Answer:
d. 629 nm
Explanation:
slit separation d = .16 x 10⁻³ m
distance of screen D = 1.4 m
distance of second bright band = 11 x 10⁻³
distance of second bright band = 2 x band width
= 2 x λ D /d
Putting the values given ,
11 x 10⁻³ = 2 x λ x 1.4 / .16 x 10⁻³
λ = 1.76 x 10⁻⁶ / 2.8
= .6285 x 10⁻⁶
= 628.5 x 10⁻⁹
= 629 nm approx .
A 1.2-m length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x= 5.0m on x-axis.
a. 1.6 nt in the negative z direction
b. 1.6 nt in the positive z direction
c. 2.4 T in the positive z direction
d. 2.4 nt in the negative z direction
e. None of the above
Answer:
None of the above
Explanation:
The formula of the magnetic field of a point next to a wire with current is:
B = 2×10^(-7) × ( I /d)
I is the intensity of the current.
d is the distance between the wire and the point.
● B = 2*10^(-7) × (20/5) = 8 ×10^(-7) T
Please help!
Much appreciated!
Answer:
F = 2.7×10¯⁶ N.
Explanation:
From the question given:
F = (9×10⁹ Nm/C²) (3.2×10¯⁹ C × 9.6×10¯⁹ C) /(0.32)²
Thus we can obtain the value value of F by carrying the operation as follow:
F = (9×10⁹) (3.2×10¯⁹ × 9.6×10¯⁹) /(0.32)²
F = 2.7648×10¯⁷ / 0.1024
F = 2.7×10¯⁶ N.
Therefore, the value of F is 2.7×10¯⁶ N.
Does the moon light originate from the moon only
Answer:
No
Explanation:
Moon has no light of its own. It just shines because its surface reflects light from the sun and that's what we see.
:-)
"When red light in vacuum is incident at the Brewster angle on a certain glass slab, the angle of refraction is"
Complete Question
When red light in vacuum is incident at the Brewster angle on a certain glass slab, the angle of refraction is [tex]36.0 ^o[/tex] . What are
(a) the index of refraction of the glass and
(b) the Brewster angle?
Answer:
a
[tex]n_r = 1.376[/tex]
b
[tex]i = 54^o[/tex]
Explanation:
From the question we are told that
The angle of refraction is [tex]r = 36.0 ^o[/tex]
Generally according Brewster law
[tex]i + r = 90[/tex]
Here [tex]i[/tex] is the angle of incidence which is also the Brewster angle
So
[tex]i + 36.0 = 90[/tex]
[tex]i = 54^o[/tex]
Now the refractive index is mathematically represented as
[tex]n_r = tan (i)[/tex]
substituting values
[tex]n_r = tan (54)[/tex]
[tex]n_r = 1.376[/tex]
Suppose a 1300 kg car is traveling around a circular curve in a road at a constant
9.0 m/sec. If the curve in the road has a radius of 25 m, then what is the
magnitude of the unbalanced force that steers the car out of its natural straight-
line path?
Answer:
F = 4212 N
Explanation:
Given that,
Mass of a car, m = 1300 kg
Speed of car on the road is 9 m/s
Radius of curve, r = 25 m
We need to find the magnitude of the unbalanced force that steers the car out of its natural straight- line path. The force is called centripetal force. It can be given by :
[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N[/tex]
So, the force has a magnitude of 4212 N
What is temperature?
O A. The force exerted on an area
B. A measure of mass per unit volume
O C. The net energy transferred between two objects
OD. A measure of the movement of atoms or molecules within an
object
Answer:
The net energy transferred between two objects
Explanation:
The physical property of matter that expresses hot or cold is called temperature. It demonstrates the thermal energy. A thermometer is used to measure temperature. It defines the rate to which the chemical reaction occurs. It tells about the thermal radiation emitted from an object.
The correct option that defines temperature is option C.
