Answer:
Explanation:
dV = k (dq)/d
dV = k (sigma da)/d
dV = k (5 C r (r dr d(phi)))/d
d = sqrt(r^2 + x^2)
dV = 5 k C (r^2/sqrt(r^2+x^2)) dr d(phi)
==> V = int{5 k C (r^2/sqrt(r^2+x^2)) dr d(phi)} ; from r=0 to r=R and phi=0 to phi=2pi
==> V = 5 k C int{(r^2/sqrt(r^2+x^2)) dr d(phi)} ; from r=0 to r=R and phi=0 to phi=2pi
==> V = 5 k C (2 pi) int{(r^2/sqrt(r^2+x^2)) dr} ; from r=0 to r=R
==> V = 5 k C (2 pi) (1/2) (R sqrt(R^2+x^2) - x^2 ln(sqrt(r^2+x^2) + r) + x^2 Ln(x))
If the velocity of a pitched ball has a magnitude of 47.0 m/s and the batted ball's velocity is 56.5 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Answer:
Change in momentum =15.01kgm/s
Impulse applied by bat = 15.01Ns
Explanation:
Please see attachment below.
A 0.60-kg particle has a speed of 2.0 m/s at point A and a kinetic energy of 7.5 J at point B. What is (a) its kinetic energy at A? (b) Its speed at point B ? (c) The total work done on the particle as it moves from A to B ?
Answer:
Explanation:
note:
solution is attached due to error in mathematical equation. please find the attachment
Answer:
(a) 1.2 J
(b) 5 m/s
(c) 6.3 J
Explanation:
(a) Kinetic energy at A
Ek = 1/2mv².................. Equation 1
Where Ek = Kinetic energy at A, m = mass of the particle, v = velocity at A.
Given: m = 0.6 kg, v = 2.0 m/s
Substitute into equation 1
Ek = 1/2(0.6)(2²)
Ek = 0.6(2)
Ek = 1.2 J.
(b)
Speed at point B
Ek' = 1/2mv'²............... Equation 2
Make v' the subject of the equation,
v' = √(2Ek'/m).................. Equation 3
Where, Ek' = kinetic energy at B, v' = velocity at B.
Given: Ek' = 7.5 J, m = 0.6 kg,
Substitute into equation 3
v' = √[(2×7.5)/0.6]
v' =√(15/0.6)
v' = √25
v' = 5 m/s.
(c)
Wt = Δ kinetic energy from A to B
Where Wt = total work done as the particle moves from A to B.
Wt = 1/2m(v'²-v²)
Wt = 1/2(0.6)(5²-2²)
Wt = 0.3(25-4)
Wt = 0.3(21)
Wt = 6.3 J
A current of 8 A exists in a copper (Cu) wire which has a diameter of 5 mm. What is the current density? Each atom of copper contributes one conduction electron, and the average thermal speed r k T m of an electron is 2.4 × 106 m/s . The mass
Explanation:
Below is an attachment containing the solution.
To model a tornado, superimpose the stream function for a vortex (Eq. 6.91) and a source (Eq. 6.83). Assume that the circulation is 8000 m2 /s and the pressure at the radius of 40 m is 2 kPa less than atmospheric
Answer:
The stream function for this potential flow is -4696.8
Explanation:
Given that,
Circulation = 8000 m²/s
Radius = 40 m
Pressure = 2 kPa
Suppose the determine the stream function for this potential flow
We need to calculate the stream function
Using formula of stream function
[tex]\Psi=-(\dfrac{\Gamma}{2\pi})ln(r)[/tex]
Where, Γ = circulation
r = radius
Put the value into the formula
[tex]\Psi=-\dfrac{8000}{2\pi}\times ln(40)[/tex]
[tex]\Psi=-4696.8[/tex]
Hence, The stream function for this potential flow is -4696.8
Two long parallel wires 20 cm apart carry currents of 5.0 A and 8.0 A in the same direction. Is there any point between the two wires where the magnetic field is zero?
Answer:
x= 0.077 m from the wire carrying 5.0 A current.
Explanation:
If the wire can be approximated as an infinite one, and we can neglect the diameter of the wire, we can find the magnetic field B at a distance d from the wire, with the following expression:[tex]B =\frac{\mu_{0} * I}{2*\pi*d}[/tex]
Due to the currents are in the same direction, this means that the magnetic field lines (taking the shape of circumferences) will have opposite directions between the wires.So, if we assume that at some distance from both wires, the magnetic field will be 0, we can write the following equation:[tex]\frac{\mu_{0} * I_{1}}{2*\pi*x} - \frac{\mu_{0} * I_{2} }{2*\pi*(d-x)} = 0[/tex]
where I₁ = 5.0A, I₂= 8.0A and d = 0.2 mSimplifying common terms, we can solve for x, as follows:[tex]\frac{I_{1} }{x} = \frac{I_{2} }{(d-x)} \\ \frac{5.0A}{x(m)} = \frac{8.0A}{(0.2m-x(m))}[/tex]
⇒ [tex]x =\frac{1m}{13} = 0.077 m = 7.7 cm[/tex]
A racquet ball with mass m = 0.238 kg is moving toward the wall at v = 12.4 m/s and at an angle of θ = 31° with respect to the horizontal. The ball makes a perfectly elastic collision with the solid, frictionless wall and rebounds at the same angle with respect to the horizontal. The ball is in contact with the wall for t = 0.078 s. 1)What is the magnitude of the initial momentum of the racquet ball?
