Answer:
Explanation:
At the point midway between wires
magnetic field due to wire having current 2I₀
= 10⁻⁷ x 2 x2I₀ / r where 2r is the distance between wires .
magnetic field due to wire having current I₀
= 10⁻⁷ x 4 I₀ / r
magnetic field due to wire having current I₀
= 10⁻⁷ x 2I₀ / r
= 10⁻⁷ x 2 I₀ / r where 2r is the distance between wires .
these fields are in opposite direction as direction of current is same in both .
net magnetic field = (4 - 2 )x 10⁻⁷ x I₀ / r
= 2 x 10⁻⁷ x I₀ / r
At point A net magnetic field = 2 x 10⁻⁷ x I₀ / r
At point B , we shall calculate magnetic field
magnetic field due to nearer wire having current 2 I₀ = 10⁻⁷ x 4 I₀ / r
magnetic field due to wire far away = 10⁻⁷ x 2 I₀ / 3r
These magnetic fields act in the same direction so they will add up
net magnetic field = [ (4 I₀ / r) + (2 I₀ / 3r) ] x 10⁻⁷
= (14 I₀ / 3r ) x 10⁻⁷
Magnetic field at point B = (14 I₀ / 3r ) x 10⁻⁷
Ratio of field at A and B
= 3 / 7 . Ans
The ratio of the magnitude of the magnetic field at point A to point B is :
3 / 7
Given data :
Magnitude of the left current is 2I₀
Magnitude of the right current is I₀
First step : Determine the magnetic field at point A
The magnetic field due to the left current ( 2I₀ )
10⁻⁷ * 2 * 2I₀ / r ( 2r = distance between wires )
The magnetic field due to the right current ( I₀ )
10⁻⁷ * 2 I₀ / r
From the expressions above the magnetic fields are in opposite direction
∴ Net magnetic field = (4 - 2 )* 10⁻⁷ * I₀ / r = 2 * 10⁻⁷ * I₀ / r
Hence The magnetic field at point A = 2 * 10⁻⁷ * I₀ / r
Next step : determine the magnetic field at point B
Magnetic field due to the closest wire to point B ( i.e.2I₀ ) = 10⁻⁷ * 4 I₀ / r
Magnetic field due to the wire away from point A = 10⁻⁷ * 2 I₀ / 3r
Since the fields acts in the same directions
The net magnetic field = (4 I₀ / r) + (2 I₀ / 3r) ] * 10⁻⁷ = ( 14 I₀ / 3r ) * 10⁻⁷
Hence The magnetic field at point A = ( 14 I₀ / 3r ) * 10⁻⁷
Therefore the ratio of the magnitude of the magnetic field at point A to point B = 3/ 7
Hence we can conclude that the ratio of the magnitude of the magnetic field at point A to point B = 3 / 7
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how does a system naturally change over time
Answer:
The movement of energy and matter in a system differs from one system to another. On the other hand, in open system both the matter and energy move into and out of the system. Therefore, matter and energy in a system naturally change over time will decrease in entropy.
Explanation:
Answer:
Decrease in entropy
Explanation:
Various systems which exist in nature possess energy and matter that move through these system continuously. The movement of energy and matter in a system differs from one system to another.
In a closed system for example, only energy flows in and out of the system while matter does not enter or leave the system.
On the other hand, in open system both the matter and energy move into and out of the system.
A metal sphere A of radius a is charged to potential V. What will be its potential if it is enclosed by a spherical conducting shell B of radius b and the two are connected by a wire?
Answer:
The potential will be Va/b
Explanation:
So Let sphere A charged Q to potential V.
so, V= KQ/a. ....(1
Thus, spherical shell B is connected to the sphere A by a wire, so all charge always reside on the outer surface.
therefore, potential will be ,
V ′ = KQ/b = Va/b... That is from .....(1), KQ=Va]
You want the current amplitude through a 0.450 mH inductor (part of the circuitry for a radio receiver) to be 1.50 mA when a sinusoidal voltage with an amplitude of 13.0 V is applied across the inductor. What frequency is required?
