Answer:
Rise in temperature is given as
[tex]\Delta T = 8.4 \times 10^{-3} ^0C[/tex]
Explanation:
Initial kinetic energy of the block is given as
[tex]KE = \frac{1}{2}mv^2[/tex]
here we will have
m = 3.3 kg
v = 3.2 m/s
now we will have
[tex]KE = \frac{1}{2}mv^2[/tex]
now we will have
[tex]KE = \frac{1}{2}(3.3)(3.2)^2[/tex]
[tex]KE = 17 J[/tex]
now we know that 74% of initial kinetic energy is absorbed as internal energy of the block
so the rise in temperature of the block is given as
[tex]KE = ms\Delta T[/tex]
[tex]0.74 \times 17 J = (3.3)(452)\Delta T[/tex]
[tex]12.5 = 1491.6 \Delta T[/tex]
[tex]\Delta T = 8.4 \times 10^{-3} ^0C[/tex]
Answer:
The temperature increases 0.0084ºC
Explanation:
Please look at the solution in the attached Word file.
You’re on a team performing a high-magnetic-field experiment. A conducting bar carrying 4.1 kA will pass through a 1.3-m-long region containing a 12-T magnetic field, making a 60° angle with the field. A colleague proposes resting the bar on wooden blocks. You argue that it will have to be clamped in place, and to back up your argument you claim that the magnetic force will exceed 10,000 pounds. Are you right?
Answer:
Yes we are right as the force on wire is approx 12500 Lb
Explanation:
Magnetic force on a current carrying bar is given by the equation
[tex]\vec F = i(\vec L \times \vec B)[/tex]
here we know that
L = 1.3 m
B = 12 T
[tex]\theta = 60^0[/tex]
i = 4.1 kA
now from above formula we have
[tex]F = iLBsin60[/tex]
[tex]F = (4.1\times 10^3)(1.3 )(12)sin60[/tex]
[tex]F = 55391 N[/tex]
So this is equivalent to 12500 Lb force
Therefore, the magnetic force exerted on the conducting bar is approximately 994.5 pounds.
"The magnetic force exerted on the conducting bar can be calculated using the formula for the force on a current-carrying conductor in a magnetic field, which is given by [tex]\( F = ILB \sin(\theta) \),[/tex] where [tex]\( I \)[/tex] is the current, \( [tex]L \)[/tex] is the length of the conductor, [tex]\( B \)[/tex]is the magnetic field strength, and [tex]\( \theta \)[/tex] is the angle between the current direction and the magnetic field.
Given:
[tex]- \( I = 4.1 \times 10^3 \) A (since 4.1 kA = 4.1 \(\times\) 10^3 A) - \( L = 1.3 \) m\\ - \( B = 12 \) T\\ \( \theta = 60^\circ \)\[/tex]
Converting the angle from degrees to radians (since the sine function in physics formulas typically uses radians), we have[tex]\( \theta = 60^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{3} \) radians.[/tex]
Now, we can calculate the force:
[tex]\( F = ILB \sin(\theta) \)[/tex]
[tex]\( F = (4.1 \times 10^3 \text{ A}) \times (1.3 \text{ m}) \times (12 \text{ T}) \times \sin\left(\frac{\pi}{3}\right) \)[/tex]
Since[tex]\( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)[/tex], we can substitute this value into the equation:
[tex]\( F = (4.1 \times 10^3) \times (1.3) \times (12) \times \frac{\sqrt{3}}{2} \)[/tex]
Calculating the force in newtons:
[tex]\( F = 4.1 \times 10^3 \times 1.3 \times 12 \times \frac{\sqrt{3}}{2} \)[/tex]
[tex]\( F \approx 4.1 \times 1.3 \times 12 \times 0.866 \times 10^3 \)[/tex]
[tex]\( F \approx 5107.2 \times 0.866 \)[/tex]
[tex]\( F \approx 4424.4 \) N[/tex]
To convert the force from newtons to pounds, we use the conversion factor [tex]\( 1 \text{ N} \approx 0.22481 \text{ lb} \):[/tex]
[tex]\( F \approx 4424.4 \text{ N} \times 0.22481 \frac{\text{lb}}{\text{N}} \)[/tex]
[tex]\( F \approx 994.5 \text{ lb} \)[/tex]
Therefore, the magnetic force exerted on the conducting bar is approximately 994.5 pounds.
In conclusion, the claim that the magnetic force will exceed 10,000 pounds is incorrect. The actual force is approximately 994.5 pounds, which is less than 10,000 pounds. The conducting bar should still be clamped in place to prevent movement due to the magnetic force, but the initial claim overestimated the force."
The force of attraction between a star of mass M and a planet of mass m (where т «М) is: ЗЫ? a F where is the angular momentum of the planet and a,b are both positive constants. [Note: black hole, in the non- this does approximate the force of attraction between a planet and a relativistic limit, with a = GMm. (a) 15 points Under what conditions is a stable circular orbit possible? Give the radius of the stable circular orbit in terms of the given parameters (M, m,a, b, l). function of a (b) 15 pointsWhat is the smallest radius possible for any circular orbit as a and b, allowing for arbitrary l? (Hint: this occurs in the limit of very large .) Is this circular orbit stable or unstable? (c) [10 points] If the planet travels in a slightly non-circular orbit about a stable radius, find expression for the angular frequency of small radial oscillations
Answer:
2b2t
Explanation:
2b2t
An electron is released from rest in a uniform electric field. The electron accelerates, travelling 6.20 m in 4.50 µs after it is released. What is the magnitude of the electric field in N/C?
