Answer:
The speed of the proton relative to the laboratory frame is 0.981c
Explanation:
Given that,
Speed of electron v= 0.90c
Speed of proton u= 0.70c
We need to calculate the speed of the proton relative to the laboratory frame
Using formula of speed
[tex]u'=\dfrac{u+v}{1+\dfrac{uv}{c^2}}[/tex]
Where, u = speed of the proton relative to the electron
v = speed of electron relative to the laboratory frame
Put the value into the formula
[tex]u'=\dfrac{0.70+0.90}{1+\dfrac{0.70\times0.90\times c^2}{c^2}}[/tex]
[tex]u'=c\dfrac{0.70+0.90}{1+(0.70\times0.90)}[/tex]
[tex]u'=0.981c[/tex]
Hence, The speed of the proton relative to the laboratory frame is 0.981c
Final answer:
The speed of the proton relative to the laboratory frame, when it moves to the right with a speed of 0.70c relative to an electron moving at 0.90c, is found to be approximately 0.98c using relativistic velocity addition.
Explanation:
To find the speed of the proton relative to the laboratory frame when an electron moves to the right with a speed of 0.90c and a proton moves to the right with a speed of 0.70c relative to the electron, we use the formula for relativistic velocity addition. This formula is given by:
V = (v + u) / (1 + vu/c²),
where V is the velocity of the proton relative to the lab, v is the velocity of the electron relative to the lab (0.90c), u is the velocity of the proton relative to the electron (0.70c), and c is the speed of light. Plugging in these values, we get:
V = (0.90c + 0.70c) / (1 + (0.90*0.70)) = 0.98c.
Therefore, the speed of the proton relative to the laboratory frame is approximately 0.98c, close to the speed of light, demonstrating the relativistic effects when velocities approach the speed of light.
Logan observes a paramecium under a microscope. The eyepiece of the microscope has a horizontal scale marked in mm. The paramecium starts at the 65 mm mark and ends up at the 39 mm mark.
What is the paramecium's displacement? Express your answer with appropriate units.
Answer:
d = -26 mm
Explanation:
given,
initial position of paramecium, x = 65 mm
final position of paramecium, y = 39 mm
displacement of the paramecium's = ?
displacement = final position - initial position
d = y - x
d = 39 - 65
d = -26 mm
The paramecium's displacement comes out to be -26 mm
Very short pulses of high-intensity laser beams are used to repair detached portions of the retina of the eye. The brief pulses of energy absorbed by the retina welds the detached portion back into place. In one such procedure, a laser beam has a wavelength of 810 nm and delivers 250 mW of power spread over a circular spot 510 μm in diameter. The vitreous humor (the transparent fluid that fills most of the eye) has an index of refraction of 1.34.
A) If the laser pulses are each 1.50 ms long, how much energy is delivered to the retina with each pulse? Express your answer with the appropriate units
B) What average pressure would the pulse of the laser beam exert at normal incidence on a surface in air if the beam is fully absorbed? Express your answer with the appropriate units.
C) What are the wavelength of the laser light inside the vitreous humor of the eye? Express your answer with the appropriate units.
D) What is the frequency of the laser light inside the vitreous humor of the eye? Express your answer with the appropriate units.
E) What is the maximum value of the electric field in the laser beam? Express your answer with the appropriate units.
F) What is the maximum value of the magnetic field in the laser beam?
The procedures involves using short pulses of high intensity laser beams to repair detached parts of the retina. The energy delivered per pulse, average pressure exerted by laser, new wavelength and frequency of laser light inside the eye, and the maximum electric and magnetic field in the laser beam are calculated. The details of these calculations have been given.
Explanation:A) The energy delivered to the retina with each pulse can be calculated using the formula Energy = Power x Time. Here, power is 250mW or 0.250 joules/second and time is 1.50 ms or 0.00150 seconds. So, the energy is 0.250 joules/second x 0.00150 seconds = 0.000375 joules or 375 μJ per pulse.
B) The average pressure exerted by the laser beam can be calculated using the formula Pressure = Power/Area where the area of the laser spot is the area of a circle of diameter 510 μm. The laser light is completely absorbed upon hitting a surface, therefore the pressure the beam exerts is 2* Power/Area of beam.
C) Inside the vitreous humor of the eye, the wavelength of the laser light decreases as it is inversely proportional to the refractive index of the medium. So the new wavelength in the vitreous humor is 810 nm/ 1.34 = 604.48 nm.
D) The frequency of light doesn't change when it enters a different medium so it remains the same in the vitreous humor as it was when in air. It can be calculated using the formula Frequency = Speed of Light / Wavelength.
E) The maximum electric field in the laser beam can be computed using the formula for energy density.
F) The maximum value of the magnetic field in the laser beam is proportional to the maximum electric field.
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The energy delivered to the retina per pulse is 0.375 mJ. The pressure exerted is approximately 3.68 Pa, and the wavelength inside the vitreous humor is 604.5 nm. The frequency is 3.70 × 10¹⁴ Hz, with maximum electric and magnetic field values of 1.53 × 10⁵ V/m and 5.10 × 10⁻⁴ T, respectively.
A) Energy Delivered to the Retina Per Pulse
The power of the laser beam is 250 mW, which is 0.250 W. The duration of each pulse is 1.50 ms, which is 1.50 × 10⁻³ s.
Calculate the energy delivered during each pulse using the formula:
Energy (J) = Power (W) × Time (s).
Energy = 0.250 W × 1.50 × 10⁻³ s
= 3.75 × 10⁻⁴ J (or 0.375 mJ).
B) Average Pressure Exerted by the Pulse
If the beam is fully absorbed, the pressure exerted is given by:
Pressure (P) = Power (P) / Area (A) × c,
where c is the speed of light (approximately 3 × 10⁸ m/s).
Calculate the area of the circular spot:
Area (A) = π (d/2)²
where diameter
d = 510 μm = 510 × 10⁻⁶ m.
Area = π (510 × 10⁻⁶ / 2)²
≈ 2.04 × 10⁻⁷ m².
P = 0.250 W / (2.04 × 10⁻⁷ m²) × 3 × 10⁸ m/s
≈ 3.68 N/m² (Pa).
