Answer:
(a) 6.283 Wb (b) 69.11 Wb (c) I = 0.628 A
Explanation:
Given that,
The diameter of the loop, d = 40 cm
Radius, r = 20 cm
Initial magnetic field, B = 5 mT
Final magnetic field, B' = 55 mT
Initial magnetic flux,
[tex]\phi_i=BA\\\\=5\times 10^{-3}\times \pi \times 20^2\\\\=6.283\ Wb[/tex]
Final magnetic flux,
[tex]\phi_f=B'A\\\\=55\times 10^{-3}\times \pi \times 20^2\\\\=69.11\ Wb[/tex]
Due to change in magnetic field an emf will be generated in the loop. It is given by :
[tex]\epsilon=-\dfrac{d\phi}{dt}\\\\=\phi_f-\phi_i\\\\=69.11-6.283\\\\=62.827\ V[/tex]
Let I be the current in the loop. We can find it using Ohm's law such that,
[tex]\epsilon=IR\\\\I=\dfrac{\epsilon}{R}\\\\I=\dfrac{62.827}{100}\\\\=0.628\ A[/tex]
Hence, this is the required solution.
large roll of fabric
Costume Rendering
Scissors
Pattern
Bolt
Answer:
Do we have to choose out of those three?
Explanation:
I WILL REPORT YOU IF YOU ANSWER WITH LINK!!! WILL GIVE BRAINLIEST
You have two identical bowling balls that are 1.00 m apart as measured from their centers. There is an attractive force of gravity between them due to their mass. When a net charge of +0.40 nC is placed on each bowling ball, the force exerted by the electrostatic forces exactly balances the force of gravity on the bowling balls resulting in a net charge of zero. The formula for the electrostatic force is similar to the formula for the gravitational force. There is a constant that is multiplied by the magnitude of each of the two charges and divided by the square of the distance between them. What two conclusions can you draw from this information?
CHOOSE 2 ANSWERS
Answer:
I think is is
Explanation:
B and C why because i have a gut feeling
During a soccer game, a player grabs and holds an opponent's shirt outside of the penalty box. After the foul is called, what kick is awarded to put the ball back into play?
a
Penalty Kick
b
Indirect Free Kick
c
Kickoff
d
Direct Free Kick
Calculate the force of an object that has a mass of 10kg and an acceleration of 4m/s²
A-40N
B-8N
C-2.5N
D-0.4N
PLEASE HELP
A problem says a plane is accelerating
3.42 m/s2 northeast. Which one of these
tables includes that information correctly?
Answer:
The answer is C, I just guessed and got it right lol
Explanation:
⦁ A certain resistor is required to dissipate 0.25 W, what standard rating should be used?
Answer:A
Explanation:
What is another example of convection
Answer:
boiling water - When water heats, the heat from the burner is transferred to the container, heating the water at the bottom. This hot water rises, creating a circular motion, as colder water descends to replace it.
Radio waves travel at the speed of light. What is the wavelength of a radio signal with a frequency of 9.45 x 10^7 Hz?
The wavelength of this radio signal is equal to 3.18 meters.
Given the following data:
Frequency = [tex]9.45 \times10^7[/tex] Hz.Speed of light = [tex]3 \times 10^8[/tex] m/s.What is wavelength?Wavelength can be defined as the distance between two (2) successive crests (troughs) of a wave.
How to calculate wavelength.Mathematically, the wavelength of a wave is given by this formula:
[tex]\lambda = \frac{V}{F}[/tex]
Where:
F is the frequency of a wave.V is the speed of a sound wave.[tex]\lambda[/tex] is the wavelength of a sound wave.Substituting the given parameters into the formula, we have;
[tex]\lambda = \frac{3 \times 10^8}{9.45 \times10^7}[/tex]
Wavelength = 3.18 meters.
Find more information on waves here: brainly.com/question/23460034
The wavelength of the radio signal travel at speed of light is 3.17m.