Answer:
3.067MHzExplanation:
The formula for calculating the voltage across an inductor is expressed as
[tex]V_l = IX_l\\\\Since\ X_l = 2\pi fL\\V_l = I(2\pi fL)[/tex]
Given parameters
current amplitude I = 1.50mA = 1.5*10⁻³A
inductance L = 0.450mH = 0.450*10⁻³H
Voltage across the inductor [tex]V_l[/tex] = 13.0V
Required
frequency f
Substituting the given parametres into the formula, we have;
[tex]V_l = I(2\pi fL)\\\\13 = 1.50*10^{-3}(2*3.14*f*0.450*10^{-3})\\\\13 = 4.239*10^{-6}f\\\\f = \frac{13}{4.239*10^{-6}} \\\\f = 3,066,761 Hertz\\\\f = 3.067MHz[/tex]
Hence, the frequency required is 3.067MHz
A rectangular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil with its north pole pointing down. What is the direction of the induced current in the coil?
Answer:
There is no induced current on the coil.
Explanation:
Current is induced in a coil or a circuit, when there is a break of flux linkage. A break in flux linkage is caused by a changing magnetic field, and must be achieved by a relative motion between the coil and the magnet. Holding the magnet above the center of the coil will cause no changing magnetic filed since there is no relative motion between the coil and the magnet.
A bungee cord with a spring constant of 800 StartFraction N over m EndFraction stretches 6 meters at its greatest displacement. How much elastic potential energy does the bungee cord have? The bungee cord has J of elastic potential energy.
Explanation:
EE = ½ kx²
EE = ½ (800 N/m) (6 m)²
EE = 14,400 J
Answer:
14,400 J
Explanation:
Its the answer
An airplane flies 1,592 miles east from Phoenix, Arizona, to Atlanta, Georgia, in 3.68 hours.
What is the average velocity of the airplane? Round your answer to the nearest whole number.
Answer:
433
Explanation:
A stationary coil is in a magnetic field that is changing with time. Does the emf induced in the coil depend
Answer:
Explanation:
The e.m.f induced in the coil depend on the following :
(a) No. of turns in the coil
(b) Cross-sectional Area of the coil
(c) Magnitude of Magnetic field
(d) Angular velocity of the coil
Electrons are accelerated through a voltage difference of 270 kV inside a high voltage accelerator tube. What is the final kinetic energy of the electrons?
Each electron winds up with kinetic energy of
(270 keV)
plus
(whatever KE it had when it started accelerating).
The AB rope is fixed to the ground at its A end, and forms 30º with the vertical. Its other end is connected to two ropes by means of the B-ring of negligible weight. The vertical rope supports the E block and the other rope passes through the grounded articulated pulley C to join at its end to the 80 N weight block D. The inclined section of the BD rope forms 60º with the vertical one; determine the weight of the E block necessary for the balance of the system and calculate the tension in the AB rope.
Answer:
T = 80√3 N ≈ 139 N
W = 160 N
Explanation:
Sum of forces on B in the x direction:
∑F = ma
80 N sin 60° − T sin 30° = 0
T = 80 N sin 60° / sin 30°
T = 80√3 N
T ≈ 139 N
Sum of forces on B in the y direction:
∑F = ma
80 N cos 60° + T cos 30° − W = 0
W = 80 N cos 60° + T cos 30°
W = 40 N + 120 N
W = 160 N
a 1010 W radiant heater is constructed to operate at 115 V. (a) What is the current in the heater when the unit is operating?
Answer:
8.78 AmpsExplanation:
Given data:
power rating of the heater P= 1010 W
voltage of the heater V= 115 volts
current taken by the heater I= ?
We can apply the power formula to solve for the current in the heater
i.e P= IV
Making I the current subject of formula we have
I= P/V
Substituting our given data into the expression for I we have
I=1010/115= 8.78 A
Hence the current when the unit/heater is operating is 8.78 AmpTwo blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of static friction are μA = 0.16 and μB = 0.23. Determine the incline angle θ for which both blocks begin to slide. Also find the required stretch or compression in the connecting spring for this to occur. The spring has a stiffness of k = 2.1 lb/ft .
Answer:
[tex]\theta=10.20^{\circ}[/tex]
[tex]\Delta l=0.10 ft[/tex]
Explanation:
First of all, we analyze the system of blocks before starting to move.