Answer:
[tex]||\vec p || = 2.951\,\frac{kg\cdot m}{s}[/tex]
Explanation:
The initial momentum of the racquet ball is:
[tex]||\vec p || = (0.238\,kg)\cdot (12.4\,\frac{m}{s} )[/tex]
[tex]||\vec p || = 2.951\,\frac{kg\cdot m}{s}[/tex]
Bulb ܣ is rated for 20W at 12V. Bulb ܤ is rated for 20W at 120V. The two bulbs are mistakenly mixed up so that bulb ܣ is connected to a 120V line and bulb ܤ is connected to a 12V line. a. Determine the power dissipated by Bulb ܣ while connected to the 120Vline. b. Determine the power dissipated by Bulb ܤ while connected to the 12Vline. c. Which bulb, if any, is more likely to burn out and why?
Answer:
a. 2000W
b.0.2W
c. Bulb 1
Explanation:
Data given:
bulb 1 rating =20w,12v
bulb 2 rating =20w,120v.
we calculate the resistance for each bulb
[tex]bulb 1: R=\frac{V^2}{P}\\ bulb 1: R=\frac{12^2}{20}\\ R=7.2 ohms[/tex]
for bulb 2
[tex]bulb 2: R=\frac{V^2}{P}\\ bulb 2: R=\frac{120^2}{20}\\ R=720 ohms[/tex]
when miss connection occur we need to calculate the dissipated power from wash bulb
a. for bulb 1 with R=7.2ohms and V=120v,
the power is calculated as
[tex]P=\frac{v^2}{R}\\ P=\frac{120^2}{7.2}\\ P=2000W[/tex]
b. for bulb 2 with R=720 ohms and V=12v,
the power is calculated as
[tex]P=\frac{v^2}{R}\\ P=\frac{12^2}{720}\\ P=0.2W\\[/tex]
c. Bulb 1 is more likely to burn out because it is operating above the rated power.
Consider a sphere, an infinitely long cylinder, and a plane of infinite length and width (a, b and c below). Imagine that you can hover above each one in your own personal helicopter. In which case do you have the most freedom to move about without your view of the object changing? In other words, for each case consider if there are directions that you can move in without the objects distance or orientation, relative to you, changing.
Answer:
A plane of infinite length and width.
Explanation:
For the case of the sphere, we should consider spherical coordinates.
You can move around the sphere without changing your distance from the center of the sphere, however you can alter your azimuthal angle and polar angle, since they are symmetric in spherical coordinate system.
r --> Cannot change
Φ --> Free to change
θ --> Free to change
For the case of the infinite cylinder, we should consider cylindrical coordinates.
You can change your height and angular coordinate, but you cannot change your distance from the axis of the cylinder.
r --> Cannot change
θ --> Free to change
z --> Free to change
For the case of the infinite plane, we should consider cartesian coordinates.
Since the length and width of the plane is infinite, we cannot recognize whether we are getting closer or further away.
x --> Free to move
y --> Free to move
z --> Free to move
Therefore, in the case of infinite plane you have the most freedom to move about without your view of the object changing.
Final answer:
An infinite plane allows the most freedom of movement without changing the view, due to its infinite symmetry. In contrast, a sphere and a cylinder have more limited symmetrical properties, resulting in a restricted freedom of movement to keep the view unchanged.
Explanation:
When comparing the freedom to move above a sphere, an infinitely long cylinder, and an infinite plane, the plane provides the most freedom without changing your view of the object. On a sphere, any move results in a different orientation or distance from the surface. With the cylinder, movement along the axis does not change the view, but any other direction does. However, the infinite plane allows movement in any direction on a two-dimensional plane without changing the viewed geometry or orientation.
From a mathematical perspective, this question revolves around symmetry and geometric invariance under transformation. The infinite plane exhibits infinite symmetry; therefore, no matter how you move parallel to its surface, the view remains unchanged. Conversely, a sphere has rotational symmetry around its center, and a cylinder along its axis, limiting the directions in which one can move without altering the perceived distance or orientation.