Answer:
A measure of the movement of atoms or molecules within an object
Explanation:
Process of elimination
When the current in a toroidal solenoid is changing at a rate of 0.0200 A/s, the magnitude of the induced emf is 12.7 mV. When the current equals 1.50 A, the average flux through each turn of the solenoid is 0.00458 Wb. How many turns does the solenoid have?
Answer:
[tex]N = 208 \ turns[/tex]
Explanation:
From the question we are told that
The rate of current change is [tex]\frac{di }{dt} = 0.0200 \ A/s[/tex]
The magnitude of the induced emf is [tex]\epsilon = 12.7 \ mV = 12.7 *10^{-3} \ V[/tex]
The current is [tex]I = 1.50 \ A[/tex]
The average flux is [tex]\phi = 0.00458 \ Wb[/tex]
Generally the number of turns the number of turn the solenoid has is mathematically represented as
[tex]N = \frac{\epsilon_o * I}{ \phi * \frac{di}{dt} }[/tex]
substituting values
[tex]N = \frac{ 12.7*10^{-3} * 1.50 }{ 0.00458 * 0.0200 }[/tex]
[tex]N = 208 \ turns[/tex]
help... Please help!!!!!!!!!!!
Answer:
a) 6.8--5.10 thats equal 11.9
b) m=ris/run +10 equal 0.06/8 =7.5*10^-3
A lamp has the shape of a parabola when viewed from the side. The lamp is centimeters wide and centimeters deep. How far is the light source from the bottom of the lamp if the light source is placed at the focus
The question is not complete so i have attached it.
Answer:
The light source is 2 cm from the bottom of the lamp
Explanation:
From the attached image, we can see that the parabola opens up with its vertex at the origin.
Now, the standard form of equation for a parabola is:
x² = 4ay
Looking at the parabola in the attachment, the top right edge of the lamp has a coordinate of (12,18)
Thus, we have;
12² = 4a(18)
144 = 72a
a = 144/72
a = 2
Looking at the parabola again, the line of symmetry is at x = 0
Thus, axis of symmetry is at x = 0.
Thus, focus is at (0, 2)
So, if the light source is placed at the focus, the distance of the light source from the bottom of the lamp is 2 cm
The distance of the light source from the bottom of the lamp is 2 cm.
The given parameters;
the top right edge of the lamp has a coordinate of (12,18)Apply standard parabola equation to determine the distance of the light source from the bottom of the lamp;
[tex]x^2 = 4ay\\\\12^2 = 4a(18)\\\\144 = 72 a\\\\a = \frac{144}{72} \\\\a = 2 \ cm[/tex]
Thus, the distance of the light source from the bottom of the lamp is 2 cm.
"Your question is not complete, it seems to be missing the following information";
the top right edge of the lamp has a coordinate of (12,18)
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A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, and 130 cm. The loop is in a uniform magnetic field of magnitude 75.0 mT whose direction is parallel to the current in the 130 cm side of the loop. What is the magnitude of the magnetic force on the
Given that,
Current = 4 A
Sides of triangle = 50.0 cm, 120 cm and 130 cm
Magnetic field = 75.0 mT
Distance = 130 cm
We need to calculate the angle α
Using cosine law
[tex]120^2=130^2+50^2-2\times130\times50\cos\alpha[/tex]
[tex]\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}[/tex]
[tex]\alpha=\cos^{-1}(0.3846)[/tex]
[tex]\alpha=67.38^{\circ}[/tex]
We need to calculate the angle β
Using cosine law
[tex]50^2=130^2+120^2-2\times130\times120\cos\beta[/tex]
[tex]\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}[/tex]
[tex]\beta=\cos^{-1}(0.923)[/tex]
[tex]\beta=22.63^{\circ}[/tex]
We need to calculate the force on 130 cm side
Using formula of force
[tex]F_{130}=ILB\sin\theta[/tex]
[tex]F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0[/tex]
[tex]F_{130}=0[/tex]
We need to calculate the force on 120 cm side
Using formula of force
[tex]F_{120}=ILB\sin\beta[/tex]
[tex]F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63[/tex]
[tex]F_{120}=0.1385\ N[/tex]
The direction of force is out of page.