Answer:
E= 3.4893 N/C
Explanation:
Given s=6.20 m , t=2.50μs, m=9.11*10^-31 Kg , q= 1.6*10^-19 C
the distance traveled by the electron in time t is
s=ut+0.5at^2
here, u is the initial velocity of the electron, t is time taken and
a is acceleration.
Since the electron is initially at rest u=0
now s=0.5at^2
Therefore a=2s/t^2
also. we know that strength of electric field is
E=ma/q
[tex]E= \frac{2ma}{qt^2}[/tex]
now puting the values we get
[tex]E=\frac{9.11\times 10^-31\times 2\times 6.20}{1.6\times 10^-19\times (4.5\times 10^-6)^2}[/tex]
therefore, E= 3.4865 N/C
The magnitude of the electric field is calculated by first determining the acceleration of the electron and then using the electric force equation to find the electric field. The resulting electric field is 34.8 N/C.
To find the magnitude of the electric field, we first need to calculate the acceleration of the electron. Given that the electron travels a distance of 6.20 m in a time of 4.50 µs (4.50 × 10-6 s), we can use the equations of motion.
Initial velocity, u = 0 (since the electron is released from rest)
Time, t = 4.50 × 10-6 s
Distance, s = 6.20 m
Using the equation of motion: s = ut + 0.5at2
Substitute the values: 6.20 = 0 + 0.5a(4.50 × 10-6)2
6.20 = 0.5a(20.25 × 10-12)
a = 6.20 / (0.5 × 20.25 × 10-12)
a = 6.20 / (10.125 × 10-12)
a = 6.12 × 1011 m/s2
Now, we calculate the electric field using Newton's Second Law, F = ma, and the electric force equation, F = eE, where e is the charge of the electron (1.60 × 10-19 C) and E is the electric field.
ma = eE
(9.11 × 10-31 kg)(6.12 × 1011 m/s2) = (1.60 × 10-19 C)E
(5.57 × 10-19 N) = (1.60 × 10-19 C)E
E = 5.57 × 10-19 N / 1.60 × 10-19 C
E = 3.48 × 101 N/C
Thus, the magnitude of the electric field is 34.8 N/C.
he "escape velocity from Earth (the speed required to escape Earth's gravity) is 2.5 x 10 miles per hour. What is this speed in m/s? (1 mile 1609 m)
Answer: [tex]11.17\ \text{ m/s}[/tex]
Explanation:
Given : The escape velocity : [tex]v=2.5\times10\text{ miles per hour}[/tex]
We know that 1 mile = 1609 meters (approx)
and 1 hour= 3600 seconds
To convert escape velocity 2.5 x 10 miles per hour into m/s , we need to multiply it by 1609.34 and divide it by 3600.
Thus, the escape velocity in m/s is given by :-
[tex]v=2.5\times10\times\dfrac{1609}{3600}\\\\=11.1736111111\approx11.17\text{ m/s}[/tex]
Hence, the speed in m/s = 11.17
A caterpillar tries to climb straight up a wall a meter high, but for every 2 cm up it climbs, it slides down 1 cm. Eventually, it reaches the top. When it reaches the top, it does not pull itself over so it will slide down 1 cm. What is the total displacement traveled? (Include direction, whether up, down, or n/a.)
Answer:
Total displacement traveled = 298
Explanation:
According to the given information, to actually climb for 1 cm, the caterpillar has to travel for 3 cm (2 cm upwards and 1 cm downwards).
So in order to climb straight up a one meter (100 cm) high wall, it needs to travel for 99 × 3 = 297 cm.
Then after a little it can travel up another cm to reach the top.
Therefore, the total displacement traveled = 297 + 1 = 298 cm
"Displacement" means the distance and direction from start to finish, regardless of what happened in between.
The caterpillar's displacement is 99 cm straight up.
Suppose you apply a force of 40 N to a 0.25-meter-long wrench attached to a bolt in a direction perpendicular to the bolt. Determine the magnitude of the torque when the force is applied at an angle of 135degrees to the wrench.
Answer:
Torque = 7.07 N.m
Explanation:
It is given that,
Force acting on the wrench, F = 40 N
Length of wrench, l = 0.25 meters
It is attached perpendicular to the bolt such that the force is applied at an angle of 135 degrees to the wrench. The formula for torque is given by :
[tex]\tau=r\times F[/tex]
[tex]\tau=rF\ sin\theta[/tex]
[tex]\tau=0.25\ m\times 40\ N\ sin(135)[/tex]
[tex]\tau=7.07\ N.m[/tex]
So, the magnitude of torque applied to the wrench is 7.07 N-m. Hence, this is the required solution.
To calculate the magnitude of the torque, use the formula τ = r * F * sin(θ), where r is the lever arm length (0.25 m), F is the force applied (40 N), and θ is the angle between force and lever arm (135°). The sine of 135° provides the necessary component of the force that contributes to the torque.
Explanation:The question deals with the concept of torque in physics, particularly how torque is influenced by the angle at which a force is applied. Torque (τ) is the product of the force (F) applied, the distance (r) from the pivot point to the point where the force is applied, and the sine of the angle (θ) between the force vector and the lever arm, which can be represented as τ = r * F * sin(θ). Given that a force of 40 N is applied to the 0.25-meter-long wrench at a 135-degree angle to the wrench, the magnitude of the torque can be calculated using this formula.