If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wavelengths on a screen 1.0 m away
Answer:
0.82 mm
Explanation:
The formula for calculation an [tex]n^{th}[/tex] bright fringe from the central maxima is given as:
[tex]y_n=\frac{n \lambda D}{d}[/tex]
so for the distance of the second-order fringe when wavelength [tex]\lambda_1[/tex] = 745-nm can be calculated as:
[tex]y_2 = \frac{n \lambda_1 D}{d}[/tex]
where;
n = 2
[tex]\lambda_1[/tex] = 745-nm
D = 1.0 m
d = 0.54 mm
substituting the parameters in the above equation; we have:
[tex]y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}[/tex]
[tex]y_2[/tex] = 0.00276 m
[tex]y_2[/tex] = 2.76 × 10 ⁻³ m
The distance of the second order fringe when the wavelength [tex]\lambda_2[/tex] = 660-nm is as follows:
[tex]y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}[/tex]
[tex]y^'}_2[/tex] = 1.94 × 10 ⁻³ m
So, the distance apart the two fringe can now be calculated as:
[tex]\delta y = y_2-y^{'}_2[/tex]
[tex]\delta y[/tex] = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m
[tex]\delta y[/tex] = 10 ⁻³ (2.76 - 1.94)
[tex]\delta y[/tex] = 10 ⁻³ (0.82)
[tex]\delta y[/tex] = 0.82 × 10 ⁻³ m
[tex]\delta y[/tex] = 0.82 × 10 ⁻³ m [tex](\frac{1.0mm}{10^{-3}m} )[/tex]
[tex]\delta y[/tex] = 0.82 mm
Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm
An ideal gas is compressed isothermally to one-third of its initial volume. The resulting pressure will be
A) three times as large as the initial value.
B) less than three times as large as the initial value.
C) more than three times as large as the initial value.
D) equal to the initial value.
Answer:
A three times as large as the initial value
a carbon steel ball with a 30mm diameter is pressed against a flat carbon steel plate with a force of 20n. calculate the diameter of the circular contact area and the maximum pressure that occurs at the center of the contact area
Final answer:
The diameter of the circular contact area is 30 mm and the maximum pressure at the center of the contact area is approximately 0.14 N/mm^2.
Explanation:
To calculate the diameter of the circular contact area, we need to determine the radius of the contact area first. The radius can be found by dividing the diameter of the ball by 2. In this case, the radius is 15 mm.
The area of the circular contact area can be calculated using the formula A = πr^2, where A is the area and r is the radius. The maximum pressure at the center of the contact area can be found by dividing the force applied on the plate by the area of the contact area.
Using these formulas, the diameter of the circular contact area is 30 mm and the maximum pressure at the center of the contact area is approximately 0.14 N/mm^2.
A magnetic field has a magnitude of 1.2 \times 10^{-3} T, and an electric field has a magnitude of 4.6 \times 10^{3}N/C. Both fields point in the same direction. A positive 1.8 \mu C charge moves at a speed of 3.1 \times 10^6 m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge.
Answer: F = 113.4.[tex]10^{-3}[/tex]N
Explanation: Net Force is the total forces acting in an object. In this case, there are two forces acting on the charge: one due to magnetic field (Fm) and another due to electric field (Fe). So, net force is
F = Fe + Fm
Force due to electric field
To determine this force:
Fe = q.E, where q is the charge and E is electric field.
Calculating:
Fe = q.E
Fe = 1.8.[tex]10^{-6}[/tex].4.6.[tex]10^{3}[/tex]
Fe = 8.28.[tex]10^{-3}[/tex]N
Force due to magnetic field: It can only happens when the charge is in movement, so
Fm = q.(v×B), where v represents velocity and B is magnetic field
The cross product indicates that force is perpendicular to the velocity and the field.
Calculating:
Fm = q.v.B.senθ
As θ=90°,
Fm = q.v.B
Fm = 1.8.[tex]10^{-6}[/tex].3.1.[tex]10^{6}[/tex].1.2.[tex]10^{-3}[/tex]
Fm = 6.696.[tex]10^{-3}[/tex]N
F, Fm and Fe make a triangle. So, using Pythagorean theorem:
F = [tex]\sqrt{Fe^{2} + Fm^{2} }[/tex]
F = [tex]\sqrt{(8.28.10^{-3} )^{2} +(6.696.10^{-3} )^{2} }[/tex]
F = 113.4.[tex]10^{-3}[/tex]N
The net force acting on the charge is F = 113.4.[tex]10^{-3}[/tex]N
The top of a swimming pool is at ground level. If the pool is 3.00 m deep, how far below ground level does the bottom of the pool appear to be located for the following conditions? (The index of refraction of water is 1.333.)
(a) The pool is completely filled with water.
______m below ground level
(b) The pool is filled halfway with water.
______m below ground level
Answer:
a) 2.25 m
b) 2.625 m
Explanation:
Refraction is the name given to the phenomenon of the speed of light changing the the boundary when it moves from one physical medium to the other.
Refractive index is the ratio of the speed of light in empty vacuum (air is an appropriate substitution) to the speed of light in the medium under consideration.
In terms of real and apparent depth, the refractive index is given by
η = (real depth)/(apparent depth)
a) Real depth = 3.00 m
Apparent depth = ?
Refractive index, η = 1.333
1.333 = 3/(apparent depth)
Apparent depth = 3/1.3333 = 2.25 m.
Hence the bottom of the pool appears to be 2.25 m below the ground level.
b) Real depth = 1.5 m
Apparent depth = ?
Refractive index, η = 1.333
1.3333 = 1.5/(apparent depth)
Apparent depth = 1.5/1.3333 = 1.125 m
But the pool is half filled with water, there is a 1.5 m depth on top of the pool before refraction starts.