C) Wavelength of Laser Light Inside the Vitreous Humor
The wavelength inside the medium is given by: λ' = λ / n, where λ is the wavelength in air (810 nm) and n is the refractive index (1.34).
λ' = 810 nm / 1.34
≈ 604.5 nm.
D) Frequency of the Laser Light Inside the Vitreous Humor
The frequency of light remains constant through different media and is given by:
f = c / λ,
where c is the speed of light
λ is the wavelength in air.
Frequency f = 3 × 10⁸ m/s / 810 × 10⁻⁹ m
≈ 3.70 × 10¹⁴ Hz.
E) Maximum Value of the Electric Field in the Laser Beam
The maximum electric field E₀ is given by:
E₀ = √(2I/ε₀c),
where I is the intensity,
ε₀ is the permittivity of free space,
c is the speed of light.
Intensity I = P / A
= 0.250 W / 2.04 × 10⁻⁷ m²
≈ 1.23 × 10⁶ W/m².
Using ε₀ = 8.85 × 10⁻¹² F/m: E₀
≈ √(2 × 1.23 × 10⁶ W/m² / 8.85 × 10⁻¹² F/m × 3 × 10⁸ m/s)
≈ 1.53 × 10⁵ V/m.
F) Maximum Value of the Magnetic Field in the Laser Beam
The magnetic field B₀ is related to the electric field E₀ by:
B₀ = E₀ / c.
B₀ = 1.53 × 10⁵ V/m / 3 × 10⁸ m/s
≈ 5.10 × 10⁻⁴ T.
These calculations help us understand how laser pulses interact with biological tissues, aiding in precision medical procedures.
Therefore, the energy delivered to the retina per pulse is 0.375 mJ. The pressure exerted is approximately 3.68 Pa, and the wavelength inside the vitreous humor is 604.5 nm. The frequency is 3.70 × 10¹⁴ Hz, with maximum electric and magnetic field values of 1.53 × 10⁵ V/m and 5.10 × 10⁻⁴ T, respectively.
A car starts from rest at a stop sign. It accelerates at 3.8 m/s 2 m/s2 for 6.0 s, coasts for 1.6 ss , and then slows down at a rate of 3.3 m/s 2 m/s2 for the next stop sign.
How far apart are the stop signs?
To solve this problem we will start by calculating the distance traveled while relating the first acceleration in the given time. From that acceleration we will calculate its final speed with which we will calculate the distance traveled in the second segment. With this speed and the acceleration given, we will proceed to calculate the last leg of its route.
Expression for the first distance is
[tex]s_1 = ut +\frac{1}{2} at^2[/tex]
[tex]s_1 = 0+\frac{1}{2} (3.8)(6)^2[/tex]
[tex]s_1 = 68.4m[/tex]
The expression for the final speed is
[tex]v = v_0 +at[/tex]
[tex]v = 0+(3.8)(6)[/tex]
[tex]v = 22.8m/s[/tex]
Then the distance becomes as follows
[tex]s_2 = vt[/tex]
[tex]s_2 = (22.8)(1.6)[/tex]
[tex]s_2 = 36.48m[/tex]
The expression for the distance at last sop is
[tex]v_1^2=v_0^2 +2as_3[/tex]
[tex]22.8^2 = 0+2(3.3)s_3[/tex]
[tex]s_3 =78.7636m[/tex]
Therefore the required distance between the signs is,
[tex]S = s_1+s_2+s_3[/tex]
[tex]S = 68.4+36.48+78.76[/tex]
[tex]S = 183.64m[/tex]
Therefore the total distance between signs is 183.54m
A 0.46-kg cord is stretched between two supports, 7.2 m apart. When one support is struck by a hammer, a transverse wave travels down the cord and reaches the other support in 0.74 s.
What is the tension in the cord? (Express your answer to two significant figures and include the appropriate units.
Answer:
T = 6.0 N
Explanation:
given,
mass of the cord = 0.46 Kg
length of the supports = 7.2 m
time taken to travel = 0.74 s
tension in the chord = ?
using formula for tension calculation
[tex]T = \dfrac{v^2.m}{l}[/tex]
[tex]v = \dfrac{l}{s}[/tex]
[tex]v = \dfrac{7.2}{0.74}[/tex]
v = 9.73 m/s
now, calculation of tension
[tex]T = \dfrac{9.73^2\times 0.46}{7.2}[/tex]
T = 6.0 N
The tension in the cord is equal to 6.0 N.
When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to make this force twice as strong, the distance would have to be changed to
A) √2d
B) d/√2
C) d/4
D) 2d
E) d/2
Answer:b
Explanation:
Given
Force of attraction is F when charges are d distance apart.
Electrostatic force is given by
[tex]F=\frac{kq_1q_2}{d^2}---1[/tex]
where k=constant
[tex]q_1[/tex] and [tex]q_2[/tex] are charges
d=distance between them
In order to double the force i.e. 2F
[tex]2F=\frac{kq_1q_2}{d'^2}----2[/tex]
divide 1 and 2 we get
[tex]\frac{F}{2F}=\frac{d'^2}{d^2}[/tex]
[tex]d'=\frac{d}{\sqrt{2}}[/tex]
The energy required to dissociate KF into neutral atoms is 498 kJ/mol. Given that the first ionization energy for K is 418 kJ/mol, calculate the electron affinity (in kJ/mol) for F. Show your work for all calculations.
Answer: Electron affinity of F equals
275.8kJ/mol
Explanation: Electron affinity is the energy change when an atom gains an electron.
Let's first calculate the energy required -E(r)to dissociate KF into ions not neutral atom which is given.
E(r) = {z1*z2*e²}/{4π*permitivity of space*r}
z1 is -1 for flourine
z2 is +1 for potassium
e is magnitude of charge 1.602*EXP{-9}C
r is ionic bond length of KF(is a constant for KF 0.217nm)
permitivity of free space 8.854*EXP{-12}.
