Answers
Answer:
298rad/s and 116.96 ohms
Explanation:
Given an L-R-C series circuit where
L = 0.450 H,
C=2.50×10^−5F, and resistance R= 0
In this situation we have a simple LC circuit with angular frequency
Wo = 1√LC
= 1/√(0.450)(2.50×10^-5)
= 1/√0.00001125
= 1/0.003354
= 298rad/s
B) Now we need to find the value of R such that it gives a 10% decrease in angular frequency.
Wi/W° = (100-10)/100
Wi/W° = 90/100
Wi/W° = 0.90 ............... 1
Angular frequency of oscillation
The complete aspect of the solution is attached, please check.
a. The angular frequency of the circuit when R = 0 Ohms is 294.12 rad/s.
b. The value R must have to give a decrease in angular frequency of 10.0 % compared to the initial value is equal to 116.96 Ohms.
Given the following data:
Inductance, L = 0.450 HenryCapacitance, C = [tex]2.50\times 10^{-5}[/tex] Faradsa. To determine the angular frequency of the circuit when R = 0 Ohms:
Mathematically, the angular frequency of a LC circuit is given by the formula:
[tex]\omega = \frac{1}{\sqrt{LC} } \\\\\omega =\frac{1}{\sqrt{0.450 \times 2.50\times 10^{-5}}} \\\\\omega =\frac{1}{\sqrt{1.125 \times 10^{-5}}} \\\\\omega = \frac{1}{0.0034} \\\\\omega = 294.12\;rad/s[/tex]
b. To find the value R must have to give a decrease in angular frequency of 10.0 % compared to the value calculated above:
The mathematical expression is given as follows:
[tex]\frac{\omega_f}{\omega_i} = \frac{100-10}{100} \\\\\frac{\omega_f}{\omega_i} =\frac{90}{100} \\\\\frac{\omega_f}{\omega_i} =0.9[/tex]
[tex](\frac{\omega_f}{\omega_i})^2 = 1 - \frac{R^2C}{4L} \\\\0.90^2=1 - \frac{R^2C}{4L}\\\\R=\sqrt{\frac{4L(1-0.81)}{C}} \\\\R=\sqrt{\frac{4\times 0.450 \times (0.19)}{2.50\times 10^{-5}}}\\\\R = \sqrt{\frac{0.342}{2.50\times 10^{-5}} }\\\\R =\sqrt{13680}[/tex]
R = 116.96 Ohms.
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Related Questions
Two protons, A and B, are next to an infinite plane of positive charge. Proton B is twice as far from the plane as proton A. Which proton has the larg
Answers
Answer:
They both have the same acceleration
Can anyone provide me the answer with explanation?
Answers
Answer:
the answer to your question us c honey
Answer:
C
Explanation:
This is so because different materials vary in resistance and conductance of current, heat. Metals are good conductors while none metals like rubber, plastic, glass etc are good insulators or resistors.
A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately what? Group of answer choices
Answers
Answer:
0.05cos10t
Explanation:
X(t) = Acos(wt+φ)
The oscillation angular frequency can be calculated using below formula
w = √(k/M)
Where K is the spring constant
But we were given body mass of 5.0 kg
We know acceleration due to gravity as 9.8m)s^2
The lenghth of spring which stretches =10 cm
Then we can calculate the value of K
k = (5.0kg*9.8 m/s^2)/0.10 m
K= 490 N/m
Then if we substitute these values into the formula above we have
w = √(k/M)
w = √(490/5)
= 9.90 rad/s=10rads/s(approximately)
Its position as a function of time can be calculated using the below expresion
X(t) = Acos(wt+φ)
We were given amplitude of 5 cm , if we convert to metre = 0.05m
w=10rads/s
Then if we substitute we have
X(t)=0.05cos(10×t)
X(t)= 0.05cos10t
Therefore,Its position as a function of time=
X(t)= 0.05cos10t
You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and hang a 1.8 g mass from it. This stretches the spring to a length of 5.2 cm . You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.8 cm .
Required:
What is the magnitude of the charge (in nC) on each bead?
Answers
Answer:
The magnitude of the charge is 54.9 nC.