So, apparent depth of the pool = 1.5 + 1.125 = 2.625 m below the ground level
The apparent depth of a swimming pool is measured by considering the water's refractive index. With the pool completely filled, the bottom appears to be 2.25m deep. When halfway filled, the pool appears to be 2.625m deep.
Explanation:When light travels from a medium with a high refractive index to one with a lower refractive index, the light is refracted, or bent, making objects appear closer than they actually are. We can calculate this apparent depth by using the formula d' = d / n, where d' is the apparent depth, d is the actual depth, and n is the refractive index.
(a) If the pool is completely filled with water, for a person looking from above ground, the bottom of the pool appears to be closer than it actually is. Substituting the given values into the formula, we get the apparent depth: d' = 3.00m / 1.333 = 2.25 m below ground level.
(b) If the pool is halfway filled with water, the apparent depth of the water is calculated in the same way. However, the depth beneath the water is below the refractive index border and is not subject to refraction. Therefore, the apparent total depth of the partially filled pool is the sum of the actual depth of the air part (1.50m) and the apparent depth of the water part (1.50m / 1.333 = 1.125m). This gives us 2.625m.
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Two coils have the same number of circular turns and carry the same current. Each rotates in a magnetic field acting perpendicularly to its axis of rotation. Coil 1 has a radius of 5.6 cm and rotates in a 0.24-T field. Coil 2 rotates in a 0.44-T field. Each coil experiences the same maximum torque. What is the radius (in cm) of coil 2?
Answer:
4.14 cm.
Explanation:
Given,
For Coil 1
radius of coil, r₁ = 5.6 cm
Magnetic field, B₁ = 0.24 T
For Coil 2
radius of coil, r₂ = ?
Magnetic field, B₂ = 0.44 T
Using formula of maximum torque
[tex]\tau_{max}= NIAB[/tex]
Since both the coil experience same maximum torques
now,
[tex] NIA_1B_1 = NIA_2B_2[/tex]
[tex]A_1B_1 = A_2B_2[/tex]
[tex] r_1^2 B_1 = r_2^2 B_2[/tex]
[tex] 5.6^2\times 0.24= r_2^2\times 0.44[/tex]
[tex]r_2 = 4.14\ cm[/tex]
Radius of the coil 2 is equal to 4.14 cm.
Use the exact values you enter to make later calculations.You measure the potential difference across a capacitor at different times while it's charging and record the following results.Voltage (V) Time (s)1.4790 0.015843.0000 0.036894.5210 0.066516.0210 0.11700
The final voltage the capacitor reaches after you go get some coffee is 6.500 V.
(a) Determine the time constant from the slope.
Answer:
0.0800 is time constant for slope.
Explanation:
See attached pictures for explanation.
To find the time constant from the slope, use the formula RC = -1/slope, where R is the resistance and C is the capacitance. Calculate the slope by taking the ratio of voltage change to time change between two data points.
Explanation:To determine the time constant from the slope, we can use the formula:
RC = -1/slope
where R is the resistance in the circuit and C is the capacitance of the capacitor.
In this case, since we are only given the voltage and time data, we need to find the slope by taking the ratio of the change in voltage to the change in time between any two points.
Let's take the first and second data points:
Slope = (V2 - V1) / (t2 - t1)
= (3.0000 V - 1.4790 V) / (0.036894 s - 0.015843 s)
Now, calculate the slope:
Slope = (1.5210 V) / (0.021051 s)
Slope ≈ 72.27 V/s
Once we have the slope, we can plug it into the formula RC = -1/slope to find the time constant:
RC = -1 / (72.27 V/s)
Calculate RC:
RC ≈ -0.0138 s/V
So, the time constant (τ) is approximately 0.0138 seconds per volt (s/V). This value represents the product of resistance and capacitance in the circuit.
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You are to connect resistors R1 and R2, with R1 > R2, to a battery, first individually, then in series, and then in parallel. Rank those arrangements according to the amount of current through the battery, greatest first.
Answer:
Parallel, R2, R1, series
Explanation:
From Ohm's law, V=IR hence making current the subject of the formula then [tex]I=\frac {V}{R}[/tex]
Since R1>R2 eg 4>2 then current for individual connection of R2 will be greater than that for R1 for example, assume V is 8 then if R2 we will have 8/2=4 A but for R1 we shall have 8/4=2 A.
When in parallel, the equivalent resistance will be given by [tex]\frac {1}{R1}+\frac {1}{R2}[/tex] for example here it will be 1/4+1/2=3/4. Still taking V of 8 then I= 8/(3/4)=10.667 A
When in series connection, equivalent resistance is given by adding R1 and R2 hence using the same figures we shall have 4+2=6 hence I=8/6=1.33A
We can conclude that the arrangement of current from greatest will be parallel, R2, R1, series
According to the amount of current, the arrangement will be "Parallel, R2, R1, Series".