Now let's solve
E(r)= {(-1)*(1)*(1.602*EXP{-9})²} /
{4*3.142*8.854*EXP(-12)*0.217*EXP(-9)
E(r) = - 1.063*EXP{-18}J
But the energy is released out that is exothermic so we find - E(r)
Which is +1.603*EXP{-18}J
Let's now convert this into kJ/mol
By multiplying by Avogadro constant 6.022*EXP(23) for the mole and diving by 1000 for the kilo
So we have,
1.603*EXP(-18) *6.022*EXP(23)/1000
-E(r) = 640.2kJ/mol.
Now let's obtain our electron affinity for F
We use this equation
Energy of dissociation (nuetral atom)= electron affinity of F +(-E(r)) + ionization energy of K.
498kJ/mol
=e affinity of F + 640.2kJ/mol
+(-418kJ/mol)
(Notice the negative sign in ionization energy for K. since it ionize by losing an electron)
Making electron affinity of F subject of formula we have
Electron affinity (F)=498+418-640.2
=275.8kJ/mol.
A 3.0 cm × 3.0 cm parallel-plate capacitor has a 3.0 mm spacing. The electric field strength inside the capacitor is 1.2×105 V/m . What is the potential difference across the capacitor? How much charge is on each plate?
To solve this problem it is necessary to apply the concepts related to the voltage depending on the electric field and the distance, as well as the load depending on the capacitance and the voltage. For the first part we will use the first mentioned relationship, for the second part, we will not only define the load as the capacitance by the voltage but also place it in terms of the Area, the permittivity in free space, the voltage and the distance.
PART A ) Voltage in function of electric field and distance can be defined as,
[tex]V = Ed[/tex]
Our values are,
[tex]E = 1.2*10^5 V/m[/tex]
[tex]d = 3.0mm = 3*10^{-3}[/tex]
Replacing,
[tex]V = (1.2*10^5)(3*10^{-3})[/tex]
[tex]V = 360v[/tex]
Therefore the potential difference across the capacitor is 360V
PART B) The charge can be defined as,
[tex]Q = CV = \frac{\epsilon AV}{d}[/tex]
Here,
[tex]\epsilon = 8.85*10^{-12} F/m[/tex], Permittivity of free space
[tex]A = s^2[/tex], area of each capacitor plate
s = Length of capacitor plate
Replacing,
[tex]Q = \frac{\epsilon AV}{d}[/tex]
[tex]Q = \frac{(8.85*10^{-12})(0.03)^2(240)}{2.0*10^{-8}m}[/tex]
[tex]Q = 9.558*10^{-10}C[/tex]
Therefore the charge on each plate is [tex]9.558*10^{-10}C[/tex]
The potential difference across the parallel-plate capacitor is 360 V. The charge on each plate of the capacitor is approximately 0.95 x 10^-8 C.
The potential difference across a parallel-plate capacitor is calculated using the formula V = Ed, where E is the electric field strength and d is the distance (or spacing) between the plates. As given, E is 1.2 x 10^5 V/m, and d is 3.0 mm (or 3.0 x 10^-3 m). Therefore, the potential difference V across the plates is given by V = 1.2 x 10^5 V/m * 3.0 x 10^-3 m = 360 V.
The amount of charge Q on each plate of the capacitor can be found using the formula Q = εEA, where ε is the permittivity of free space (ε = 8.85 x 10^-12 F/m), E is the electric field strength, and A is the area of the plate. Substituting the values given, we have A = 3.0 cm * 3.0 cm = 9 cm^2 = 9 x 10^-4 m^2, E = 1.2 x 10^5 V/m, and ε = 8.85 x 10^-12 F/m. Therefore, Q = εEA = 8.85 x 10^-12 F/m * 1.2 x 10^5 V/m * 9 x 10^-4 m^2 ≈ 0.95 x 10^-8 C.
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A large centrifuge is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries. At what angular velocity is the centripetal acceleration 10 g if the rider is 15.0 m from the center of rotation
Answer:
ω = 2.55 rad/sec
Explanation:
Assuming no other external forces acting in the horizontal plane, the only force keeping the rider in a circular path of a radius equal to his distance to the center of rotation, is the centripetal force.
According to Newton's 2nd law, in the horizontal direction, we have:
F = Fc = m*a = m*ω²*r
We know that a = ac = 10*g = 98.0 m/s², and that r = 15.0 m.
Replacing these values in (1), and solving for ω, we get:
ω = √98.0m/s²/ 15.0 m = 2.55 rad/sec
The centrifugal force in a rocket.
The centrifugal force is the force that is related to the outwards or away from the body. The larger force is used to expose the aspiring astronomers to accelerations that are the same as those experienced by the rockets that are launched and the air reentries.
Thus the answer is ω = 2.55 rad/sec
The centrifugal force is used to measure the acceleration of the astronauts when they are launched in the air or atmosphere. The angular velocity of the centripetal force that accelerates to 10g if the rider is 15.0 meters from the center.Taking no external forces acting on the plane,As per the Newton's 2nd law, the formulae.F= Fc = m*a = m*ω²*r know that a = ac = 10*g = 98.0 m/s², and that r = 15.0 m. On replacing these values in (1), and solving for ω, we get: ω equal to 2.55 rad/sec.Learn more about the centrifuge is used.
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A -4.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.
1. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.
2. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 1.20 m.
To find the net electric force on an electron due to two point charges at the origin and at x=0.8m, one has to first calculate the electric field due to each charge at the location of the electron. After finding the total electric field it's multiplied with the charge of the electron to obtain the force.
Explanation:The subject of this question is Physics, specifically the concept of Electric Force. In this problem, we need to find the net force on an electron located on the x-axis at two different points because of two point charges at the origin and at x=0.8m.
Step 1: Calculate the electric field at the given point due to each point charge by using the formula E= KQ/r² where k = 9 x 10⁹ N m²/C²(Q is the charge and r is the distance from the charge to the point in question).
Step 2: Once the electric field due to each charge is found, sum these together to get the total electric field at the point.
Step 3: Electric force that an electron experiences in the field can be found using the electric field (E) and the charge of an electron (e) by the formula F= eE where e is 1.602 x 10⁻¹⁹ Coulomb.
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The net electric force on an electron at x = 0.200 m is [tex]1.66 * 10^{-17}[/tex] N towards the origin, due to both charges. At x = 1.20 m, the net force is [tex]6.95 * 10^{-18}[/tex] N towards q1, away from the origin.