Explanation:
The charge on each bead can be found using Coulomb's law:
[tex] F_{e} = \frac{k*q_{1}q_{2}}{r^{2}} [/tex]
Where:
q₁ and q₂ are the charges, q₁ = q₂
r: is the distance of spring stretching = 4.8x10⁻² m
[tex]F_{e}[/tex]: is the electrostatic force
[tex] F_{e} = \frac{k*q^{2}}{r^{2}} \rightarrow q = \sqrt{\frac{F_{e}}{k}}*r [/tex]
Now, we need to find [tex]F_{e}[/tex]. To do that we have that Fe is equal to the spring force ([tex]F_{k}[/tex]):
[tex] F_{e} = F_{k} = -kx [/tex]
Where:
k is the spring constant
x is the distance of the spring = 4.8 - 4.0 = 0.8 cm
The spring constant can be found by equaling the sping force and the weight force:
[tex] F_{k} = -W [/tex]
[tex] -k*x = -m*g [/tex]
where x is 5.2 - 4.0 = 1.2 cm, m = 1.8 g and g = 9.81 m/s²
[tex] k = \frac{mg}{x} = \frac{1.8 \cdot 10^{-3} kg*9.81 m/s^{2}}{1.2 \cdot 10^{-2} m} = 1.47 N/m [/tex]
Now, we can find the electrostatic force:
[tex] F_{e} = F_{k} = -kx = -1.47 N/m*0.8 \cdot 10^{-2} m = -0.0118 N [/tex]
And with the magnitude of the electrostatic force we can find the charge:
[tex]q = \sqrt{\frac{F_{e}}{k}}*r = \sqrt{\frac{0.0118 N}{9 \cdot 10^{9} Nm^{2}/C^{2}}}*4.8 \cdot 10^{-2} m = 54.9 \cdot 10^{-9} C = 54.9 nC[/tex]
Therefore, the magnitude of the charge is 54.9 nC.
I hope it helps you!
The magnitude of the charge (in nC) on each bead is equal to 55.21 nC.
Given the following data:
Original length = 4.0 cm to m = 0.04 mMass = 1.8 grams to kg = 0.0018New length = 5.2 cm to m = 0.052.Final length = 4.8 cm to m = 0.048 m.Extension, e = [tex]0.052 - 0.048[/tex] = 0.012 m
Scientific data:
Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]Coulomb's constant = [tex]8.99 \times 10^9\; Nm^2/C^2[/tex]To calculate the magnitude of the charge (in nC) on each bead, we would apply Coulomb's law:
First of all, we would determine the spring constant of this lightweight spring by using this formula:
[tex]W = mg = Ke \\\\K=\frac{mg}{e} \\\\K=\frac{0.0018 \times 9.8}{0.012} \\\\K=\frac{0.01764}{0.012}[/tex]
Spring constant, K = 1.47 N/m.
For the electrostatic force:
[tex]F = ke\\\\F = 1.47 \times 0.08[/tex]
F = 0.01176 Newton.
Coulomb's law of electrostatic force.
Mathematically, the charge in an electric field is given by this formula:
[tex]q = \sqrt{\frac{F}{k} } \times r[/tex]
Substituting the given parameters into the formula, we have;
[tex]q = \sqrt{\frac{0.01176 }{8.99 \times 10^9} } \times 0.048\\\\q=\sqrt{1.3228 \times 10^{-12}} \times 0.048\\\\q=1.1502 \times 10^{-6} \times 0.048\\\\q= 5.521 \times 10^{-8}\;C[/tex]
Note: 1 nC = [tex]1 \times 10^{-9}\;C[/tex]
Charge, q = 55.21 nC.
Read more on electric field here: https://brainly.com/question/14372859
The capacitor is originally charged. How does the current I in the ammeter behave as a function of time after the switch is closed?
1. I = 0 always.
2. I = constant, not equal to 0.
3. I increases, then is constant.
4. I instantly jumps up, then slowly decreases.
5. None of the above.
Answers
Answer:
The current in the ammeter is zero.
(1) is correct option.
Explanation:
Given that,
The capacitor is charged.
We need find the current after closed switched
We know that,
When switch is closed then the capacitor behave as a short circuit, and the all current flows through it. the current is zero.
Then, the ammeter reads zero.
Hence, The current in the ammeter is zero.
(1) is correct option.
The cylinder is displaced 0.17 m downward from its equilibrium position and is released at time t = 0. Determine the displacement y and the velocity v when t = 3.1 s. The displacement and velocity are positive if downward, negative if upward. What is the magnitude of the maximum acceleration?
