0.25069 \)[/tex] m
Explanation:
To determine the distance L that the block will travel up the incline before coming to rest, we need to use the conservation of energy principle. The mechanical energy conserved will be the initial potential energy stored in the spring when compressed and the final kinetic energy of the block up the slope, taking into account the work done against friction.
Initially, the spring's potential energy (Us) is given by Us = 1/2 k d^2, where k is the spring constant and d is the compression distance. As the spring pushes the block up the incline, the block gains gravitational potential energy (Ug = mgh), does work against friction (Wf), and could have some residual kinetic energy (which is zero at the highest point).
The work done against friction can be found by Wf = μk N L, where μk is the coefficient of kinetic friction, N is the normal force (N=m*g*cos(θ)), and L is the distance traveled. Since we are considering the point where the block comes to rest, we set the total mechanical energy at this position equal to the initial energy.
Equating the initial and final energies we get: (1/2 k d^2) = mgh + μk m*g*cos(θ)L. We can now solve for L by plugging in the known values: m = 2.5 kg, k = 940 N/m, d = 0.13 m, μk = 0.11, g = 9.8 m/s^2, and θ = 29 degrees.
Substituting these values, we solve for L:
L = [(1/2) * 940 N/m * (0.13 m)^2 - 2.5 kg * 9.8 m/s^2 * sin(29 degrees)] / [2.5 kg * 9.8 m/s^2 * cos(29 degrees)* 0.11]
let's break it down step by step.
Given:
[tex]- Spring constant: \(k = 940 \, \text{N/m}\)\\- Displacement: \(x = 0.13 \, \text{m}\)\\- Mass: \(m = 2.5 \, \text{kg}\)\\- Gravitational acceleration: \(g = 9.8 \, \text{m/s}^2\)\\- Angle: \(\theta = 29^\circ\)\\- Length: \(l = 0.11 \, \text{m}\)[/tex]
Let's calculate it step by step:
1. Calculate the potential energy stored in the spring using the formula:
[tex]\[ U_{\text{spring}} = \frac{1}{2} k x^2 \][/tex]
Substitute the given values:
[tex]\[ U_{\text{spring}} = \frac{1}{2} \times 940 \, \text{N/m} \times (0.13 \, \text{m})^2 \]\[ U_{\text{spring}} = \frac{1}{2} \times 940 \times 0.0169 \, \text{N} \cdot \text{m}^2 \]\[ U_{\text{spring}} = 7.963 \, \text{J} \][/tex]
2. Calculate the gravitational potential energy using the formula:
[tex]\[ U_{\text{gravitational}} = mgh \]\\where \( h \) is the vertical height.\[ h = l \sin(\theta) \]\[ h = 0.11 \, \text{m} \times \sin(29^\circ) \]\[ h \approx 0.055 \, \text{m} \][/tex]
Substitute the values into the gravitational potential energy formula:
[tex]\[ U_{\text{gravitational}} = 2.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.055 \, \text{m} \]\[ U_{\text{gravitational}} = 1.3525 \, \text{J} \][/tex]
3. Now, plug these values into the given expression:
[tex]\[ L = \frac{(7.963 \, \text{J} - 1.3525 \, \text{J})}{(2.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \cos(29^\circ) \times 0.11 \, \text{m})} \][/tex]
Let's evaluate the denominator first:
[tex]\[ 2.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \cos(29^\circ) \times 0.11 \, \text{m} \]\[ = 2.5 \times 9.8 \times \cos(29^\circ) \times 0.11 \]\[ \approx 26.368 \][/tex]
Now, let's plug it back into the expression:
[tex]\[ L = \frac{(7.963 \, \text{J} - 1.3525 \, \text{J})}{26.368} \]\[ L = \frac{6.6105}{26.368} \]\[ L \approx 0.25069 \][/tex]
Therefore, the value of L is approximately [tex]\( 0.25069 \)[/tex] m.
2 m3 of an ideal gas are compressed from 100 kPa to 200 kPa. As a result of the process, the internal energy of the gas increases by 10 kJ, and 150 kJ of heat is transferred to the surroundings. How much work was done by the gas during the process?