We need to calculate the force on 50 cm side
Using formula of force
[tex]F_{50}=ILB\sin\alpha[/tex]
[tex]F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38[/tex]
[tex]F_{50}=0.1385\ N[/tex]
The direction of force is into page.
Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.
a. The magnitude of the magnetic force on the 130 cm side is 0 Newton.
b. The magnitude of the magnetic force on the 120 cm side is 0.1385 Newton.
c. The magnitude of the magnetic force on the 50 cm side is 0.1385 Newton.
Given the following data:
Current = 4.00 Amperes.Magnetic field strength = 75.0 mT = [tex]7.5 \times 20^{-3}\;T[/tex]Length = 130 cm to m = 1.3 mHypotenuse = 130 cmOpposite side = 120 cmAdjacent side = 50 cmLet us assume the current is flowing in a counterclockwise direction in the right-angle triangle.
First of all, we would determine the angles by using cosine rule:
[tex]C^2=A^2 +B^2 - 2ABCos\alpha \\\\120^2=130^2 +50^2 - 2(130)(50)Cos\alpha\\\\14400 = 16900 + 2500 -13000Cos\alpha\\\\13000Cos\alpha=19400-14400 \\\\Cos\alpha=\frac{5000}{13000} \\\\\alpha = Cos^{-1}(0.3846)\\\\\alpha =67.38^\circ[/tex]
[tex]C^2=A^2 +B^2 - 2ABCos\beta \\\\50^2=120^2 +130^2 - 2(120)(130)Cos\beta \\\\2500 = 14400 + 16900 -31200Cos\beta\\\\31200Cos\alpha=31300-2500 \\\\Cos\beta=\frac{28800}{31200} \\\\\beta = Cos^{-1}(0.9231)\\\\\beta =22.62^\circ[/tex]
a. To the determine the magnitude of the magnetic force on the 130 cm side:
Mathematically, the force acting on a current in a magnetic field is given by the formula:
[tex]F = BILsin\theta[/tex]
Where:
B is the magnetic field strength.I is the current flowing through a conductor.L is the length of conductor.[tex]\theta[/tex] is the angle between a conductor and the magnetic field.Substituting the given parameters into the formula, we have;
[tex]F_{130}=7.5 \times 20^{-3}\times 4 \times 1.3 \times sin(0)\\\\F_{130}=7.5 \times 20^{-3}\times 4 \times 1.3 \times0\\\\F_{130}=0\;Newton[/tex]
b. To the determine the magnitude of the magnetic force on the 120 cm side:
[tex]F_{120}=BILsin\beta[/tex]
[tex]F_{120}=7.5 \times 20^{-3}\times 4 \times 1.2 \times sin(22.62)\\\\F_{120}=7.5 \times 20^{-3}\times 4 \times 1.2 \times0.3846\\\\F_{120}=0.1385\;Newton[/tex]
c. To the determine the magnitude of the magnetic force on the 50 cm side:
[tex]F_{50}=BILsin\alpha[/tex]
[tex]F_{50}=7.5 \times 20^{-3}\times 4 \times 0.5 \times sin(67.38)\\\\F_{50}=7.5 \times 20^{-3}\times 4 \times 1.2 \times0.9231\\\\F_{50}=0.1385\;Newton[/tex]
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where c is the speed of light and G is the universal gravitational constant. RBH gives the radius of the event horizon of a black hole with mass ????. In other words, it gives the radius to which some amount of mass ???? would need to be compressed in order to form a black hole. The mass of the Sun is about 1.99×1030 kg. What would be the radius of a black hole with this mass?
Answer:
The radius of the black hole will be 2949.6 m.
