Answer:
The force that the wire exerts on the electron is [tex]-4.128\times10^{-20}i-6.88\times10^{-20}j+0k[/tex]
Explanation:
Given that,
Current = 8.60 A
Velocity of electron [tex]v= (5.00\times10^{4})i-(3.00\times10^{4})j\ m/s[/tex]
Position of electron = (0,0.200,0)
We need to calculate the magnetic field
Using formula of magnetic field
[tex]B=\dfrac{\mu I}{2\pi d}(-k)[/tex]
Put the value into the formula
[tex]B=\dfrac{4\pi\times10^{-7}\times8.60}{2\pi\times0.200}[/tex]
[tex]B=0.0000086\ T[/tex]
[tex]B=-8.6\times10^{-6}k\ T[/tex]
We need to calculate the force that the wire exerts on the electron
Using formula of force
[tex]F=q(\vec{v}\times\vec{B}[/tex]
[tex]F=1.6\times10^{-6}((5.00\times10^{4})i-(3.00\times10^{4})j\times(-8.6\times10^{-6}) )[/tex]
[tex]F=(1.6\times10^{-19}\times3.00\times10^{4}\times(-8.6\times10^{-6}))i+(1.6\times10^{-19}\times5.00\times10^{4}\times(-8.6\times10^{-6}))j+0k[/tex]
[tex]F=-4.128\times10^{-20}i-6.88\times10^{-20}j+0k[/tex]
Hence, The force that the wire exerts on the electron is [tex]-4.128\times10^{-20}i-6.88\times10^{-20}j+0k[/tex]
Part A) Components of the Force The force components on the electron are: [tex]\[F_x = -8.26 \times 10^{-20} \, \text{N}, \quad F_y = -1.38 \times 10^{-19} \, \text{N}, \quad F_z = 0 \, \text{N}\][/tex]
Part B) Magnitude of the Force The magnitude of the force is:[tex]\[F \approx 1.60 \times 10^{-19} \, \text{N}\][/tex]
Part A: Calculate the force components
The force on a moving charge in a magnetic field is given by the Lorentz force equation:
[tex]\[\vec{F} = q \vec{v} \times \vec{B}\][/tex]
First, we need to find the magnetic field [tex]\(\vec{B}\)[/tex] produced by the wire at the position of the electron. The magnetic field due to a long, straight current-carrying wire is given by:
[tex]\[B = \frac{\mu_0 I}{2 \pi r}\][/tex]
where:
- [tex]\(\mu_0 = 4 \pi \times 10^{-7} \, \text{T} \cdot \text{m/A}\)[/tex] (the permeability of free space)
- [tex]\(I = 8.60 \, \text{A}\)[/tex] (the current through the wire)
- [tex]\(r = 0.200 \, \text{m}\)[/tex] (the distance from the wire to the electron)
Calculating [tex]\(B\)[/tex]:
[tex]\[B = \frac{4 \pi \times 10^{-7} \times 8.60}{2 \pi \times 0.200} = \frac{4 \times 10^{-7} \times 8.60}{0.200} = \frac{3.44 \times 10^{-6}}{0.200} = 1.72 \times 10^{-5} \, \text{T}\][/tex]
The direction of [tex]\(\vec{B}\)[/tex] follows the right-hand rule. Since the current flows in the [tex]\(-x\)[/tex]-direction, at the point [tex]\((0, 0.200, 0)\)[/tex], the magnetic field [tex]\(\vec{B}\)[/tex] is directed into the page (negative [tex]\(z\)[/tex]-direction):
[tex]\[\vec{B} = -1.72 \times 10^{-5} \hat{k} \, \text{T}\][/tex]
Now we use the Lorentz force equation with:
[tex]\[q = -1.60 \times 10^{-19} \, \text{C} \quad (\text{charge of an electron})\][/tex]
[tex]\[\vec{v} = (5.00 \times 10^4 \hat{i} - 3.00 \times 10^4 \hat{j}) \, \text{m/s}\][/tex]
[tex]\[\vec{B} = -1.72 \times 10^{-5} \hat{k} \, \text{T}\][/tex]
The cross product [tex]\(\vec{v} \times \vec{B}\)[/tex]:
[tex]\[\vec{v} \times \vec{B} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\5.00 \times 10^4 & -3.00 \times 10^4 & 0 \\0 & 0 & -1.72 \times 10^{-5}\end{vmatrix}= \hat{i}( (-3.00 \times 10^4)(-1.72 \times 10^{-5}) - 0) - \hat{j}( (5.00 \times 10^4)(-1.72 \times 10^{-5}) - 0)\][/tex]
[tex]\[= \hat{i}( 5.16 \times 10^{-1}) - \hat{j}( -8.60 \times 10^{-1})\][/tex]
[tex]\[= 0.516 \hat{i} + 0.860 \hat{j} \, \text{N/C}\][/tex]
Now, multiply by the charge of the electron:
[tex]\[\vec{F} = q \vec{v} \times \vec{B} = -1.60 \times 10^{-19} (0.516 \hat{i} + 0.860 \hat{j})\][/tex]
[tex]\[\vec{F} = -0.516 \times 1.60 \times 10^{-19} \hat{i} - 0.860 \times 1.60 \times 10^{-19} \hat{j}\][/tex]
[tex]\[\vec{F} = -8.26 \times 10^{-20} \hat{i} - 1.376 \times 10^{-19} \hat{j} \, \text{N}\][/tex]
So, the components of the force are:
[tex]\[F_x = -8.26 \times 10^{-20} \, \text{N}, \quad F_y = -1.376 \times 10^{-19} \, \text{N}, \quad F_z = 0 \, \text{N}\][/tex]
Part B: Calculate the magnitude of the force
The magnitude of the force is given by:
[tex]\[F = \sqrt{F_x^2 + F_y^2 + F_z^2}\][/tex]
[tex]\[F = \sqrt{(-8.26 \times 10^{-20})^2 + (-1.376 \times 10^{-19})^2}\][/tex]
[tex]\[F = \sqrt{(6.82 \times 10^{-39}) + (1.89 \times 10^{-38})}\][/tex]
[tex]\[F = \sqrt{2.57 \times 10^{-38}}\][/tex]
[tex]\[F \approx 1.60 \times 10^{-19} \, \text{N}\][/tex]
So, the magnitude of the force is approximately [tex]\(1.60 \times 10^{-19} \, \text{N}\).[/tex]
The complete question is attached here:
An electron is moving in the vicinity of a long, straight wire that lies along the z-axis. The wire has a constant current of 8.60 A in the -z-direction. At an instant when the electron is at point (0, 0.200 m, 0) and the electron's velocity is (5.00 x 104 m/s) -(3.00 x 104 m/s).
