Answer:
FALSE
Explanation:
As we know that equation for charging of a capacitor is given by
[tex]q = CV(1 - e^{-\frac{t}{\tau}})[/tex]
now at t = 0 we can say there is no charge on the capacitor
now as the time will increase the value of charge will increase on the plates of capacitor
now after t = 1 time constant
we will have
[tex]q = CV(1 - e^{-1})[/tex]
[tex]q = 0.63 CV[/tex]
so after one time constant the capacitor will charged upto 63% of its maximum value.
A single loop of wire with an area of 0.0900 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendicular to the plane of the loop, and is decreasing at a constant rate of 0.190 T s. (a) What emf is induced in this loop? (b) If the loop has a resistance of find the current induced in the loop
Answer:
(a) 0.0171 V
Explanation:
A = 0.09 m^2, dB/dt = 0.190 T/s
(a) According to the law of electromagntic induction
e = dФ / dt
e = A dB / dt
e = 0.09 x 0.190 = 0.0171 V
(b)
as we know
i = e / R
we can find induced current by dividing induced emf by resistance
(a) The emf induced in the loop is 0.0171 V.
(b) The current in the loop is determined from the ratio of induced emf to resistance in the loop.
emf induced in the loopThe emf induced in the loop is calculated by applying Faradays law as follows;
emf = dФ/dt
emf = A dB/dt
where;
A is the area of the loopdB/dt is the change in magnetic field with timeemf = 0.09 x 0.19
emf = 0.0171 V
Current in the loopThe current in the loop is determined by applying Ohm's law;
I = emf/R
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A worker stands still on a roof sloped at an angle of 35° above the horizontal. He is prevented from slipping by static frictional force of 560 N. Find the mass of the worker.
Answer:
99.63 kg
Explanation:
From the force diagram
N = normal force on the worker from the surface of the roof
f = static frictional force = 560 N
θ = angle of the slope = 35
m = mass of the worker
W = weight of the worker = mg
W Cosθ = Component of the weight of worker perpendicular to the surface of roof
W Sinθ = Component of the weight of worker parallel to the surface of roof
From the force diagram, for the worker not to slip, force equation must be
W Sinθ = f
mg Sinθ = f
m (9.8) Sin35 = 560
m = 99.63 kg
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
A windmill is turning at 12.5 rad/s in a clockwise direction. The wind suddenly shifts, producing an angular acceleration of 0.75 rad/s2 in a counterclockwise direction. After 12.0 s, how many revolutions has the windmill blades turned?
Answer:
Windmill will complete 0.27 revolutions.
Explanation:
We have equation of motion s = ut + 0.5at²
Initial speed, u = 12.5 rad/s
Time, t = 12 seconds.
Angular acceleration, a = -0.75 rad/s²
Substituting
s = 12.5 x 12 - 0.5 x 0.75 x 12² = 96 rad
1 revolution = 360 rad
[tex]96rad=\frac{96}{360}=0.27revolution[/tex]
So, windmill will complete 0.27 revolutions.
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."
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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The angular velocity of the rear wheel of a stationary exercise bike is 5.40 rad/s at time t = 0.000 sec, and its angular acceleration is constant and equal to 1.50 rad/s^2. A particular spoke coincides with the +x axis at time t = 0.000 sec. (a) What angle (in rev) does this spoke make with the +x axis at time t = 4.00 s? (b) What is the angular velocity (rev/s) at this time?
Answer:
The angle and angular velocity at time 4.00 are 5.35 rev and 1.81 rev/s.
Explanation:
Given that,
Angular velocity = 5.40 rad/s
Time t = 0.000 sec
Angular acceleration = 1.50 rad/s^2
(a). We need to calculate the angle at time t = 4.00 s
Using formula for angle
[tex]\theta=\omega_{0}t+\dfrac{1}{2}\alpha t^2[/tex]
Where, [tex]\omega_{0}[/tex]=angular velocity
[tex]\alpha[/tex]=angular acceleration
t = time
Put the value into the formula
[tex]\theta=5.40\times4.00+\dfrac{1}{2}\times1.50\times(4.00)^2[/tex]
[tex]\theta=33.6\ rad[/tex]
[tex]\theta=\dfrac{33.6}{2\pi}\ rad[/tex]
[tex]\theta=5.35\ rev[/tex]
(b). We nee to calculate the angular velocity at 4.00 s
Using formula of angular velocity
[tex]\omega=\omega_{0}+\alpha t[/tex]
[tex]\omega =5.40+1.50\times4.00[/tex]
[tex]\omega=11.4\ rad/s[/tex]
[tex]\omega=\dfrac{11.4}{2\pi}\ rad/s[/tex]
[tex]\omega=1.81\ rev/s[/tex]
Hence, The angle and angular velocity at time 4.00 are 5.35 rev and 1.81 rev/s.