Explanation:
The radius of this black hole will be the Schwarzschild radius of the mass of the sun
[tex]r_{s}[/tex] = [tex]\frac{2GM}{c^{2} }[/tex]
where
G is the gravitational constant = 6.67 x 10^-11 m^3⋅kg^-1⋅s^-2
M is the mass of the sun = 1.99×10^30 kg
c is the speed of light = 3 x 10^8 m/s
substituting values into the equation, we have
[tex]r_{s}[/tex] = [tex]\frac{2*6.67*10^{-11}*1.99*10^{30} }{(3*10^{8} )^{2} }[/tex] = 2949.6 m
Which of the following regions of the electromagnetic spectrum have longer wavelengths than visible light? 1. infrared radiation 2. ultraviolet radiation 3. microwave radiation
Answer:infrared radiation
Explanation:
Infrared radiation and microwave radiation of the electromagnetic spectrum have longer wavelengths than visible light.
What is electromagnetic wave?EM waves are another name for electromagnetic waves. When an electric field interacts with a magnetic field, electromagnetic waves are created. These electromagnetic waves make up electromagnetic radiations. It is also possible to say that electromagnetic waves are made up of magnetic and electric fields that are oscillating. The basic equations of electrodynamics, Maxwell's equations, have an answer in electromagnetic waves.
If we arrange electromagnetic wave with decrease in wavelength, we get:
Radio waves > microwave > Infrared > Visible light > Ultraviolet > X-rays > Gamma radiation.
Hence, Infrared radiation and microwave radiation of the electromagnetic spectrum have longer wavelengths than visible light.
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If mirror M2 in a Michelson interferometer is moved through 0.233 mm, a shift of 792 bright fringes occurs. What is the wavelength of the light producing the fringe pattern?
Answer:
The wavelength is [tex]\lambda = 589 nm[/tex]
Explanation:
From the question we are told that
The distance of the mirror shift is [tex]k = 0.233 \ mm = 0.233*10^{-3} \ m[/tex]
The number of fringe shift is n = 792
Generally the wavelength producing this fringes is mathematically represented as
[tex]\lambda = \frac{ 2 * k }{ n }[/tex]
substituting values
[tex]\lambda = \frac{ 2 * 0.233*10^{-3} }{ 792 }[/tex]
[tex]\lambda = 5.885 *10^{-7} \ m[/tex]
[tex]\lambda = 589 nm[/tex]
Calculate the density of the following material.
1 kg helium with a volume of 5.587 m³
700 kg/m³
5.587 kg/m³
0.179 kg/m³
Answer:
[tex]density \: = \frac{mass}{volume} [/tex]
1 / 5.587 is equal to 0.179 kg/m³
Hope it helps:)
Answer:
The answer is
0.179 kg/m³Explanation:
Density of a substance is given by
[tex]Density \: = \frac{mass}{volume} [/tex]
From the
mass = 1 kg
volume = 5.583 m³
Substitute the values into the above formula
We have
[tex]Density \: = \frac{1 \: kg}{5.583 \: {m}^{3} } [/tex]
We have the final answer as
Density = 0.179 kg/m³Hope this helps you
The Milky Way has a diameter (proper length) of about 1.2×105 light-years. According to an astronaut, how many years would it take to cross the Milky Way if the speed of the spacecraft is 0.890 c?
Answer:
t = 134834.31 years
Explanation:
First we find the speed of the ship:
v = 0.890 c
where,
v = speed of the ship = ?
c = speed of light = 3 x 10⁸ m/s
Therefore, using the values, we get:
v = (0.89)(3 x 10⁸ m/s)
v = 2.67 x 10⁸ m/s
Now, we find the distance in meters:
Distance = s = (1.2 x 10⁵ light years)(9.461 x 10¹⁵/1 light year)
s = 11.35 x 10²⁰ m
Now, for the time we use the following equation:
s = vt
t = s/v
t = (11.35 x 10²⁰ m)/(2.67 x 10⁸ m/s)
t = (4.25 x 10¹² s)(1 h/3600 s)(1 day/24 h)(1 year/365 days)
t = 134834.31 years
An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length? Aluminum has a linear expansion coefficient of 25 × 10-6 C-1.