Using the provided equation:
Torque = radius * force * sin(angle)Torque= 0.25 m * 40 N * sin(135°) = 0.25 m * 40 N * sin(135°)
Here, sin(135°) is a positive value since 135° is in the second quadrant where sine values are positive. It is important to note that the angle must be converted to radians or the correct sine value must be used if the calculator is set to degrees. The calculated torque will have the unit of Newton-meters (N.m).
The world energy consumption was about 6*10^22 J. How much area must a parallel plate capacitor need to store this energy Assume we maintain the capacitor at delta V= 5 volts for safety reasons, and have a plate separation distance of 1 meter. 5 *10^32 m^2 5 *10^61 m^2 9.10^8 m^2 4*10^14 m^2 2 *10^23 m^2
Answer:
[tex]A = 5 \times 10^{32} m^2[/tex]
Explanation:
As we know that the energy stored in the capacitor is given as
[tex]Q = \frac{1}{2}CV^2[/tex]
here we know that
[tex]Q = 6 \times 10^{22} J[/tex]
also we know that
[tex]V = 5 Volts[/tex]
now we have
[tex]6 \times 10^{22} = \frac{1}{2}C(5^2)[/tex]
[tex]C = 4.8 \times 10^{21} F[/tex]
now we know the formula of capacitance
[tex]C = \frac{\epsilon_0 A}{d}[/tex]
[tex]4.8 \times 10^{21} = \frac{(8.85 \times 10^{-12})(A)}{1}[/tex]
[tex]A = 5 \times 10^{32} m^2[/tex]
The maximum wavelength for photoelectric emission in tungsten is 230 nm. What wavelength of light must be used inorder for electrons with a maximum energy of 1.5 eV to be ejected ?
Answer:
180.04 nm
Explanation:
λ₀ = maximum wavelength for photoelectric emission in tungsten = 230 x 10⁻⁹ m
E₀ = maximum energy of ejected electron = 1.5 eV = 1.5 x 1.6 x 10⁻¹⁹ J
λ = wavelength of light used = ?
Using conservation of energy
Energy of the light used = Maximum energy required for photoelectric emission + Energy of ejected electron
[tex]\frac{hc}{\lambda }=\frac{hc}{\lambda_{o} } + E_{o}[/tex]
[tex]\frac{(6.63 \times 10^{-34})(3 \times 10^{8})}{\lambda }=\frac{(6.63 \times 10^{-34})(3 \times 10^{8})}{230 \times 10^{-9} } + 1.5 \times 1.6 \times 10^{-19}[/tex]
λ = 180.04 x 10⁻⁹ m
λ = 180.04 nm
An eagle is flying horizontally at a speed of 3m/s when the fish in her talons wiggles loose and falls into the lake 10m below. Calculate the velocity of the fish relative to the water when it hits the water.
Answer:
Velocity of the fish relative to the water when it hits the water = 14.32 m/s along 77.91° below horizontal.
Explanation:
Vertical motion of fish:
Initial speed, u = 0
Acceleration, a = 9.81 m/s²
Displacement, s = 10 m
We have equation of motion, v² = u² + 2as
Substituting
v² = 0² + 2 x 9.81 x 10 = 196.2
v = 14 m/s
Final vertical speed = 14 m/s
Final horizontal speed = initial horizontal speed = 3 m/s
Final velocity = 3 i - 14 j m/s
Magnitude
[tex]v=\sqrt{3^2+(-14)^2}=14.32m/s[/tex]
Direction
[tex]\theta =tan^{-1}\left ( \frac{-14}{3}\right )=-77.91^0[/tex]
Velocity of the fish relative to the water when it hits the water = 14.32 m/s along 77.91° below horizontal.
According to the quantum theory of radiation,
A. radiation goes through only one of the two slits in the double-slit experiment with radiation.
B. an electromagnetic field can have only certain values of total energy and no other values.
C. electromagnetic fields are really made of particles.
D. electromagnetic fields must vibrate only at one particular frequency.
E. the total quantity of energy that an electromagnetic field can contain must be less than a certain maximum value
Answer:
Option B
Explanation:
According to quantum theory of radiation electromagnetic field or electromagnetic radiation( like light) produce by accelerated charge object and the quantum of EM radiations is photon which has discrete energy. So, EM field can have only certain values of total energy and no other value due the discrete nature of the energy of photon. Hence option B is correct
Suppose a plane accelerates from rest for 32.3, achieving a takeoff speed of 47.1 m/s after traveling a distance of 607 m down the runway. A smaller plane with the same acceleration has a takeoff speed of 28.2 m/s. Starting from rest, after what distance will this smaller plane reach its takeoff speed?
Answer:
217.28 m/s
Explanation:
u = 0, t 32.3 s, v = 47.1 m/s, s = 607 m
Let a be the acceleration.
Use third equation of motion.
v^2 = u^2 + 2 a s
47.1 x 47.1 = 0 + 2 a x 607
a = 1.83 m/s^2
For small plane
a = 1.83 m/s^2 , v = 28.2 m/s, u = 0, Let teh distance be s.