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
The momentum of an isolated system is conserved A) only in inelastic collisions. B) only in elastic collisions. C) in both elastic and inelastic collisions
Answer:
It is actually in both.
Explanation:
The momentum of an isolated system is conserved for both elastic and inelastic collision. OPtion C is correct
Collision is the process by which two bodies come into contact with the release of energy.
Collision can be elastic or inelastic.
Foe elastic collision, both momentum, and energy is conserved while for inelastic collision only momentum is conserved.
From the explanation above, we can see that momentum of an isolated system is conserved for both elastic and inelastic collision
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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
40-turn circular coil (radius = 4.0 cm, total resistance = 0.20 ) is placed in a uniform magnetic field directed perpendicular to the plane of the coil. The magnitude of the magnetic field varies with time as given by B = 50 sin(10 t) mT where t is measured in s. What is the magnitude of the induced current in the coil at 0.10 s?
To find the magnitude of the induced current in the coil at 0.10 s, we need to calculate the magnitude of the magnetic flux through the coil and use Faraday's law of electromagnetic induction.
Explanation:To find the magnitude of the induced current in the coil at 0.10 s, we need to first calculate the magnitude of the magnetic flux through the coil. The magnetic flux is given by Φ = B * A, where B is the magnetic field and A is the area of the coil. The area of the coil is calculated as A = π * r^2, where r is the radius of the coil.
Next, we can use Faraday's law of electromagnetic induction, which states that the induced emf is equal to the rate of change of magnetic flux. Mathematically, this can be written as emf = -dΦ/dt. Since the coil has resistance, we can use Ohm's Law to find the magnitude of the induced current, which is given by I = emf/R. Plugging in the values and calculating, we can find the magnitude of the induced current at 0.10 s.
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To find the magnitude of the induced current in the coil, we need to use Faraday's Law of Electromagnetic Induction. The magnetic field B is given as B = 50sin(10πt) mT. At t = 0.1s, the magnetic field is B = 50sin(10π(0.1)) mT.
Explanation:To find the magnitude of the induced current in the coil, we need to use Faraday's Law of Electromagnetic Induction, which states that the induced emf (voltage) is equal to the negative rate of change of magnetic flux through the coil.
The magnetic field B is given as B = 50sin(10πt) mT, where t is measured in seconds.
The magnetic flux through the coil is given by Φ = B.A, where A is the area of the coil.
Substituting the given values, we have A = πr^2 = π(0.04^2) = 0.005 cm^2.
At t = 0.1s, the magnetic field is B = 50sin(10π(0.1)) mT.
We can now calculate the magnetic flux through the coil using Φ = B.A.
Finally, we can use Faraday's Law to find the induced emf and divide it by the resistance of the coil to find the magnitude of the induced current.
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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 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.
What is the value of (8 104)2, written with the correct mumber of significant figures? A) 206.324 B) 206.323 C)2063 D) 206 E) 200
Answer:
B) 206.323
Explanation:
The value of (8 104)2, written with the correct mumber of significant figures is 206.323.
Answer:
B. 206.323
Explanation:
As a radioactive material undergoes nuclear disintegration, a) b) c) d) e) Its atomic number always remains the same Its mass number never increases; Its mass number never decreases; It always emits an alpha particle; None of the preceding statements is correct.
The differential equation below models the temperature of a 91°C cup of coffee in a 21°C room, where it is known that the coffee cools at a rate of 1°C per minute when its temperature is 71°C. Solve the differential equation to find an expression for the temperature of the coffee at time t. (Let y be the temperature of the cup of coffee in °C, and let t be the time in minutes, with t = 0 corresponding to the time when the temperature was 91°C.) dy/dt =(− 1/50)(y − 21)
Answer:
[tex]T=21+70e^{\frac{-t}{50} }[/tex]
Explanation:
We start from:
[tex]\frac{dT}{dt}=\frac{-1}{50}(T-21)[/tex]
Separating variables:
[tex]-50dT=(T-21)dt[/tex]
[tex]-50\frac{dT}{T-21}=dt[/tex]
Integrating with initial conditions:
[tex]-50\int\limits^{T}_{91} {\frac{1}{T-21} } \, dT= \int\limits^t_{0} {} \, dt[/tex]
[tex]-50ln(\frac{T-21}{91-21})=t[/tex]
[tex]ln(\frac{T-21}{71})=\frac{-t}{50}[/tex]
Isolating T:
[tex]\frac{T-21}{70} =e^\frac{-t}{50} }[/tex]
[tex]T=21+70e^{\frac{-t}{50} }[/tex]
You may note that when t is zero the temperature is 91 ºC, as is specified by the problem. As well, when t is bigger (close to infinite), the temperature tends to be the room temperature (21 ºC)
The given differential equation can be solved to yield the function y(t) = 70 e^(-t/50) + 21, which describes the temperature of the coffee as a function of time.