A -4.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.
1. Find the net electric force that the two charges would exert on an electron placed at a point on the x-axis at x = 0.200 m.
To find the net electric force on an electron at x = 0.200 m, we will use Coulomb's Law:
[tex]F = (k * |q_1 * q_2| )/ r^2[/tex]
Here, [tex]q_1[/tex] and [tex]q_2[/tex] are the charges, r is the distance between them, and k is Coulomb's constant,[tex]8.99 * 10^9 N.m^2/C^2.[/tex]
The electron at x = 0.200 m is 0.200 m from q1 and 0.600 m from [tex]q_2[/tex]. Both forces are attractive since all charges are negative.
Force due to [tex]q_1[/tex]:
[tex]F_1 = k * |e * q_1| / (0.200)^2 = 8.99 * 10^9 N.m^2/C^2 * |-1.6 x 10^-{19} C * -4.00 * 10^{-9} C| / (0.200 m)^2 = 1.44 * 10^{-17} N[/tex]
Force due to [tex]q_2[/tex]:
[tex]F_2 = k * |e * q_2| / (0.600)^2 = 8.99 * 10^9 N.m^2/C^2 * |-1.6 x 10^{-19} C * -5.50 x 10^{-9} C| / (0.600 m)^2 = 2.20 * 10^{-18} N[/tex]
The net force will be the sum of these forces, taking directions into account.
Net force = F1 (towards -x) + F2 (towards -x) [tex]= 1.44 * 10^{-17} N + 2.20 * 10^{-18} N = 1.66 * 10^{-17} N[/tex]
This force is towards the origin.
2. Find the net electric force that the two charges would exert on an electron placed at a point on the x-axis at x = 1.20 m.
We repeat the procedure for the electron at x = 1.20 m:
The electron at x = 1.20 m is 1.20 m from q1 and 0.400 m from q2.
Force due to q1:
[tex]F1 = k * |e * q1| / (1.20)^2 = 8.99 * 10^9 N.m^2/C^2 * |-1.6 x 10^{-19} C * -4.00 * 10^-9 C| / (1.20 m)^2 = 2.00 * 10^{-18} N[/tex]
Force due to q2:
[tex]F2 = k * |e * q2| / (0.400)^2 = 8.99 * 10^9 N.m^2/C^2 * |-1.6 x 10^{-19} C * -5.50 * 10^{-9} C| / (0.400 m)^2 = 4.95 * 10^{-18} N[/tex]
The net force is:
Net force = F1 (towards +x) - F2 (towards -x) [tex]= 2.00 * 10^{-18} N + 4.95 * 10^{-18} N = 6.95 * 10^{-18}[/tex] N towards q1, away from the origin
What must be the distance between point charge q1 = 28.0 μC and point charge q2 = −57.0 μC for the electrostatic force between them to have a magnitude of 5.05 N?
Answer:
1.686 m
Explanation:
From coulomb's law,
F = kq1q2/r² ...................................... Equation 1
Where F = electrostatic force between the two charges, q1 = first charge, q2 = second charge, r = distance between the charges.
making r the subject of the equation,
r = √(kq1q2/F).......................... Equation 2
Given: F = 5.05 N, q1 = 28.0 μC = 28×10⁻⁶ C, q2 = 57.0 μC = 57.0×10⁻⁶ C
Constant: k = 9.0×10⁹ Nm²/C².
Substituting into equation 2
r = √(9.0×10⁹×28×10⁻⁶×57.0×10⁻⁶/5.05)
r = √(14364×10⁻³/5.05)
r = √(14.364/5.05)
r = √2.844
r = 1.686 m
r = 1.686 m.
Thus the distance must be 1.686 m
A horizontal disk with a radius of 10 cm rotates about a vertical axis through its center. The disk starts from rest at t = 0 and has a constant angular acceleration of 2.1 rad/s^2. At what value of t will the radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude?
Answer:
0.69s
Explanation:
10 cm = 0.1 m
Let t be the time that radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude. At that time we have the angular velocity would be
[tex]\omega = \alpha t = 2.1 t[/tex]
And so the radial acceleration is
[tex]a_r = \omega^2 r = (2.1t)^2 r = 2.1^2 t^2 * 0.1= 0.441 t^2 m/s^2[/tex]
The tangential acceleration is always the same since angular acceleration is constant:
[tex]a_t = \alpha * r = 2.1 * 0.1 = 0.21 m/s^2[/tex]
For these 2 quantities to be the same
[tex]a_r = a_t[/tex]
[tex]0.441 t^2 = 0.21[/tex]
[tex]t^2 = 0.21/0.441 = 0.4762[/tex]
[tex]t = \sqrt{0.4762} = 0.69 s[/tex]
The value of t whereby the radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude is; t = 0.69 s
Calculating Radial and Tangential Acceleration
Let the time that radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude be denoted as t. Thus, angular velocity at time (t) is;
ω = αt
where;
α is angular acceleration
We are given;
radius; r = 10 cm = 0.1 mconstant angular acceleration; α = 2.1 rad/s²Thus, ω = 2.1t
Now, we can find the radial acceleration from the formula;
α_r = ω²r
Thus;
α_r = (2.1t)² × 0.1
α_r = 0.441 t² m/s²
The tangential acceleration is gotten from the formula is;
α_t = α × r
α_t = 2.1 × 0.1
α_t = 0.21 m/s²
The condition in the question implies that the tangential acceleration is equal to the radial acceleration. Thus;
α_t = α_r
0.21 = 0.441 t²
t = √(0.21/0.441)
t = 0.69 s
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An eight-turn coil encloses an elliptical area having a major axis of 40.0 cm and a minor axis of 30.0 cm. The coil lies in the plane of the page and carries a clockwise current of 6.20 A. If the coil is in a uniform magnetic field of 1.98 10-4 T directed toward the left of the page, what is the magnitude of the torque on the coil? Hint: The area of an ellipse is A = ?ab, where a and b are, respectively, the semimajor and semiminor axes of the ellipse.
