Answer:
Explanation:
Give that,
Spring constant (k)=40N/m
Force applied =75N
Since the force is applied to the right, we don't know if it is compressing or stretching the spring
So let assume it compress
Using hooke's law
F=-ke
e=-F/k
Then, e=-75/40
e=-1.875m
The deformation is 1.875m.
Let assume it stretch
Using hooke's law
-F=-ke
e=F/k
Then, e=75/40
e=1.875m
The elongation is 1.875m
The deformation or elongation of a spring with a force constant of 400 N/m when a force of 75 N is applied in the rightward direction is 18.75 cm.
Explanation:To determine the deformation or elongation of a spring, we can use Hooke's Law, which states that the force exerted on a spring is directly proportional to its deformation. In this case, the force constant (k) of the spring is given as 400 N/m and the applied force (F) is 75 N. We can use the formula F = kx, where x is the deformation or elongation of the spring.
Substituting the given values, we have 75 N = 400 N/m * x. Solving for x, we find that the deformation of the spring is x = 0.1875 m or 18.75 cm.
Therefore, when a force of 75 N is applied to the spring, it will deform or elongate by 18.75 cm in the rightward direction.
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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.
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).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
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
The most soaring vocal melody is in Johann Sebastian Bach's Mass in B minor. In one section, the basses, tenors, altos, and sopranos carry the melody from a low D to a high A. In concert pitch, these notes are now assigned frequencies of 146.8 Hz and 880.0 Hz. (Use 343 m/s as the speed of sound, and 1.20 kg/m3 as the density of air.)
Find the wavelength of the initial note.
Answer:
2.33651226158 m
Explanation:
From the question the required data is as follows
f = Frequency of the initial note = 146.8 Hz
v = Velocity of sound in air = 343 m/s
The wavelength of a wave is given by
[tex]\lambda=\dfrac{v}{f}[/tex]
[tex]\Rightarrow \lambda=\dfrac{343}{146.8}[/tex]
[tex]\Rightarrow \lambda=2.33651226158\ m[/tex]
The wavelength of the initial note is 2.33651226158 m
Under what circumstances are two circuits considered equivalent? A. their input values are the same B. the output of one is the inverse of the output of the other C. their output values are the same for all possible input combinations D. their output values are always 1 E. the input of one matches the output of the other
Two circuits are considered equivalent when their output values are the same for all possible input combinations.
A circuit equivalent to another is one that meets the same conditions, (eg same current), under a different configuration.
The equivalent circuit made in this way is not the same as the original one, but if the total voltages, the total currents and the total resistance of the circuit will be equal, which will be the equivalent.The concept of equivalent circuit is used when saying, for example, that a real generator is equivalent to an ideal one with its internal resistance in series.Therefore, we can conclude that two circuits are considered equivalent when their output values are the same for all possible input combinations.
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Final answer:
Two circuits are considered equivalent if they produce the same output for all possible input combinations, as can be demonstrated using a truth table in digital logic. The correct option is C.
Explanation:
Two circuits are considered equivalent when their output values are the same for all possible input combinations. This means that regardless of the input values, both circuits will produce the same output across all scenarios. For instance, in digital logic, two circuits using different logic gates such as AND, OR, NAND, or XOR would be equivalent if they yield the same output for each combination of inputs, as represented in a truth table. An example of this is the use of De Morgan's laws where an AND gate followed by a NOT gate can be equivalent to a NAND gate, producing the same outputs. Understanding these principles is essential for fields like electronics, computer engineering, and programming. Hence, Option C is correct.
) A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coefficient of kinetic friction between the stone and the surface
Answer:
-0.3
Explanation:
F' = μmg ........... Equation 1
Where F' = Frictional force, μ = coefficient of kinetic friction, m = mass of the stone, g = acceleration due to gravity.
But,
F' = ma ............ Equation 2
Where a = acceleration of the stone.
Substitute equation 2 into equation 1
ma = μmg
dividing both side of the equation by m
a = μg
make μ the subject of the equation
μ = a/g............... Equation 3
From the equation of motion,
v² = u²+2as................. Equation 4
Where v and u are the final and the initial velocity respectively, s = distance.
Given: v = 0 m/s (to rest), u = 8.0 m/s, s = 11 m.
Substitute into equation 4
0² = 8² + 2×11×a
22a = -64
a = -64/22
a = -32/11 m/s² = -2.91 m/s²
substitute the values of a and g into equation 3
μ = -2.91/9.8
μ = -0.297
μ ≈ -0.3
Final answer:
The coefficient of kinetic friction is found using the work-energy principle by equating the initial kinetic energy of the stone to the work done by friction. Given that the stone travels 11 meters and comes to rest, the coefficient of kinetic friction is calculated as approximately 0.296.
Explanation:
To find the coefficient of kinetic friction between the stone and the surface, we need to use the work-energy principle, which states that the work done by all the forces acting on an object is equal to the change in its kinetic energy. Since the stone comes to rest, all of its initial kinetic energy has been converted into work done against friction.
