Explanation:
According to Ohms Law :
V = I * R
(A) R (Resistance) = 0.022 / 0.75 = 0.03 Ohms
Also,
[tex]r = \alpha \frac{length}{area} = \alpha \frac{5.8}{3.14 \times 0.001 \times 0.001} [/tex]
(B)
[tex] \alpha(resistivity) = 1.62 \times {10}^{ - 8} [/tex]
Drift speed is missing. It is given as;
1.7 × 10^(-5) m/s
A) R = 0.0293 ohms
B) ρ = 1.589 × 10^(-8)
C) n = 8.8 × 10^(28) electrons
This is about finding, resistance and resistivity.
We are given;Length; L = 5.8 m
Diameter; d = 2mm = 0.002 m
Radius; r = d/2 = 0.001 m
Voltage; V = 22 mv = 0.022 V
Current; I = 750 mA = 0.75 A
Area; A = πr² = 0.001²π
Drift speed; v_d = 1.7 × 10^(-5) m/s
A) Formula for resistance is;R = V/I
R = 0.022/0.75
R = 0.0293 ohms
B) formula for resistivity is given by;ρ = RA/L
ρ = (0.0293 × 0.001²π)/5.8
ρ = 1.589 × 10^(-8)
C) Formula for current density is given by;J = n•e•v_d
Where;
J = I/A = 0.75/0.001²π A/m² = 238732.44 A/m²
e is charge on an electron = 1.6 × 10^(-19) C
v_d = 1.7 × 10^(-5) m/s
n is number of free electrons per unit volume
Thus;
238732.44 = n(1.6 × 10^(-19) × 1.7 × 10^(-5))
238732.44 = (2.72 × 10^(-24))n
n = 238732.44/(2.72 × 10^(-24))
n = 8.8 × 10^(28)
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You stand 17.5 m from a wall holding a softball. You throw the softball at the wall at an angle of 38.5∘ from the ground with an initial speed of 27.5 m/s. At what height above its initial position does the softball hit the wall? Ignore any effects of air resistance.
The ball's horizontal position in the air is
[tex]x=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t[/tex]
It hits the wall when [tex]x=17.5\,\mathrm m[/tex], which happens at
[tex]17.5\,\mathrm m=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t\implies t\approx0.813\,\mathrm s[/tex]
Meanwhile, the ball's vertical position is
[tex]y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ t-\dfrac g2t^2[/tex]
where [tex]g[/tex] is the acceleration due to gravity, 9.80 m/s^2.
At the time the ball hits the wall, its vertical position (relative to its initial position) is
[tex]y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ(0.813\,\mathrm s)-\dfrac g2(0.813\,\mathrm s)^2\approx\boxed{10.7\,\mathrm m}[/tex]
A series LR circuit contains an emf source of 19 V having no internal resistance, a resistor, a 22 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4 s after the switch is closed, what is the resistance of the resistor
Answer: R = 394.36ohm
Explanation: In a LR circuit, voltage for a resistor in function of time is given by:
[tex]V(t) = \epsilon. e^{-t.\frac{L}{R} }[/tex]
ε is emf
L is indutance of inductor
R is resistance of resistor
After 4s, emf = 0.8*19, so:
[tex]0.8*19 = 19. e^{-4.\frac{22}{R} }[/tex]
[tex]0.8 = e^{-\frac{88}{R} }[/tex]
[tex]ln(0.8) = ln(e^{-\frac{88}{R} })[/tex]
[tex]ln(0.8) = -\frac{88}{R}[/tex]
[tex]R = -\frac{88}{ln(0.8)}[/tex]
R = 394.36
In this LR circuit, the resistance of the resistor is 394.36ohms.
Which unbalanced force accounts for the direction of the net force of the rocket?
a. Air resistance
b. Friction
c. Gravity
d. Thrust of rocket engine
It depends on what stage of the mission you're talking about.
==> While it's sitting on the pad before launch, the forces on the rocket are balanced, so there's no net force on it.