Answers
Complete Question
The image of this question is shown on the first uploaded image
Answer:
a
[tex]d =0.161 \ m[/tex]
b
[tex]v = - 0.054 \ m/s[/tex]
c
[tex]a = 6.12 \ m/s^2[/tex]
Explanation:
From the question we are told that
The maximum displacement is A = 0.17 m
The time considered is [tex]t = 3.1 \ s[/tex]
The spring constant is [tex]k = 137 \ N \cdot m[/tex]
The mass is [tex]m = 3.8 \ kg[/tex]
Generally given that the motion which the cylinder is undergoing is a simple harmonic motion , then the displacement is mathematically represented as
[tex]d = A cos (w t )[/tex]
Where [tex]w[/tex] is the angular frequency which is mathematically evaluated as
[tex]w = \sqrt{\frac{k}{m} }[/tex]
substituting values
[tex]w = \sqrt{\frac{137}{ 3.8} }[/tex]
[tex]w =6[/tex]
So the displacement is at t
[tex]d = 0.17 cos (6 * 3.1 )[/tex]
[tex]d =0.161 \ m[/tex]
Generally the velocity of a SHM(simple harmonic motion) is mathematically represented as
[tex]v = - Asin (wt)[/tex]
substituting values
[tex]v = - 0.17 sin ( 6 * 3.1 )[/tex]
[tex]v = - 0.054 \ m/s[/tex]
Generally the maximum acceleration is mathematically represented as
[tex]a = w^2 * A[/tex]
substituting values
[tex]a_{max} = 6^2 * (0.17)[/tex]
substituting values
[tex]a = 6^2 * (0.17)[/tex]
[tex]a = 6.12 \ m/s^2[/tex]
The AB rope is fixed to the ground at its A end, and forms 30º with the vertical. Its other end is connected to two ropes by means of the B-ring of negligible weight. The vertical rope supports the E block and the other rope passes through the grounded articulated pulley C to join at its end to the 80 N weight block D. The inclined section of the BD rope forms 60º with the vertical one; determine the weight of the E block necessary for the balance of the system and calculate the tension in the AB rope.
Answers
Answer:
T = 80√3 N ≈ 139 N
W = 160 N
Explanation:
Sum of forces on B in the x direction:
∑F = ma
80 N sin 60° − T sin 30° = 0
T = 80 N sin 60° / sin 30°
T = 80√3 N
T ≈ 139 N
Sum of forces on B in the y direction:
∑F = ma
80 N cos 60° + T cos 30° − W = 0
W = 80 N cos 60° + T cos 30°
W = 40 N + 120 N
W = 160 N
You simultaneously shine two light beams, each of intensity I0, on an ideal polarizer. One beam is unpolarized, and the other beam is polarized at an angle of exactly 30.0∘ to the polarizing axis of the polarizer. Find the intensity of the light that emerges from the polarizer. Express your answer in term of I0 .
Answers
Answer:
The emerging intensity is equal to 0.75[tex]I_{o}[/tex]
Explanation:
The initial intensity of the light = [tex]I_{o}[/tex]
The angle of polarization β = 30°
We know that the polarized light intensity is related to the initial light intensity by
[tex]I[/tex] = [tex]I_{0} cos^{2}\beta[/tex]
where [tex]I[/tex] is the emerging polarized light intensity
inserting values gives
[tex]I[/tex] = [tex]I_{0} cos^{2}[/tex] 30°
[tex]cos^{2}[/tex] 30° = [tex](cos 30)^{2}[/tex] = [tex](\frac{\sqrt{3} }{2} )^{2}[/tex] = 0.75
[tex]I[/tex] = 0.75[tex]I_{o}[/tex]
Consider 2 converging lenses of focal lengths 5 mm (objective) and 50 mm.(eyepiece) An object 0.1 mm in size is placed a distance of 5.2 mm from the objective.
1. What is the size and location of the image from the objective? What is the linear magnification of this objective?
2. Treat the image from the objective as an object for the eyepiece. If the eyepiece creates an image at infinity, how far apart are the two lenses?
3. What is the angular magnification of the pair of lenses?
Answers
Answer:
1) q₁ = 12.987 cm , b) L = 17.987 cm , c) m = 179.87
Explanation:
We can solve the geometric optics exercises with the equation of the constructor
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distance to the object and the image respectively.