Part A:What is the force that the wire exerts on the electron?
Part B:Calculate the magnitude of this force.
A parallel-plate capacitor is constructed from two aluminum foils of 1 square centimeter area each placedon both sides of a rubber square of the same size. The rubber dielectric is 2.5 mm thick, hasr2.5, andbreakdown field strength of 25 megavolts per meter. Find the voltage rating of the capacitor using a safetyfactor of 10.
Answer:
The voltage will be 0.0125V
Explanation:
See the picture attached
Singly charged positive ions are kept on a circular orbit in a cyclotron. The magnetic field inside the cyclotron is 1.833 T. The mass of the ions is 2.00×10-26 kg, and speed of the ions is 2.05 percent of the speed of the light. What is the diameter of the orbit? (The speed of the light is 3.00×108 m/s.)
Answer:
The diameter is 0.8376 m
Explanation:
Magnetic force is the force that is associated with the magnetic field, the magnitude of the magnitude force can be obtained using equation 1;
F = q v B .....................................1,
where q is the magnitude of the charge of the particle =;
v is the velocity and;
B is the magnetic field = 1.833 T ;
Here, the path of the charge is circular so the force can be also considered as the centripetal force which is represented in equation 2;
[tex]F_{c}[/tex] = m [tex]v^{2}[/tex] / r ....................................2,
m is the particle's mass = 2.00 x[tex]10^{-26}[/tex] kg
v is the speed of ion = 2.05% of 3.00 x [tex]10^{8}[/tex] = 0.0205 x 3.00 x [tex]10^{8}[/tex]
= 6.15 x [tex]10^{6}[/tex] m/s ;
Singly charged ion has a charge equal to the electron charge and the magnitude = 1.60217646 ×[tex]10^{-19}[/tex] C and;
r is the radius of the circular path.
to get the diameter of the orbit we equate equation 1 to 2 and isolate r in equation 2.
q v B = m [tex]v^{2}[/tex] / r
r = m v/q B.....................................3
r = (2.00 x[tex]10^{-26}[/tex] kg) x (6.15 x [tex]10^{6}[/tex] m/s) / (1.60217646 ×[tex]10^{-19}[/tex] C) x (1.833 T)
r = 0.4188 m
The diameter is r x 2
D = 0.4188 m x 2
D = 0.8376 m
Therefore the diameter is 0.8376
A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right so that it slides along the surface. Let R = 1.45 ft and let the angle at which the sphere separates from the cylinder be θs = 34°. The sphere was placed in motion at the very top of the cylinder. Determine the sphare;s initial speed.
Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
A 62.0 kg skier is moving at 6.90 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high.
How fast is the skier moving when she gets to the bottom ofthe hill?
Final answer:
The skier is moving at 11.1 m/s when she gets to the bottom of the hill. This solution is derived using the principles of energy conservation and work done by friction.
Explanation:
A 62.0 kg skier is moving at 6.90 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high. To determine how fast the skier is moving when she gets to the bottom of the hill, we analyze the problem using principles of energy conservation and work done by friction.
We start with the equation that equates the final kinetic and potential energies to the initial energies along with the work done by friction. This equation is 0.5 mv² + 0 = 0.5 mu² + μmgl + mgh, where v is the final velocity, μ is the coefficient of friction, g is the acceleration due to gravity (9.8 m/s²), l is the length of the rough patch, and h is the height of the hill.
Plugging in the given values,
we have: 0.5 v² = 0.5 x 6.9 x 6.9 + 0.3 x 9.8 x 4.5 + 9.8 x 2.50. S
implifying, we get v² = 123.07, which leads to v = 11.1 m/s.
Thus, the skier's speed at the bottom of the hill is 11.1 m/s.
Two identical black holes collide head-on. Each of them has a mass equivalent to 37 solar masses. (The sun has a mass of about 2×1030 kg.) As the black holes collide, they merge, forming a single, larger black hole and additional gravitational waves that carry momentum out of the system. Before the collision, one black hole is moving with a speed of 56 km/s, while the other one is moving at 69 km/s. After the collision the larger black hole moves with speed 4 km/s. How much momentum was carried away by gravitational waves?
Answer:
[tex]3.7\times10^{35}\text{ kg m/s}[/tex]
Explanation:
The system of the colliding bodies is ideally isolated, so no external forces act on it. By the principle of conservation of linear momentum, the total initial momentum is equal to the total final momentum.
Both bodies had a head-on collision. We take the direction of the faster body as the positive direction. Because they have the same mass, let's call this mass m.
Hence, we have for the initial momentum
[tex]69m - 56m = 13m[/tex]
The final momentum is
[tex](m+m) \times4 =[/tex]8m
The difference in both momenta is the momentum carried by the gravitational waves.