Answer:
the new length is 17.435cm
Explanation:
the new length is 17.435cm
pls give brainliest
The new length of aluminum rod is 17.435 cm.
The linear expansion coefficient is given as,
[tex]\alpha=\frac{L_{1}-L_{0}}{L_{0}(T_{1}-T_{0})}[/tex]
Given that, An aluminum rod 17.400 cm long at 20°C is heated to 100°C.
and linear expansion coefficient is [tex]25*10^{-6}C^{-1}[/tex]
Substitute, [tex]L_{0}=17.400cm,T_{1}=100,T_{0}=20,\alpha=25*10^{-6}C^{-1}[/tex]
[tex]25*10^{-6}C^{-1} =\frac{L_{1}-17.400}{17.400(100-20)}\\\\25*10^{-6}C^{-1} = \frac{L_{1}-17.400}{1392} \\\\L_{1}=[25*10^{-6}C^{-1} *1392}]+17.400\\\\L_{1}=17.435cm[/tex]
Hence, The new length of aluminum rod is 17.435 cm.
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A charge of 15 is moving with velocity of 6.2 x17 which makes an angle of 48 degrees with respect to the magnetic field. If the force on the particle is 4838 N, find the magnitude of the magnetic field.
a. 06.0T.
b. 08.0T.
c. 07.0T.
d. 05.0 T.
Complete question:
A charge of 15C is moving with velocity of 6.2 x 10³ m/s which makes an angle of 48 degrees with respect to the magnetic field. If the force on the particle is 4838 N, find the magnitude of the magnetic field.
a. 0.06 T
b. 0.08 T
c. 0.07 T
d. 0.05 T
Answer:
The magnitude of the magnetic field is 0.07 T.
Explanation:
Given;
magnitude of the charge, q = 15C
velocity of the charge, v = 6.2 x 10³ m/s
angle between the charge and the magnetic field, θ = 48°
the force on the particle, F = 4838 N
The magnitude of the magnetic field can be calculated by applying Lorentz force formula;
F = qvBsinθ
where;
B is the magnitude of the magnetic field
B = F / vqsinθ
B = (4838) / (6.2 x 10³ x 15 x sin48)
B = 0.07 T
Therefore, the magnitude of the magnetic field is 0.07 T.
Select the situation for which the torque is the smallest.
a. A 200 kg piece of silver is placed at the end of a 2.5 m tree branch.
b. A 20 kg piece of marble is placed at the end of a 25 m construction crane arm.
c. A 8 kg quartz rock is placed at the end of a 62.5 m thin titanium rod.
d. The torque is the same for two cases.
e. The torque is the same for all cases.
Answer:
e. The torque is the same for all cases.
Explanation:
The formula for torque is:
τ = Fr
where,
τ = Torque
F = Force = Weight (in this case) = mg
r = perpendicular distance between force an axis of rotation
Therefore,
τ = mgr
a)
Here,
m = 200 kg
r = 2.5 m
Therefore,
τ = (200 kg)(9.8 m/s²)(2.5 m)
τ = 4900 N.m
b)
Here,
m = 20 kg
r = 25 m
Therefore,
τ = (20 kg)(9.8 m/s²)(25 m)
τ = 4900 N.m
c)
Here,
m = 8 kg
r = 62.5 m
Therefore,
τ = (8 kg)(9.8 m/s²)(62.5 m)
τ = 4900 N.m
Hence, the correct answer will be:
e. The torque is the same for all cases.
Seismic attenuation and how spherical spreading affect amplitude, can anyone explain this please!
Answer:
Hey there!
This can be a confusing topic, so it's totally fine if you get confused...
First, Seismic Attenuation is how seismic waves lose energy as they expand and spread.
Secondly, when distance increases, amplitude decreases. This is because the distance (spherical spreading would mean radius) is inversely proportional to amplitude.