Let's apply this equation to our case
1) f = 5mm = 0.5 cm
p₁ = 5.2 mm = 0.52 cm
h = 0.1 mm = 0.01 cm
1 / q₁ = 1 / f- 1 / p
1 / q₁ = 1 / 0.5 - 1 / 0.52 = 2 - 1.923
1 / q₁ = 0.077
q₁ = 12.987 cm
2) in this part they tell us that the eyepiece creates an image at infinity, therefore the object that comes from being at the focal length of the eyepiece
p₂ = 5 cm
The absolute thing that goes through the two lenses is
L = q₁ + p₂
L = 12.987 +5
L = 17.987 cm
3) This lens configuration forms the so-called microscope, whose expression for the magnifications
m = -L / f_target 25 cm / f_ocular
m = - 17.987 / 0.5 25 / 5.0
m = 179.87
Experiments are performed with ultracold neutrons having velocities of 7.54 m/s. (a) What is the wavelength (in nm) of such a neutron
Answers
Answer:
λ = 52.5 nm
Explanation:
De Broglie's duality principle states that all matter has wave and particle characteristics, being related by the expression
p = h / λ
where the moment
p = mv
λ = h / mv
let's calculate
λ = 6.63 10⁻³⁴ / 1.675 10⁻²⁷ 7.54
λ = 5.25 10⁻⁸ m
Let's reduce anm
λ = 5.25 10⁻⁸ m (10⁹ nm / 1m)
λ = 52.5 nm
"Two waves of the same frequency have amplitudes 1.00 and 2.00. They interfere at a point where their phase difference is 60.0°. What is the resultant amplitude?"
Answers
Answer:
The resultant amplitude of the two waves is 2.65.
Explanation:
Given;
amplitude of the first wave, A₁ = 1
amplitude of the second wave, A₂ = 2
phase difference of the two amplitudes, θ = 60.0°.
The resultant amplitude of two waves after interference is given by;
[tex]A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2Cos \theta} \\\\A = \sqrt{1^2 + 2^2 + 2(1)(2)Cos 60} \\\\A= 2.65[/tex]
Therefore, the resultant amplitude of the two waves is 2.65.
An electron moving at 3.94 103 m/s in a 1.23 T magnetic field experiences a magnetic force of 1.40 10-16 N. What angle does the velocity of the electron make with the magnetic field? There are two answers between 0° and 180°. (Enter your answers from smallest to largest.)
Answers
Answer:
10.4⁰ and 169.6⁰Explanation:
The force experienced by the moving electron in the magnetic field is expressed as F = qvBsinθ where;
q is the charge on the electron
v is the velocity of the electron
B is the magnetic field strength
θ is the angle that the velocity of the electron make with the magnetic field.
Given parameters
F = 1.40*10⁻¹⁶ N
q = 1.6*10⁻¹⁹C
v = 3.94*10³m/s
B = 1.23T
Required
Angle that the velocity of the electron make with the magnetic field
Substituting the given parameters into the formula:
1.40*10⁻¹⁶ = 1.6*10⁻¹⁹ * 3.94*10³ * 1.23 * sinθ
1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁹⁺³sinθ
1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁶sinθ
sinθ = 1.40*10⁻¹⁶/7.75392 * 10⁻¹⁶
sinθ = 1.40/7.75392
sinθ = 0.1806
θ = sin⁻¹0.1806
θ₁ = 10.4⁰
Since sinθ is positive in the 1st and 2nd quadrant, θ₂ = 180-θ₁
θ₂ = 180-10.4
θ₂ = 169.6⁰
Hence, the angle that the velocity of the electron make with the magnetic field are 10.4⁰ and 169.6⁰
10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ????? How far will this pulse travel in the same time if the tension is doubled?
Answers
Answer: Tension = 47.8N, Δx = 11.5×[tex]10^{-6}[/tex] m.
Tension = 95.6N, Δx = 15.4×[tex]10^{-5}[/tex] m
Explanation: A speed of wave on a string under a tension force can be calculated as:
[tex]|v| = \sqrt{\frac{F_{T}}{\mu} }[/tex]
[tex]F_{T}[/tex] is tension force (N)
μ is linear density (kg/m)
Determining velocity:
[tex]|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }[/tex]
[tex]|v| = \sqrt{0.00874 }[/tex]
[tex]|v| =[/tex] 0.0935 m/s
The displacement a pulse traveled in 1.23ms:
[tex]\Delta x = |v|.t[/tex]
[tex]\Delta x = 9.35.10^{-2}*1.23.10^{-3}[/tex]
Δx = 11.5×[tex]10^{-6}[/tex]
With tension of 47.8N, a pulse will travel Δx = 11.5×[tex]10^{-6}[/tex] m.
Doubling Tension:
[tex]|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }[/tex]
[tex]|v| = \sqrt{2.0.00874 }[/tex]
[tex]|v| = \sqrt{0.01568}[/tex]
|v| = 0.1252 m/s
Displacement for same time:
[tex]\Delta x = |v|.t[/tex]
[tex]\Delta x = 12.52.10^{-2}*1.23.10^{-3}[/tex]
[tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex]
With doubled tension, it travels [tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex] m
3. What conclusion can you make about the electric field strength between two parallel plates? Explain your answer referencing Photo 2.