==> When the engines ignite, their thrust (d) is greater than the force of gravity. So the net force on the rocket is upward, and the spacecraft accelerates upward.
==> After the engines shut down, the net force acting on the rocket is due to Gravity (c).
. . . If the rocket has enough vertical speed, it escapes the Earth completely, and just keeps going.
. . . If it has enough horizontal speed, it enters Earth orbit.
. . . If it doesn't have enough vertical or horizontal speed, it falls back to Earth.
A rocket will preserve to speed up so long as there's a resultant pressure upwards resulting from the thrust of the rocket engine.
What unbalanced force bills for the course of the internet pressure of the rocket?A rocket launches whilst the pressure of thrust pushing it upwards is greater than the burden force because of gravity downwards. This unbalanced pressure reasons a rocket to accelerate upwards. A rocket will maintain to hurry up so long as there's a resultant force upwards resulting from the thrust of the rocket engine.
What's the net pressure of unbalanced?
If the forces on an item are balanced, the net pressure is zero. If the forces are unbalanced forces, the results do not cancel each difference. Any time the forces acting on an object are unbalanced, the net pressure is not 0, and the movement of the item modifications.
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Two identical rooms in a house are connected by an open doorway. The temperatures in the two rooms are maintained at different values. Which room contains more air
Answer:
The room with the lower temperature
Explanation:
Using
PV=nRT
Since both the rooms same volume and are connected, so they will have same pressure
PV=nRT=constant
nT=Constant/R=constant
If T is more n has to be less
Thus, lower the temperature, more the number molecules.
A simple pendulum is 3.00 m long. (a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at 3.00 m/s2? s (b) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating downward at 3.00 m/s2? s (c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 3.00 m/s2? s
Answer:
a,)3.042s
b)4.173s
c)3.281s
Explanation:
For a some pendulum the period in seconds T can be calculated using below formula
T=2π√(L/G)
Where L = length of pendulum in meters
G = gravitational acceleration = 9.8 m/s²
Then we are told to calculate
(a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at 3.00 m/s2?
Since oscillations for this pendulum is located in the elevator that is accelerating upward at 3.00 then
use G = 9.8 + 3.0 = 12.8 m/s²
Period T=2π√(L/G)
T= 2π√(3/12.8)
T=3.042s
b) (b) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating downward at 3.00 m/s2?
G = 9.8 – 3.0 = 6.8 m/s²
T= 2π√(3/6.8)
T=4.173s
C)(c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 3.00 m/s2?
Net acceleration is
g'= √(g² + a²)
=√(9² + 3²)
Then period is
T=2π√(3/11)
T=3.281s
Coherent light that contains two wavelengths, 660 nm (red) and 470 nm (blue), passes through two narrow slits that are separated by 0.310 mm. Their interference pattern is observed on a screen 4.40 m from the slits. What is the disatnce on the screen between the first order bright fringe for each wavelength?
Answer:
0.002699 m or 2.699 mm
Explanation:
y = Fringe distance
d= Distance between slits = 0.310mm
L = Screen distance = 4.40m
λ= Wavelength
Given from question
λ₁= 660 nm = 6.6 x 10^-9 m
λ₂= 470 nm = 4.7 x 10^-9 m
d = 0.340 mm = 3.4 x 10^-3 m
L = 4.40 m
In the case of constructive interference, we use below formula
y/L = mλ/d
For first order wavelength
(y₁/4.40) =(1×660x10⁻⁹)/(0.310*10⁻³)
y₁= (0.310*10⁻³)×(4.40)/(0.310*10⁻³)
y₁=0.00937m
(y2/4.40) =(1×470x10⁻⁹)/(0.310*10⁻³)
y2= =(1×470x10⁻⁹)×(4.40)/(0.310*10⁻³)
y2=0.00667m
distance between the fringes is given by (y₁ -y2)
=0.00937-0.00667=0.002699m
Therefore, distance on the screen between the first-order bright fringes for the two wavelengths is 0.002699 m or 2.699 mm
the atomic number of a nucleus increases during which nuclear reactions
Answer:
Answer A : Fusion followed by beta decay (electron emission)
Explanation:
Notice that you want the Atomic number to increase, that is the number of protons in a nucleus. So if all four cases given experience the same fusion of nuclei, the only one that net increases the number of protons in the last stage, is the reaction that undergoes a beta decay (with emission of an electron) thus leaving a positive imbalance of positive charge (proton generated in the beta decay of a neutron).