[tex]13 m - 8m = 5m[/tex]
Converting to the appropriate units and using the actual value of m (37 × a solar mass), we have
[tex]5\times10^3 \text{ m/s}\times37\times2\times10^{30} \text{ kg} = 3.7\times10^{35}\text{ kg m/s}[/tex]
Cylinder A has a mass of 2kg and cylinder B has a mass of 10kg. Determinethe velocity of A after it has displaced 2m from its original starting position. Neglect the mass of the cable and pulleys and assume that both cylinders start at res
The velocity of A is 5.16m/s²
Explanation:
Given-
mass of cylinder A, mₐ = 2kg
mass of cylinder B, mb = 10kg
Distance, s = 2m
Velocity of A, v = ?
Let acceleration due to gravity, g = 10m/s²
We know,
[tex]a = \frac{mb * g - ma * g}{ma + mb} \\\\a = \frac{10 * 10 - 2 * 10}{ 2 + 10} \\\\a = \frac{80}{12} \\\\a = 6.67m/s^2[/tex]
We know,
[tex]v = \sqrt{2as}[/tex]
[tex]v = \sqrt{2 X 6.67 X 2} \\\\v = \sqrt{26.68} \\\\v = 5.16m/s^2[/tex]
Therefore, the velocity of A is 5.16m/s²
The driving force for fluid flow is the pressure difference, and a pump operates by raising the pressure of a fluid (by converting the mechanical shaft work to flow energy). A gasoline pump is measured to consume 3.8 kW of electric power when operating, If the pressure differential between-the outlet and inlet of the pump is measured to be 7 kPa and the changes in velocity and elevation are negligible, determine the maximum possible volume flow rate of gasoline.
Answer:
[tex]\dot V = 0.542 \frac{m^{3}}{s}[/tex]
Explanation:
The power needed for the pump to raise the pressure of gasoline is defined by following equation. The maximum possible volume flow rate is isolated and then calculated:
[tex]\dot W = \dot V \cdot \Delta P\\\dot V = \frac{\dot W}{\Delta P}\\\dot V = \frac{3.8 kW}{7 kPa}\\\dot V = 0.542 \frac{m^{3}}{s}[/tex]
Explanation:
Below is an attachment containing the solution.
Two 3.0 μC charges lie on the x-axis, one at the origin and the other at What is the potential (relative to infinity) due to these charges at a point at on the x-axis?
Complete Question:
Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)
Answer:
The potential due to these charges is 11250 V
Explanation:
Potential V is given as;
[tex]V =\frac{Kq}{r}[/tex]
where;
K is coulomb's constant = 9x10⁹ N.m²/C²
r is the distance of the charge
q is the magnitude of the charge
The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:
[tex]V =\frac{9X10^9 X3X10^{-6}}{6} =4500 V[/tex]
The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:
[tex]V =\frac{9X10^9 X3X10^{-6}}{4} =6750 V[/tex]
Total potential due to this charges = 4500 V + 6750 V = 11250 V
The potential due to two 3.0 μC charges on the x-axis at different distances from a point can be calculated using Coulomb's Law.
Explanation:The potential due to two point charges can be found using Coulomb's Law. The potential, V, at a point on the x-axis is the sum of the potentials from each charge. The potential due to a point charge can be calculated using the formula V = k * (Q / r), where k is the electrostatic constant, 9 x 10^9 Nm^2/C^2, Q is the charge, and r is the distance between the charge and the point. In this case, since the charges are on the x-axis, the distance between the origin and the point is x, and the distance between the other charge and the point is (6-x). So, the potential at the point is V = k * (3.0 x 10^-6 / x) + k * (3.0 x 10^-6 / (6-x)) relative to infinity.
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An airplane is flying horizontally with a speed of 103 km/hr (278 m/s) when it drops a payload. The payload hits the ground 30 s later. (Neglect air drag and the curvature of the Earth. Take g = 10 m/s².)
At what altitude H is the airplane flying?
Answer:
H = 4500 m
Explanation:
Once dropped, the payload moves along a trajectory, that can be decomposed along two directions independent each other.Just by convenience, we choose these directions to be coincident with the horizontal (-x) and vertical (y) axes.As both movements are independent each other due to both are perpendicular, in the vertical direction, the initial speed is 0.So, in order to find the vertical displacement at any point in time, we can use the following kinematic equation, where a=-g., and H = -Δy.[tex]H = \frac{1}{2}*g*t^{2} = \frac{1}{2} * 10 m/s2*(30s)^{2} = 4500 m[/tex]
The airpane was flying at a 4500 m altitude.If the frequency of the radio station is 88.1MHz(8.81 •10^7Hz), what is the wavelength of the wave used by the radio station for its broadcast? The answer should have three significant figures
Answer:
3.41m
Explanation:
The following were obtained from the question:
f (frequency) = 8.81x10^7Hz
V (velocity of electromagnetic wave) = 3x10^8 m/s
λ (wavelength) =?
Velocity, frequency and wavelength of a wave are related with the equation below:
V = λf
λ = V/f
λ = 3x10^8 /8.81x10^7
λ = 3.41m
Therefore, the wavelength of the radio wave is 3.41m
Answer:
Answer: 3.41
Explanation:
Edge 2020 (E2020)
You are generating traveling waves on a stretched string by wiggling one end. If you suddenly begin to wiggle more rapidly without appreciably affecting the tension, you will cause the waves to move down the string a. faster than before.b. at the same speed as before.c. slower than before.