Answer:
9.25 x 10^-4 Nm
Explanation:
number of turns, N = 8
major axis = 40 cm
semi major axis, a = 20 cm = 0.2 m
minor axis = 30 cm
semi minor axis, b = 15 cm = 0.15 m
current, i = 6.2 A
Magnetic field, B = 1.98 x 10^-4 T
Angle between the normal and the magnetic field is 90°.
Torque is given by
τ = N i A B SinФ
Where, A be the area of the coil.
Area of ellipse, A = π ab = 3.14 x 0.20 x 0.15 = 0.0942 m²
τ = 8 x 6.20 x 0.0942 x 1.98 x 10^-4 x Sin 90°
τ = 9.25 x 10^-4 Nm
thus, the torque is 9.25 x 10^-4 Nm.
The magnitude of the torque on the coil is 9.25 x 10⁻⁴ Nm
Torque:According to the question we have the following data:
Number of turns of the coil N = 8
Semi-major axis of the ellipse a = 40/2 cm = 0.2 m
Semi-minor axis, b = 30/2 cm = 0.15 m
Current in the coil, i = 6.2 A
Magnetic field, B = 1.98 x 10⁻⁴ T
The angle between the normal and the magnetic field is 90°.
So the torque on the coil is given by:
τ = NiABsinθ
Now, the area of ellipse:
A = πab
A = 3.14 x 0.20 x 0.15 = 0.0942 m²
Thus,
τ = 8 x 6.20 x 0.0942 x 1.98 x 10⁻⁴ x sin 90°
τ = 9.25 x 10⁻⁴ Nm
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If He gas has an average kinetic energy of 5930 J/mol under certain conditions, what is the root mean square speed of F2 gas molecules under the same conditions
To solve this problem we will apply the concept related to kinetic energy based on the ideal gas constant and temperature. From there and with the given values we will find the temperature of the system. As the temperature is the same it will be possible to apply the root mean square speed formula that is dependent on the element's molar mass, the ideal gas constant and the temperature, this would be:
[tex]KE = \frac{3}{2} RT[/tex]
Where,
KE = Average kinetic energy of an ideal gas
[tex]R = 8.314JK^{-1}mol^{-1}[/tex]= Ideal gas constant
T = Temperature
Replacing we have,
[tex]KE = \frac{3}{2} RT[/tex]
[tex]5930J/mol = \frac{3}{2}(8.314JK^{-1}mol^{-1})T[/tex]
[tex]T = 475.503K[/tex]
Therefore the temperature is 475.5K
RMS velocity of [tex]F_2[/tex] gas is
[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]
Where,
M = Molar mass of [tex]F_2[/tex]
[tex]M = 38.00g/mol[/tex]
[tex]M = 38.00*10^{-3} kg/mol[/tex]
[tex]T = 475.5K[/tex]
[tex]R = 8.314JK^{-1}mol^{-1}[/tex]
Replacing we have,
[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]
[tex]v_{rms} = \sqrt{\frac{3(8.314JK^{-1}mol^{-1})(475.5K )}{38.00*10^{-3} kg/mol}}[/tex]
[tex]v_{rms} = 558.662m/s[/tex]
Therefore, the RMS velocity of [tex]F_2[/tex] gas is 558.6m/s
An oscillator creates periodic waves on two strings made ofthe same material. The tension is the same in both strings.If the strings have different thicknesses,which of the following parameters, if any, will be different in thetwo strings?Check all that apply.a. wave frequencyb. wave speedc. wavelengthd. none of the above
The wave speed and wavelength will differ across strings of different thicknesses held at the same tension; the wave frequency remains the same.
When two strings of different thicknesses are made of the same material and held at the same tension, the wave speed and wavelength will differ, whereas the wave frequency will remain the same. This is because the speed of a wave on a string is determined by the tension (T) and the linear mass density (μ), where v = (T/μ)¹/2; as the tension is constant and the material is the same, it is the difference in thickness (hence, different densities) that causes a variance in wave speed. Since wave frequency (f) is related to the speed (v) and wavelength (λ) by the equation v = fλ, and the oscillator creates waves at a fixed frequency, a change in wave speed inherently impacts the wavelength.
You are given vectors Aâ = 4.8 i^â 7.0 j^ and Bâ = - 4.0 i^+ 7.3 j^. A third vector Câ lies in the xy-plane. Vector Câ is perpendicular to vector Aâ and the scalar product of Câ with Bâ is 14.0. Find the X and Y components of vector C
Answer:
Vector C = -1.56i^ +1.07j^
This question requires that we use the properties of the scalar product of two vectors to find the required x and y component.
The dot product of two perpendicular vectors is equal to and the product of two vectors that are not parallel is equal to a nonzero value.
These are the properties that have been used in solving this problem alongside solving the simultaneous questions generator.
Explanation:
The full solution can be found in the attachment below.
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where in your environment can you observe some of the general properties of wave motion
Answer:
Air, water, rock and soil
Explanation:
The environment where the general properties of wave motion can be observed is the wave medium which is any substance or particle that carries the wave, or through which the wave travels.
ocean waves are carried by watersound waves are carried by airthe seismic waves of an earthquake are carried by rock and soilThe system below uses massless pulleys and ropes. The coefficient of friction is μ. Assume that M1 and M2 are sliding. Gravity is directed downwards. Find the tension in the rope.
Explanation:
Using Newtons second law on each block
F = m*a
Block 1
[tex]T_{1} - u*g*M_{1} = M_{1} *a \\\\T_{1} = M_{1}*(a + u*g) ... Eq1[/tex]
Block 2
[tex]T_{2} - u*g*M_{2} = M_{2} *a \\\\T_{2} = M_{2}*(a + u*g) ... Eq2[/tex]
Block 3
[tex]- (T_{1} + T_{2} ) + g*M_{3} = M_{3} *a \\\\T_{1} + T_{2} = M_{3}*( -a + g) ... Eq3[/tex]
Solving Eq1,2,3 simultaneously
Divide 1 and 2
[tex]\frac{T_{1} }{T_{2}} = \frac{M_{1}*(a+u*g)}{M_{2}*(a+u*g)} \\\\\frac{T_{1} }{T_{2}} = \frac{M_{1} }{M_{2} }\\\\ T_{1} = \frac{M_{1} *T_{2} }{M_{2} } .... Eq4[/tex]
Put Eq 4 into Eq3
[tex]T_{2} = \frac{M_{3}*(g-a) }{1+\frac{M_{1} }{M_{2} } } ...Eq5[/tex]
Put Eq 5 into Eq2 and solve for a
[tex]a = \frac{M_{3}*g -u*g*(M_{1} + M_{2}) }{M_{1} + M_{2} + M_{3} } .... Eq6[/tex]
Substitute back in Eq2 and use Eq4 and solve for T2 & T1
[tex]T_{2} = M_{2}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\T_{1} = M_{1}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\[/tex]
A liquid has a specific gravity of 1.1 at room temperature. What is its (a) Density at room temperature in kg/m3 (b) Specific volume at room temperature in ft3/lbm? (c) If the liquid is placed in a 2 L bottle that has a mass of 157 g, how much will the full bottle weigh?