Answers
Answer:
From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases
Explanation:
I cannot find any attached photo, but we can proceed anyways theoretically.
The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point
i.e
[tex]E=\frac{F}{Q}[/tex]
But the force F
[tex]F= \frac{kQ1Q2}{r^2}[/tex]
But the electric field intensity due to a point charge Q at a distance r meters away is given by
[tex]E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }[/tex]
From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases
If Superman really had x-ray vision at 0.12 nm wavelength and a 4.1 mm pupil diameter, at what maximum altitude could he distinguish villains from heroes, assuming that he needs to resolve points separated by 5.4 cm to do this?
Answers
Answer:
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
Explanation:
Given:
wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m
Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m
Separation distance (D) = 5.4 cm = 0.054 m
Find:
Maximum altitude to see(L)
Computation:
Resolving power = 1.22(λ / d)
D / L = 1.22(λ / d)
0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]
0.054 / L = 1.22 [0.03 × 10⁻⁶]
L = 0.054 / 1.22 [0.03 × 10⁻⁶]
L = 0.054 / [0.0366 × 10⁻⁶]
L = 1.47 × 10⁶
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
Radio station WCCO in Minneapolis broadcasts at a frequency of 830 kHz. At a point some distance from the transmitter, the magnetic-field amplitude of the electromagnetic wave from WCCO is 4.82×10-11 T.A) Calculate the wavelength.B) Calculate the wave number.C) Calculate the angular frequency.
D) Calculate the electric-field amplitude.
Answers
Answer:
A
[tex]\lambda = 361.45 \ m[/tex]
B
[tex]k = 0.01739 \ rad/m[/tex]
C
[tex]w = 5.22 *10^{6} \ rad/s[/tex]
D
[tex]E = 0.01446 \ N/C[/tex]
Explanation:
From the question we are told that
The frequency is [tex]f = 83 0 \ kHz = 830 *10^{3} \ Hz[/tex]
The magnetic field amplitude is [tex]B = 4.82*10^{-11} \ T[/tex]
Generally wavelength is mathematically represented as
[tex]\lambda = \frac{c}{f}[/tex]
where c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]
=> [tex]\lambda = \frac{3.0*10^{8}}{ 830 *10^{3}}[/tex]
=> [tex]\lambda = 361.45 \ m[/tex]
Generally the wave number is mathematically represented as
[tex]k = \frac{2 \pi }{\lambda }[/tex]
=> [tex]k = \frac{2 * 3.142 }{ 361.45 }[/tex]
=> [tex]k = 0.01739 \ rad/m[/tex]
Generally the angular frequency is mathematically represented as
[tex]w = 2 * \pi * f[/tex]
=> [tex]w = 2 * 3.142 * 830*10^{3}[/tex]
=> [tex]w = 5.22 *10^{6} \ rad/s[/tex]
The the electric-field amplitude is mathematically represented as
[tex]E = B * c[/tex]
=> [tex]E = 4.82 *10^{-11} * 3.0*10^{8}[/tex]
=> [tex]E = 0.01446 \ N/C[/tex]
This question involves the concepts of wavelength, frequency, wave number, and electric field.
a) The wavelength is "361.44 m".
b) The wave number is "0.0028 m⁻¹".
c) The angular frequency is "5.22 x 10⁶ rad/s".
d) The electric field amplitude is "0.0145 N/C".
a)
The wavelength can be given by the following formula:
[tex]c=f\lambda[/tex]
where,
c = speed of light = 3 x 10⁸ m/s
f = frequency = 830 KHz = 8.3 x 10⁵ Hz
λ = wavelength = ?