Use third equation of motion
28.2^2 = 0 + 2 x 1.83 x s
s = 217.28 m/s
A solenoid of length 0.700 m having a circular cross-section of radius 5.00 cm stores 6.00 µJ of energy when a 0.400-A current runs through it. What is the winding density of the solenoid? (μ 0 = 4π × 10-7 T · m/A)
Answer:
N/l = 104
Explanation:
Energy stored in the inductor is given by the formula
[tex]U = \frac{1}{2}Li^2[/tex]
now we have
[tex]6\times 10^{-6} = \frac{1}{2}L(0.400)^2[/tex]
now we have
[tex]L = 7.5 \times 10^{-5}[/tex]
now we have
[tex]L = \frac{\mu_0 N^2 \pi r^2}{l}[/tex]
[tex]7.5 \times 10^{-5} = \frac{4\pi \times 10^{-7} N^2 \pi(0.05)^2}{0.7}[/tex]
[tex]N = 73 turns[/tex]
now winding density is turns per unit length
[tex]N/l = 104[/tex]
The winding density of the given solenoid is 104 turns per meter.
Energy stored in inductorThe formula for energy stored in the inductor can be used to determine the inductance of the solenoid as follows.
U = ¹/₂LI²
6 x 10⁻⁶ = ¹/₂ (L) x (0.4)²
6 x 10⁻⁶ = 0.0.8L
L = 7.5 x 10⁻⁵
Number of turns of the solenoidThe number of turns of the solenoid is calculated as follows;
[tex]L = \frac{\mu N^2\pi r^2}{l} \\\\7.5 \times 10^{-5} = \frac{(4\pi \times 10^{-7} ) \times N^2 \times \pi(0.05)^2}{0.7} \\\\N = 73 \ turns[/tex]
Winding densityThe winding density if the number of turns per length
N/l = 73/0.7
N/l = 104
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4) Block A has a mass of 3kg and velocity of 13m/s, catching up with a second block B that has a mass of 3kg and is moving with 5m/s in the same direction. They suffer an elastic collision and block B is now moving with 8m/s. What is the speed of block A after collision? 10m/s
8m/s
5m/s
13m/s
18m/s
Answer:
Option A is the correct answer.
Explanation:
Here momentum is conserved.
That is [tex]\left (m_Av_A+m_Bv_B \right )_{initial}=\left (m_Av_A+m_Bv_B \right )_{final}[/tex]
Substituting values
[tex]3\times 13+3\times 5=3v_A+3\times 8\\\\3v_A=39+15-24\\\\3v_A=30\\\\v_A=10m/s[/tex]
Speed of block A after collision = 10 m/s
Option A is the correct answer.
Answer:
8m/s
Explanation:
What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 535 nm ? The index of refraction of the film is 1.33, and there is air on both sides of the film.
Explanation:
It is given that,
The thinnest soap film appears black when illuminated with light with a wavelength of 535 nm, [tex]\lambda=5.35\times 10^{-7}\ m[/tex]
Refractive index, [tex]\mu=1.33[/tex]
We need to find the thickness of soap film. The soap film appear black means there is an destructive interference. The condition for destructive interference is given by :
[tex]2t=m\dfrac{\lambda}{\mu}[/tex]
t = thickness of film
m = 0,1,2....
[tex]\mu[/tex] = refractive index
[tex]t=m\dfrac{\lambda}{2\mu}[/tex]
[tex]t=\dfrac{\lambda}{2\mu}[/tex]
For thinnest thickness, m = 1
[tex]t=1\times \dfrac{5.35\times 10^{-7}\ m}{2\times 1.33}[/tex]
[tex]t=2.01\times 10^{-7}\ m[/tex]
Hence, this is the required solution.
Final answer:
The thinnest soap film that appears is approximately 50.47 nm due to destructive interference.
Explanation:
The question is asking for the thinnest soap film that appears black when illuminated with light of a specific wavelength, in this case, 535 nm. The phenomenon described is known as thin film interference, which occurs when light waves reflected off the top and bottom surfaces of a film interfere with each other.
The film appears black at the thinnest point where destructive interference occurs, which means the reflected light waves are out of phase and cancel each other out.
To find the thinnest film thickness that appears black, we can use the formula for destructive interference in thin films, taking into account the phase shift that occurs upon reflection from a medium with a lower index of refraction to a higher one. This formula is:
[tex]2nt = (m + \(\frac{1}{2}\))\(\lambda\),[/tex]where n is the index of refraction, t is the thickness of the film, \(\lambda\) is the wavelength of light in vacuum, and m is an integer representing the order of the interference.
For the thinnest film, we use m = 0.
Therefore, the thinnest soap film thickness t is calculated as:
[tex]2nt = \(\frac{1}{2}\)\(\lambda\)2nt = \(\frac{1}{2}\) \\times 535 nm / 1.33t = \(\frac{535 nm}{(2 \\times 1.33 \\times 2)}\)t = 100.94 nm[/tex]
However, regarding physical film thickness, we should only consider the distance that light travels within the film, which is factored by n to give the optical path length. Thus, the actual thickness would be half of 100.94 nm, yielding a value of approximately 50.47 nm.
A cylindrical specimen of some metal alloy 11.2 mm (0.4409 in.) in diameter is stressed elastically in tension. A force of 14100 N (3170 lbf) produces a reduction in specimen diameter of 7 × 10-3 mm (2.756 × 10-4 in.). Compute Poisson's ratio for this material if its elastic modulus is 100 GPa (14.5 × 106 psi).