To solve this problem we will proceed to find the density from the specific gravity. Later we will find the specific volume as the inverse of the density. Finally with the data obtained we will find the total weight in the bottle.
a) [tex]\rho = \gamma * 1000[/tex]
Here,
[tex]\rho[/tex] = Density
[tex]\gamma[/tex] = Specific gravity
[tex]\rho = 1.1 * 1000[/tex]
[tex]\rho = 1100 kg/m3[/tex]
b)
[tex]\text{Specific volume}= \frac{1}{\rho}[/tex]
[tex]\upsilon = \frac{1}{1100}[/tex]
[tex]\upsilon = 0.00090909 m^3/kg[/tex]
From the equivalences of meters to feet and kilograms to pounds, we have to
[tex]1m = 3.280839895 ft[/tex]
[tex]1 kg = 2.2046 lbm[/tex]
Converting the previous value to British units:
[tex]\upsilon = 0.00090909 m^3/kg (\frac{3.280839895^3 ft^3}{1m^3} )(\frac{1kg}{2.2046 lbm})[/tex]
[tex]\upsilon= 0.0145757 ft^3 / lbm[/tex]
c)
[tex]V = 2*10^{-3} m^3[/tex]
Mass of the liquid in bottle is
[tex]m = V\rho[/tex]
[tex]m= (2*10^{-3} m^3 )(1100kg/m^3)[/tex]
[tex]m = 2.2kg = 2200g[/tex]
Therefore the Total weight
[tex]W= 157 + 2200 = 2357 g[/tex]
An electric dipole is in a uniform electric field of magnitude 8.50×104N/C. The charges in the dipole are separated by 1.10×10−10m, and the torque on the dipole when its dipole moment is perpendicular to the electric field is 6.60×10−26N⋅m. Calculate the magnitude of the charges that make up the dipole.
Answer: q = 7.06 × 10^-21 C
the magnitude of the charges that make up the dipole is 7.06 × 10^-21 C
Explanation:
Given:
Torque, τ = 6.60×10^−26N⋅m
Angle made by p with a uniform electric field, θ = 90° (perpendicular)
Electric field, E = 8.50×10^4N/C
Length between dipole r = 1.10×10^−10 m
Torque acting on the dipole is given by the relation,
τ = pE sinθ....1
But,
p = qr .....2
Substituting equation 1 to 2
τ= qrEsinθ ....3
Making q the subject of formula
q = τ/rEsinθ .....4
Where;
q = magnitude of the charges that make up the dipole.
Substituting the given values into equation 4:
q = 6.60×10^−26N⋅m/(1.10×10^−10 m × 8.50×10^4N/C × sin90°)
q = 0.70588 × 10^-20 C
q = 7.06 × 10^-21 C
The magnitude of the charge making up the electric dipole, given a torque of 6.60×10−26N⋅m when the dipole moment is perpendicular to an electric field of 8.50×104N/C, and a separation of 1.10×10−10m between the charges, is approximately 7.05×10−21 C.
Explanation:The torque (τ) on a dipole in a uniform electric field is given by the equation τ = pEsinθ, where p is the dipole moment, E is the electric field strength, and θ is the angle between the dipole moment and the electric field. In this case, the dipole moment is perpendicular to the electric field, so θ= 90 degrees, and sinθ= 1. The dipole moment, p, is the product of the magnitude of the charge (q) and the separation (d) between the charges, so p = qd.
Given τ = 6.60×10−26N⋅m, E = 8.50×104N/C, and d = 1.10×10−10m, we can first use the torque equation to find the dipole moment: p = τ / E = 6.60×10−26N⋅m / 8.50×104N/C = 7.76×10−31 C⋅m. Then, use p = qd to calculate the charge: q = p / d = 7.76×10−31 C⋅m / 1.10×10−10m = 7.05×10−21 C.
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Suppose that F3 = 310 N Determine the magnitude of the resultant force F′=F2+F3.
Answer:
F' = 169.45N
This is a vector addition involving two vectors. In order to do this correctly, we need to resolve each of those forces into their vertical and horizon components and sum them up accordingly (all vertical components summed together and all horizontal components summed together). Then the magnitude of the summation is found by taking the square root of the sun of the squares of the summations along the vertical and the horizontal.
Explanation:
See the attachment below for the full solution to the problem.
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A sinusoidal voltage is given by the expression ????(????)=20cos(5π×103 ????+60°) V. Determine its (a) frequency in hertz, (b) period in milliseconds, (c) amplitude in volts, and (d) phase angle in degrees
There are some placeholders in the expression, but they can be safely assumed
Answer:
(a) [tex]f=1617.9\ Hz[/tex]
(b) [tex]T=0.618\ ms[/tex]
(c) [tex]A=20 \ Volts[/tex]
(d) [tex]\varphi=60^o[/tex]
Explanation:
Sinusoidal Waves
An oscillating wave can be expressed as a sinusoidal function as follows
[tex]V(t)&=A\cdot \sin(2\pi ft+\varphi )[/tex]
Where
[tex]A=Amplitude[/tex]
[tex]f=frequency[/tex]
[tex]\varphi=Phase\ angle[/tex]
The voltage of the question is the sinusoid expression
[tex]V(t)=20cos(5\pi\times 103t+60^o)[/tex]
(a) By comparing with the general formula we have
[tex]f=5\pi\times 103=1617.9\ Hz[/tex]
[tex]\boxed{f=1617.9\ Hz}[/tex]
(b) The period is the reciprocal of the frequency:
[tex]\displaystyle T=\frac{1}{f}[/tex]
[tex]\displaystyle T=\frac{1}{1617.9\ Hz}=0.000618\ sec[/tex]
Converting to milliseconds
[tex]\boxed{T=0.618\ ms}[/tex]
(c) The amplitude is
[tex]\boxed{A=20 \ Volts}[/tex]
(d) Phase angle:
[tex]\boxed{\varphi=60^o}[/tex]
Light from a helium-neon laser (λ=633nm) passes through a circular aperture and is observed on a screen 4.0 m behind the aperture. The width of the central maximum is 2.5 cm .What is the diameter (in mm) of the hole?