Therefore,
[tex]3\ x\ 10^8\ m/s=(8.3\ x\ 10^5\ Hz)\lambda\\\\\lambda=\frac{3\ x\ 10^8\ m/s}{8.3\ x\ 10^5\ Hz}\\\\[/tex]
λ = 361.44 m
b)
The wave number can be given by the following formula:
[tex]wave\ number = \frac{1}{\lambda} = \frac{1}{361.44\ m}[/tex]
wave number = 0.0028 m⁻¹
c)
The angular frequency is given as follows:
[tex]\omega = 2\pi f = (2)(\pi)(8.3\ x\ 10^5\ Hz)[/tex]
ω = 5.22 x 10⁶ rad/s
d)
The electric field amplitude can be given by the following formula:
[tex]\frac{E}{B} = c\\\\c(B)=E\\\\E = (3\ x\ 10^8\ m/s)(4.82\ x\ 10^{-11}\ T)\\[/tex]
E = 0.0145 N/C
Learn more about wavelength and frequency here:
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The Bohr model pictures a hydrogen atom in its ground state as a proton and an electron separated by the distance a0 = 0.529 × 10−10 m. The electric potential created by the electron at the position of the proton is
Answers
Answer:
E = -8.23 10⁻¹⁷ N / C
Explanation:
In the Bohr model, the electric potential for the ground state corresponding to the Bohr orbit is
E = k q₁ q₂ / r²
in this case
q₁ is the charge of the proton and q₂ the charge of the electron
E = - k e² / a₀²
let's calculate
E = - 9 10⁹ (1.6 10⁻¹⁹)² / (0.529 10⁻¹⁰)²
E = -8.23 10⁻¹⁷ N / C
how does a system naturally change over time
Answers
Answer:
The movement of energy and matter in a system differs from one system to another. On the other hand, in open system both the matter and energy move into and out of the system. Therefore, matter and energy in a system naturally change over time will decrease in entropy.
Explanation:
Answer:
Decrease in entropy
Explanation:
Various systems which exist in nature possess energy and matter that move through these system continuously. The movement of energy and matter in a system differs from one system to another.
In a closed system for example, only energy flows in and out of the system while matter does not enter or leave the system.
On the other hand, in open system both the matter and energy move into and out of the system.
When water freezes, it expands about nine percent. What would be the pressure increase inside your automobile engine block if the water in there froze? The bulk modulus of ice is 2.0 × 109 N/m2, and 1 ATM = 1.01 × 105 N/m2.
Answers
Answer:
The pressure increase inside the automobile engine block is 1782.18 ATM
Explanation:
Given;
the change in volume of water, ΔV = 9%
the bulk modulus of ice, K = 2 x 10⁹ N/m²
Bulk modulus is given by;
[tex]K = -V\frac{dP}{dV}[/tex]
for pressure increase in the automobile engine block, when the water in there froze;
[tex]dP = K(\frac{dV}{V} )\\\\dP = K(\frac{0.09V}{V} )\\\\dP = 0.09K\\\\dP = 0.09 (2*10^9)\\\\dP = 1.8 *10^{8} \ N/m^2\\\\dP = 1782.18 \ ATM[/tex]
Therefore, the pressure increase inside the automobile engine block is 1782.18 ATM
The pressure increase inside your automobile engine block will be 1782.18 atm. The force involved vertical to the surface of an object per unit area is pressure.
What is pressure?The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.
It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.
The given data in the problem is;
ΔV is the change in volume of water = 9%
K is the bulk modulus of ice = 2 x 10⁹ N/m²
dP is the change in the presure=?
The bulk modulus is found as;
[tex]\rm K=-V \frac{dp}{dv} \\\\[/tex]
The change in the presure is obtained as;
[tex]\rm dP = K\frac{dV}{V} \\\\ \rm dP = K\frac{0.09V}{V} \\\\ \rm dP = 0.09 K \\\\ \rm dP = 0.09 \times 2 \times 10^9 \\\\ \rm dP = 1.8 \times 10^8 \\\\ \rm dP =1782.18 \ atm[/tex]
Hence the pressure increase inside your automobile engine block will be 1782.18 atm.
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please help !!!!!!!!!!
Answers
Answer:
Lighthouse 1 during the day will be warmer, lighthouse 2 during the night will be warmer.
Explanation:
As the paragraph stated land absorbs heat and heats up faster than water. So during the day the lighthouse farthest away from the water will be hotter. But then the converse is true also land losses heat faster than water at night. So the water retains the heat from the day better making the lighthouse by the water warmer at night.
A brick weighs 50.0 N, and measures 30.0 cm × 10.0 cm × 4.00 cm. What is the maximum pressure it can exert on a horizontal surface due to its weight?
Answers
Answer:
Pressure, P = 1250 Pa
Explanation:
Given that,
Weight of a brick, F = 50 N
Dimension of the brick is 30.0 cm × 10.0 cm × 4.00 cm
We need to find the maximum pressure it can exert on a horizontal surface due to its weight. Pressure is equal to the force acting per unit area. Pressure exerted is inversely proportional to the area of cross section. So, we need to minimize area. Taking to smaller dimensions.
A = 40 cm × 10 cm = 400 cm² = 0.04 m²
So,
Pressure,
[tex]P=\dfrac{50\ N}{0.04\ m^2}\\\\P=1250\ Pa[/tex]
So, the maximum pressure of 1250 Pa it can exert on a horizontal surface.
The maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.
What is pressure?The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure. It is denoted by P.
The given data in the problem is;
W is the weight of a brick = 50 N
The dimension of the brick = 30.0 cm × 10.0 cm × 4.00 cm
A is the area,
The area is found as;
A=40 cm × 10 cm = 400 cm² = 0.04 m²
The pressure is the ratio of the force and area
[tex]\rm P = \frac{F}{A} \\\\ \rm P = \frac{50}{0.04} \\\\ \rm P =1250 \ Pascal[/tex]
Hence the maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.
To learn more about the pressure refer to the link;
https://brainly.com/question/356585
n ultraviolet light beam having a wavelength of 130 nm is incident on a molybdenum surface with a work function of 4.2 eV. How fast does the electron move away from the metal
Answers
Answer:
The speed of the electron is 1.371 x 10⁶ m/s.
Explanation:
Given;
wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m
the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J
The energy of the incident light is given by;
E = hf
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
f = c / λ
[tex]E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J[/tex]
Photo electric effect equation is given by;
E = W₀ + K.E
Where;
K.E is the kinetic energy of the emitted electron
K.E = E - W₀
K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J
K.E = 8.563 x 10⁻¹⁹ J
Kinetic energy of the emitted electron is given by;
K.E = ¹/₂mv²
where;
m is mass of the electron = 9.11 x 10⁻³¹ kg
v is the speed of the electron
[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s[/tex]
Therefore, the speed of the electron is 1.371 x 10⁶ m/s.
Light of wavelength 519 nm passes through two slits. In the interference pattern on a screen 4.6 m away, adjacent bright fringes are separated by 5.2 mm in the general vicinity of the center of the pattern. What is the separation of the two slits?
Answers
Answer:
The separation of the two slits is 0.456 mm.
Explanation:
Given the wavelength of light = 519 nm
The indifference pattern = 4.6 m
Adjacent bright fringes = 5.2 mm
In the interference, the equation required is Y = mLR/d
Here, d sin theta = mL
L = wavelgnth
For bright bands, m is the order = 1,2,3,4
For dark bands, m = 1.5, 2.5, 3.5, 4.5
R = Distance from slit to screen (The indifference pattern)
Y = Distance from central spot to the nth order fringe or fringe width
Thus, here d = mLR/Y
d = 1× 519nm × 4.6 / 5.2mm
d = 0.459 mm
You plan to take your hair blower to Europe, where the electrical outlets put out 240 V instead of the 120 V seen in the United States. The blower puts out 1700 W at 120 V.Required:a. What could you do to operate your blower via the 240V line in Europe? which one is it?b. What current will your blower draw from a European outlet?c. What resistance will your blower appear to have when operated at 240 ?
Answers
Answer:
a) Connect a series resistance of 8,47 ohms
b)14,16 [A]
c) r = 10,96 ohms
Explanation:
My blower requires 120 (v) then, I have to connect a series resistor to make the nominal 240 (v) of the European voltage outlet drop to 120 (V) but at the same time keep the level of current to operate my blower
In America
P = V*I
1700 (w) = 120*I
I = 1700/120 [A]
I = 14,16 [A] current needed for the blower
In Europe
120 (v) (the drop of voltage I need) when a current of 14,16 passes through to series resistor is
V = I*R 120 = 14,16* R R = 8,47 ohms
c) P = I*r²
1700 (w) = 14,16 (A) * r²
r² = 120,06
r = 10,96 ohms
An isolated system consists of two masses. The first, m1, has a mass of 1.90 kg, and is initially traveling to the east with a speed of 6.71 m/s. The second, m2, has a mass of 2.94 kg, and is initially traveling to the west with an unknown initial speed. The two masses collide head-on in a completely inelastic collision that stops them both. Calculate the initial kinetic energy of m2.
Answers
Answer:
m1v1=m2v2, v2=4.3m/s KE=(0.5)(2.94)(4.3)=6.2J
A mass m = 0.6 kg is released from rest at the top edge of a hemispherical bowl with radius = 1.1 meters. The mass then slides without friction down the inner surface toward the bottom of the bowl. At a certain point of its path the mass achieves a speed v = 3.57 m/s. At this point, what angle \theta\:θ ( in degrees) does the mass make with the top of the bowl?