[tex]\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0[/tex]
[tex]\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]
[tex]16sin(\theta)-2.91cos(\theta)=0[/tex]
[tex]tan(\theta)=0.18[/tex]
[tex]\theta=arctan(0.18)[/tex]
[tex]\theta=10.20^{\circ}[/tex]
Hence, the incline angle θ for which both blocks begin to slide is 10.20°.
Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.
[tex]P_{A}sin(\theta)-F_{fA}-F_{spring}=0[/tex]
Where:
[tex]F_{spring} = k\Delta l=2.1\Delta l[/tex]
[tex]P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0[/tex]
[tex]\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}[/tex]
[tex]\Delta l=0.10 ft[/tex]
Therefore, the required stretch or compression in the connecting spring is 0.10 ft.
I hope it helps you!
(a) The inclined angle for which both blocks begin to slide is 10.3⁰.
(b) The compression of the spring is 0.22 ft.
The given parameters;
mass of block A, = 11 lbmass of block B, = 5 lbcoefficient of static friction for A, = 0.16coefficient of static friction for B, = 0.23 spring constant, k = 2.1 lb/ftThe normal force on block A and B:
[tex]F_n_A = m_Agcos \ \theta\\\\F_n_B = m_Bgcos \ \theta[/tex]
The frictional force on block A and B:
[tex]F_f_A = \mu_s_AF_n_A \\\\F_f_B = \mu_s_BF_n_A[/tex]
The net force on the blocks when they starts sliding;
[tex](m_Ag sin \theta+ m_Bgsin\theta) - (F_f_A + F_f_B) = 0\\\\m_Ag sin \theta+ m_Bgsin\theta = F_f_A + F_f_B\\\\m_Ag sin \theta+ m_Bgsin\theta = \mu_Am_Agcos\theta \ + \ \mu_Bm_Bgcos\theta\\\\gsin\theta(m_A + m_B) = gcos\theta (\mu_Am_A + \mu_Bm_B)\\\\\frac{sin\theta}{cos \theta} = \frac{\mu_Am_A\ + \ \mu_Bm_B}{m_A\ + \ m_B} \\\\tan\theta = \frac{(0.16\times 11) \ + \ (0.23 \times 5)}{11 + 5} \\\\tan\theta = 0.1819\\\\\theta = tan^{-1}(0.1819)\\\\\theta = 10.3 \ ^0[/tex]
The change in the energy of the blocks is the work done in compressing the spring;
[tex]\Delta E = W\\\\F_A (sin \theta )d- \mu F_n d= \frac{1}{2} kd^2\\\\F_A sin\theta \ - \ \mu F_A cos\theta = \frac{1}{2} kd\\\\d = \frac{2F_A(sin\theta - \mu cos \theta) }{k} \\\\d = \frac{2\times 11(sin \ 10.3\ - \ 0.16\times cos \ 10.3) }{2.1} \\\\d = 0.22 \ ft[/tex]
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Two football teams, the Raiders and the 49ers are engaged in a tug-of-war. The Raiders are pulling with a force of 5000N. Which of the following is an accurate statement?
A. The tension in the rope depends on whether or not the teams are in equilibrium.
B. The 49ers are pulling with a force of more than 5000N because of course they’d be winning.
C. The 49ers are pulling with a force of 5000N.
D. The tension in the rope is 10,000N.
E. None of these statements are true.
Answer:
E. None of these statements are true.
Explanation:
We can't say the exact or approximate amount of tension on the rope, since we do know for sure from the statement who is winning.
for A, the tension on the rope does not depend on if both teams pull are in equilibrium.
for B, the 49ers would be pulling with a force more than 5000 N, if they were winning. The problem is that we can't say with all confidence that they'd be winning.
for C, we don't know how much tension exists on the rope, and its direction, so we can't work out how much tension the 49ers are pulling the rope with.
for D, just as for C above, we can't work out how much tension there is on the rope, since we do not know how much force the 49ers are pulling with.
we go with option E.
A 1.25-kg ball begins rolling from rest with constant angular acceleration down a hill. If it takes 3.60 s for it to make the first complete revolution, how long will it take to make the next complete revolution?