Let me know if this helps :)
"A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if"
Answer:
A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if
the dispersion is great
You need to repair a broken fence in your yard. The hole in your fence is
around 3 meters in length and for whatever reason, the store you go to
has oddly specific width 20cm wood. Each plank of wood costs $16.20,
how much will it cost to repair your fence? (Hint: 1 meter = 100 cm) *
Answer:
cost = $ 243.00
Explanation:
This exercise must assume that it uses a complete table for each piece, we can use a direct ratio of proportions, if 1 table is 0.20 m wide, how many tables will be 3.00 m
#_tables = 3 m (1 / 0.20 m)
#_tables = 15 tables
Let's use another direct ratio, or rule of three, for cost. If a board costs $ 16.20, how much do 15 boards cost?
Cost = 15 (16.20 / 1)
cost = $ 243.00
Light with an intensity of 1 kW/m2 falls normally on a surface with an area of 1 cm2 and is completely absorbed. The force of the radiation on the surface is
Answer:
The force of the radiation on the surface is 3.33 x 10⁻¹⁰ N
Explanation:
Given;
intensity of light, I = 1kw/m² = 1000 W/m²
area of the surface, A = 1 cm² = 1 x 10⁻⁴ m²
Since the light is completely absorbed, the force of the radiation is given by;
F = P/c
where;
c is the speed of light = 3 x 10⁸ m/s
But P = IA
F = IA /c
F = (1000 X 1 X 10⁻⁴) / 3 x 10⁸
F = 3.33 x 10⁻¹⁰ N
Therefore, the force of the radiation on the surface is 3.33 x 10⁻¹⁰ N
The force of radiation will be "3.33 × 10⁻¹⁰ N"
Intensity and ForceAccording to the question,
Intensity of force, I = 1 kW/m² or,
= 1000 W/m²
Area of surface, A = 1 cm² or,
= 1 × 10⁻⁴ m²
Speed of light, c = 3 × 10³ m/s
As we know the relation,
→ F = [tex]\frac{P}{c}[/tex]
or,
P = IA
or,
F = [tex]\frac{IA}{c}[/tex]
By substituting the values, we get
= [tex]\frac{1000\times 1\times 10^{-4}}{3\times 10^3}[/tex]
= 3.33 × 10⁻¹⁰ N
Thus the response above is correct.
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You simultaneously shine two light beams, each of intensity I0, on an ideal polarizer. One beam is unpolarized, and the other beam is polarized at an angle of exactly 30.0∘ to the polarizing axis of the polarizer. Find the intensity of the light that emerges from the polarizer. Express your answer in term of I0 .
Answer:
The emerging intensity is equal to 0.75[tex]I_{o}[/tex]
Explanation:
The initial intensity of the light = [tex]I_{o}[/tex]
The angle of polarization β = 30°
We know that the polarized light intensity is related to the initial light intensity by
[tex]I[/tex] = [tex]I_{0} cos^{2}\beta[/tex]
where [tex]I[/tex] is the emerging polarized light intensity
inserting values gives
[tex]I[/tex] = [tex]I_{0} cos^{2}[/tex] 30°
[tex]cos^{2}[/tex] 30° = [tex](cos 30)^{2}[/tex] = [tex](\frac{\sqrt{3} }{2} )^{2}[/tex] = 0.75
[tex]I[/tex] = 0.75[tex]I_{o}[/tex]
A circular loop in the plane of a paper lies in a 0.45 T magnetic field pointing into the paper. The loop's diameter changes from 17.0 cm to 6.0 cm in 0.53 s.
A) Determine the direction of the induced current.
B) Determine the magnitude of the average induced emf.
C) If the coil resistance is 2.5 Ω, what is the average induced current?
Answer:
(A). The direction of the induced current will be clockwise.
(B). The magnitude of the average induced emf 16.87 mV.
(C). The induced current is 6.75 mA.