Therefore, answer A is the correct one.
Answer:
A : Fusion followed by beta decay (electron emission)
Explanation:
Ap3x
g Two point sources emit sound waves of 1.0-m wavelength. The source 1 is at x = 0 and source 2 is at x = 2.0 m along x-axis. The sources, 2.0 m apart, emit waves which are in phase with each other at the instant of emission. Where, along the line between the sources, are the waves out of phase with each other by π radians?
Answer:
constructive interferencia 0, 1 , 2 m
destructive inteferencia 1/4, 3/4. 5/4, 7/4 m
Explanation:
This exercise is equivalent to the double slit experiment, the two sources are in phase and separated by a distance, therefore the waves observed in the line between them have an optical path difference and a phase difference, given by the expression
Δr / λ = Φ / 2π
Δr = Φ/2π λ
let's apply this expression to our case
λ = 1 m
Δr = Φ 1 / 2π
We have constructive interference for angle of Φ = 0, 2π, ...
let's find the values where they occur
Φ Δr
0 0
2π 1
4π 2
Destructive interference occurs by Φ = π /2, 3π / 2, ...
Φ Δr
π/2 ¼ m
3π /2 ¾ m
5π /2 5/4 m
7π /2 7/4 m
An atom in the ground state has a collision with an electron, then emits a photon with a wavelength of 1240 nm. What conclusion can you draw about the initial kinetic energy of the electron
Answer:
attached below is the free body diagram of the missing illustration
Initial kinetic energy of the electron = 3 eV
Explanation:
The conclusion that can be drawn about the kinetic energy of the electron is
[tex]E_{e} = E_{3} - E_{1}[/tex]
E[tex]_{e}[/tex] = initial kinetic energy of the electron
E[tex]_{1}[/tex] = -4 eV
E[tex]_{3}[/tex] = -1 eV
insert the values into the equation above
[tex]E_{e}[/tex] = -1 -(-4) eV
= -1 + 4 = 3 eV
A race-car drives around a circular track of radius RRR. The race-car speeds around its first lap at linear speed v_iv i v, start subscript, i, end subscript. Later, its speed increases to 4v_i4v i 4, v, start subscript, i, end subscript. How does the magnitude of the car's centripetal acceleration change after the linear speed increases
Answer:
The magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.
Explanation:
The initial centripetal acceleration, a of the race-car around the circular track of radius , R with a linear speed v is a = v²/R.
When the linear speed of the race-car increases to v' = 4v, the centripetal acceleration a' becomes a' = v'²/R = (4v)²/R = 16v²/R.
So the centripetal acceleration, a' = 16v²/R.
To know how much the magnitude of the car's centripetal acceleration changes, we take the ratio a'/a = 16v²/R ÷ v²/R = 16
a'/a = 16
a' = 16a.
So the magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.
You have three resistors: R1 = 1.00 Ω, R2 = 2.00 Ω, and R3 = 4.00 Ω in parallel. Find the equivalent resistance for the combination
Answer:
4 / 7
Explanation:
1/total resistance = 1/1 + 1/2 + 1/4
= 1¾
total resistance = 1 ÷ 1¾
= 4/7
Find the current through a person and identify the likely effect on her if she touches a 120 V AC source in the following circumstances. (Note that currents above 10 mA lead to involuntarily muscle contraction.)