Answer:
The time taken is [tex]\Delta t = 1.5 \ s[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m = 1.25 \ kg[/tex]
The time taken to make the first complete revolution is t= 3.60 s
The displacement of the first complete revolution is [tex]\theta = 1 rev = 2 \pi \ radian[/tex]
Generally the displacement for one complete revolution is mathematically represented as
[tex]\theta = w_i t + \frac{1}{2} * \alpha * t^2[/tex]
Now given that the stone started from rest [tex]w_i = 0 \ rad / s[/tex]
[tex]2 \pi =0 + 0.5* \alpha *(3.60)^2[/tex]
[tex]\alpha = 0.9698 \ s[/tex]
Now the displacement for two complete revolution is
[tex]\theta_2 = 2 * 2\pi[/tex]
[tex]\theta_2 = 4\pi[/tex]
Generally the displacement for two complete revolution is mathematically represented as
[tex]4 \pi = 0 + 0.5 * 0.9698 * t^2[/tex]
=> [tex]t^2 = 25.9187[/tex]
=> [tex]t= 5.1 \ s[/tex]
So
The time taken to complete the next oscillation is mathematically evaluated as
[tex]\Delta t = t_2 - t[/tex]
substituting values
[tex]\Delta t = 5.1 - 3.60[/tex]
[tex]\Delta t = 1.5 \ s[/tex]
The time for the ball to complete the next revolution is 1.5 s.
The given parameters;
mass of the ball, m = 1.25 kgtime of motion, t = 3.6 sone complete revolution, θ = 2πThe constant angular acceleration of the ball is calculated as follows;
[tex]\theta = \omega t \ + \ \frac{1}{2} \alpha t^2\\\\2\pi = 0 \ + \ 0.5(3.6)^2 \alpha\\\\2\pi = 6.48 \alpha \\\\\alpha = \frac{2 \pi }{6.48} \\\\\alpha = 0.97 \ rad/s^2[/tex]
The time to complete the next revolution is calculated as follows;
[tex]4\pi = 0 + \frac{1}{2} (0.97)t^2\\\\8\pi = 0.97t^2\\\\t^2 = \frac{8\pi }{0.97} \\\\t^2 = 25.91\\\\t = \sqrt{ 25.91} \\\\t = 5.1 \ s[/tex]
[tex]\Delta t = 5.1 \ s \ - \ 3.6 \ s \\\\\Delta t = 1.5 \ s[/tex]
Thus, the time for the ball to complete the next revolution is 1.5 s.
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A sinusoidal sound wave moves through a medium and is described by the displacement wave function s(x, t) = 1.99 cos(15.2x − 869t) where s is in micrometers, x is in meters, and t is in seconds. (a) Find the amplitude of this wave. µm (b) Find the wavelength of this wave. cm (c) Find the speed of this wave. m/s (d) Determine the instantaneous displacement from equilibrium of the elements of the medium at the position x = 0.050 9 m at t = 2.94 ms. µm (e) Determine the maximum speed of a element's oscillatory motion. mm/s
Answer:
a) A = 1.99 μm , b) λ = 0.4134 m , c) v = 57.2 m / s , d) s = - 1,946 nm ,
e) v_max = 1,739 mm / s
Explanation:
A sound wave has the general expression
s = s₀ sin (kx - wt)
where s is the displacement, s₀ the amplitude of the wave, k the wave vector and w the angular velocity, in this exercise the expression given is
s = 1.99 sin (15.2 x - 869 t)
a) the amplitude of the wave is
A = s₀
A = 1.99 μm
b) wave spectrum is
k = 2π /λ
in the equation k = 15.2 m⁻¹
λ = 2π / k
λ = 2π / 15.2
λ = 0.4134 m
c) the speed of the wave is given by the relation
v = λ f
angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 869 / 2π
f = 138.3 Hz
v = 0.4134 138.3
v = 57.2 m / s
d) To find the instantaneous velocity, we substitute the given distance and time into the equation
s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)
s = 1.99 sin (0.77368 - 2.55486)
remember that trigonometry functions must be in radians
s = 1.99 (-0.98895)
s = - 1,946 nm
The negative sign indicates that it shifts to the left