Explanation:
Given that,
Magnetic field = 0.45 T
The loop's diameter changes from 17.0 cm to 6.0 cm .
Time = 0.53 sec
(A). We need to find the direction of the induced current.
Using Lenz law
If the direction of magnetic field shows into the paper then the direction of the induced current will be clockwise.
(B). We need to calculate the magnetic flux
Using formula of flux
[tex]\phi_{1}=BA\cos\theta[/tex]
Put the value into the formula
[tex]\phi_{1}=0.45\times(\pi\times(8.5\times10^{-2})^2)\cos0[/tex]
[tex]\phi_{1}=0.01021\ Wb[/tex]
We need to calculate the magnetic flux
Using formula of flux
[tex]\phi_{2}=BA\cos\theta[/tex]
Put the value into the formula
[tex]\phi_{2}=0.45\times(\pi\times(3\times10^{-2})^2)\cos0[/tex]
[tex]\phi_{2}=0.00127\ Wb[/tex]
We need to calculate the magnitude of the average induced emf
Using formula of emf
[tex]\epsilon=-N(\dfrac{\Delta \phi}{\Delta t})[/tex]
Put the value into t5he formula
[tex]\epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})[/tex]
[tex]\epsilon=0.016867\ V[/tex]
[tex]\epsilon=16.87\ mV[/tex]
(C). If the coil resistance is 2.5 Ω.
We need to calculate the induced current
Using formula of current
[tex]I=\dfrac{\epsilon}{R}[/tex]
Put the value into the formula
[tex]I=\dfrac{0.016867}{2.5}[/tex]
[tex]I=0.00675\ A[/tex]
[tex]I=6.75\ mA[/tex]
Hence, (A). The direction of the induced current will be clockwise.
(B). The magnitude of the average induced emf 16.87 mV.
(C). The induced current is 6.75 mA.
A loop of wire in the shape of a rectangle rotates with a frequency of 143 rotation per minute in an applied magnetic field of magnitude 2 T. Assume the magnetic field is uniform. The area of the loop is A = 2 cm2 and the total resistance in the circuit is 7 Ω.
1. Find the maximum induced emf.
e m fmax =
2. Find the maximum current through the bulb.
Imax
Answer:
1. e m fmax = 0.00598 Volt
2. Imax = 0.000854 Amp
Explanation:
1. Find the maximum induced emf.
e m fmax =
Given that e m fmax = N*A*B*w
N = 1
A = 2 cm^2 = 0.0002 m^2
f = 143 rotation per minute = 143/min
f = (143/min) * (1 min/60 sec) = 2.38/sec
w = 2Πf = 2 * Π * 2.38 = 14.95 rad/sec
B = 2T
e m fmax = N*A*B*w
e m fmax = 1 * 0.0002 * 2 * 14.95
e m fmax = 0.00598 Volt.
2. Find the maximum current through the bulb.
Imax = e m fmax / R
Where R is the total resistance in the circuit is 7 Ω.
Imax = 0.00598/7 = 0.000854 Amp.
Imax = 0.000854 Amp
1) The maximum induced EMF in the loop of wire is; EMF_max = 9.52 × 10^(-4) V
2) The maximum current through the bulb is;
I_max = 1.36 × 10^(-4) A
We are given;
Number of turns; N = 1
Magnitude of magnetic field; B = 2 T
Area; A = 2 cm² = 0.0002 m²
Angular frequency; ω = 143 /min = 2.38 /s
Resistance; R = 7 Ω.