Answers
Answer:
The angle is [tex]\theta = 36.24 ^o[/tex]
Explanation:
From the question we are told that
The mass is [tex]m = 0.6 \ kg[/tex]
The radius is [tex]r = 1.1 \ m[/tex]
The speed is [tex]v = 3.57 \ m /s[/tex]
According to the law of energy conservation
The potential energy of the mass at the top is equal to the kinetic energy at the bottom i.e
[tex]m * g * h = \frac{1}{2} * m * v^2[/tex]
=> [tex]h = \frac{1}{2 g } * v^2[/tex]
Here h is the vertical distance traveled by the mass which is also mathematically represented as
[tex]h = r * sin (\theta )[/tex]
So
[tex]\theta = sin ^{-1} [ \frac{1}{2* g* r } * v^2][/tex]
substituting values
[tex]\theta = sin ^{-1} [ \frac{1}{2* 9.8* 1.1 } * (3.57)^2][/tex]
[tex]\theta = 36.24 ^o[/tex]
A father and his young son get on a teeter-totter. The son sits 2 m fromthe center, but the father has to sit closer to balance. Where does the father have to sit to balance the teeter-totter if he weighs 4 times as much as his son?
Answers
Answer:
The distance of the father from the center is [tex]d_f = \frac{1}{2} \ m[/tex]
Explanation:
From the question we are told that
The distance of the son from the center is [tex]d_s = 2 \ m[/tex]
Let the mass of the son be [tex]m_s[/tex]
then the mass of the father is [tex]m_f = 4m_s[/tex]
Now for the teeter-totter to be balanced the torque due to the weight of the father must be equal to the torque due to the weight the son, this is mathematically represented as
[tex]\tau_s = \tau_f[/tex]
Where [tex]\tau_s[/tex] is the torque of the son which is mathematically represented as
[tex]\tau_ s = m_s * d_s * g[/tex]
while [tex]\tau_f[/tex] is the torque of the father which is mathematically represented as
[tex]\tau_f = m_f * d_f * g[/tex]
=> [tex]\tau_f = 4 m_s * d_f * g[/tex]
So
[tex]4 m_s * d_f * g = m_s * d_s * g[/tex]
substituting values
[tex]4 * d_f * = 2[/tex]
=> [tex]d_f = \frac{1}{2} \ m[/tex]
A stone is dropped from the bridge, it takes 4s to reach the water. what's the height of the bridge?
Answers
Explanation:
Using Equations of Motion :
[tex]s = ut + \frac{1}{2} g {t}^{2} [/tex]
Height = 0 * 4 + 4.9 * 16
Height = 78.4 m
In a double-slit experiment the distance between slits is 5.0 mm and the slits are 1.4 m from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength 450 nm, and the other due to light of wavelength 590 nm. What is the separation in meters on the screen between the m = 5 bright fringes of the two interference patterns?
Answers
Answer:
Δy = 1 10⁻⁴ m
Explanation:
In double-slit experiments the constructive interference pattern is described by the equation
d sin θ = m λ
In this case we have two wavelengths, so two separate patterns are observed, let's use trigonometry to find the angle
tan θ = y / L
as the angles are small,
tan θ = sin θ / cos θ = sin θ
substituting
sin θ = y / L
d y / L = m λ
y = m λ / d L
let's apply this formula for each wavelength
λ = 450 nm = 450 10⁻⁹ m
m = 5
d = 5.0 mm = 5.0 10⁻³ m
y₁ = 5 450 10⁻⁹ / (5 10⁻³ 1.4)
y₁ = 3.21 10⁻⁴ m
we repeat the calculation for lam = 590 nm = 590 10⁻⁹ m
y₂ = 5 590 10⁻⁹ / (5 10⁻³ 1.4)
y₂= 4.21 10⁻⁴ m
the separation of these two lines is
Δy = y₂ - y₁
Δy = (4.21 - 3.21) 10⁻⁴ m
Δy = 1 10⁻⁴ m
A fish in a flat-sided aquarium sees a can of fish food on the counter. To the fish's eye, the can looks to be 43 cmcm outside the aquarium.
Required:
What is the actual distance between the can and the aquarium?
Answers
Answer:
The actual distance is [tex]d_a = 0.3233\ m[/tex]
Explanation:
From the question we are told that
The distance of the can is d = 43 cm = 0.43 m
Generally the actual distance is mathematically represented as
[tex]d_a = [\frac{ n_a }{n_w} ]* d[/tex]
Where [tex]n_a , n_w[/tex] are the refractive index of air and water and their value is
[tex]n_a = 1 , \ \ \ n_w = 1.33[/tex]
So
[tex]d_a = [\frac{ 1 }{1.33} ]* 0.43[/tex]
[tex]d_a = 0.3233\ m[/tex]