(a) if she is standing on a rubber mat and offers a total resistance of 300kΩ
(b) if she is standing barefoot on wet grass and has a resistance of only 4000kΩ
Answer:
A) 0.4 mA
B) 0.03 mA
Explanation:
Given that
voltage source, V = 120 V
to solve this question, we would be using the very basic Ohms Law, that voltage is proportional to the current and the resistance passing through the circuit, if temperature is constant.
mathematically, Ohms Law, V = IR
V = Voltage
I = Current
R = Resistance
from question a, we were given 300kΩ, substituting this value of resistance in the equation, we have
120 = I * 300*10^3 Ω
making I the subject of the formula,
I = 120 / 300000
I = 0.0004 A
I = 0.4 mA
Question said, currents above 10 mA causes involuntary muscle contraction, this current is way below 10 mA, so nothing happens.
B, we have Resistance, R = 4000kΩ
Substituting like in part A, we have
120 = I * 4000*10^3 Ω
I = 120 / 4000000
I = 0.00003 A
I = 0.03 mA
This also means nothing happens, because 0.03 mA is very much lesser compared to in the 10 mA
The current through a person will be:
a) 0.4 mA
b) 0.03 mA
Given:
Voltage, V = 120 V
Ohm's Law:It states that the voltage or potential difference between two points is directly proportional to the current or electricity passing through the resistance, and directly proportional to the resistance of the circuit.
Ohms Law, V = I*R
where,
V = Voltage
I = Current
R = Resistance
a)
Given: Resistance= 300kΩ
[tex]120 = I * 300*10^3 ohm\\\\I = 120 / 300000\\\\I = 0.0004 A[/tex]
Thus, current will be, I = 0.4 mA
b)
Given: R = 4000kΩ
[tex]120 = I * 4000*10^3 ohm\\\\I = 120 / 4000000\\\\I = 0.00003 A[/tex]
Thus, current will be, I = 0.03 mA
From calculations, we observe that nothing happens, because 0.03 mA is very much lesser compared to in the 10 mA.
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g In the atmosphere, the shortest wavelength electromagnetic waves are called A. infrared waves. B. ultraviolet waves. C. X-rays. D. gamma rays. E.
Answer:gamma ray
Explanation:
The temperature difference between the inside and the outside of a house on a cold winter day is 33°F. (a) Express this difference on the Celsius scale. 0.55 Incorrect: Your answer is incorrect. °C (b) Express this difference on the Kelvin scale. 273.7 Incorrect: Your answer is incorrect. K
Answer:
a) 0.56°C
b) 273.56 K
Explanation:
If we want to convert from Fahrenheit scale to Celcius scale we use the formula;
T(°C) = (T(°F) - 32) × 5/9
Where T(°F) = 33°F
Hence;
T(°C) = (33°F - 32) × 5/9
T°C = 0.56°C
b)
T(K) = T°C + 273
T(K) = 0.56 + 273
T(K) = 273.56 K
The actual depth of a shallow pool 1.00 m deep is not the same as the apparent depth seen when you look straight down at the pool from above. How deep (in cm) will it appear to be
Answer:
d' = 75.1 cm
Explanation:
It is given that,
The actual depth of a shallow pool is, d = 1 m
We need to find the apparent depth of the water in the pool. Let it is equal to d'.
We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,
[tex]n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{1\ m}{1.33}\\\\d'=0.751\ m[/tex]
or
d' = 75.1 cm
So, the apparent depth is 75.1 cm.
Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plates in Capacitor B are separated by twice the separation of the plates of Capacitor A. If Capacitor A has a capacitance of CA-17.8nF, what is the capacitance of Capacitor? .
Answer:
CB = 4.45 x 10⁻⁹ F = 4.45 nF
Explanation:
The capacitance of a parallel plate capacitor is given by the following formula:
C = ε₀A/d
where,
C = Capacitance
ε₀ = Permeability of free space
A = Area of plates
d = Distance between plates
FOR CAPACITOR A:
C = CA = 17.8 nF = 17.8 x 10⁻⁹ F
A = A₁
d = d₁
Therefore,
CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F ----------------- equation 1
FOR CAPACITOR B:
C = CB = ?