e) the speed of the oscillating part is
v = ds / dt)
v = - s₀(-w) cos (kx -wt)
the maximum speed occurs when the cosines is 1
v_maximo = s₀w
v_maximum = 1.99 869
v_maximo = 1739.31 μm / s
let's reduce to mm / s
v_maxio = 1739.31 miuy / s (1 mm / 103 mu)
v_max = 1,739 mm / s
a) A is = 1.99 μm , b) λ is = 0.4134 m , c) v is = 57.2 m / s , d) s is = - 1,946 nm, e) v_max is = 1,739 mm / s
Calculation of Wavelength
When A sound wave has the general expression is:
Then, s = s₀ sin (kx - wt)
Now, where s is the displacement, Then, s₀ is the amplitude of the wave, k the wave vector, and w the angular velocity, Now, in this exercise the expression given is
s is = 1.99 sin (15.2 x - 869 t)
a) When the amplitude of the wave is
A is = s₀
Thus, A = 1.99 μm
b) When the wave spectrum is
k is = 2π /λ
Now, in the equation k = 15.2 m⁻¹
Then, λ = 2π / k
After that, λ = 2π / 15.2
Thus, λ = 0.4134 m
c) When the speed of the wave is given by the relation is:
Then, v = λ f
Now, the angular velocity and frequency are related is:
w is = 2π f
Then, f = w / 2π
After that, f = 869 / 2π
Now, f = 138.3 Hz
Then, v = 0.4134 138.3
Thus, v = 57.2 m / s
d) Now, To find the instantaneous velocity, When we substitute the given distance and time into the equation
Then, s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)
After that, s = 1.99 sin (0.77368 - 2.55486)
Then remember that trigonometry functions must be in radians
After that, s = 1.99 (-0.98895)
Thus, s = - 1,946 nm
When The negative sign indicates that it shifts to the left
e) When the speed of the oscillating part is
Then, v = ds / dt)
Now, v = - s₀(-w) cos (kx -wt)
When the maximum speed occurs when the cosines is 1
Then, v_maximo = s₀w
After that, v_maximum = 1.99 869
v_maximo = 1739.31 μm / s
Now, let's reduce to mm / s
Then, v_maxio = 1739.31 miuy / s (1 mm / 103 mu)
Therefore, v_max = 1,739 mm / s
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what is the average flow rate in of gasoline to the engine of a plane flying at 700 km/h if it averages 100.0 km/l
Answer:
1.94cm³/s
Explanation:
1L = 1000cm³
Ihr = 3600s
So
Using
Average flow rate
Fr= 1L/100Km x 700Km/1hr x 1hr/3600s x 1000cm³/ 1L
= 1.94cm³/s
if a 1-m diameter sewer pipe is flowing at a depth of 0.4 m and has a flow rate of 0.15 m^3/s, what will be the flow rate when the pipe flows full?
Answer:
0.35 m³/s
Explanation:
When the pipe's depth is 0.4 m, the area of the circular segment is:
A = ½ R² (θ − sin θ)
The depth of the water is:
h = R (1 − cos(θ/2))
Solving for θ:
0.4 = 0.5 (1 − cos(θ/2))
0.8 = 1 − cos(θ/2)
cos(θ/2) = 0.2
θ/2 = acos(0.2)
θ = 2 acos(0.2)
θ ≈ 2.74 rad
The area is therefore:
A = ½ (0.5 m)² (2.74 − sin 2.74)
A = 0.338 m²
The cross-sectional area when the pipe is full is:
A = π (0.5 m)²
A = 0.785 m²
The flow velocity is constant:
v = v
Q / A = Q / A
(0.15 m³/s) / (0.338 m²) = Q / (0.785 m²)
Q = 0.35 m³/s
A mechanic wants to unscrew some bolts. She has two wrenches available: one is 35 cm long, and one is 50 cm long. Which wrench makes her job easier and why?
Answer:
50 cm long
When 35cm long wrench is compared to 50cm long wrench, we find that the 50cm long wrench produces more turning effect of force because it has longer distance between fulcrum and line of action of force. At conclusion, the more the turning effect of force the more it is easy to unscrew bolts.
A stone is dropped from the bridge, it takes 4s to reach the water. what's the height of the bridge?