Answer:
Option as B is correct At the same speed as before
Explanation:
As we know the relation between speed of the wave and tension in string
The speed of wave in stretched string
ν = [tex]\sqrt{\frac{T}{\mu} }[/tex]
speed of wave is the directly proportional to the square root of tension as mentioned in question tension of string is unaffected when in linear mass density is constant, so we can say that the speed of wave will be the same
Option as B is correct At the same speed as before
If you suddenly begin to wiggle more rapidly without appreciably affecting the tension, you will cause the waves to move down the string at the same speed as before (Option b).
What is a wave?A wave can be defined as a type of disturbance that contains energy independently of particle motion.
The wave can move at a velocity (frequency) that is directly proportional to the tension.In this case, tension is constant, thereby velocity of the wave will remain constant.In conclusion, if you suddenly begin to wiggle more rapidly without appreciably affecting the tension, you will cause the waves to move down the string at the same speed as before (Option b).
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What is the voltage across six 1.5-V batteries when they are connected (a) in series, (b) in parallel, (c) three in parallel with one another and this combination wired in series with the remaining three?
The resultant voltage depends on the arrangement of the batteries. For series configuration, the voltage sums up to 9V. For parallel, it remains 1.5V. And for combined series and parallel, it sums up to 3V.
Explanation:The voltage across batteries depends on how they are connected.
When the batteries are connected in series, the voltages add up. So, for six 1.5-V batteries, the total voltage is 6 * 1.5V = 9V. When the batteries are connected in parallel, the voltage remains the same as one battery, which is 1.5V, no matter how many batteries are connected. If three batteries are connected in parallel with each other and then in series with the remaining three also organized in parallel, the voltage would be 1.5V (parallel group) + 1.5V (parallel group) = 3V.Learn more about Voltage here:https://brainly.com/question/31347497
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A particle of mass m is confined to a box of length`. Its initial wave function is identical to that of the displacement of the string in the problem above, Boas Ch. 13, Sec. 4, #4.Find the solution of the Schrodinger equation
Answer:
φ = √2/L sin (kx), E = (h² / 8 mL²) n²
Explanation:
The Schrödinger equation for a particle in a box is, described by a particle within a potential for simplicity with infinite barrier
V (x) = ∞ x <0
0 0 <x <L
∞ x> L
This means that we have a box of length L
We write the equation
(- h’² /2m d² / dx² + V) φ = E φ
h’= h / 2π
The region of interest is inside the box, since being the infinite potential there can be no solutions outside the box. The potential is zero
- h’² /2m d²φ/ dx² = E φ
The solution for this equation is a sine wave,
Because it is easier to work with exponentials, let's use the reaction between the sine function and cook with the exponential
[tex]e^{ikx}[/tex] = cos kx + i sin kx
Let's make derivatives
dφ / dx = ika e^{ikx}
d²φ / dx² = (ik) e^{ikx} = - k² e^{ikx}
Let's replace
- h'² / 2m (-k² e^{ikx}) = E e^{ikx}
E = h'² / 2m k²
To have a solution this expression
Now let's work on the wave function, as it is a second degree differential bond, two solutions must be taken
φ = A e^{ikx} + B e^{-ikx}
This is a wave that moves to the right and the other to the left.
Let's impose border conditions
φ (0) = 0
φ (L) = 0
For being the infinite potential
With the first border condition
0 = A + B
A = -B
They are the second condition
0 = A e^{ikL}+ B e^{-ikL}
We replace
0 = A (e^{ikL} - e^{-ikL})
We multiply and divide by 2i, to use the relationship
sin kx = (e^{ikx} - e^{-ikx}) / i2
0 = A 2i sin kL
Therefore kL = nπ
k = nπ / L
The solution remains
φ = A sin (kx)
E = (h² / 8 mL²) n²
To find the constant A we must normalize the wave function
φ*φ = 1
A² ∫ sin² kx dx = 1
We change the variable
sin² kx = ½ (1 - cos 2kx)
A =√ 2 / L
The definitive function is
φ = √2/L sin (kx)
A remote-controlled car’s wheel accelerates at 22.7 rad/s2 . If the wheel begins with an angular speed of 10.3 rad/s, what is the wheel’s angular speed after exactly twenty full turns
Explanation:
Below is an attachment containing the solution.
The element hydrogen has the highest specific heat of all elements. At a temperature of 25°C, hydrogen’s specific heat capacity is 14300J/(kg K). If the temperature of a .34kg sample of hydrogen is to be raised by 25 K, how much heat will have to be transferred to the hydrogen?
Answer:
121550 J
Explanation:
Parameters given:
Mass, m = 0.34kg
Specific heat capacity, c = 14300 J/kgK
Change in temperature, ΔT = 25K
Heat gained/lost by an object is given as:
Q = mcΔT
Since ΔT is positive in this case and also because we're told that heat was transferred to the hydrogen sample, the hydrogen sample gained heat. Therefore, Q:
Q = 0.34 * 14300 * 25
Q = 121550J or 121.55 kJ
block A with a mass of 10 kg rests on a 30 degree incline. the coefficient of kinetic friction is 0.20. theattatched string is parallel to the incline and passes over amassless frictionless pulley at the top. block B with a massof 8.0kg is attached to the dangling end of the string. theacceleration of B is:
a. 0.69 up
b. 0.69 down
c. 2.6 up
d. 2.6 down
e. 0
Answer:
Please find attached
Explanation:
The acceleration of the block B is 0.69 m/s² downwards in the direction of block B.