Parallel and Series connectionBy using Ohm's law
→ Voltage (V) = Current (I) × Resistance (R)
or,
→ I = [tex]\frac{V}{R}[/tex]
Since R1 > R2
Let, Current (V) = 8 then the resistance will be:
R2 = [tex]\frac{8}{2}[/tex] = 4 A
and,
R1 = [tex]\frac{8}{2}[/tex] = 4 A
When the resistance is in parallel connection,
= [tex]\frac{1}{R1} + \frac{1}{R2}[/tex]
= [tex]\frac{1}{4} +\frac{1}{2}[/tex]
By taking L.C.M, we get
= [tex]\frac{1+2}{4}[/tex] = [tex]\frac{3}{4}[/tex] and,
The current be:
→ I = [tex]\frac{8}{\frac{3}{4} }[/tex] = 10.667 A
When the resistance is in series connection,
= R1 + R2
= 4 + 2
= 6 and,
The current be:
→ I = [tex]\frac{8}{6}[/tex] = 1.33 A
Thus the response above is appropriate.
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A circular coil that has N = 230 turns and a radius of r = 10.0 cm lies in a magnetic field that has a magnitude of B 0 = 0.0650 T directed perpendicular to the plane of the coil. What is the magnitude of the magnetic flux Φ B through the coil?
Answer:
Φ = 0.469 Wb
Explanation:
Given,
N = 230 turns
Radius, r = 10 cm
Magnetic field, B = 0.0650 T
Magnetic flux = ?
now,
Φ = NBA
Φ = 230 x 0.0650 x π x r²
Φ = 230 x 0.0650 x π x 0.1²
Φ = 0.469 Wb
Hence, the magnetic flux is equal to Φ = 0.469 Wb
The magnitude of the magnetic flux ΦB through the coil is approximately 0.0496 T·m². when A circular coil that has N = 230 turns and a radius of r = 10.0 cm.
Given:
N = 230 turns
r = 10.0 cm = 0.1 m
B₀ = 0.0650 T
The magnetic flux through a coil can be calculated using the formula:
ΦB = B₀ × A × N,
The area of the coil, we can use the formula for the area of a circle:
A = π × r²,
Let's calculate the magnetic flux:
A = π × r² = 3.14159 × (0.1)² = 0.0314159 m²
ΦB = B₀ × A × N = 0.0650 × 0.0314159 × 230 = 0.04958735 T·m²
Therefore, the magnitude of the magnetic flux ΦB through the coil is approximately 0.0496 T·m².
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"A 0.15 kg ball moving at 40 m/s is struck by a bat. The bat reverses the ball's direction and gives it a speed of 50 m/s. What average force does the bat apply to the ball if they are in contact for 6.0 ×10 -3 s?"
Final answer:
The average force exerted by the bat on the ball is 2250 N.
Explanation:
To calculate the average force exerted by the bat on the ball, we can use the impulse-momentum principle. The impulse experienced by the ball is equal to the change in its momentum. We can calculate the initial momentum by multiplying the mass of the ball by its initial velocity, and similarly, we can calculate the final momentum using the mass and final velocity. By subtracting the initial momentum from the final momentum, we get the change in momentum. Finally, dividing the change in momentum by the time of contact gives us the average force.
Using the given values, the initial momentum of the ball is (0.15 kg) × (-40 m/s) = -6 kg·m/s, and the final momentum is (0.15 kg) × (50 m/s) = 7.5 kg·m/s. The change in momentum is 7.5 kg·m/s - (-6 kg·m/s) = 13.5 kg·m/s. Dividing this by the time of contact, 6.0 × 10^-3 s, gives us an average force of 2250 N.
3. A uniformly charged ring of radius 10.0 cm has a total T charge of 75.0 mC. Find the electric field on the axis of the ring at (a) 1.00 cm, (b) 5.00 cm, (c) 30.0 cm, and (d) 100 cm from the center of the ring.
To solve this problem we will apply the concepts related to the electric field in a ring. This concept is already standardized in the following mathematical expression, which relates the coulomb constant, the distance to the axis, the distance of the two points. Mathematically it is described as,
[tex]E = \frac{k_exq}{(x^2+r^2)^{(3/2)}} \hat{i}[/tex]
Here,
[tex]k_e[/tex] = Coulomb constant
q = Charge
x = Distance to the axis
r = Distance between the charges
Our values are given as,
[tex]r = 10.0cm = 10*10^{-2} m[/tex]
[tex]q = 75\mu C = 75*10^{-6} C[/tex]
[tex]x_a = 1.00cm[/tex]
[tex]x_b = 5.00cm[/tex]
[tex]x_c = 30cm[/tex]
[tex]x_d = 100cm[/tex]
So applying this to our 4 distances, we have
PART A)
[tex]E_a = \frac{(9*10^9)(1.00*10^{-2})(75*10^{-6})}{((1.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]
[tex]E_a = 6.64*10^6N/C \hat{i}[/tex]
PART B)
[tex]E_b = \frac{(9*10^9)(5.00*10^{-2})(75*10^{-6})}{((5.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]
[tex]E_b = 24.1*10^6N/C \hat{i}[/tex]
PART C)
[tex]E_c = \frac{(9*10^9)(30.00*10^{-2})(75*10^{-6})}{((30.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]
[tex]E_c = 6.39*10^6N/C \hat{i}[/tex]
PART D)
[tex]E_d = \frac{(9*10^9)(100.00*10^{-2})(75*10^{-6})}{((100.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]
[tex]E_d = 0.664*10^6N/C \hat{i}[/tex]
Old cannons were built on wheeled carts, both to facilitate moving the cannon and to allow the cannon to recoil when fired. When a 150 kg cannon and cart recoils at 1.5 m/s, at what velocity would a 10 kg cannonball leave the cannon?