Answer:
d = 0.247 mm
Explanation:
given,
λ = 633 nm
distance from the hole to the screen = L = 4 m
width of the central maximum = 2.5 cm
2 y = 0.025 m
y = 0.0125 m
For circular aperture
[tex]sin \theta = 1.22\dfrac{\lambda}{d}[/tex]
using small angle approximation
[tex]\theta = \dfrac{y}{D}[/tex]
now,
[tex]\dfrac{y}{D} = 1.22\dfrac{\lambda}{d}[/tex]
[tex]y = 1.22\dfrac{\lambda\ D}{d}[/tex]
[tex]d = 1.22\dfrac{\lambda\ D}{y}[/tex]
[tex]d = 1.22\dfrac{633\times 10^{-9}\times 4}{0.0125}[/tex]
d =0.247 x 10⁻³ m
d = 0.247 mm
the diameter of the hole is equal to 0.247 mm
The diameter of the hole is approximately 0.0201 mm.
Explanation:The width of the central maximum in a diffraction pattern can be determined using the formula:
w = (2 * λ * D) / x
Where w is the width of the central maximum, λ is the wavelength of the light, D is the distance between the aperture and the screen, and x is the diameter of the hole. Rearranging the formula, we can solve for x:
x = (2 * λ * D) / w
Plugging in the given values, we get:
x = (2 * 633 * 10^-9 * 4.0) / 0.025
x ≈ 0.0201 mm
A proton is traveling horizontally to the right at 4.20×10^6m/s.Part A:Find (a)the magnitude and (b) direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.50cm.Part B: counterclockwise from the left directionPart C:How much time does it take the proton to stop after entering the field?Part D:What minimum field ((a)magnitude and (b)direction) would be needed to stop an electron under the conditions of part (a)?
Answer:
2630250 N/C, horizontally left
0
[tex]1.67\times 10^{-8}\ s[/tex]
1434.825 N/C, horizontally left
Explanation:
m = Mass of particle
u = Initial velocity = [tex]4.2\times 10^6\ m/s[/tex]
v = Final velocity = 0
t = Time taken
s = Displacement = 3.5 cm
q = Charge of particle = [tex]1.6\times 10^{-19}\ C[/tex]
Force is given by
[tex]F=qE[/tex]
Acceleration is given by
[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times E}{1.67\times 10^{-27}}\\\Rightarrow a=95808383.23353E[/tex]
[tex]v^2-u^2=2as\\\Rightarrow v^2-u^2=2\times 95808383.23353Es\\\Rightarrow E=\dfrac{0^2-(4.2\times 10^6)^2}{2\times 95808383.23353\times 0.035}\\\Rightarrow E=-2630250\ N/C[/tex]
Magnitude of electric field is 2630250 N/C
Direction is horizontally to the left
The angle counterclockwise from left is zero.
[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times -2630250}{1.67\times 10^{-27}}\\\Rightarrow a=-2.52\times 10^{14}\ m/s^2[/tex]
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-4.2\times 10^6}{-2.52\times 10^{14}}\\\Rightarrow t=1.67\times 10^{-8}\ s[/tex]
The time taken is [tex]1.67\times 10^{-8}\ s[/tex]
Acceleration is given by
[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times E}{9.11\times 10^{-31}}\\\Rightarrow a=175631174533.4797E[/tex]
[tex]v^2-u^2=2as\\\Rightarrow v^2-u^2=2\times 175631174533.4797Es\\\Rightarrow E=\dfrac{0^2-(4.2\times 10^6)^2}{2\times 175631174533.4797\times 0.035}\\\Rightarrow E=-1434.825\ N/C[/tex]
Magnitude of electric field is 1434.825 N/C
Direction is horizontally to the left
A car starting from rest accelerates at a rate of 2 m/sec2 . Find the average speed of this object in the first 10 seconds.
Answer:
The final speed of the car in the first 10 seconds will be 20 m/s.
Explanation:
Given that,
Initial speed of the car, u = 0
Acceleration of the car, [tex]a=2\ m/s^2[/tex]
Time, t = 10 s
We need to find the speed of the car in the first 10 seconds. It can be calculated using first equation of motion. It is given by :
[tex]v=u+at[/tex]
v is the final speed of the car
[tex]v=at[/tex]
[tex]v=2\ m/s^2\times 10\ s[/tex]
v = 20 m/s
So, the final speed of the car in the first 10 seconds will be 20 m/s. Hence, this is the required solution.
Which of the following statements about electric field lines due to static charges are true? (Select all that apply.)
a. Electric field lines can never cross each other.
b. Electric field lines point from negative charges toward positive charges.
c. Static charges can create closed loop electric field lines that do not begin or end on a charge.
d. Wider spacing between electric field lines indicates a lower magnitude of electric field.
Answer:
a. Electric field lines can never cross each other.
d. Wider spacing between electric field lines indicates a lower magnitude of electric field.
Explanation:
a. The electric field lines cannot be crossed, since this would mean that there would be more than one electric field vector for the same point at the place where the crossing occurs.
d. The space between the field lines is inversely proportional to the intensity of the electric field.
Answer: "a" and "d" are correct
Explanation:
(a) Field lines can never cross. Since a field line represents the direction of the field at a given point, if two field lines crossed at some point, that would imply that the electric field was pointing in two different directions at a single point.