Answer:
work done is -150 kJ
Explanation:
given data
volume v1 = 2 m³
pressure p1 = 100 kPa
pressure p2 = 200 kPa
internal energy = 10 kJ
heat is transferred = 150 kJ
solution
we know from 1st law of thermodynamic is
Q = du +W ............1
put here value and we get
-140 = 10 + W
W = -150 kJ
as here work done is -ve so we can say work is being done on system
Final answer:
The work done by the gas during the compression where 2 m³ of an ideal gas is compressed from 100 kPa to 200 kPa, with an internal energy increase of 10 kJ, and 150 kJ of heat transferred to the surroundings, is -160 kJ.
Explanation:
The student has asked how much work was done by the gas during a compression process in which 2 m³ of an ideal gas is compressed from 100 kPa to 200 kPa, the internal energy increases by 10 kJ, and 150 kJ of heat is transferred to the surroundings. To find the work done by the gas, we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W). The formula is ΔU = Q - W.
In this scenario, we have ΔU as +10 kJ (because the internal energy increases) and Q as -150 kJ (because heat is transferred to the surroundings, meaning it is leaving the system, thus it's a negative value). Plugging these values into the first law of thermodynamics gives us:
10 kJ = -150 kJ - W
When we rearrange the equation to solve for W, it becomes:
W = -150 kJ - 10 kJ
W = -160 kJ
Since work done by the system is a negative value in this case, it indicates that 160 kJ of work has been done on the gas by the surroundings during the compression. Thus, the work done by the gas itself is -160 kJ.
A very long uniform line of charge has charge per unit length 4.54 μC/m and lies along the x-axis. A second long uniform line of charge has charge per unit length -2.58 μC/m and is parallel to the x-axis at y1 = 0.384 m .
What is the magnitude of the net electric field at point y2 = 0.204 m on the y-axis?
Answer:
The magnitude of the net electric field is [tex]6.57\times10^{5}\ N/C[/tex]
Explanation:
Given that,
Charge density [tex]\lambda = 4.54\ \mu C/m[/tex]
Charge density [tex]\lambda' = -2.58\ \mu C/m[/tex]
Distance [tex]y_{1}= 0.384\ m[/tex]
Distance [tex]y_{2}= 0.204\ m[/tex]
We need to calculate the magnitude of the net electric field
Using formula of electric field
[tex]E=E_{1}+E_{2}[/tex]
[tex]E=\dfrac{1}{2\pi\epsilon_{0}}(\dfrac{\lambda}{r}+\dfrac{\lambda'}{r'})[/tex]
Put the value into the formula
[tex]E=\dfrac{1}{2\pi\times8.85\times10^{-12}}(\dfrac{4.54\times10^{-6}}{0.204}+\dfrac{2.58\times10^{-6}}{0.384-0.204})[/tex]
[tex]E=6.57\times10^{5}\ N/C[/tex]
Hence, The magnitude of the net electric field is [tex]6.57\times10^{5}\ N/C[/tex]
Place several E-Field Sensors at a few points on different equipotential lines, and look at the relationship between the electric field and the equipotential lines. Which statement is true?
1-At any point, the electric field is perpendicular to the equipotential line at that point, and it is directed toward lines of higher voltages.
2-At any point, the electric field is perpendicular to the equipotential line at that point, and it is directed toward lines of lower voltages.
3-At any point, the electric field is parallel to the equipotential line at that point.
Answer:
2.
Explanation:
If a charge is moved along a equipotential line, no work is done on the charge.If we remember that the work done by an external force, is just the product of the component of the force parallel to the displacement, if the force produces no work, this means that is perpendicular to the displacement.So, as the electric field is just the force per unit charge, and has the same direction as the force (for a positive charge), it must be perpendicular to any equipotential line.As the electric field (by convention) has the same direction as it would be taken by a positive test charge, and positive charges move from higher voltages to lower ones, the electric field is directed toward lines of lower voltages (like it happens between the plates of a capacitor).1 kg ball rolls off a 33 m high cliff, and lands 23 m from the base of the cliff. Express the displacement and the gravitational force in terms of vectors and calculate the work done by the gravitational force. Note that the gravitational force is <0, , 0>, where is a positive number (+9.8 N/kg). (Let the origin be at the base of the cliff, with the direction towards where the ball lands, and the direction taken to be upwards.)