Answer:
The Poisson's ratio for this material is 0.4370.
Explanation:
Given that,
Diameter of metal = 11.2 mm
Force = 14100 N
Reduction diameter [tex]d=7\times10^{-3}\ mm[/tex]
Elastic modulus = 100 GPa
We need to calculate the change in length
Using formula of modulus elasticity
[tex]E=\dfrac{FL}{A\Delta L}[/tex]
The change in length is
[tex]\Delta L=\dfrac{FL}{AE}[/tex]
[tex]\dfrac{\Delta L}{L}=\dfrac{14100}{\pi\times\dfrac{(11.2\times10^{-3})^2}{4}100\times10^{9}}[/tex]
[tex]\dfrac{\Delta L}{L}=0.00143[/tex]
We need to calculate the Poisson's ratio
Using formula of Poisson's ratio
[tex]\nu=\dfrac{longitudinal\ strain}{Transverse strain}[/tex]
[tex]\nu=\dfrac{-\dfrac{\Delta d}{d}}{-\dfrac{\Delta L}{L}}[/tex]
Put the value into the formula
[tex]\nu=\dfrac{\dfrac{7\times10^{-6}}{11.2\times10^{-3}}}{0.00143}[/tex]
[tex]\nu=0.4370[/tex]
Hence, The Poisson's ratio for this material is 0.4370.
A series LR circuit consists of a 2.0-H inductor with negligible internal resistance, a 100-ohm resistor, an open switch, and a 9.0-V ideal power source. After the switch is closed, what is the maximum power delivered by the power supply?
Answer:
The maximum power delivered by the power supply is 0.81 W.
Explanation:
Given that,
Inductance L= 2.0 H
Resistance R = 100 ohm
Voltage = 9.0 V
We need to calculate the power
Using formula of power
[tex]P = \dfrac{V^2}{R}[/tex]
Where, P = power
V = voltage
R = resistance
Put the value into the formula
[tex]P = \dfrac{(9.0)^2}{100}[/tex]
[tex]P =0.81\ W[/tex]
Hence, The maximum power delivered by the power supply is 0.81 W.
The maximum power delivered by the power supply in a series LR circuit can be calculated using the maximum current, the resistance, and the voltage of the power supply.
Explanation:In a series LR circuit, the power delivered by the power supply is maximized when the current is at its maximum value. Initially, when the switch is closed, the current rises exponentially with time and eventually reaches its maximum value. The time constant of the circuit is T = L/R, where L is the inductance and R is the resistance. Therefore, the maximum power delivered by the power supply can be calculated using the formula P = (I_max)^2 * R, where I_max is the maximum current.
In this case, the initial current can be calculated using the formula I(0) = V/R, where V is the voltage of the power supply. The time constant can be calculated using the given inductance and negligible internal resistance. Plug these values into the formulas to find the maximum power delivered.
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Who is Janna Levin:
Professionally?
Race?
Contribution to Science?
Provide your Reference
Answer:
Janna levin is a cosmologist and professor at physics.
She is an american by race
She was the presenter of Nova feature Black hole Apocalypse and has writtenany science non-fiction books
www.jannalevin.com is her own page where u get her correct info and bio
A rod of 2.0-m length and a square (2.0 mm × 2.0 mm) cross section is made of a material with a resistivity of 6.0 × 10−8 Ω ⋅ m. If a potential difference of 0.50 V is placed across the ends of the rod, at what rate is heat generated in the rod?
The rate of heat generation in the rod is calculated by first determining the resistance using the formula R = ρL/A, then using that resistance value in the power formula P = V²/R. Using the provided values, the rate of heat generation in the rod under a potential difference of 0.50 V is 8.33 Watts.
Explanation:The subject of this question is about the rate of heat generation in a rod under a potential difference, which is a topic in Physics. To solve for this, we first need to compute for the resistance of the rod using the formula R = ρL/A, where R is the resistance, ρ is the resistivity of the material, L is the length, and A is the cross-sectional area of the rod. Given that ρ = 6.0 × 10−8 Ω ⋅ m, L = 2.0 m, and A = (2.0 mm × 2.0 mm) = 4.0 × 10-6 m², we get R = (6.0 × 10−8 Ω ⋅ m * 2.0 m) / 4.0 × 10-6 m² = 0.03 Ω.
Next, we use the formula P = V²/R to calculate the rate of heat generation (power). Here, P is the power, V is the potential difference, and R is the resistance. With V = 0.50 V and R = 0.03 Ω, after substituting the values we get P = (0.50 V)² / 0.03 Ω = 8.33 Watts. Therefore, the rate at which heat is generated in the rod is 8.33 Watts.
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The rate at which heat is generated in the rod is found by calculating the current and using Joule's Law. The final rate of heat generation is 8.33 Watts.