First, let's calculate the initial kinetic energy (KE) of the stone:
KE = (1/2)mv²KE = (1/2)(8.0 m/s)²KE = 32 J (joules)This energy is equal to the work done by friction (Wf):
Wf = Frictional force (f) x Distance (d)
Since the frictional force is equal to the kinetic friction coefficient (µk) multiplied by the normal force (N), and the normal force is equal to the weight of the stone (mg, where g is the acceleration due to gravity), we can express the work done by friction as:
Wf = µkmgd
Setting the work done by friction equal to the initial kinetic energy gives us:
32 J = µkmg(11 m)
Solving for the coefficient of kinetic friction:
µk = 32 J / (mg x 11 m)
Now, assuming the acceleration due to gravity (g) is 9.8 m/s²:
µk = 32 J / (m x 9.8 m/s² x 11 m)
Since the mass (m) cancels out, we don't need to know it:
µk = 32 J / (9.8 m/s² x 11 m)
µk = 0.296
Therefore, the coefficient of kinetic friction between the stone and the level surface is approximately 0.296.
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.
A large block is being pushed against a smaller block such that the smaller block remains elevated while being pushed. The mass of the smaller block is m = 0.45 kg. It is found through repeated experimentation that the blocks need to have a minimum acceleration of a = 13 m / s 2 in order for the smaller block to remain elevated and not slide down. What is the coefficient of static friction between the two blocks?
Explanation:
According to the free body diagram a block of mass m will have expression for force as follows.
N = ma
and, [tex]f_{c} - mg[/tex] = 0
[tex]\mu_{s}N - mg[/tex] = 0
[tex]\mu_{s} = \frac{mg}{N}[/tex] = [tex]\frac{mg}{ma}[/tex]
= [tex]\frac{g}{a}[/tex]
= [tex]\frac{9.8}{13}[/tex]
= 0.75
Therefore, we can conclude that the value of coefficient of static friction between the two blocks is 0.75.
The coefficient of static friction between the large block and smaller block is equal to 0.754.
Given the following data:
Mass of smaller block (m) = 0.45 kg.Acceleration (a) = [tex]13 \;m/s^2[/tex]Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]
To determine the coefficient of static friction between the large block and smaller block:
A force of static friction can be defined as the frictional force that resists the relative motion of two (2) surfaces.
Hence, a force of static friction is a frictional force that keeps an object at rest or stationary rather than being in relative motion.
Mathematically, the force of static friction is given by the formula;
[tex]Fs = uFn[/tex]
Where;
Fs represents the force of static friction.μ represents the coefficient of friction.Fn represents the normal force.For these block systems, the forces acting on them is given by:
[tex]uma - mg = 0\\\\uma = mg\\\\ua =g\\\\u=\frac{g}{a}[/tex]
Substituting the parameters into the formula, we have;
[tex]u=\frac{9.8}{13}[/tex]
u = 0.754
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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
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 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]
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.
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.
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
A rigid body rotates about a fixed axis with a constant angular acceleration. Which one of the following statements is true concerning the tangential acceleration of any point on the body?
a. The tangential acceleration is zero m/s^2.
b. The tangential acceleration depends on the change in the angular velocity.
c. The tangential acceleration is constant in both magnitude and direction.
d. The tangential acceleration depends on the angular velocity.
e. The tangential acceleration is equal to the centripetal
The tangential acceleration of a point on a rotating rigid body with constant angular acceleration depends on the change in the angular velocity. It's represented by the formula a_t = r * α, thus can change if the radius changes, even if the angular acceleration is constant.
Explanation:In this case, a rigid body rotates about a fixed axis with constant angular acceleration. The tangential acceleration of any point on the body would depend on the change in the angular velocity, making the correct answer (b). This is because the tangential acceleration is directly proportional to the angular acceleration and the distance from the axis of rotation, as represented by the formula a_t = r * α, where a_t is the tangential acceleration, r is the radius, and α is the angular acceleration.
Therefore, if the angular acceleration is constant, the tangential acceleration can change if the radius changes. However, if the radius is also constant, then the tangential acceleration will be constant in magnitude, but its direction will change as the direction of the tangent to the motion changes.
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The tangential acceleration of a point on a rotating body with constant angular acceleration depends on the change in the angular velocity. This is due to the relationship defined by the equation for tangential acceleration at = α x r.
Explanation:In the context of a rigid body rotating about a fixed axis with constant angular acceleration, the correct statement in relation to the tangential acceleration of any point on the body would be: b. The tangential acceleration depends on the change in the angular velocity.
This is because, in physics, tangential acceleration is a measure of how the tangential velocity of a point at a certain radius changes with time. It is directly proportional to the angular acceleration (α) and the radius (r), expressed by the equation at = α x r. Therefore, the tangential acceleration will change as the angular velocity changes, provided there is angular acceleration.