Answer:
d = <23, 33, 0> m , F_W = <0, -9.8, 0> , W = -323.4 J
Explanation:
We can solve this exercise using projectile launch ratios, for the x-axis the displacement is
x = vox t
Y Axis
y = [tex]v_{oy}[/tex] t - ½ g t²
It's displacement is
d = x i ^ + y j ^ + z k ^
Substituting
d = (23 i ^ + 33 j ^ + 0) m
Using your notation
d = <23, 33, 0> m
The force of gravity is the weight of the body
W = m g
W = 1 9.8 = 9.8 N
In vector notation, in general the upward direction is positive
W = (0 i ^ - 9.8 j ^ + 0K ^) N
W = <0, -9.8, 0>
Work is defined
W = F. dy
W = F dy cos θ
In this case the force of gravity points downwards and the displacement points upwards, so the angle between the two is 180º
Cos 180 = -1
W = -F y
W = - 9.8 (33-0)
W = -323.4 J
The magnitude of the magnetic field that a long and extremely thin current-carrying wire produces at a distance of 3.0 µm from the center of the wire is 2.0 × 10-3 T. How much current is flowing through the wire?
Answer:
Current (I) = 3 x 10^-2 A
Explanation:
As we know, [tex]B = 4\pi 10^-7 *l/ 2\pi r[/tex]
By putting up the values needed from the data...
Current (I) = 2 x 3.14 x (3.0 x 10^-6) (2.0 x 10^-3) / 4 x 3.14 x 10^-7 = 3 x 10^-2 A
Answer: 0.03002A
Explanation: The formulae that relates the magnetic field strength B at a point (r) away from the center of a conductor carrying a current of value (I) is given below as
B = Uo×I/2πr
From our question, B =2.0×10^-3 T, r = 3.0×10^-6m
I =?, Uo = permeability of free space = 1.256×10^-6 mkg/s²A².
By substituting the parameters, we have that
2×10^-3 = 1.256×10^-6 × I/2π(3.0×10^-6)
2×10^-3 × 2π(3.0×10^-6) = 1.256×10^-6 × I
3.77×10^-8 = 1.256×10^-6 × I
I = 3.77×10^-8/ 1.256×10^-6
I = 3.002×10^-2 = 0.03002A
A transformer is to be used to provide power for a computer disk drive that needs 6.4 V (rms) instead of the 120 V (rms) from the wall outlet. The number of turns in the primary is 405, and it delivers 500 mA (the secondary current) at an output voltage of 6.4 V (rms). Find the current in the primary.
Answer:
The current in the primary is 0.026 A
Explanation:
Using the formula
I1 = (V1/V2)*I2
we have
I1 = (6.4/120)*0.500
I1 = 0.026 A
To find the current in the primary coil of a transformer, we use the power equivalence between the primary and secondary sides, and solve for the primary current using the given voltages and secondary current. This results in a primary current of 26.7 mA.
Calculating the Primary Current in a Transformer
To determine the primary current in a transformer, we need to use the relationship between the power in the primary and secondary circuits. The power delivered by an ideal transformer is the same on both sides, which means Powerprimary = Powersecondary. This can be written as:
Vprimary × Iprimary = Vsecondary × IsecondaryGiven parameters are:
Vprimary = 120 V (rms)Vsecondary = 6.4 V (rms)Isecondary = 500 mA = 0.5 AWe first calculate the power on the secondary side:
Powersecondary = Vsecondary × Isecondary = 6.4 V × 0.5 A = 3.2 WFor an ideal transformer, this power must be equal to the power on the primary side:
[tex]Powerprimary = Vprimar *Iprimary[/tex]Thus, we have:
3.2 W = 120 V × [tex]Iprimary[/tex]Solving for [tex]Iprimary[/tex], we get:
[tex]Iprimary[/tex] = 3.2 W / 120 V = 0.0267 A = 26.7 mATherefore, the current in the primary coil is 26.7 mA.