Given the data in the question;
Frequency of the radio wave; [tex]f = 9.45 * 10^{7}Hz = 9.45 * 10^{7} s^{-1}[/tex]Wavelength of a radio signal; [tex]\lambda = \ ?[/tex]WavelengthWavelength the spatial period of a periodic wave. That is to say, when the shapes of waves are Wavelength , the distance over which they are repeated is called wavelength. Wavelength is expressed as;
[tex]\lambda = \frac{v}{f}[/tex]
Where [tex]\lambda[/tex] is wavelength, f is the frequency of the wave and c is the velocity or speed of light ( [tex]c = 3*10^8m/s[/tex] )
We substitute our values into the expression above.
[tex]\lambda = \frac{c}{ f}\\ \\\lambda = \frac{3*10^8m/s}{9.45*10^7s^{-1}} \\\\\lambda = \frac{3*10^8ms/s}{9.45*10^7}\\\\\lambda = 3.17m[/tex]
Therefore, the wavelength of the radio signal travel at speed of light is 3.17m.
Learn more about wavelength: brainly.com/question/16776907
Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius RRR which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2cm2. How large does the radius RRR of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/mmV/m at the receiver
Answer:
R₁ = 0.126 m
Explanation:
Let's use the definition of intensity which is the power per unit area
I = P / A
the generated power is constant
P = I A
power is
P = E / t
if we perform the calculations for a given time, the wave energy is
E = q V
we substitute
P = [tex]\frac{q V\ A}{t}[/tex]
we can write this equation for two points, point 1 the antenna and point 2 the receiver
V₁A₁ = V₂A₂
A₁ = [tex]\frac{V_2}{V_1} \ A_2[/tex]
A₁ = 0.1 10⁻³ 5 10⁻⁴ /V₁
A₁ = 5 10⁻⁸ /V₁
In general, the electric field on the antenna is very small on the order of micro volts, suppose V₁ = 1 10⁻⁶ V
let's calculate
A₁ = 5 10⁻⁸ / 1 10⁻⁶
A₁ = 5 10⁻² m²
the area of a circle is
A = π r²
we substitute
π R1₁²= 5 10⁻²
R₁ = [tex]\sqrt{ \frac{5 \ 10^{-2} }{\pi } }}[/tex]
R₁ = 0.126 m
Electron spin: Radio astronomers can detect clouds of hydrogen too cool to radiate optical wavelengths of light by means of the 21 cm spectral line corresponding with the flipping of the electron in a hydrogen atom from having its spin parallel to the proton spin to having it antiparallel. From this wavelength, and thus E between states, find the magnetic field experienced by the electron in a hydrogen atom
Answer:
the magnetic field experienced by the electron is 0.0511 T
Explanation:
Given the data in the question;
Wavelength λ = 21 cm = 0.21 m
we know that Bohr magneton μ[tex]_B[/tex] is 9.27 × 10⁻²⁴ J/T
Plank's constant h is 6.626 × 10⁻³⁴ J.s
speed of light c = 3 × 10⁸ m/s
protein spin causes magnetic field in hydrogen atom.
so
Initial potential energy = -μ[tex]_B[/tex]B × cos0°
= -μ[tex]_B[/tex]B × 1
= -μ[tex]_B[/tex]B
Final potential energy = -μ[tex]_B[/tex]B × cos180°
= -μ[tex]_B[/tex]B × -1
= μ[tex]_B[/tex]B
so change in energy will be;
ΔE = μ[tex]_B[/tex]B - ( -μ[tex]_B[/tex]B )
ΔE = 2μ[tex]_B[/tex]B
now, difference in energy levels will be;
ΔE = hc/λ
2μ[tex]_B[/tex]B = hc/λ
2μ[tex]_B[/tex]Bλ = hc
B = hc / 2μ[tex]_B[/tex]λ
so we substitute
B = [(6.626 × 10⁻³⁴) × (3 × 10⁸)] / [2(9.27 × 10⁻²⁴) × 0.21 ]
B = [ 1.9878 × 10⁻²⁵ ] / [ 3.8934 × 10⁻²⁴ ]
B = 510556326.09
B = 0.0511 T
Therefore, the magnetic field experienced by the electron is 0.0511 T
list some application of atmospheric pressure?