A = A₁/2
d = 2 d₁
Therefore,
CB = ε₀(A₁/2)/2d₁
CB = (1/4)(ε₀A₁/d₁)
using equation 1:
CB = (1/4)(17.8 X 10⁻⁹ F)
CB = 4.45 x 10⁻⁹ F = 4.45 nF
A wire along the z axis carries a current of 4.9 A in the z direction Find the magnitude and direction of the force exerted on a 3.3 cm long length of this wire by a uniform magnetic field pointing in the x direction having a magnitude 0.43T
Answer:
0.069 N, in the X directionExplanation:
According to Flemming's left hand rule, it sates that if the first three fingers of the left hand are held mutually at right angles to one another, the fore finger will point in the direction of magnetic field, the middle finger will point in direction of current, while the thumb will point to the direction of force.
Mathematically the law is stated as
F= BIL
given data
Magnetic field B= 0.43T
Current I= 4.9 A
length of conductor L= 3.3cm to meter , 3.3/100= 0.033 m
Applying the formula the force is calculated as
F= 0.43*4.9* 0.033= 0.069 N
According to Flemming's rule the direction of all parameters are mutually perpendicular to one another, then the Force is in the X direction
A long solenoid of radius 3 cm has 1100 turns per meter. If the solenoid carries a current of 1.5 A, then calculate the magnetic field at the center of the solenoid.a. 2.1E^-3T b. 1.0E^-3 T c. 1.7E^-4T d. 7.0E^-2 T
Answer:
The magnetic field at the center of the solenoid is 2.1 × 10⁻³ T
Explanation:
The magnetic field B at the center of the solenoid is given by
B = μ₀ni where μ₀ = permeability of free space = 4π × 10⁻⁷H/m, n = number of turns per unit length of the solenoid = 1100 turns per meter and i = current in the solenoid = 1.5 A.
So B = μ₀ni
= 4π × 10⁻⁷H/m × 1100 × 1.5 A
= 4π × 10⁻⁷H/m × 1650 A-turns/m
= 20734.5 × 10⁻⁷T
= 2.07345 × 10⁻³ T
≅ 2.1 × 10⁻³ T
So the magnetic field at the center of the solenoid is 2.1 × 10⁻³ T
Four friends push on the same block in different directions. Allie pushes on the block to the north with a force of 18 N. Bill pushes on the block to the east with a force of 14 N. Chris pushes on the block to south with a force of 23 N. Debra pushes on the block to the west with a force of 20 N. Assuming it does not move vertically, in which directions will the block move? north and west south and east south and west north and east
Answer:
South and West
Explanation:
Those people are pushing the hardest. It will move south faster than it moves west.
When using a crowbar to remove a nail, the person should hold it at which of the following spots to increase the IMA of the lever? nearest the end prying out the nail furthest from the end prying out the nail right in the middle the location where the person holds it will not affect the IMA
Answer: the furthest from the end prying out the nail
Answer:
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Explanation:
Two parallel slits are illuminated with monochromatic light of wavelength 567 nm. An interference pattern is formed on a screen some distance from the slits, and the fourth dark band is located 1.83 cm from the central bright band on the screen. (a) What is the path length difference corresponding to the fourth dark band? (b) What is the distance on the screen between the central bright band and the first bright band on either side of the central band? (Hint: The angle to the fourth dark band and the angle to the first bright band are small enough that tan θ ≈ sin θ.)
Answer:
a)1984.5nm
b)523mm
Explanation:
A)A destructive interference can be explained as when the phase shifting between the waves is analysed by the path lenght difference
θ=(m+0.5)λ where m= 1,2.3....
Where given from the question the 4th dark Fringe which will take place at m= 3
θ=7/2y
Where y= 567nm
= 7/2(567)=1984.5nm
But
B)tan θ ≈ y/d
And sinθ = mλ/d
y=mλd when m= 1 which is the first bright we have
Then y=(1× 567.D)/d
But the distance from Central to the 4th dark Fringe is 1.83cm then
y= 7λD/2d= 1.83cm
D/d=(2)×(1.83×10^-2)/(7×567×10^-9)
=92221.5
y= (567×10^-9)× (92221.5)
=0.00523m
Therefore, the distance between the first and center is y1-y0= 523mm
An electron moves through a uniform electric field E = (2.60i + 5.90j) V/m and a uniform magnetic field B= 0.400k in m/s^2.) T.