Heat Generated in a Rod Due to Electric Current
To determine the rate at which heat is generated in the rod, we start by calculating the resistance of the rod. Given:
Length of the rod (L) = 2.0 m
Cross-sectional area (A) = (2.0 mm × 2.0 mm) = (2.0 × [tex]10^{-3}[/tex] m) × (2.0 × [tex]10^{-3}[/tex] m) = 4.0 × [tex]10^{-6}[/tex] m²
Resistivity of the material (ρ) = 6.0 × [tex]10^{-8}[/tex] Ω·m
Potential difference (V) = 0.50 V
The resistance (R) of the rod can be calculated using the formula:
R = ρ(L/A)
Substituting the given values:
R = (6.0 × [tex]10^{-8}[/tex] Ω·m) × (2.0 m / 4.0 × [tex]10^{-6}[/tex] m²) = 3.0 × [tex]10^{-2}[/tex] Ω
Next, we calculate the current (I) flowing through the rod using Ohm's Law:
I = V / R
Substituting the values:
I = 0.50 V / 3.0 × [tex]10^{-2}[/tex] Ω = 16.67 A
The rate of heat generated (P) in the rod is given by Joule's Law:
P = I²R
Substituting the calculated values:
P = (16.67 A)² × 3.0 ×[tex]10^{-2}[/tex] Ω = 8.33 W
Therefore, the rate at which heat is generated in the rod is 8.33 Watts.
A circular loop with a 0.03-m radius and 18-Ω resistance is facing away from this page. A magnetic field that points into the page was 0.35 T one second ago and is now 0.25 T. What current (in which sense of circulation) appears in this loop, if any?
Answer:
15.7 x 10⁻⁶ A
Explanation:
r = radius of the circular loop = 0.03 m
Area of the loop is given as
A = πr² = (3.14) (0.03)² = 0.002826 m²
R = Resistance of the resistor = 189 Ω
ΔB = Change in magnetic field = 0.25 - 0.35 = - 0.10 T
Δt = time interval = 1 sec
Current is given as
[tex]i = - A\left ( \frac{\Delta B}{\Delta t} \right )[/tex]
[tex]i = \left ( \frac{A}{R} \right )\left ( \frac{\Delta B}{\Delta t} \right )[/tex]
[tex]i = \left ( \frac{-0.002826}{18} \right )\left ( \frac{- 0.10}{1} \right )[/tex]
i = 15.7 x 10⁻⁶ A
Ocean waves of wavelength 22 m are moving directly toward a concrete barrier wall at 4.0 m/s . The waves reflect from the wall, and the incoming and reflected waves overlap to make a lovely standing wave with an antinode at the wall. (Such waves are a common occurrence in certain places.) A kayaker is bobbing up and down with the water at the first antinode out from the wall. A) How far from the wall is she?B) What is the period of her up and down motion?
If ocean waves of wavelength 22 m are moving directly toward a concrete barrier wall at 4.0 m/s .
A. The kayaker is 11 meters away from the wall.
B. The period of the kayaker's up and down motion is 5.5 seconds.
What is the distance?
A) The distance from the kayaker to the wall is:
Distance = 1/2 × Wavelength
Distance = 1/2 × 22 m
Distance = 11 m
Therefore the kayaker is 11 meters away from the wall.
B) The period can be calculated using the formula:
Period = 1 / Frequency
Use this formula:
Frequency = Velocity / Wavelength
Frequency = 4.0 m/s / 22 m
Frequency ≈ 0.182 Hz
Period:
Period = 1 / 0.182 Hz
Period ≈ 5.49 seconds
Period ≈ 5.5 seconds
Therefore, the period of the kayaker's up and down motion is approximately 5.5 seconds.
A 0.43-kg object mass attached to a spring whose spring constant is 561 N/m executes simple harmonic motion. If its maximum speed is 8 m/s, what is the amplitude of its oscillation (in m)? Round your answer to the nearest tenth.
Answer:
0.22 m
Explanation:
m = 0.43 kg, K = 561 N/m
Vmax = 8 m/s
Let the amplitude of the oscillations be A.
The formula for the angular frequency of oscillation sis given by
[tex]\omega = \sqrt{\frac{K}{m}}[/tex]
[tex]\omega = \sqrt{\frac{561}{0.43}}[/tex]
ω = 36.1 rad/s
The formula for the maximum velocity is given by
Vmax = ω x A
A = Vmax / ω
A = 8 / 36.1 = 0.22 m
Salt water has greater density than fresh water. A boat floats in both fresh water and in salt 12) water. Where is the buoyant force greater on the boat?
Final answer:
The buoyant force on a boat is the same in both salt water and fresh water, as it depends on the weight of the fluid displaced, not on the fluid's density.
Explanation:
According to the principle discovered by Archimedes, a boat experiences buoyant force which is equal to the weight of the water it displaces. This buoyant force is not dependent on the weight of the boat but rather on the density of the fluid and the volume of the fluid displaced.
As the density of salt water is higher than that of fresh water, a boat will displace less volume of salt water to float. Therefore, the weight of the salt water displaced is equal to the weight of fresh water displaced when the boat floats.
This means that the buoyant force on the boat is the same in both salt water and fresh water.
The elementary particle called a muon is unstable and decays in about 2.20μs2.20μs , as observed in its rest frame, into an electron, a neutrino, and an antineutrino. What lifetime do you observe for muons approaching you at 0.9270.927 the speed of light?
Answer:
5.865 μs
Explanation:
t₀ = Time taken to decay a muon = 2.20 μs
c = Speed of Light in vacuum = 3×10⁸ m/s
v = Velocity of muon = 0.927 c
t = Lifetime observed
Time dilation
[tex]t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{1-\frac{(0.927c)^2}{c^2}}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{1-0.927^2}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{0.140671}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{0.3750}\\\Rightarrow t=5.865\times 10^{-6}\ seconds[/tex]
∴Lifetime observed for muons approaching at 0.927 the speed of light is 5.865 μs
Use Gauss' law to find the E field of an infinite solid cylinder of charge of radius R and charge density lambda per unit length. As usual, draw everything on the diagrams and label them.