Answer:
Drinking straw, syringe, Dropper, vacuum, etc.
Explanation:
Why did the “yielders” conform in Asch’s experiment?
Answer:
Asch's experiment showed that about 75% of people were "yielders" who conformed and 25% were "independent" who didn't conform. Asch concludes that people ignored reality and gave an incorrect answer in order to follow the rest of the group.
A particle of ink in a ink-jet printer carries a charge of -8x 10^-13 C and is deflected onto paper force of 3.2x10^-4. Find the strength of the electric field
o reduce the drag coefficient and thus to improve the fuel efficiency, the frontal area of a car is to be reduced. Determine the amount of fuel and money saved per year as a result of reducing the frontal area from 18 to 14 ft2. Assume the car is driven 12,000 mi a year at an average speed of 55 mi/h. Take the density and price of gasoline to be 50 lbm/ft3 and $3.10/gal, respectively; the density of air to be 0.075 lbm/ft3, the heating value of gasoline to be 20,000 Btu/lbm; and the overall efficiency of the engine to be 30 percent. Take the drag coefficient as CD
Complete question
To reduce the drag coefficient and thus to improve the fuel efficiency, the frontal area of a car is to be reduced. Determine the amount of fuel and money saved per year as a result of reducing the frontal area from 18 to 14 ft2. Assume the car is driven 12,000 mi a year at an average speed of 55 mi/h. Take the density and price of gasoline to be 50 lbm/ft3 and $3.10/gal, respectively; the density of air to be 0.075 lbm/ft3, the heating value of gasoline to be 20,000 Btu/lbm; and the overall efficiency of the engine to be 30 percent. Take the drag coefficient as CD=0.3 for a passenger car.
Answer:
22.22%
$57
Explanation:
From the question we are told that
Initial area of frontal area [tex]a_1=18ft^2[/tex]
Final area of frontal area [tex]a_2=14ft^2[/tex]
Distance covered a year [tex]D=12000mile[/tex]
Average speed a year [tex]V_{avg}=55mile/h[/tex]
Density [tex]\rho=50 lbm/ft3[/tex]
Price [tex]P= $3.10/gal[/tex]
Density of air [tex]\rho_{air} 0.075 lbm/ft3[/tex]
Heating value of gasoline [tex]Q=20,000 Btu/lbm[/tex]
Efficiency [tex]\eta=30\%[/tex]
Drag coefficient [tex]CD=0.3[/tex]
[tex]\triangle A=18-14ft^2=4ft^2[/tex]
Generally the equation for drag force [tex]C_D[/tex] is mathematically given as
[tex]F_D=\frac{C_DAPV^2}{2}[/tex]
[tex]F_D=\frac{0.3*18*0.075*(80.685^2)*A}{2}[/tex]
[tex]F_D=73.24Alb[/tex]
where [tex]v=55mil/h*1.467=80.685ft/s[/tex]
Generally the equation for work done W is mathematically given as[tex]W=F_D*L[/tex]
where [tex]L=12000mile*5280[/tex]
[tex]L=63360000[/tex]
[tex]W=73.24A*63360000[/tex]
[tex]W=4.6*10^9A[/tex]
Generally the equation for overall efficiency [tex]\eta[/tex] is mathematically given as
where
[tex]W_{req}=required\ gasoline\ power\ efficiency[/tex]
[tex]\eta=\frac{W}{W_{req}}[/tex]
[tex]W_{req}=\frac{W}{\eta}[/tex]
[tex]W_{req}=\frac{4.6*10^9}{0.3}[/tex]
[tex]W_{req}=1.55*10^{10}A[/tex]
Generally the equation for reduction fee with change in frontal area [tex]\triangle M[/tex] is mathematically given as
[tex]\triangle M =\triangle Vgasoline*cost[/tex]