Required:
a. Determine the acceleration of the electron when it has a velocity v= 8.0i m/s.
b. What If? For the electron moving along the x-axis in the fields in part (a), what speed (in m/s) would result in the electron also experiencing an acceleration directed along the x-axis?
A) The acceleration of the electron along the x -axis is ; 4.57 * 10⁻¹¹ m /s²
B) The speed that would result in the electron experiencing an acceleration along the x-axis is 4.57 * 10⁻¹¹ * time m/s
Given Data :
Electric field ( E ) = ( 2.60i + 5.90j ) V/m
Magnetic field ( B ) = 0.400 k T
Velocity ( v ) = 8.0i m/s
A) Determine the acceleration of the electronApplying Lorentz force
F = q ( E + ( v * B ) )
= 1.6 * 10⁻¹⁹ ( 2.60 i + 5.90 j + ( 8.0 i * 0.4 k ) ) N
= 1.6 * 10⁻¹⁹ ( 2.60 i + 5.90 j + ( 3.2 ( -j ) ) N
= 1.6 * 10⁻¹⁹ ( 2.60 i + 2.70 j ) N
Ax = 4.57 * 10⁻¹¹ m /s²
B) The speed of the electron moving along the x-axisAx = Fx / Mc
= ( 1.6 * 10⁻¹⁹ * 2.60 ) / 9.1 * 10⁻³¹
= ( 4.16 * 10⁻¹⁹ ) / 9.1 * 10⁻³¹
= 0.457 * 10¹²
= 4.57 * 10⁻¹¹ m /s²
Therefore The speed that would result in the electron experiencing an acceleration along the x-axis is 4.57 * 10⁻¹¹ * time
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Which notation is better to use? (Choose between 4,000,000,000,000,000 m and 4.0 × 1015 m)
Answer:
4 x 10¹⁵
Explanation:
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm^2, separated by a distance of 1.70 mm. A 25.0-V potential difference is applied to these plates. Calculate: a. the electric field between the plates b. the surface charge density c. the capacitance d. the charge on each plate.
Answer:
(a) 1.47 x 10⁴ V/m
(b) 1.28 x 10⁻⁷C/m²
(c) 3.9 x 10⁻¹²F
(d) 9.75 x 10⁻¹¹C
Explanation:
(a) For a parallel plate capacitor, the electric field E between the plates is given by;
E = V / d -----------(i)
Where;
V = potential difference applied to the plates
d = distance between these plates
From the question;
V = 25.0V
d = 1.70mm = 0.0017m
Substitute these values into equation (i) as follows;
E = 25.0 / 0.0017
E = 1.47 x 10⁴ V/m
(c) The capacitance of the capacitor is given by
C = Aε₀ / d
Where
C = capacitance
A = Area of the plates = 7.60cm² = 0.00076m²
ε₀ = permittivity of free space = 8.85 x 10⁻¹²F/m
d = 1.70mm = 0.0017m
C = 0.00076 x 8.85 x 10⁻¹² / 0.0017
C = 3.9 x 10⁻¹²F
(d) The charge, Q, on each plate can be found as follows;
Q = C V
Q = 3.9 x 10⁻¹² x 25.0
Q = 9.75 x 10⁻¹¹C
Now since we have found other quantities, it is way easier to find the surface charge density.
(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e
σ = Q / A
σ = 9.75 x 10⁻¹¹ / 0.00076
σ = 1.28 x 10⁻⁷C/m²
At a rock concert, a dB meter registered 131 dB when placed 2.6 m in front of a loudspeaker on the stage. The intensity of the reference level required to determine the sound level is 1.0×10−12W/m2.
a) What was the power output of the speaker, assuming uniform spherical spreading of the sound and neglecting absorption in the air?
b) How far away would the sound level be 86 dB?