Answer:
[tex]E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex]
Explanation:
As we know that electric field due to long cylinder on a cylindrical Gaussian surface must be constant
so on the Gaussian surface we will have
[tex]\int E. dA = \frac{q_{en}}{\epsilon_0}[/tex]
now the electric field is passing normally through curved surface area of the cylinder
so we will have
[tex]E (2\pi rL) = \frac{q_{en}}{\epsilon_0}[/tex]
here enclosed charge in the cylinder is given as
[tex]q_{en} = \lambda L[/tex]
from above equation
[tex]E(2\pi rL) = \frac{\lambda L}{\epsilon_0}[/tex]
[tex]E = \frac{\lambda}{2\pi \epsilon_0 r}[/tex]
An electron is released from rest in a uniform electric field of 418 N/C near a particle detector. The electron arrives at the detector with a speed of 3.70 x 106 m/s (a) What was the uniform acceleration of the electron? (Enter the magnitude.) m/s (b) How long did the electron take to reach the detector? (c) What distance was traveled by the electron? cm (d) What is the kinetic energy of the electron when it reaches the detector?
Answer:
a) 7.35 x 10¹³ m/s²
b) 5.03 x 10⁻⁸ sec
c) 9.3 cm
d) 6.23 x 10⁻¹⁸ J
Explanation:
E = magnitude of electric field = 418 N/C
q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
m = mass of the electron = 9.1 x 10⁻³¹ kg
a)
acceleration of the electron is given as
[tex]a = \frac{qE}{m}[/tex]
[tex]a = \frac{(1.6\times 10^{-19})(418)}{(9.1\times 10^{-31})}[/tex]
a = 7.35 x 10¹³ m/s²
b)
v = final velocity of the electron = 3.70 x 10⁶ m/s
v₀ = initial velocity of the electron = 0 m/s
t = time taken
Using the equation
v = v₀ + at
3.70 x 10⁶ = 0 + (7.35 x 10¹³) t
t = 5.03 x 10⁻⁸ sec
c)
d = distance traveled by the electron
using the equation
d = v₀ t + (0.5) at²
d = (0) (5.03 x 10⁻⁸) + (0.5) (7.35 x 10¹³) (5.03 x 10⁻⁸)²
d = 0.093 m
d = 9.3 cm
d)
Kinetic energy of the electron is given as
KE = (0.5) m v²
KE = (0.5) (9.1 x 10⁻³¹) (3.70 x 10⁶)²
KE = 6.23 x 10⁻¹⁸ J
The space shuttle travels at a speed of about 7.6times10^3 m/s. The blink of an astronaut's eye lasts about 110 ms. How many football fields (length=91.4 m) does the shuttle cover in the blink of an eye? The three-toed sloth is the slowest-moving land mammal. On the ground, the sloth moves at an average speed of 0.033 m/s, considerably slower than the giant tortoise have gone relative to the sloth? Do not put units in your answer.
Answer:
It covers distance of 9.15 football fields in the said time.
Explanation:
We know that
[tex]Distance=Speed\times Time[/tex]
Thus distance covered in blinking of eye =
[tex]Distance=7.6\times 10^{3}m/s\times 110\times 10^{-3}s\\\\Distance=836 meters[/tex]
Thus no of football fields=[tex]\frac{936}{91.4}=9.15Fields[/tex]
The acceleration of a particle is given by a = 6t - 28, where a is in meters per second squared and t is in seconds. Determine the velocity and displacement as functions of time. The initial displacement at t = 0 is s0= -8 m, and the initial velocity is v0 = 6 m/s. After you have the general expressions, evaluate these expressions at the indicated times.
Answers:
At t = 5.8 s, s = m, v = m/s
At t = 12.7 s, s = m, v = m/s
Answer:
Velocity, V = 3t²- 28t+6
Displacement, s = t³ - 14t²+6t -8
At t = 5.8 s
s = -249.05 m
v = -55.48 m/s
At t = 12.7 s
s = -141.48 m
v = 134.27 m/s
Explanation:
We have acceleration of a particle is given by a = 6t - 28
Velocity
[tex]v=\int adt=\int (6t - 28)dt=3t^2-28t+C[/tex]
At t = 0 we have v₀ = 6 m/s
v₀ = 6 = 3 x 0 ²-28 x 0+C
C = 6
So velocity, V = 3t² - 28t+6
Displacement
[tex]s=\int vdt=\int (3t^2-28t+6)dt=t^3-14t^2+6t+C[/tex]
At t = 0 we have s₀ = -8 m
s₀ = -8 = 0³ + 14 x 0² + 6 x 0 + C
C = -8
So displacement, s = t³ - 14t²+6t -8
At t = 5.8 s
s = 5.8³ - 14 x 5.8²+6 x 5.8 - 8 = -249.05 m
v = 3 x 5.8² - 28 x 5.8 + 6 = -55.48 m/s
At t = 12.7 s
s = 12.7³ - 14 x 12.7²+6 x 12.7 - 8 = -141.48 m
v = 3 x 12.7² - 28 x 12.7 + 6 = 134.27 m/s
A solenoid having N turns and carrying a current of 2.000 A has a length of 34 00 cm. If the magnitude of the magnetic field generated at the center of the solenoid is 9.000 mT what is the value of N? (μo = 4π x10^-7 T. m/A) A) 2318 B) 1218 C) 860.0 D) 3183 E) 1591
Answer:
B) 1218
Explanation:
N = Total number of turns in the solenoid
L = length of the solenoid = 34.00 cm = 0.34 m
B = magnetic field at the center of the solenoid = 9 mT = 9 x 10⁻³ T
i = current carried by the solenoid = 2.000 A
Magnetic field at the center of the solenoid is given as
[tex]B = \frac{\mu _{o}N i}{L}[/tex]
[tex]9\times 10^{-3} = \frac{(4\pi\times 10^{-7} )N (2)}{0.34}[/tex]
N = 1218
The value of N is about B) 1218
[tex]\texttt{ }[/tex]
Further explanationLet's recall magnetic field strength from current carrying wire and from center of the solenoid as follows:
[tex]\boxed {B = \mu_o \frac{I}{2 \pi d} } [/tex]
B = magnetic field strength from current carrying wire (T)
μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)
I = current (A)
d = distance (m)
[tex]\texttt{ }[/tex]
[tex]\boxed {B = \mu_o \frac{I N}{L} } [/tex]
B = magnetic field strength at the center of the solenoid (T)
μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)
I = current (A)
N = number of turns
L = length of solenoid (m)
Let's tackle the problem now !