1) Formula for maximum induced EMF is;
EMF_max = NAωB
Plugging in the relevant values gives;
EMF_max = 1 × 0.0002 × 2.38 × 2
EMF_max = 9.52 × 10^(-4) V
2) Formula for maximum current through the bulb is given as;
I_max = EMF_max/R
Plugging in the relevant values;
I_max = (9.52 × 10^(-4))/7
I_max = 1.36 × 10^(-4) A
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The roller coaster car reaches point A of the loop with speed of 20 m/s, which is increasing at the rate of 5 m/s2. Determine the magnitude of the acceleration at A if pA
Answer and Explanation:
Data provided as per the question is as follows
Speed at point A = 20 m/s
Acceleration at point C = [tex]5 m/s^2[/tex]
[tex]r_A = 25 m[/tex]
The calculation of the magnitude of the acceleration at A is shown below:-
Centripetal acceleration is
[tex]a_c = \frac{v^2}{r}[/tex]
now we will put the values into the above formula
= [tex]\frac{20^2}{25}[/tex]
After solving the above equation we will get
[tex]= 16 m/s^2[/tex]
Tangential acceleration is
[tex]= \sqrt{ac^2 + at^2} \\\\ = \sqrt{16^2 + 5^2}\\\\ = 16.703 m/s^2[/tex]
Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging, the blood flow and heartbeat of the child can be measured using an echolocation technique similar to that used by bats. For the purposes of these questions, please use 1500 m/s as the speed of sound in tissue. I need help with part B and C
To clearly see an image, the wavelength used must be at most 1/4 of the size of the object that is to be imaged. What frequency is needed to image a fetus at 8 weeks of gestation that is 1.6 cm long?
A. 380 kHz
B. 3.8 kHz
C. 85 kHz
D. 3.8 MHz
Answer:
380 kHz
Explanation:
The speed of sound is taken as 1500 m/s
The length of the fetus is 1.6 cm long
The condition is that the wavelength used must be at most 1/4 of the size of the object that is to be imaged.
For this 1.6 cm baby, the wavelength must not exceed
λ = [tex]\frac{1}{4}[/tex] of 1.6 cm = [tex]\frac{1}{4}[/tex] x 1.6 cm = 0.4 cm =
0.4 cm = 0.004 m this is the wavelength of the required ultrasonic sound.
we know that
v = λf
where v is the speed of a wave
λ is the wavelength of the wave
f is the frequency of the wave
f = v/λ
substituting values, we have
f = 1500/0.004 = 375000 Hz
==> 375000/1000 = 375 kHz ≅ 380 kHz
Which examination technique is the visualization of body parts in motion by projecting x-ray images on a luminous fluorescent screen?
Answer:
Fluoroscopy
Explanation:
A Fluoroscopy is an imaging technique that uses X-rays to obtain real-time moving images of the interior of an object. In its primary application of medical imaging, a fluoroscope allows a physician to see the internal structure and function of a patient, so that the pumping action of the heart or the motion of swallowing, for example, can be watched.
A skull believed to belong to an ancient human being has a carbon-14 decay rate of 5.4 disintegrations per minute per gram of carbon (5.4 dis/min*gC). If living organisms have a decay rate of 15.3 dis/min*gC, how old is this skull
Answer:
9.43*10^3 year
Explanation:
For this question, we ought to remember, or know that the half life of carbon 14 is 5730, and that would be vital in completing the calculation
To start with, we use the formula
t(half) = In 2/k,
if we make k the subject of formula, we have
k = in 2/t(half), now we substitute for the values
k = in 2 / 5730
k = 1.21*10^-4 yr^-1
In(A/A•) = -kt, on rearranging, we find out that
t = -1/k * In(A/A•)
The next step is to substitite the values for each into the equation, giving us
t = -1/1.21*10^-4 * In(5.4/15.3)
t = -1/1.21*10^-4 * -1.1041
t = 0.943*10^4 year
g To decrease the intensity of the sound you are hearing from your speaker system by a factor of 36, you can
Answer:
Increase the distance by a factor of 6.
Explanation:
The intensity at a distance r is given by :
[tex]I=\dfrac{P}{4\pi r^2}[/tex]
Here,
P is power emitted
r is distance from source
It means that the intensity is inversely proportional to the distance from the source.
To decrease the intensity of the sound you are hearing from your speaker system by a factor of 36, we can increase the distance by a factor of 6. Hence, this is the required solution.