Answer:
22.5 m/s
Explanation:
Applying Newton's third law of motion
Momentum of the cannon and cart = momentum of the cannonball
MV = mv..................... Equation 1
Where M = mass of the cannon and the cart, V = Recoil velocity of the cannon and the cart, m = mass of the cannonball, v = velocity of the cannonball
make v the subject of the equation
v = MV/m............. Equation 2
Given: M = 150 kg, V = 1.5 m/s, m = 10 kg
Substitute into equation 2
v = 150(1.5)/10
v = 22.5 m/s
Hence the cannonball leave the cannon with a velocity of 22.5 m/s
Answer:
v2 = 22.5 m/s
Explanation:
Momentum is how hard to stop or turn a moving object . Generally, momentum measures mass in motion. Momentum is a vector quantity. Mathematically,
p = mass × velocity
The total momentum of an isolated system of bodies remains constant.
mometum before = 0
mass of the canon (m1) = 150 kg
mass of the ball (m2) = 10 kg
velocity of the ball (v2) = ?
velocity of the cannon(v1) = 1.5 m/s
momentum after = momentum before
m2v2 + m1v1 = 0
10v2 = 150 × 1.5
10v2 = 225
divide both sides by 10
v2 = 225/10
v2 = 22.5 m/s
Incandescent light bulbs contain a metallic filament inside. Metallic systems have allowed energy levels in a continuous range of energy, so electrons can make transitions of any energy within that range. In our lab, we will connect a light bulb to a variable AC voltage source (a Variac), which can deliver 0-140 V to the filament. The higher the voltage, the hotter we make the temperature, and the more energy we are giving the electrons in the metal, Suppose that at 20 V the filament is a dull red color, which means that most of the photons being emitted are red. When we increase the voltage, what color of light do you expect the filament to emit?
Answer:
Since at 20V most of the photons released are red then when the voltage keeps increasing the hotter the filament will be, therefore the color of light will be bright red.
Explanation:
The higher the energy the more the electrons in the molecules of the object will be excited, and when they de-excite to their ground states they release energy in the form of infrared light. The increase in voltage and higher temperatures make the object release brighter color and sometimes at the highest temperatures +1400 degrees Celsius, the color glows hot white.
In a typical Van de Graaff linear accelerator, protons are accelerated through a potential difference of 20 MV. What is their kinetic energy if they started from rest? Give your answer in (a) eV, (b) keV,(c) MeV, (d) GeV, and (e) joules.
Answer:
a) 2 x10^7 eV
b) 2 x10^4 keV
c) 20 MeV
d) 0.02 Gev
e) 3.2 x 10^-12J
Explanation:
The potential difference = 20 x 10^6 V
The charge on the proton = 1.6 x10^-19
The work done to move the proton will be basically the proton will acquire if it accelerates.
Kinetic energy gained = ΔVq = 20 x10^6 x 1.6 x 10^-19
=3.2 x 10^-12J or 2 x10^7 eV
2 x10^7 eV = 2 x10^4 keV = 20 MeV = 0.02 Gev
Explanation:
Below is an attachment containing the solution.
A typical wall outlet voltage in the United States is 120 volts. Personal MP3 players require much smaller voltages, typically 487.0 mV 487.0 mV . If the number of turns in the primary coil is 2464 2464 , calculate the number of turns on the secondary coil of the adapter transformer.
Answer:
Number of turns on the secondary coil of the adapter transformer is 10.
Explanation:
For a transformer,
[tex]\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }[/tex]
where [tex]V_{s}[/tex] is the voltage induced in the secondary coil
[tex]V_{p}[/tex] is the voltage in the primary coil
[tex]N_{s}[/tex] is the number of turns of secondary coil
[tex]N_{p}[/tex] is the number of turns of primary coil
From the given question,
[tex]\frac{487*10^{-3} }{120}[/tex] = [tex]\frac{N_{s} }{2464}[/tex]
⇒ [tex]N_{s}[/tex] = [tex]\frac{2462*487*10^{-3} }{120}[/tex]
= 9.999733
∴ [tex]N_{s}[/tex] = 10 turns
Answer:
607,145 turns
Explanation:
Output voltage, that is secondary voltage,Es = 120 volts
Input voltage, that is primary voltage, Ep = 487/1000 = 0.487 volts
Number of turns in secondary = Ns
Number of turns in primary, Np = 2464
∴ Es/Ep = Ns/Np
Ns = Es * Np/Ep = 1`20 X 2464/0.487 = 607,145 turns ( Step up transformer)
A blood-flow meter emits a 1.1-MHz ultrasound pulse to measure the speed of blood moving directly away from the meter. The meter’s sensor detects the pulse reflected back from the blood at a frequency 21 Hz lower than the emitted frequency. Take the speed of sound in the tissues to be 1475 m/s.