Explanation:
Using Equations of Motion :
[tex]s = ut + \frac{1}{2} g {t}^{2} [/tex]
Height = 0 * 4 + 4.9 * 16
Height = 78.4 m
An electrostatic paint sprayer contains a metal sphere at an electric potential of 25.0 kV with respect to an electrically grounded object. Positively charged paint droplets are repelled away from the paint sprayer's positively charged sphere and towards the grounded object. What charge must a 0.168-mg drop of paint have so that it will arrive at the object with a speed of 18.8 m/s
Answer:
The charge is [tex]Q = 2.177 *10^{-9} \ C[/tex]
Explanation:
From the question we are told that
The electric potential is [tex]V = 25.0 \ kV = 25.0 *10^{3}\ V[/tex]
The mass of the drop is [tex]m = 0.168 \ m g = 0.168 *10^{-3} \ g = 0.168 *10^{-6}\ kg[/tex]
The speed is [tex]v = 18.8 \ m/s[/tex]
Generally the charge on the paint drop due to the electric potential which will give it the speed stated in the question is mathematically represented as
[tex]Q = \frac{m v^2 }{ 2 * V }[/tex]
Substituting values
[tex]Q = \frac{0.168 *10^{-6} (18)^2 }{ 2 * 25*10^3 }[/tex]
[tex]Q = 2.177 *10^{-9} \ C[/tex]
In a two-slit experiment, the slit separation is 3.34 ⋅ 10 − 5 m. The interference pattern is created on a screen that is 3.30 m away from the slits. If the 7th bright fringe on the screen is 29.0 cm away from the central fringe, what is the wavelength of the light?
Answer:
The wavelength is [tex]\lambda = 419 \ nm[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 3.34 *10^{-5} \ m[/tex]
The distance of the screen is [tex]D = 3.30 \ m[/tex]
The order of the fringe is n = 7
The distance of separation of fringes is y = 29.0 cm = 0.29 m
Generally the wavelength of the light is mathematically represented as
[tex]\lambda = \frac{y * d }{ n * D}[/tex]
substituting values
[tex]\lambda = \frac{0.29 * 3.34*10^{-5} }{ 7 * 3.30}[/tex]
[tex]\lambda = 4.19*10^{-7}\ m[/tex]
[tex]\lambda = 419 \ nm[/tex]
An array of solar panels produces 9.35 A of direct current at a potential difference of 195 V. The current flows into an inverter that produces a 60 Hz alternating current with Vmax = 166V and Imax = 19.5A.
A) What rms power is produced by the inverter?
B) Use the rms values to find the power efficiency Pout/Pin of the inverter.
Answer:
(A). 1620 watt.
(B).0.8885.
Explanation:
So, we are given the following data or parameters or information which is going to assist or help us in solving this particular Question or problem. So, we have;
Current = 9.35A, direct current at a potential difference of 195 V, frequency of the inverter = 60 Hz alternating current, alternating current with Vmax = 166V and Imax = 19.5A.
(A). The rms power is produced by the inverter = (19.5 /2 ) × 166 = 1620 watt(approximately).
(B). the rms values to find the power efficiency Pout/Pin of the inverter.
P(in) = 195 × 9.35 = 1823.3 watt.
Thus, the rms values to find the power efficiency Pout/Pin of the inverter = 1620/1823.3 = 0.88852324146441793 = 0.8885.
A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the slit width is doubled
Answer:
y ’= y / 2
thus when the slit width is doubled the pattern width is halved
Explanation:
The diffraction of a slit is given by the expressions
a sin θ = m λ
where a is the width of the slit, λ is the wavelength and m is an integer that determines the order of diffraction.
sin θ = m λ / a
If this equation
a ’= 2 a
we substitute
2 a sin θ'= m λ
sin θ'= (m λ / a) 1/2
sin θ ’= sin θ / 2
We can use trigonometry to find the width
tan θ = y / L
as the angle is small
tan θ = sin θ / cos θ = sin θ
sin θ = y / L
we substitute
y ’/ L = y/L 1/2
y ’= y / 2
thus when the slit width is doubled the pattern width is halved
With the same block-spring system from above, imagine doubling the displacement of the block to start the motion. By what factor would the following change?
A. Kinetic energy when passing through the equilibrium position.
B. Speed when passing through the equilibrium position.
Answer:
A) K / K₀ = 4 b) v / v₀ = 4
Explanation:
A) For this exercise we can use the conservation of mechanical energy
in the problem it indicates that the displacement was doubled (x = 2xo)
starting point. At the position of maximum displacement
Em₀ = Ke = ½ k (2x₀)²
final point. In the equilibrium position
[tex]Em_{f}[/tex] = K = ½ m v²
Em₀ = Em_{f}
½ k 4 x₀² = K
(½ K x₀²) = K₀
K = 4 K₀
K / K₀ = 4
B) the speed value
½ k 4 x₀² = ½ m v²
v = 4 (k / m) x₀
if we call
v₀ = k / m x₀
v = 4 v₀
v / v₀ = 4
The metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off. This simple
process is which kind of a change?