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Two insulated wires, each 4.42 m long, are taped together to form a two-wire unit that is 4.42 m long. One wire carries a current of 7.33 A; the other carries a smaller current I in the opposite direction. The two-wire unit is placed at an angle of 69.4o relative to a magnetic field whose magnitude is 0.547 T. The magnitude of the net magnetic force experienced by the two-wire unit is 2.24 N. What is the current I
Answer:
Current I=6.34A
Explanation:
Given data
L=L₁=L₂=4.42 m
I₁=7.33A
Angle α=69.4°
B=0.547T
Force F=2.24N
Required
Current I
Solution
The length of each wire ,the magnetic field B,and the angle are same for both wires.
As we know that Force is:
[tex]F_{net}=I_{1}LBSin\alpha -I_{2}LBSin\alpha\\F_{net}=(I_{1}-I_{2})LBSin\alpha\\I_{2}=I_{1}-\frac{F_{net}}{LBSin\alpha}\\I_{2}=7.33A-\frac{2.24N}{(4.42m)(0.547T)Sin(69.4)} \\I_{2}=6.34A[/tex]
Current I=6.34A
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 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]
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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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
X rays of 25 keV are received with an X - ray intensifying screen that produces light photons at 425 nm. If the conversion effi ciency of intensifying screen is 20%, calculate how many light photons will be generated by an X - ray photon.
Answer:
Photons Generated= ~1715 photons
Explanation:
The detailed explanation of answer is given in attached file.
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
An initially uncharged 4.07 μF capacitor and a 7.71 k Ω resistor are connected in series to a 1.50 V battery that has negligible internal resistance. What is the initial current in the circuit, expressed in milliamperes?
Answer:
I₀ = 0.2 mA
Explanation:
Just after being connected, as the voltage between plates of a capacitor can't change instantanously, the initial voltage through the capacitor must be zero, so it presents like a short to the battery.So, in these circumstances, as the battery only "sees"a resistor, the initial current can be found applying Ohm's law to the resistor, as follows:[tex]V = I_{0}*R \\\\ I_{0} = \frac{V}{R} = \frac{1.50V}{7.71e3\Omega} = 0.2 mA[/tex]
The initial current (that will be diminishing as the capacitor charges), is 0.2 mA.0.19mA
Explanation:Given;
Capacitor of capacitance = 4.07 μF
Resistor of resistance = 7.71 kΩ = 7.71 x 1000Ω = 7710Ω
Voltage = 1.50V
Since the capacitor is initially uncharged, it behaves like a short circuit. Therefore, the only element drawing current at that instant is the resistor. This means that the initial current in the circuit is the one due to (flowing through) the resistor.
And since there is negligible internal resistance, the emf of the battery is equal to the voltage supplied by the battery and is used to supply current to the resistor. Therefore, according to Ohm's law;
V = I x R ---------------(i)
Where;
V = voltage supplied or the emf
I = current through the resistor
R = resistance of the resistor
Substitute the values of V and R into equation (i) as follows;
1.50 = I x 7710
I = [tex]\frac{1.5}{7710}[/tex]
I = 0.00019A
Multiply the result by 1000 to convert it to milliamperes as follows;
0.00019 x 1000 mA = 0.19mA
Therefore, the initial current in the circuit, expressed in milliamperes is 0.19
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.
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.
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 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 skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings her arms inward toward her body in such a way that the distance of each mass from the axis changes from 1.00 to 0.50 m. Her rate of rotation (neglecting the mass of the skater) will the final angular velocity (in rev/s) be if her moment of inertia remains approximately constant at 5kg(m2), and the distance of the masses from the axis changes from 1 m to 0.1m?a. 6b. 3c. 9d. 4e. 7
Explanation:
It is known that relation between torque and angular acceleration is as follows.
[tex]\tau = I \times \alpha[/tex]
and, I = [tex]\sum mr^{2}[/tex]
So, [tex]I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}[/tex]
= 4 [tex]kg m^{2}[/tex]
[tex]\tau_{1} = 4 kg m^{2} \times \alpha_{1}[/tex]
[tex]\tau_{2} = I_{2} \alpha_{2}[/tex]
So, [tex]I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}[/tex]
= 1 [tex]kg m^{2}[/tex]
as [tex]\tau_{2} = I_{2} \alpha_{2}[/tex]
= [tex]1 kg m^{2} \times \alpha_{2}[/tex]
Hence, [tex]\tau_{1} = \tau_{2}[/tex]
[tex]4 \alpha_{1} = \alpha_{2}[/tex]
[tex]\alpha_{1} = \frac{1}{4} \alpha_{2}[/tex]
Thus, we can conclude that the new rotation is [tex]\frac{1}{4}[/tex] times that of the first rotation rate.
Final answer:
The final angular velocity will be approximately 1 rev/s.
Explanation:
First, let's calculate the initial angular velocity using the formula:
Initial Angular Velocity = Initial Moment of Inertia * Initial Angular Velocity / Final Moment of Inertia
Given that the initial moment of inertia is 5 kg(m2), the initial angular velocity is 1 rev/s, and the final moment of inertia is approximately 5 kg(m2), we can calculate the initial angular velocity as follows: (5 kg(m2) * 1 rev/s / 5 kg(m2)) = 1 rev/s. Therefore, the final angular velocity will also be approximately 1 rev/s.
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]