At what speed, in centimeters per second, is the blood moving?
Answer:
V = 2.8cm/s
Explanation:
Please see attachment below.
This problem involves the concept of doppler effect.
Answer:
The speed at which the blood is flowing in cm/s = 1.4 cm/s
Explanation:
emitted frequency ( f ) = 1.1 * 10^6 Hz
detected frequency ( F ) = 21 Hz
speed of sound in tissues ( c ) = 1475 m/s
speed ( V ) = ?
To calculate for speed of blood flowing we apply the detected frequency formula :
F = [tex]\frac{2fV}{c}[/tex]
21 = ( 2* 1100000* V ) / 1475
therefore V = (21 * 1475) / (2 * 1100000)
V = 30975 / 2200000
V = 0.0140 m/s = 1.4 cm/sec
Two objects are moving at equal speed along a level, frictionless surface. The second object has twice the mass of the first object. They both slide up the same frictionless incline plane. Which object rises to a greater height?
a. Object 1 rises to the greater height because it weighs less.
b. Object 2 rises to the greater height because it possesses a larger amount of kinetic energy.
c. Object 2 rises to the greater height because it contains more mass.
d. Object 1 rises to the greater height because it possesses a smaller amount of kinetic energy.
e. The two objects rise to the same height.
Answer:
Option E is correct.
The two objects rise to the same height.
Explanation:
Let the mass, velocity and height the bigger object will rise to be M, V and H respectively.
Let the mass, velocity and height the object will rise to be m, v and h respectively.
Note that V = v
Using the work energy theorem
The change in kinetic energy of a body between two points is equal to the work done on the body between those two points.
Change in kinetic energy of the bigger object = (final kinetic energy) - (initial kinetic energy) = 0 - (1/2)(M)(V²) = (-MV²/2)
(The final kinetic energy = 0 J because the object comes to rest at the final point)
Work done on the bigger object = work done by all the forces acting on the body
But the only force acting on the body is the force of gravity (since the inclined plane is frictionless)
Workdone on the body = work done by the force of gravity in moving the body up a height of H = - MgH
(-MV²/2) = - MgH
H = (V²/2g)
For the small body,
Change in kinetic energy of the bigger object = (final kinetic energy) - (initial kinetic energy) = 0 - (1/2)(m)(v²) = (-mv²/2)
Workdone on the body = work done by the force of gravity in moving the body up a height of H = - mgh
(-mv²/2) = - mgh
h = (v²/2g)
Since V = v as given in the question (Both bodies have the same speeds)
H = h = (V²/2g) = (v²/2g)
Hope this Helps!!!
You create a plot of voltage (in V) vs. time (in s) for an RC circuit as the capacitor is charging, where V=V_{0} \cdot \left(1- e^{ \frac{-\left(t\right)}{RC} } \right). You curve fit the data using the inverse exponent function Y=A \cdot \left(1- e^{-\left(Cx\right)} \right)+B and LoggerPro gives the following values for A, B, and C. A = 4.211 ± 0.4211 B = 0.1699 ± 0.007211 C = 1.901 ± 0.2051 What is the time constant for and its uncertainty?
Answer:
The time constant and its uncertainty is t ± Δt = 0.526 ± 0.057 s
Explanation:
If we make a comparison we have to:
y = A*(1-e^-(C*x)) + B
If the time remains constant we have to:
t = R*C = 1/C
In this way we calculate the time constant and its uncertainty. this will be equal to:
t ± Δt = (1/1.901) ± (0.2051/1.901)*(1/1.901) = 0.526 ± 0.057 s
A typical laboratory centrifuge rotates at 3700 rpm . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation
Answer:
Explanation:
acceleration of test tube
= ω² R
= (2πn)² R
= 4π²n²R
n = no of rotation per second
= 3700 / 60
= 61.67
R = .10 m
acceleration
= 4π²n²R
= 4 x 3.14² x 61.67² x .10
= 14999 N Approx
Final answer:
The acceleration at the end of a test tube 10 cm from the axis of rotation in a centrifuge spinning at 3700 rpm is calculated to be 15051.2 m/s², demonstrating the significant centrifugal forces generated by such devices.
Explanation:
To calculate the acceleration at the end of a test tube that is 10 cm from the axis of rotation in a centrifuge spinning at 3700 rpm, we first need to convert the rotational speed to radians per second. The formula to convert revolutions per minute (rpm) to radians per second (ω in rad/s) is ω = (2π×rpm)/60. Thus, ω = (2π× 3700)/60 = 387.98 rad/s. Next, we use the formula for centripetal acceleration, a = ω²×r, where r is the radius of the circle (distance from the center of rotation to the point of interest) in meters. Given that r = 10 cm = 0.1 m, the acceleration a = (387.98)^2× 0.1 = 15051.2 m/s².
This centripetal acceleration is much larger than Earth's gravitational acceleration, indicating the extreme forces at play in a centrifuge's operation, which is crucial for its role in laboratory settings for sedimentation of materials.
At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.
Answer:
(a). The speed of the ball after collision is 2.01 m/s.
(b). The speed of the block after collision 1.11 m/s.
Explanation:
Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.