[tex]\texttt{ }[/tex]
Given:
Current = I = 2000 A
Length = L = 34.00 cm = 0.34 m
Magnetic field strength = B = 9000 mT = 9 T
Permeability of free space = μo = 4π × 10⁻⁷ T.m/A
Asked:
Number of turns = N = ?
Solution:
[tex]B = \mu_o \frac{I N}{L}}[/tex]
[tex]\frac{I N}{L} = B \div \mu_o[/tex]
[tex]IN = BL \div \mu_o[/tex]
[tex]N = BL \div (\mu_o I)[/tex]
[tex]N = ( 9 \times 0.34 ) \div ( 4 \pi \times 10^{-7} \times 2000 )[/tex]
[tex]\boxed {N \approx 1218}[/tex]
[tex]\texttt{ }[/tex]
Learn moreTemporary and Permanent Magnet : https://brainly.com/question/9966993The three resistors : https://brainly.com/question/9503202A series circuit : https://brainly.com/question/1518810Compare and contrast a series and parallel circuit : https://brainly.com/question/539204[tex]\texttt{ }[/tex]
Answer detailsGrade: High School
Subject: Physics
Chapter: Magnetic Field
The activation energy for the uncatalyzed decomposition of hydrogen peroxide at 20°C is 75.3 kJ/mol. In the presence of the enzyme catalase, the activation energy is reduced to 29.3 kJ/mol. Use the following form of the Arrhenius equation, RT ln1k1/k22 5 Ea2 2 Ea1 , to calculate how much larger the rate constant of the catalyzed reaction is.
Answer:
The rate of enzyme catalyzed reaction will increases by [tex]1.58\times 10^{8}[/tex] times.
Explanation:
According to the Arrhenius equation,
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
[tex]\log K=\log A-\frac{Ea}{2.303\times RT}[/tex]
The expression used with catalyst and without catalyst is,
[tex]\log K_1=\log A-\frac{Ea_1}{2.303\times RT}[/tex]...(1)
[tex]\log K_2=\log A-\frac{Ea_2}{2.303\times RT}[/tex]...(2)
On subtracting (2) from (1)
[tex]\log \frac{K_2}{K_1}=\frac{Ea_1-Ea_2}{2.303RT}[/tex]
where,
[tex]K_2[/tex] = rate of reaction with catalyst
[tex]K_1[/tex] = rate of reaction without catalyst
[tex]Ea_2[/tex] = activation energy with catalyst = 29.3 kJ/mol = 29300 J/mol
[tex]Ea_1[/tex] = activation energy without catalyst = 75.3 kJ/mol=75300 J/mol
R = gas constant =8.314 J /mol K
T = temperature = [tex]20^oC=273+20=293K[/tex]
Now on substituting all the values in the above formula, we get
[tex]\log \frac{K_2}{K_1}=\frac{75300 kJ/mol-29300 kJ/mol}{2.303\times 8.314 J/mol K\times 293}=1.58\times 10^{8}[/tex]
The rate of enzyme catalyzed reaction will increases by [tex]1.58\times 10^{8}[/tex] times.
The strength of the Earth’s magnetic field B at the equator is approximately equal to 5 × 10−5 T. The force on a charge q moving in a direction perpendicular to a magnetic field is given by F = q v B, where v is the speed of the particle. The direction of the force is given by the right-hand rule. Suppose you rub a balloon in your hair and your head acquires a static charge of 4 × 10−9 C. If you are at the equator and driving west at a speed of 60 m/s, what is the strength of the magnetic force on your head due to the Earth’s magnetic field? Answer in units of N
Answer:
1.2 x 10⁻¹¹ N
Explanation:
B = magnitude of magnetic field at the equator = 5 x 10⁻⁵ T
q = magnitude of charge acquired by the head = 4 x 10⁻⁹ C
v = speed of driving at the equator = 60 m/s
Magnitude of magnetic force on the head at the equator is given as
F = q v B
Inserting the above values in the equation
F = (4 x 10⁻⁹) (60) (5 x 10⁻⁵)
F = 1.2 x 10⁻¹¹ N