The normal force on each block is calculated as follows;
[tex]F_n_ A = mgcos \theta\\\\F_n_ B = m_ B g[/tex]
The frictional force on block A is calculated as;
[tex]F_f = \mu_k F_n\\\\F_f = \mu_ kg mgcos \theta[/tex]
The horizontal force on block A is given as;
[tex]F_x = mgsin\theta[/tex]
The tension on the string due to each block is given as;
[tex]T_ A = m_ A a\\\\T_ B = m_ B a[/tex]
The net force on the block B is calculated as;
[tex]m_Bg - (T_A + m_Agsin\theta + \mu mgcos\theta) = T_B\\\\m_Bg - m_Agsin\theta - \mu mgcos\theta= T_B + T_ A\\\\m_Bg - m_Agsin\theta - \mu mgcos\theta = a(m_ B+ m_ A)\\\\a = \frac{m_Bg - m_Agsin\theta - \mu mgcos\theta}{m_B + m_ A} \\\\a = \frac{(8)(9.8)\ -\ (10)(9.8)(sin30)\ -\ (0.2)(10)(9.8)(cos30)}{8 + 10} \\\\a = 0.69 \ m/s^2 \ (in -direction \ of \ block \ B)[/tex]
Thus, the acceleration of the block B is 0.69 m/s² downwards in the direction of block B.
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Air "breaks down" when the electric field strength reaches 3 × 106 N/C, causing a spark. A parallel-plate capacitor is made from two 3.0 cm × 3.0 cm electrodes.How many electrons must be transferred from one electrode to the other to create a spark between the electrodes?
Answer:
[tex]1.5\times 10^{11}}[/tex]
Explanation:
We are given that
Electric field=[tex]E=3\times 10^6 N/C[/tex]
Dimension of parallel plate capacitor=[tex]3 cm\times 3 cm[/tex]
Area of parallel plate capacitor=[tex]A=3\times 3=9 cm^2=9\times 10^{-4}m^2[/tex]
[tex]1 cm^2=10^{-4} m^2[/tex]
We have to find the number of electrons must be transferred from one electrode to the other to create a spark between the electrodes.
[tex]E=\frac{Q}{\epsilon_0A}[/tex]
Where [tex]\epsilon_0=8.85\times 10^{-12}C^2/Nm^2[/tex]
Substitute the values
[tex]3\times 10^6=\frac{Q}{8.85\times 10^{-12}\times 9\times 10^{-4}}[/tex]
[tex]Q=3\times 10^6\times 8.85\times 10^{-12}\times 9\times 10^{-4}}[/tex]
[tex]Q=2.4\times 10^{-8} C[/tex]
We know that
[tex]Q=ne=n\times 1.6\times 10^{-19} [/tex]
Where e=[tex]1.6\times 10^{-19} C[/tex]
[tex]n=\frac{Q}{e}=\frac{2.4\times 10^{-8}}{1.6\times 10^{-19}}=1.5\times 10^{11}}[/tex]
A SMA wire in the un-stretched condition is then given an initial strain of εo (to preload the wire) at room temperature (RT). Its ends are then rigidly fixed. What force is developed in the wire?
Answer: tensional force
Explanation:
Tension force on a material occurs when two equal forces act on a material in an opposite direction away from the ends of the material.
Pre-tensing a wire material increases its load bearing capacity and reduces its flexure.
Analyzing the Light Bulb: You should have noticed that the light bulb doesn't have a single well-defined "resistance," since the current vs. voltage plot is nonlinear. Nevertheless, one can define a "voltage-dependent resistance" as R(V)=V/I(V)as the ratio of voltage to current.1Basic Behavior: According to your data, does this resistance increase or decrease with voltage? A reasonable (and correct) thought is that the impact is really with temperature, as the light bulb heats up with more power going into it. How does your data imply resistance varies with temperature?Thermal Expansion: One hypothesis you might have is that the reason is that the resistor expands slightly with increased temperature (since most materials do), and hence the cross-sectional area and length of the resistor change.Supposing the resistor increases in size by the same factor in every direction, what direction does the resistance change? (I.e., does the resistance get larger or smaller?) Is this the direction that you expect based on your answer to the previous part?
Answer:
Resistance increases with increase in temperature which depends on power supplied which also depends on voltage.
Thermal expansion will make resistance larger.
Explanation:
Light bulb is a good example of a filament lamp. If we plot the graph of voltage against current we will notice that resistance is constant at constant temperature.
The filament heats up when an electric current passes through it, and produces light as a result.
The resistance of a lamp increases as the temperature of its filament increases. The current flowing through a filament lamp is not directly proportional to the voltage across it.
tensile stress begins to appear in resistor as the temperature rises. Thus, the resistance value increases as the temperature rises. Resistance value can only decrease as the temperature rises in case of thin film resistor with aluminium substrate.
In case of a filament bulb, the resistance will increase as increase in length of the wire. The thermal expansion in this regard is linear expansivity in which resistance is proportional to length of the wire.
Resistance therefore get larger.
A 20.0-kg cannonball is fired from a cannon with muzzle speed of 1 000 m/s at an angle of 37.08 with the horizontal. A second ball is fired at an angle of 90.08. Use the isolated system model to find (a) the maximum height reached by each ball and (b) the total mechanical energy of the ball–Earth system at the maximum height for each ball. Let y5 0 at the cannon.