Explanation:In your given differential equation, dy/dt = (− 1/50)(y − 21), y represents the temperature of the coffee at time t, and the equation describes how the temperature changes over time. This is a type of first-order linear differential equation, which can be solved using an integrating factor. The general solution of such equation is given by y(t) = [integral(t, e^(-t/50)*(-1/50)dt] + C e^(t/50), where C is a constant. To solve for C, we use the initial condition: at t = 0, y = 91°C, which yields C = (91 - 21), or C = 70. Substituting back into the original equation provides the final formula for the temperature of the coffee at given time t: y(t) = 70 e−t/50 + 21. It states that, the temperature of coffee decreases over time from its initial temperature, and it will eventually cool to the same temperature as the room (21°C).
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Acharged particle moving through a magnetic field at right angles to the field with a speed of 36.2 m/s experiences a magnetic force of 7.38x104 N. Determine the magnetic force on an identical particle when it travels through the same magnetic field with a speed of 12.1 m/s at an angle of 17° relative to the magnetic field. Express your answer in microNewtons.
Answer:
7212.3 N
Explanation:
F = 7.38 x 10^4 N, v = 36.2 m/s
Let b be the strength of magnetic field and charge on the particle is q.
F = q v B Sin theta
Here theta = 90 degree
7.38 x 10^4 = q x 36.2 x B x Sin 90
q B = 2038.7 .....(1)
Now, theta = 17 degree, v = 12.1 m/s
F = q v B Sin theta
F = 2038.7 x 12.1 x Sin 17 ( q v = 2038.7 from equation (1)
F = 7212.3 N
A metal wheel 50 cm in radius rotates on a desk anti-clockwise at a constant rate of 800 rev/min about its central axis. Determine (a) its angular velocity (not speed!!!),
(b) its linear speed at a point 30 cm from its center of rotation,
(c) the radial acceleration of a point on the rim, and
(d) the total distance a point on the rim moves in 120 seconds.
Note: You must draw a diagram and pick appropriate axes to aid your calculation.
Write down the detail process.
Answer:
(a) 83.73 rad/s
(b) 251.2 m/s
(c) 3505.4 rad/s^2
(d) 5024 m
Explanation:
R = 50 cm = 0.5 m, f = 800 rpm = 800 / 60 rps
(a) Angular velocity, w = 2 x 3.14 x f = 2 x 3.14 x 800 / 60 = 83.73 rad / s
(b) The relation between linear speed and the angular speed is
V = r w
Here, r = 30 cm = 0.3 m
V = 0.3 x 83.73 = 25.12 m/s
(c) Radial acceleration = R w^2 = 0.5 x 83.73 x 83.73 = 3505.4 rad/s^2
(d) Time period T = 2 x 3.14 / w
T = 2 x 3.14 / 83.73 = 0.075 sec
In 0.075 second, angle turn = 360 degree
In 120 second, the angle turn = 360 x 120 / 0.075 = 576000 degree
In 360 degree, the distance traveled = 2 x pi x R
In 576000 degree, the distance traveled = 2 x 3.14 x 0.5 x 576000 / 360
= 5024 m
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]
We are given a piece of copper of mass m=0.82 g to form a cylindrical wire of resistance R=0.87.2. What should be the length / of this wire?: 2.16 m. What would be the diameter d of this wire?: d 0.232 mm. To make things certain, let us use copper's mass density of 8.96 g/cm3 and resistivity p= 1.7 x 10-8 2.m.
Answer:
Length of wire=2.16 m
Diameter of wire=0.232 mm
Explanation:
m= mass of copper wire= 0.82 g
R= Resistance of copper wire= 0.87 ohms
D= Density of copper= 8.96 g/cm^3
ρ= Resistivity= 1.7×10^-8 Ωm
[tex]Density=\frac{mass}{volume}\\\Rightarrow volume=\frac{mass}{density}\\\Rightarrow volume=\frac {0.82}{8.96}\\\Rightarrow volume=0.091\ cm^3\\ volume = \pi r^2 l\\\Rightarrow \pi r^2=\frac{volume}{l}\\ \Rightarrow \pi r^2=\frac {0.091}{l}\\[/tex]
[tex]\rho=R\frac{A}{l}\\\Rightarrow 1.7\times 10^{-8}=0.87 \frac{\pi r^2}{l}\\\Rightarrow 1.7\times 10^{-8}=0.87 \frac{0.091\times 10^{-6}}{l^2}\\\Rightarrow l^2=\frac {0.87\times 0.091\times 10^{-6}}{1.7\times 10^{-8}}\\\Rightarrow l^2=0.046\times 10^2\\\Rightarrow length=2.16\ m[/tex]
[tex]\pi r^2=\frac {0.091}{l}\\\Rightarrow r^2=\frac {0.091\times 10^{-6}}{2.16 \pi}\\\Rightarrow r^2=1.34\times 10^{-8}\\\Rightarrow r=0.00011\ m\\\Rightarrow d=0.000232\ m\\\therefore diameter=0.232\ mm[/tex]
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.