Where
[tex]\triangle Vgasoline= volume\ reduction\ of\ gasoline[/tex]
[tex]\triangle Vgasoline=\frac{E_{req}}{H*P}[/tex]
[tex]\triangle Vgasoline=\frac{1.55*10^{10}A}{20000*778.169*32.2*50}[/tex]
if
[tex]20000btu/ibm=20000*778.169*32.2(1bm.ft^2/s)[/tex]
[tex]\triangle Vgasoline=\frac{1.55*10^{10}A}{20000*778.169*32.2*50}[/tex]
[tex]\triangle Vgasoline=0.61859A[/tex]
Therefore
[tex]\triangle M =0.61859A*3.10[/tex]
if [tex]1ft^2=7.48gal[/tex]
[tex]\triangle M =0.61859(4)*3.10*7.48[/tex]
[tex]\triangle M = \$ 57.671[/tex]
Generally the equation for reduction of fuel [tex]F_r[/tex]is mathematically given as
[tex]F_r=\frac{\triangle A}{\triangle i}*100[/tex]
where
[tex]\triangle i=18ft^2[/tex]
[tex]F_r=\frac{4}{18}*100[/tex]
Fuel reduction price by reducing front area is 22.22%
Money saved per year is $57
Aliens from the planet Mars can have long or short antennas. The allele for a long antenna is DOMINANT, and the allele for a short antenna is RECESSIVE. Which of the following is the pair of alleles an alien would have to inherit to have a short antenna?
Recessive, dominant
Dominant, dominant
Recessive, recessive
Dominant, recessive
Answer:
Recessive, dominant
Explanation:
umm thats the answer
form
bonds with each other.
There are many kinds of mixtures. Some mixtures are
chunky like a mixture of peanuts and raisins. These
mixtures are called
I
mixtures.
Answer:
Homogeneous mixtures
Explanation:
I think so because homogeneous means mixed mixtures
which process of the method a neutral object obtains an. electrical charge
As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length of your window is Lw. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity g. Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward). If the bottom of your window is a height hb above the ground, what is the velocity vground of the pot as it hits the ground? You may introduce the new variable vb, the speed at the bottom of the window, defined by
vb = Lwt + gt2.
Answer:
[tex]\mathbf{v_{ground} = \sqrt{{v^2+2ghb}}}[/tex]
Explanation:
From the information given:
The avg. velocity post the window is;
[tex]v_{avg} = \dfrac{L_w}{t}[/tex]
[tex]v_b[/tex] = velocity located at the top of the window
[tex]v_b[/tex] = velocity situated at the bottom of the window
Using the equation of kinematics:
[tex]v_b = v_t + gt[/tex]
Hence,
[tex]v_t = v_b - gt[/tex]
To determine the average velocity as follows:
[tex]v_{avg} = \dfrac{1}{2} (v_t + v_b)\dfrac{L_w}{t}= \dfrac{1}{2}(v_b - gt +v_b) \\ \\\dfrac{L_w}{t} = v_b - \dfrac{1}{2}gt \\ \\ v_b = \dfrac{L_w}{t }+ \dfrac{1}{2} gt\\ \\ = \dfrac{1}{t} \Bigg(L_w + \dfrac{1}{2}gt^2 \Bigg) \\ \\[/tex]
where;
[tex]v_b[/tex] = velocity gained when fallen through the height h.
Similarly, using the equation of kinematics, we have;
[tex]v_b^2 = 2gh \\ \\h = \dfrac{v_b^2}{2g}[/tex]
[tex]\implies \dfrac{(L_w + \dfrac{1}{2} gt^2_^2}{2gt^2}[/tex]
Thus, the velocity at the ground is;
[tex]v^2_{grround} = v_b^2 + 2ghb[/tex]
[tex]\mathbf{v_{ground} = \sqrt{{v^2+2ghb}}}[/tex]
If a gas turned into a solid without going through the liquid state and how do you reverse it?