Answer:
Explanation:
A) 131 dB = 10*log(I / 1e-12W/m²)
where I is the intensity at 2.6 m away.
13.1 = log(I / 1e-12W/m²
1.25e13= I / 1e-12W/m²
I = 1.25 x10^1W/m²
power = intensity * area
P = I * A = 12.5W/m² * 4π(2.6m)² =1061 W ◄
B) 86 dB = 10*log(I / 1e-12W/m²)
8.6 = log(I / 1e-12W/m²)
3.98e8 = I / 1e-12W/m²
I = 3.98e-4 W/m²
area A = P / I = 1061W / 3.98e-4W/m² = 2.66e6 m²
A = 4πr²
2.66e6 m² = 4πr²
r = 14.5m ◄
A 6.7 cm diameter circular loop of wire is in a 1.27 T magnetic field. The loop is removed from the field in 0.16 ss . Assume that the loop is perpendicular to the magnetic field.
Required:
What is the average induced emf?
Answer:
The induced emf is [tex]\epsilon = 0.0280 \ V[/tex]
Explanation:
From the question we are told
The diameter of the loop is [tex]d = 6.7 cm = 0.067 \ m[/tex]
The magnetic field is [tex]B = 1.27 \ T[/tex]
The time taken is [tex]dt = 0.16 \ s[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = - N * \frac{\Delta \phi}{dt}[/tex]
Where N = 1 given that it is only a circular loop
[tex]\Delta \phi = \Delta B * A[/tex]
Where [tex]\Delta B = B_f - B_i[/tex]
where [tex]B_i[/tex] is 1.27 T given that the loop of wire was initially in the magnetic field
and [tex]B_f[/tex] is 0 T given that the loop was removed from the magnetic field
Now the area of the of the loop is evaluated as
[tex]A = \pi r^2[/tex]
Where r is the radius which is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{0.067}{2}[/tex]
[tex]r = 0.0335 \ m[/tex]
So
[tex]A = 3.142 * (0.0335)^2[/tex]
[tex]A = 0.00353 \ m^2[/tex]
So
[tex]\Delta \phi = (0- 127)* (0.00353)[/tex]
[tex]\Delta \phi = -0.00448 Weber[/tex]
=> [tex]\epsilon = - 1 * \frac{-0.00448}{0.16}[/tex]
=> [tex]\epsilon = 0.0280 \ V[/tex]
Two long straight wires carry currents perpendicular to the xy plane. One carries a current of 50 A and passes through the point x = 5.0 cm on the x axis. The second wire has a current of 80 A and passes through the point y = 4.0 cm on the y axis. What is the magnitude of the resulting magnetic field at the origin?
Answer:
450 x10^-6 T
Explanation:
We know that the magnetic of each wire is derived from
ByB= uoi/2pir
Thus B1= 4 x 3.14 x 10^-7 x50/( 2 x 3.142x 0.05)
= 0.2 x 10^ -3T
B2=
4 x 3.14 x 10^-7 x80/( 2 x 3.142x 0.04)
= 0.4 x 10^ -3T
So
(Bnet)² = (Bx)² + ( By)²
= (0.2² + 0.4²)mT
= 450 x10^-6T
The magnitude of magnetic field at the origin is required.