OA a physical change
OB. a chemical change
OC. a nuclear change
OD
an ionic change
B. A chemical change
Explanation:
I'm guessing ?
Two parallel metal plates, each of area A, are separatedby a distance 3d. Both are connected to ground and each plate carries no charge. A third plate carrying charge Qis inserted between the two plates, located a distance dfrom the upper plate. As a result, negative charge is induced on each of the two original plates. a) In terms of Q, find the amount of charge on the upper plate, Q1, and the lower plate, Q2. (Hint: it must be true that Q
Answer:
Upper plate Q/3
Lower plate 2Q/3
Explanation:
See attached file
The ceiling of your lecture hall is probably covered with acoustic tile, which has small holes separated by about 6.1 mm. Using light with a wavelength of 578 nm, how far could you be from this tile and still resolve these holes
Answer:
8.65x10^3m
Explanation:
See attached file
A car moving east at 45 km/h turns and travels west at 30 km/h. What is the
magnitude and direction of the change in velocity?
mahalle 1.11
Explanation:
Change in Velocity = final velocity - initial velocity
Change in velocity = 30km/h - (- 45km/h )
= 75 km/h due west
What is the wavelength of electromagnetic radiation which has a frequency of 3.818 x 10^14 Hz?
Answer:
7.86×10⁻⁷ m
Explanation:
Using,
v = λf.................. Equation 1
Where v = velocity of electromagnetic wave, λ = wave length, f = frequency.
make λ the subject of the equation
λ = v/f............... Equation 2
Note: All electromagnetic wave have the same speed which is 3×10⁸ m/s.
Given: f = 3.818×10¹⁴ Hz
Constant: v = 3×10⁸ m/s
Substitute these values into equation 2
λ = 3×10⁸/3.818×10¹⁴
λ = 7.86×10⁻⁷ m
Hence the wavelength of the electromagnetic radiation is 7.86×10⁻⁷ m
The wavelength of this electromagnetic radiation is equal to [tex]7.86 \times 10^{-7} \;meters[/tex]
Given the following data:
Frequency = [tex]3.818\times 10^{14}\;Hz[/tex]Scientific data:
Velocity of an electromagnetic radiation = [tex]3 \times 10^8\;m/s[/tex]
To determine the wavelength of this electromagnetic radiation:
Mathematically, the wavelength of an electromagnetic radiation is calculated by using the formula;
[tex]Wavelength = \frac{Speed }{frequency}[/tex]
Substituting the given parameters into the formula, we have;
[tex]Wavelength = \frac{3 \times 10^8}{3.818\times 10^{14}}[/tex]
Wavelength = [tex]7.86 \times 10^{-7} \;meters[/tex]
Read more wavelength on here: https://brainly.com/question/6352445
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.
Find more information about Current here:
brainly.com/question/24858512
Describe how, using a positively-charged rod and two neutral metal spheres, we canmake one sphere positive without touching it to the rod. You might want to draw adiagram to help you.
Answer:
se the principle of induction.
place the two metallic spheres together, now we bring the positively charged bar closer to the first sphere.
The charge that was induced in the sphere is distributed as infirm as possible,
At this time I separate the spheres and move the bar away, by separating the spheres the excess positive
Explanation:
For this exercise we will use that the electric charge is not created, it is not destroyed and charges of the same sign repel.
Let's use the principle of induction. We place the two metallic spheres together, one in front of the other, now we bring the positively charged bar closer to the first sphere.
Here the positive charge of the bar repels the positive charge of the sphere, but as this is mocil it moves as far away as possible, until the negative charge that remains neutralizes the positive charge of the bar.
The charge that was induced in the sphere is distributed as infirm as possible, most of it in the furthest sphere, since the Coulomb force decreases.
At this time I separate the spheres and move the bar away, by separating the spheres the excess positive charge in the last sphere cannot be neutralized, therefore this sphere remains with a positive charge.