Given that,
Mass of steel block = 2.30 kg
Mass of ball = 0.500 kg
Length of cord = 50.0 cm
We need to calculate the initial speed of the ball
Using conservation of energy
[tex]\dfrac{1}{2}mv^2=mgl[/tex]
[tex]v=\sqrt{2gl}[/tex]
Put the value into the formula
[tex]u=\sqrt{2\times9.8\times50.0\times10^{-2}}[/tex]
[tex]u=3.13\ m/s[/tex]
The initial speed of the ball [tex]u_{1}=3.13\ m/s[/tex]
The initial speed of the block [tex]u_{2}=0[/tex]
(a). We need to calculate the speed of the ball after collision
Using formula of collision
[tex]v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}[/tex]
Put the value into the formula
[tex]v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13[/tex]
[tex]v_{1}=-2.01\ m/s[/tex]
Negative sign shows the opposite direction of initial direction.
(b). We need to calculate the speed of the block after collision
Using formula of collision
[tex]v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}[/tex]
Put the value into the formula
[tex]v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0[/tex]
[tex]v_{2}=1.11\ m/s[/tex]
Hence, (a). The speed of the ball after collision is 2.01 m/s.
(b). The speed of the block after collision 1.11 m/s.
A 4.9 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. The force is given by F with arrow(x) = (2.6 − x^2) N, , where x is in meters and the initial position of the block is x=0(a) What is the kinetic energy of the block as it passes through x = 2.1 m?
(b) What is the maximum kinetic energy of the block between x = 0 and x = 2.1 m?
Answer with Explanation:
We are given that
Mass of block=m=4.9 kg
Initial velocity, u=0
[tex]F=(2.6-x^2) i N[/tex]
Initial position, x=0
a.We have to find the kinetic energy of the blocks as it passes through x=2.1 m
Work done=Kinetic energy=[tex]\int_{0}^{2.1}(2.6-x^2) dx[/tex]
Kinetic energy of the block=[tex][2.6 x-\frac{x^3}{3}]^{2.1}_{0}[/tex]
Kinetic energy of the block=[tex]2.6\times 2.1-\frac{(2.1)^3}{3}-0=2.373 J[/tex]
Kinetic energy of the block=2.373 J
b.Initial kinetic energy of block=[tex]K_i=\frac{1}{2}(4.9)(0)=0[/tex]
According to work energy theorem
[tex]W=K_f-K_i[/tex]
[tex]2.373 =k_f-0[/tex]
[tex]k_f=2.373 J[/tex]
Hence, the maximum kinetic energy of the block =2.373 J
A proton moves perpendicular to a uniform magnetic field B at a speed of 1.00 x10^7 m/s and experiences an acceleration of 2.00 x10^13 m/s^2. in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field for which the magnitude of the field is a minimium
Answer:
So, Magnitude of the field (B) = 2.09 x 10^(-2) T.
It's in the negative direction since acceleration is in positive direction.
Explanation:
First of all, since acceleration is in positive x- direction, the magnetic field must be in negative y- direction.
We know that The magnitude of the Lorentz force F is; F = qvB sinθ
So, B = F/(qvsinθ)
F = ma.
Speed(v) = 1.00 x10^(7) m/s
acceleration (a) = 2.00 x10^(13) m/s^(2)
Mass of proton = 1.673 × 10^(-27) kilograms
q(elementary charge of proton) = 1.602×10^(−19)
Since right hand thumb rule, θ= 90°
So;B = [1.673 × 10^(-27) x 2.00 x10^(13)] / [ {1.602×10^(−19)} x {1.00 x10^(7)} x sin 90]
So,B = 2.09 x 10^(-2) T.
It's in the negative direction since acceleration is in positive direction.
The magnitude of the magnetic field that the proton experiences is 2.08 x 10^-4 Tesla. The direction of the magnetic field, determined by the right-hand rule, is in the negative y-direction.
Explanation:The force experienced by a charged particle moving within a magnetic field, like the proton in your question, can be calculated using the Lorentz force law: F = qvBsinθ, where F is the force, q is the charge of the particle, v is its velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. Because the proton is moving perpendicular to the field, θ = 90°, and sinθ = 1. Therefore, F = qvB. To find the magnetic field, B, we can rearrange this formula to give B = F/qv.
Now, we know from Newton's second law that F = ma, so we can replace F in our formula with the proton's mass (m, which is 1.67 x 10^-27 kg for a proton) times the given acceleration (a), yielding B = ma/qv.
Substitute the given values for acceleration, proton's charge, and velocity into our equation, we get: B = (1.67 x 10^-27 kg)(2 x 10^13 m/s^2) / (1.60 x 10^-19 C)(1 x 10^7 m/s) = 2.08 x 10^-4 Tesla (T).
The direction of the magnetic field is given by the right-hand rule as discussed in Essential Knowledge 2.D.1. If you arrange your right hand such that your thumb points in the direction of the proton's velocity (positive z-direction) and your fingers curl in the direction of the force (positive x-direction), your palm points in the direction of the magnetic field, which would be in the negative y-direction.
Learn more about Magnetic Field here:https://brainly.com/question/36936308
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