Answer:
(a). The maximum height by first ball is [tex]1.8545\times10^{4}\ m[/tex]
The maximum height by second ball is [tex]5.1020\times10^{4}\ m[/tex]
(b). The total mechanical energy of the ball–Earth system at the maximum height for each ball is [tex]1.0\times10^{7}\ J[/tex]
Explanation:
Given that,
Mass of cannonball = 20.0 kg
Speed = 1000 m/s
Angle with horizontal= 37.08
Fired angle = 90.08
We need to calculate the speed of the ball
Using formula of speed
[tex]v_{y}=v\sin\theta_{H}[/tex]
Put the value into the formula
[tex]v_{y}=1000\times\sin37.08[/tex]
[tex]v_{y}=602.9\ m/s[/tex]
(a). We need to calculate the maximum height by first ball
Using conservation of energy
[tex]\dfrac{1}{2}mv^2=mgh[/tex]
[tex]h= \dfrac{v^2}{2g}[/tex]
Put the value into the formula
[tex]h=\dfrac{(602.9)^2}{2\times9.8}[/tex]
[tex]h=1.8545\times10^{4}\ m[/tex]
We need to calculate the maximum height by second ball
Using conservation of energy
[tex]\dfrac{1}{2}mv^2=mgh[/tex]
[tex]h= \dfrac{v^2}{2g}[/tex]
Put the value into the formula
[tex]h=\dfrac{(1000)^2}{2\times9.8}[/tex]
[tex]h=5.1020\times10^{4}\ m[/tex]
(b). We need to calculate the total mechanical energy of the ball–Earth system at the maximum height for each ball
Using formula of energy
[tex]E=\dfrac{1}{2}mv^2[/tex]
[tex]E=\dfrac{1}{2}\times20\times1000^2[/tex]
[tex]E=1.0\times10^{7}\ J[/tex]
Hence, (a). The maximum height by first ball is [tex]1.8545\times10^{4}\ m[/tex]
The maximum height by second ball is [tex]5.1020\times10^{4}\ m[/tex]
(b). The total mechanical energy of the ball–Earth system at the maximum height for each ball is [tex]1.0\times10^{7}\ J[/tex]
Three individual point charges are placed at the following positions in the x-y plane:Q3= 5.0 nC at (x, y) = (0,0);Q2= -3.0 nC at (x, y) = (4 cm, 0); and Q1= ?nC at (x, y) = (2 cm,0);What isthe magnitude, and sign, ofcharge Q1such that the net force exerted on charge Q3, exerted bycharges Q1and Q2, is zero?
Answer:
Explanation:
net force exerted on charge Q₃, exerted by charges Q₁and Q₂, will be zero
if net electric field due to charges Q₁ and Q₂ at origin is zero .
electric field due to Q₂
= 9 X 10⁹ X 3 x10⁹ / .04²
electric field due to Q₁
= 9 X 10⁹ X Q₁ / .02²
For equilibrium
9 X 10⁹ X Q₁ / .02² = 9 X 10⁹ X 3 x10⁻⁹ / .04²
Q₁ = 3 X10⁻⁹ x .02² / .04²
= 3 / 4 x 10⁻⁹
.75 x 10⁻⁹ C
The op amp in this circuit is ideal. R3 has a maximum value of 100 kΩ and σ is restricted to the range of 0.2 ≤ σ ≤ 1.0. a. Calculate the range of vO if vI = 40 mV. b. If σ is not restricted, at what value of σ will the operational amplifier saturate?
I have attached the circuit image missing in the question.
Answer:
A) The range of vo is; -6.6V≤ vo ≤-1V
B) σ = 0.1861
Explanation:
A) First of all, Let VΔ be the voltage from the potentiometer contact to the ground.
Thus; [(0 - vg)/(2000)] +[(0 - vΔ)/(50,000)] = 0
So, [(- vg)/(2000)] +[(- vΔ)/(50,000)] = 0
Simplifying further; -25 vg - vΔ = 0
From the question, vg = 40mV = 0.04 V
So - 25(0.04) = vΔ
So: vΔ = - 1 V
Now, [vΔ/(σRΔ)] + [(vΔ - 0)/(50,000)] + [(vΔ - vo)/((1 - σ)RΔ))] = 0
So, multiplying each term by RΔ to get; [vΔ/(σ)] + [(vΔ x RΔ)/(50,000)] + [(vΔ - vo)/((1 - σ))] = 0
So RΔ = 100kΩ or 100,000Ω from the question.
So, substituting for RΔ, we get,
[vΔ/(σ)] + [2vΔ] + [(vΔ - vo)/((1 - σ))] = 0
Let's put the value of - 1 for vΔ as gotten before.
So, ( - 1/σ) - 2 + [(-1 - vo)/(1 - σ)] = 0
Now let's make vo the subject of the equation to get;
-1 - vo = (1 - σ)[2 + (1/σ)]
-1 - vo = 2 - 2σ + (1/σ) - 1
-vo = 1 + 2 - 2σ + (1/σ) - 1
-vo = 2 - 2σ + (1/σ)
vo = - 1 (2 - 2σ + (1/σ))
When σ = 0.2; vo = - 1(2 - 0.4 + 5) =
- 1 x 6.6 = - 6.6V
Also when σ = 1;
vo = - 1(2 - 2 + 1) = - 1V
Therefore, the range of vo is;
- 6.6V ≤ vo ≤ - 1V
B) it will saturate at vo = - 7V
So, from;
vo = - 1 (2 - 2σ + (1/σ))
-7 = - 1 (2 - 2σ + (1/σ))
Divide both sides by (-1)
7 = (2 - 2σ + (1/σ))
Now, subtract 2 from both sides to get; 5 = - 2σ + (1/σ)
Multiply each term by α to get;
5σ = - 2σ^(2) + 1
So 2σ^(2) + 5σ - 1 = 0
Solving simultaneously and picking the positive value , we get σ to be approximately 0.1861
Final answer:
The question involves understanding an ideal operational amplifier's output behavior, focusing on its output voltage changes and saturation with respect to variations in σ (sigma). Calculations for the output voltage range given a specific input and understanding the conditions under which the op-amp will saturate are key.