The bar of Prob. 5/82 is repeated here. The ends of the 0.4-m bar remain in contact with their re- spective support surfaces. End B has a velocity of 0.5 m/s and an acceleration of 0.3 m/s2 in the di- rections shown. Determine the angular accelera- tion of the bar and the acceleration of end A.
Answer:
0.24
Explanation:
w=vt
then
w1/v1=w2/v2
0.4/0.5=w2/0.3
w2= 0.3*0.4/0.5=0.24//
The angular acceleration of the bar is 0.75 rad/s², and the acceleration of end A is 0.3 m/s² in the opposite direction to end B. These values were determined using the relationships between linear and angular velocities and accelerations.
To determine the angular acceleration of the bar and the acceleration of end A, let's go through the step-by-step process:
Determine the Angular Velocity
Given that the bar has a length of 0.4 meters and the velocity of end B is 0.5 m/s, we start by calculating the angular velocity ω. The formula relating linear velocity v and angular velocity ω is:
[tex]v = \omega * r[/tex]
where r is the length of the bar. Rearranging for ω :
[tex]\omega = v / r = 0.5 m/s / 0.4 m = 1.25 rad/s[/tex]
Determine the Angular Acceleration
Next, we use the given acceleration of end B, which is 0.3 m/s². To find the angular acceleration α, we use the formula:
[tex]a = \alpha * r[/tex]
Rearranging for α:
[tex]\alpha = a / r = 0.3 m/s^2 / 0.4 m = 0.75 rad/s^2[/tex]
Determine the Acceleration of End A
The acceleration of end A, which is located at the opposite end of the bar, can be computed using:
[tex]a_A = \alpha * r[/tex]
Since end A is on the same bar, r here will be the same:
[tex]a_A = 0.75 rad/s^2* 0.4 m = 0.3 m/s^2[/tex]
Therefore, the acceleration of end A is also 0.3 m/s², but in the opposite direction to end B.
A gun with a muzzle velocity of 100 m/s is fired horizontally from a tower. Neglecting air resistance, how far downrange will the bullet be 8 seconds later? g
Answer:
313.6 m downward
Explanation:
The distance covered by the bullet along the vertical direction can be calculated by using the equation of motion of a projectile along the y-axis.
In fact, we have:
[tex]y(t) = h +u_y t + \frac{1}{2}at^2[/tex]
where
y(t) is the vertical position of the projectile at time t
h is the initial height of the projectile
[tex]u_y = 0[/tex] is the initial vertical velocity of the projectile, which is zero since the bullet is fired horizontally
t is the time
a = g = -9.8 m/s^2 is the acceleration due to gravity
We can rewrite the equation as
[tex]y(t)-h = \frac{1}{2}gt^2[/tex]
where the term on the left, [tex]y(t)-h[/tex], represents the vertical displacement of the bullet. Substituting numbers and t = 8 s, we find
[tex]y(t)-h= \frac{1}{2}(-9.8)(8)^2 = -313.6 m[/tex]
So the bullet has travelled 313.6 m downward.
Suppose a 50 turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.325 m^2. It is squished to have no area in 0.225 s. What is the magnitude of the average induced emf in volts if the uniform magnetic field has a strength of 1.5 T?
Answer:
EMF = 108.3 Volts
Explanation:
As per Faraday's law of electromagnetic induction we know that rate of change in magnetic flux will induce EMF in closed loop
So it is given as
[tex]EMF = N\frac{d\phi}{dt}[/tex]
[tex]EMF = N\frac{BA - 0}{\Delta t}[/tex]
now we know that
N = 50 turns
B = 1.5 T
A = 0.325 m^2
[tex]\Delta t = 0.225 s[/tex]
now we have
[tex]EMF = (50)(\frac{1.5\times 0.325}{0.225})[/tex]
[tex]EMF = 108.3 Volts[/tex]
The magnitude of the average induced emf is 110.12 volts.
The magnitude of the average induced emf can be calculated using Faraday's law. The formula is given as:
emf = -N * A * (dB/dt)
Where:
N is the number of turns in the coilA is the area of the coildB/dt is the rate of change of the magnetic fieldIn this case, the number of turns is 50, the area is 0.325 m², and
the rate of change of the magnetic field is (0 T - 1.5 T) / 0.225 s = -6.67 T/s.
Substituting these values into the formula, we get:
emf = -(50)(0.325)(-6.67) = 110.12 V
The magnitude of the average induced emf is 110.12 volts.
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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
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.
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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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.