(b) Wider spacing between electric field lines indicates a lower magnitude of electric field.
A point charge Q is located a short distance from a point charge 3Q, and no other charges are present. If the electrical force on Q is F, what is the electrical force on 3Q?a. 3Fb. √3Fc. F/√3d. Fe. F/3
Answer:
d) F
Explanation:
According to columb's law:
"The magnitude of electrostatic force between two charges is directly proportional to the product of magnitude of two charges and inversly proportional to separation between them."
If q₁ and q₂ are magnitude of two charges, d is distance between them and k is dielectric constant, then force F is given by
[tex]F=\frac{kq_{1}q_{2}}{d^2}[/tex]
According to this force exerted on point charge Q is same as that of 3Q, so force point 3Q charge experience is also F
What is the magnitude of the electric force of attraction between an iron nucleus (q=+26e)(q=+26e) and its innermost electron if the distance between them is 1.5×10−12m1.5×10−12m?
The magnitude of the electric force of attraction between the iron nucleus and its innermost electron is approximately 2.645 × 10^-3 Newtons.
To calculate the magnitude of the electric force of attraction between an iron nucleus and its innermost electron, we can use Coulomb's law:
Given:
Charge of the iron nucleus, q1 = +26e
Charge of the electron, q2 = -e (where e is the elementary charge, 1.6 × 10^-19 C)
Distance between them, r = 1.5 × 10^-12 m
Convert the charge of the nucleus from elementary charges to coulombs:
q1 = +26e × 1.6 × 10^-19 C/e = +4.16 × 10^-18 C
Calculate the electric force using Coulomb's law:
F = (8.9875 × 10^9 N m^2/C^2) × |(+4.16 × 10^-18 C) × (-1.6 × 10^-19 C)| / (1.5 × 10^-12 m)^2
F ≈ (8.9875 × 10^9) × | -6.656 × 10^-37 | / (2.25 × 10^-24)
F ≈ (8.9875 × 10^9) × 2.947 × 10^-13
F ≈ 2.645 × 10^-3 N
So, the magnitude of the electric force of attraction between the iron nucleus and its innermost electron is approximately 2.645 × 10^-3 Newtons.
Ricardo, of mass 80 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 30 kg canoe. When the canoe is at rest in the placid water, they exchange seats, which are 3.0 m apart and symmetrically located with respect to the canoe's center. If the canoe moves 40 cm horizontally relative to a pier post, what is Carmelita's mass?
Answer:
m=57.65 kg
Explanation:
Given Data
Ricardo mass m₁=80 kg
Canoe mass m₂=30 kg
Canoe Length L= 3 m
Canoe moves x=40 cm
When Canoe was at rest the net total torque is zero.
Let the center of mass is at x distance from the canoe center and it will be towards the Ricardo cause. So the toque around the center of mass is given as
[tex]m_{1}(L/2-x)=m_{2}x+m_{2}(L/2-x)[/tex]
We have to find m₂.To find the value of m₂ first we need figure out the value of.As they changed their positions the center of mass moved to other side by distance 2x.
so
2x=40
x=40/2
x=20 cm
Substitute in the above equation we get
[tex]m_{x}=\frac{m_{1}(L/2-x)-m_{2}x }{L/2+x}\\m_{x}=\frac{80(\frac{3}{2}-0.2 )-30*0.2}{3/2+0.2}\\m_{x}=57.65 kg[/tex]
In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of radius 0.500 cm at the same elevation as the first tube. The pressure differs by DP between the liquid in one tube and the liquid in the second tube. (a) Find the volume flow rate as a function of DP. Evaluate the volume flow rate for (b) DP 5 6.00 kPa and (c) DP 5 12.0 kPa. 49. The Venturi tube discussed
To find the volume flow rate in ideal flow between two horizontal tubes, use the equation A1v1 = A2v2, where A1 and A2 are the cross-sectional areas of the tubes and v1 and v2 are the velocities of the liquid. To find the volume flow rate as a function of DP, substitute the values of A1 and v1 into the equation Q = A1v1. For specific values of DP, substitute the values of A1 and v1 into the equation Q = A1v1.
Explanation:For liquids, the volume flow rate can be determined using the equation A1v1 = A2v2. In this equation, A1 and A2 are the cross-sectional areas of the first and second tubes, and v1 and v2 are the velocities of the liquid flowing through the tubes. Since the first tube is larger in radius, it has a larger cross-sectional area. Therefore, the velocity of the liquid in the first tube is smaller than in the second tube. The volume flow rate can be expressed as:
Q = A1v1
where Q is the volume flow rate and is equal to the product of the cross-sectional area and velocity of the liquid in the first tube.
(a) The volume flow rate as a function of DP can be found by substituting the values of A1 and v1 into the equation Q = A1v1. Since the radii of the tubes are given, the cross-sectional areas can be calculated using the formula A = πr2.
(b) To evaluate the volume flow rate for DP = 6.00 kPa, substitute the values of A1 and v1 into the equation Q = A1v1.
(c) To evaluate the volume flow rate for DP = 12.0 kPa, substitute the values of A1 and v1 into the equation Q = A1v1.
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A guitar string is 90 cm long and has a mass of 3.7 g . The distance from the bridge to the support post is L=62cm, and the string is under a tension of 500 N . What are the frequencies of the fundamental and first two overtones?
To solve this problem we will apply the concept of frequency in a string from the nodes, the tension, the linear density and the length of the string, that is,
[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]
Here
n = Number of node
T = Tension
[tex]\mu[/tex] = Linear density
L = Length
Replacing the values in the frequency and value of n is one for fundamental overtone
[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]
[tex]f = \frac{1}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})[/tex]
[tex]\mathbf{f = 281.2Hz}[/tex]
Similarly plug in 2 for n for first overtone and determine the value of frequency
[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]
[tex]f = \frac{2}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})[/tex]
[tex]\mathbf{f = 562.4Hz}[/tex]
Similarly plug in 3 for n for first overtone and determine the value of frequency
[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]
[tex]f = \frac{3}{2(62*10^{-2})}\bigg (\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}} \bigg)[/tex]
[tex]\mathbf{f= 843.7Hz}[/tex]