Explanation:
The question revolves around an operational amplifier (op-amp) circuit, exploring its behavior under certain conditions, specifically examining output voltage (vO) variations and saturation point related to the parameter σ (sigma). The op-amp is assumed to be ideal, implying infinite gain, zero input current, and that its input terminals are at the same potential.
a. Range of vO if vI = 40 mV
Given that the op-amp is ideal, the output voltage will depend on the input voltage (vI), the gain settings (σ), and the feedback resistor (R3). In practical scenarios, the gain can be adjusted by changing the value of σ or R3. However, for an ideal op-amp, the input voltage is directly proportional to the output voltage, influenced by σ. With vI = 40 mV and σ ranging between 0.2 and 1.0, the output voltage will vary accordingly, directly proportional to these parameters.
b. Saturation point of the op-amp
Saturation in an op-amp occurs when the output voltage exceeds the power supply limits, meaning the op-amp can no longer amplify the input signal. The specific value of σ at which saturation occurs depends on the supply voltage, the op-amp's maximum output voltage capability, and the configuration of the feedback network. Without specific values for the power supply or feedback network, calculating the exact σ value for saturation is not possible. Yet, in theory, as σ approaches the op-amp's gain limit or if the gain results in an output voltage beyond what the op-amp can deliver based on its power supply, saturation will occur.
Two inclined planes A and B have the same height but different angles of inclination with the horizontal. Inclined plane A has a steeper angle of inclination than inclined plane B. An object is released at rest from the top of each of the inclined planes.
How does the speed of the object at the bottom of inclined plane A compare with that of the speed at the bottom of inclined plane B?
Answer:
It is the same.
Explanation:
Assuming no friction between the object and the surface, and no other external force acting on the object, than gravity and normal force, we can say the following:[tex]\Delta K + \Delta U = 0[/tex]
where ΔK = change in kinetic energy, and ΔU = change in gravitational potential energy.As ΔU = -m*g*h (being h the height of the plane), it will be the same for both inclined planes, as we are told that they have the same height.If the object starts from rest, the change in kinetic energy will be as follows:[tex]\Delta K = K_{f} - K_{0} = \frac{1}{2} * m*v_{f} ^{2} (1)[/tex]
[tex]\Delta K = -\Delta U = m*g*h (2)[/tex]
From (1) and (2) we see that the mass m and the height h are the same, the speed at the bottom of inclined plane A, will be the same as the one at the bottom of inclined plane B.Despite the difference in angle, the speed of the objects at the bottom of the planes will be the same because they start with the same potential energy at the top that's entirely converted to kinetic energy (the energy of motion) by the bottom, so long as energy losses are ignored.
Explanation:The subject of this question lies in the realm of Physics, specifically involving principles of mechanical energy and gravitational potential energy. In the stated scenario, the two objects start on their respective inclined planes from a state of rest. Accordingly, they possess potential energy but no kinetic energy.
As the objects slide down their respective planes, this potential energy is converted into kinetic energy—the energy of motion. Because the two planes are of identical height (thus imparting the same initial potential energy to the objects), and because all potential energy will have been converted to kinetic energy by the time the objects reach the bottom (ignoring energy losses due to friction or air resistance), both objects will possess the same kinetic energy—and thereby the same speed—at the bottom of their planes, regardless of the angle of inclination.
In practical applications, friction and other factors may have a role and might cause the object on the steeper plane (Plane A) to reach the bottom more quickly. However, that's not due to a difference in speed at the bottom; it's about the time taken to get there.
Learn more about Energy Transfer on Inclined Planes here:https://brainly.com/question/21487030
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You accidentally drop a quarter into the hot coals of a campfire. You fish out the hot quarter with a pair of pliers and drop the quarter directly on top of a large 2 kg block of ice to cool it down. In what direction does heat flow:
There is no heat flow.
From the quarter to the block of ice
Answer:
from the quarter to the block of ice
Explanation:
Heat flows from higher temperature to lower temperature until temperature of both bodies are in Equilibrium .
Since block of ice has lower temperature than that of quarter. heat transfer will take from quarter to ice until both have same temperature(in other words temperature are in Equilibrium for quarter and ice)
Final answer:
Heat will flow from the hot quarter to the block of ice as heat always transfers from a warmer object to a cooler one. The ice absorbs the heat and may melt without increasing in temperature due to the phase change.
Explanation:
When a hot quarter is dropped onto a large block of ice, heat will flow from the quarter to the block of ice. This is because heat always flows spontaneously from a hotter object to a cooler one according to thermodynamics. In this case, the hot quarter would lose heat, and the block of ice would absorb it. Even as the ice absorbs heat, it may not increase in temperature as it may be undergoing a phase change from solid to liquid at 0°C. The heat is used to break the bonds between water molecules during the melting process, thus increasing the internal potential energy instead of increasing the kinetic energy, which would raise the temperature.
An ideal gas Carnot cycle with air in a piston cylinder has a high temperature of 1000 K and a heat rejection at 400 K. During the heat addition the volume triples. Find the two specific heat transfers (q) in the cycle and the overall cycle efficiency
Answer:
W / n = - 9133 J / mol, W / n = 3653 J / mol , e = 0.600
Explanation:
The Carnot cycle is described by
[tex]e= 1 - Q_{c} / Q_{H} = 1 - T_{c} / T_{H}[/tex]
In this case they indicate that the final volume is
V = 3V₀
In the part of the heat absorption cycle from the source is an isothermal expansion
W = n RT ln (V₀ / V)
W / n = 8.314 1000 ln (1/3)
W / n = - 9133 J / mol
During the part of the isothermal compression in contact with the cold focus, as in a machine the relation of volumes is maintained in this part is compressed three times
W / n = 8.314 400 (3)
W / n = 3653 J / mol
The efficiency of the cycle is
e = 1- 400/1000
e = 0.600
A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.26.
a. What magnitude of force must the worker apply?
b. How much work is done on the crate by this force?
c. How much work is done on the crate by friction?
d. How much work is done on the crate by the normal force? By gravity?
e. What is the total work done on the crate?