Answer:
put it in a volcano
Explanation:
A small wooden block with mass 0.750 kg is suspended from the lower end of a light cord that is 1.40 m long. The block is initially at rest. A bullet with mass 0.0100 kg is fired at the block with a horizontal velocity v0. The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of 0.800 m, the tension in the cord is 5.00 N.
Required:
What was the initial speed v0 of the bullet?
Answer:
The initial speed of the bullet will be:
[tex]v_{ib}=331.36\: m/s[/tex]
Explanation:
Using the momentum conservation:
[tex]p_{i}=p_{f}[/tex]
[tex]m_{b}v_{ib}=Mv[/tex] (1)
Where:
M is the mass of the black plus the mass of the bulletv(ib) is the initial velocity of the bulletv is the velocity of the block with the bullet inside.Using the conservation of energy:
[tex]\frac{1}{2}Mv^{2}=\frac{1}{2}Mv_{2}^{2}+Mgh [/tex] (2)
Where v(2) is the speed of the system at 0.8 m of vertical height.
Using the forces acting on the system we can find v(2).
The forces can be equal to the centripetal force:
[tex]T-Mg*sin(\alpha)=M\frac{v_{2}^{2}}{L}[/tex]
α is the angle of T with respect to the horizontal, here α = 25.4°
So, v(2) will be:
[tex]T-Mg*sin(\alpha)=M\frac{v_{2}^{2}}{L}[/tex]
[tex]\frac{L}{M}(T-Mg*sin(\alpha))=v_{2}^{2}[/tex]
[tex]\frac{1.4}{0.76}(5-0.76*9.81*sin(25.4))=v_{2}^{2}[/tex]
[tex]v_{2}=1.82 \: m/s[/tex]
Using this value on equation (2) we will find v.
[tex]v^{2}=v_{2}^{2}+2gh [/tex]
[tex]v^{2}=1.82^{2}+2(9.81)(0.8) [/tex]
[tex]v=4.36\: m/s [/tex]
And finally using equation (1) we can find the initial speed of the bullet.
[tex]m_{b}v_{ib}=Mv[/tex]
[tex]0.01*v_{ib}=0.76*4.36[/tex]
[tex]v_{ib}=331.36\: m/s[/tex]
I hope it helps you!
An electron moves through a region of crossed electric and magnetic fields. The electric field E = 3059 V/m and is directed straight down. The magnetic field B = 1.14 T and is directed to the left. For what velocity v of the electron into the paper will the electric force exactly cancel the magnetic force?
Answer:
v = 2683.33 m/s
Explanation:
The magnetic force and the electric force on the electron must be the same, in order for them to cancel each other:
[tex]Electric\ Force = Magnetic\ Force\\Eq = qvBSin\theta \\\\v = \frac{E}{BSin\theta}[/tex]
where,
v = velcoity of electron = ?
E = Electric Field = 3059 V/m
B = Magnetic Field = 1.14 T
θ = Angle between velocity and magnetic field = 90°
Therefore,
[tex]v = \frac{3059\ V/m}{(1.14\ T)Sin90^o}[/tex]
v = 2683.33 m/s
So the the greater the height, the
farther something can fall, the greater
the potential energy.
True
False
PLEASE HELPPPPPP <333
Answer:
Explanation:
The answer is c. I am very sure
Answer:
i think its b
Explanation:
im not very sure
My best friend just returned from the clinic. She just found out she is HIV- positive. She invited me for dinner. What should I do? Should I go? What do I say to her and her family? Can I eat the foods she cooked or touched? Will she be all right? Imagine this letter is from a friend. Write a page response. Include reasons for your advice.
Answer:
yes, because you dont want to be mean and make her feel worse about it and hiv has something do with s3x not germs so yes go
Explanation:
6. Applying Explain how scientists see what early galaxies looked like..
Answer: The younger elliptical and lenticular galaxies had results similar to spiral galaxies like the Milky Way. The researchers found that the older galaxies have a larger fraction of low-mass stars than their younger counterparts.
Explanation:
The Image shows a magnetic field around the poles of a magnet. Identify the areas where the magnetic force is the strongest.