The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]
x = Location at x axis = 5 cm
y = Location at y axis = 4 cm
[tex]I_x[/tex] = Current at the x axis point = 50 A
[tex]I_y[/tex] = Current at the y axis point = 80 A
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi\times 10^{-7}\ \text{H/m}[/tex]
Magnitude of the magnetic field is given by
[tex]B=\dfrac{\mu_0I}{2\pi r}[/tex]
Finding the net magnetic field using the Pythagoras theorem
[tex]B^2=B_x^2+B_y^2\\\Rightarrow B^2=\left(\dfrac{\mu_0I_x}{2\pi x}\right)^2+\left(\dfrac{\mu_0I_y}{2\pi y}\right)^2\\\Rightarrow B=\dfrac{\mu_0}{2\pi}\sqrt{\left(\dfrac{I_x}{x}\right)^2+\left(\dfrac{I_y}{y}\right)^2}\\\Rightarrow B=\dfrac{4\pi\times 10^{-7}}{2\pi}\sqrt{\left(\dfrac{50}{0.05}\right)^2+\left(\dfrac{80}{0.04}\right)^2}\\\Rightarrow B=0.0004472=447.2\ \mu\text{T}[/tex]
The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]
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A 750 gram grinding wheel 25.0 cm in diameter is in the shape of a uniform solid disk. (we can ignore the small hole at the center). when it is in use, it turns at a consant 220 rpm about an axle perpendicular to its face through its center. When the power switch is turned off, you observe that the wheel stops in 45.0 s with constant angular acceleration due to friction at the axle. What torque does friction exert while this wheel is slowing down?
Answer:
Torque = 0.012 N.m
Explanation:
We are given;
Mass of wheel;m = 750 g = 0.75 kg
Radius of wheel;r = 25 cm = 0.25 m
Final angular velocity; ω_f = 0
Initial angular velocity; ω_i = 220 rpm
Time taken;t = 45 seconds
Converting 220 rpm to rad/s we have;
220 × 2π/60 = 22π/3 rad/s
Equation of rotational motion is;
ω_f = ω_i + αt
Where α is angular acceleration
Making α the subject, we have;
α = (ω_f - ω_i)/t
α = (0 - 22π/3)/45
α = -0.512 rad/s²
The formula for the Moment of inertia is given as;
I = ½mr²
I = (1/2) × 0.75 × 0.25²
I = 0.0234375 kg.m²
Formula for torque is;
Torque = Iα
For α, we will take the absolute value as the negative sign denotes decrease in acceleration.
Thus;
Torque = 0.0234375 × 0.512
Torque = 0.012 N.m
The unstretched rope is 20 meters. After getting dunked a few times the 80 kg jumper comes to rest above the water with the rope now stretched to 30 meters. What is the maximum length of the rope in meters when the jumper is being dunked?
Answer:
Therefore maximum stretch is y2 = 32.36 m
Explanation:
In this problem let's use the initial data to find the string constant, let's apply Newton's second law when in equilibrium
[tex]F_{e}[/tex] - W = 0
k Δx = mg
k = mg / Δx
k = 80 9.8 / (30-20)
k = 78.4 N / m
now let's use conservation of energy to find the velocity of the body just as the string starts to stretch y = 20 m
starting point. When will you jump
Em₀ = U = mg y
final point. Just when the rope starts to stretch
[tex]Em_{f}[/tex] = K = ½ m v²
Em₀ = Em_{f}
mg y = ½ m v²
v = √ 2g y
v = √ (2 9.8 20)
v = 19.8 m / s
now all kinetic energy is transformed into elastic energy
starting point
Em₀ = K = ½ m v²
final point
Em_{f} = [tex]K_{e}[/tex] + U = ½ k y² + m g y
Emo = Em_{f}
½ m v² = ½ k y² + mgy
k y² + 2 m g y - m v² = 0
we substitute the values and solve the quadratic equation
78.4 y² + 2 80 9.8 y - 80 19.8² = 0
78.4 y² + 1568 y - 31363.2 = 0
y² + 20 y - 400 = 0
y = [- 20 ±√ (20 2 +4 400)] / 2
y = [-20 ± 44.72] / 2
the solutions are
y₁ = 12.36 m
y₂ = 32.36 m
These solutions correspond to the maximum stretch and its rebound.
Therefore maximum stretch is y2 = 32.36 m
A fish appears to be 2.00 m below the surface of a pond when viewed almost directly above by a fisherman. What is the actual depth of the fish
Answer:
2,66
Explanation:
The refractive index= real depth/ apparent depth
real depth = refractive index * apparent depth
Let's assume index for water is 1.33
real depth = 2*1,33 = 2,66