Answer:
a. 79.1 N
b. 344 J
c. 344 J
d. 0 J
e. 0 J
Explanation:
a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force [tex]F_p[/tex] by the worker must be equal to the friction force [tex]F_f[/tex] on the crate, which is the product of friction coefficient μ and normal force N:
Let g = 9.81 m/s2
[tex]F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N[/tex]
b. The work is done on the crate by this force is the product of its force [tex]F_p[/tex] and the distance traveled s = 4.35
[tex]W_p = F_ps = 79.1*4.35 = 344 J[/tex]
c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35
[tex]W_f = F_fs = -79.1*4.35 = -344 J[/tex]
This work is negative because the friction vector is in the opposite direction with the distance vector
d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.
e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction
[tex]W_p + W_f = 344 - 344 = 0 J[/tex]
Answer:
(A) 79N
(B) W = 344J
(C) Wf= -344J
(D) W = 0J
(E) W = 0J
Explanation:
Please see attachment below.
Say you have two parallel current-carrying wires, each carrying a current of 1.0 A and with a distance of 1.0 m between them. What is the magnitude of the force per unit length experienced by each wire
Answer:
[tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]
Explanation:
Given,
current in the wire I₁ = 1 A
I₂ = 1 A
distance between them, r = 1 m
using Force per unit length formula
[tex]\dfrac{F}{l}=\dfrac{\mu_0I_1I_2}{2\pi r}[/tex]
[tex]\mu_0 = magnetic\ permeability\ of\ free\ space = 4\pi\times 10^{-7}[/tex]
[tex]\dfrac{F}{l}=\dfrac{4\pi \times10^{-7}\times 1\times 1}{2\pi\times 1}[/tex]
[tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]
Hence, the magnitude of force per unit length is equal to [tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]
Final answer:
The magnitude of the force per unit length each 1.0 A current-carrying wire experiences when separated by a distance of 1.0 m is 2 x 10⁻⁷ N/m, calculated using the formula for magnetic force between parallel conductors.
Explanation:
When discussing two parallel current-carrying wires, we're addressing the magnetic force that arises between them due to their currents. Specifically, when each wire carries a current of 1.0 A and they are placed 1.0 m apart, the force per unit length that each wire experiences can be calculated using the formula for the magnetic force between two parallel conductors. The equation is FE = (µ0 / 2π) × (I1I2 / r), where µ0 is the magnetic constant (4π x 10-7 T·m/A), I1 and I2 are the currents in the wires, and r is the distance between the wires. Given that each wire carries 1.0 A of current and the separation is 1.0 m, we can plug these values into the formula to find that FE = (4π x 10-7 T·m/A / 2π) × (1.0 A2 / 1.0 m), resulting in a force per unit length of 2 x 10-7 N/m.
If the dielectric constant is 14.1, calculate the ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge.
Answer:
[tex]\frac{Q}{Q_0}=1[/tex]
Explanation:
Capacitance is defined as the charge divided in voltage.
[tex]C=\frac{Q}{V}(1)[/tex]
Introducing a dielectric into a parallel plate capacitor decreases its electric field. Therefore, the voltage decreases, as follows:
[tex]V=\frac{V_0}{k}[/tex]
Where k is the dielectric constant and [tex]V_0[/tex] the voltage of the capacitor without a dielectric
The capacitance with a dielectric between the capacitor plates is given by:
[tex]C=kC_0[/tex]
Where k is the dielectric constant and [tex]C_0[/tex] the capacitance of the capacitor without a dielectric. So, we have:
[tex]Q=CV\\Q=kC_0\frac{V_0}{k}\\Q=C_0V_0\\Q_0=C_0V_0\\Q=Q_0\\\frac{Q}{Q_0}=1[/tex]
Therefore, a capacitor with a dielectric stores the same charge as one without a dielectric.
What determines whether the equilibrium temperature of a mixture of two amounts of water will be closer to the initially cooler or warmer water?
Final answer:
The equilibrium temperature of a mixture of two amounts of water will be closer to the initially cooler water due to the specific heat capacity of water.
Explanation:
The equilibrium temperature of a mixture of two amounts of water will be closer to the initially cooler water.
When different temperatures of water are mixed, heat is transferred between them until they reach a common equilibrium temperature. The amount of heat transferred is determined by the specific heat capacity of water, which is greater than most common substances. As a result, water undergoes a smaller temperature change for a given heat transfer. Therefore, the equilibrium temperature will be closer to the initially cooler water.
A truck runs into a pile of sand, moving 0.80 m as it slows to a stop. The magnitude of the work that the sand does on the truck is 5.5×105J. Part A Determine the magnitude of the average force that the sand exerts on the truck
Answer:
687,500 N
Explanation:
Workdone = Force × Distance
Making force the subject of the formula; we have:
Force =[tex]\frac{workdone}{distance}[/tex]
Given that:
workdone = 5.5×10⁵ J
Distance = 0.80 m
∴ Force = [tex]\frac{5.5*10^5}{0.8}[/tex]
Force = 687,500 N
Answer:
6.875×10⁵ N.
Explanation:
Force: This can be defined as the product of mass and acceleration or it can be defined as the ratio of work done and distance. The S.I unit of force is Newton.
W = F×d................. Equation 1
Where W = work done, F = force, d = distance.
make F the subject of the equation
F = W/d.................... Equation 2
Given: W = 5.5×10⁵ J, d = 0.8 m
Substitute into equation 2
F = 5.5×10⁵ /0.8
F = 6.875×10⁵ N.
Hence the force exerted on the truck by the sand = 6.875×10⁵ N.