N
Answer:
strongest are at the points of the north pole and the south pole, specifically between the red box and the letter of each pole.
Explanation:
The lines of magnetic force are drawn so that the density of lines is proportional to the intensity of the magnetic field.
Therefore, the sections where the magnetic field is strongest are at the points of the north pole and the south pole, specifically between the red box and the letter of each pole.
The volumes of two bodies are measured to be
V₁ = (10.2 ± 0.02) cm³ and V₂ = (6.4 ± 0.01) cm³. Calculate sum and difference in
volumes with error limits.
Answer:
sum of volumes = (16.6 ± 0.03) cm³
and difference of volumes = (3.8 ± 0.03) cm³
Explanation:
Here,
V₁ = (10.2 ± 0.02) cm³ and V₂ = (6.4 ± 0.01) cm³.
Now,
∆V = ± (∆V₁ + ∆V₂)
= ± (0.02 + 0.01) cm³
= ± 0.03 cm³
V₁ + V₂ = (10.2 + 6.4) cm³ = 16.6 cm³ and
V₁ - V₂ = (10.2 - 6.4) cm³ = 3.8 cm³
Thus, sum of volumes = (16.6 ± 0.03) cm³
and difference of volumes = (3.8 ± 0.03) cm³
-TheUnknownScientist
Answer:
I hope it's helpful .............
A pendulum is constructed from a heavy metal rod and a metal disk, both of uniform mass density. The center of the disk is bolted to one end of the rod, and the pendulum hangs from the other end of the rod. The rod has a mass of =1.0 kg and a length of =49.8 cm. The disk has a mass of =4.0 kg and a radius of =24.9 cm. The acceleration due to gravity is =9.8 m/s2.
The pendulum is held with the rod horizontal and then released. What is the magnitude of its angular acceleration at the moment of release?
The magnitude of the angular acceleration of the pendulum at the moment of release is; α = 18.45 rad/s²
We are given;
Mass of rod; m = 1 kg
Length of rod; L = 49.8 cm = 0.498 m
Mass of Disk; M = 4 kg
Radius of disk; r = 24.9 cm = 0.249 m
Let us first calculate the torque acting from the formula;
τ = mg(L/2) + MgL
Thus;
τ = (1 × 9.8 × (0.498/2)) + (4 × 9.8 × 0.498)
τ = 21.96 N.m
Using parallel axis theorem, we can find the moment of inertia about the given axis as;
I = (mL²/3) + ½MR² + ML²
Plugging in the relevant values gives;
I = (1 * 0.498²/3) + ½(4 * 0.249²) + (4 * 0.498²)
I = 1.19 kg.m²
The angular acceleration is given by the formula;
α = I/τ
α = 21.96/1.19
α = 18.45 rad/s²
Read more at; https://brainly.com/question/23321366
5. What is the period of a vertical mass-spring system that has an amplitude of
71.3 cm and maximum speed of 7.02 m/s? The spring constant is 12.07 N/m.
The period of the vertical mass-spring is 0.64 s.
The given parameters:Amplitude of the spring, A = 71.3 cm Maximum speed of the spring, V = 7.02 m/sSpring constant, k = 12.07 N/mThe angular speed of the vertical mass-spring is calculated as follows;
[tex]V_{max} = A \omega\\\\\omega = \frac{V_{max}}{A} \\\\\omega = \frac{7.02}{0.713} \\\\\omega = 9.85 \ rad/s[/tex]
The period of the vertical mass-spring is calculated as follows;
[tex]f = \frac{\omega }{2\pi} \\\\T = \frac{1}{f} \\\\T = \frac{2 \pi}{\omega } \\\\T = \frac{2\pi }{9.85} \\\\T = 0.64 \ s[/tex]
Thus, the period of the vertical mass-spring is 0.64 s.
Learn more about period of oscillation here: https://brainly.com/question/20070798
Question /
Which object is shown below?
A. Convex mirror
B. Convex lens
C. Concave lens
D. Concave mirror
SLIDNAT