Answer:
iek
Explanation:
Answer:
the answer to your questions above is 3 x 4 = 12 volts
Explanation:
Voltage = Current × Resistance
A microscope has an objective lens with diameter 1.04 cm. You wish to resolve an object 9.09 micrometers in size. You are using visible light of wavelength 562 nm. Using the Rayleigh criterion, what is the distance that the objective lens must be from the object to resolve it
Answer:
L = 0.1379 m = 13.79 cm
Explanation:
The Rayleigh criterion establishes that two objects are separated when the maximum of diffraction for slits coincides with the minimum of the other point, therefore the expression for the diffraction
a sin θ = m λ
the first zero occurs when m = 1
let's use trigonometry to find the angle
tan θ = y / L
as in these experiments the angles are very small
tan θ = sin θ /cos θ = sin θ
sin θ = y / L
we substitute
a y /L = λ
In the case of circular aperture the system must be solved in polar coordinates, for which a numerical constant is introduced
a y / L = 1.22 λ
L = a y / 1.22 λ
We search the magnitudes to the SI system
a = 1.04 cm = 1.04 10⁻² m
y = 9.09 10⁻⁶ m
λ = 562 10⁻⁹ m
let's calculate
L = [tex]\frac{1.04 \ 10^{-2} \ 9.09 \ 10^{-6} }{1.22 \ 562 \ 10^{-9} }[/tex]
L = 1.379 10⁻¹ m
L = 0.1379 m = 13.79 cm
A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.6 cm before reaching its new equilibrium length. The block is then pulled down slightly and released. What is the frequency of oscillation?
Answer:
f = 3.09 Hz
Explanation:
This is a simple harmonic motion exercise where the angular velocity is
w² = [tex]\frac{k}{m}[/tex]
to find the constant (k) of the spring, we use Hooke's law with the initial data
F = - kx
where the force is the weight of the body that is hanging
F = W = m g
we substitute
m g = - k x
k = [tex]- \frac{m g}{x}[/tex]
we calculate
k = [tex]- \frac{9.8 m}{- 2.6 \ 10^{-2}}[/tex]
k = 3.769 10² m
we substitute in the first equation
w² = [tex]\frac{ 3.769 \ 10^2 \ m }{m}[/tex]
w = 19.415 rad / s
angular velocity and frequency are related
w = 2πf
f = [tex]\frac{w}{2\pi }[/tex]
f = 19.415 / 2pi
f = 3.09 Hz
What happens during nuclear fission? A) Nucleus forms from two smaller nuclei. B) two nuclei give off positrons C) A nucleus gives off an alpha particle D) Two or more nuclei forms when a larger nucleus splits
Answer:
it is d, just did the quiz. all this came from the amazing com mentor who said a was wrong and it is d, thank you.
Explanation:
This splitting of the big atomic nucleus into smaller nuclei is known as nuclear fission, and further discussion is as follows:
Nuclear fission:The neutron collides with a uranium atom that splits during the fission of nuclear, producing a tremendous quantity of energy in the form of heat produced. If a uranium atom splits, additional neutrons are extricated.
It proceeds to smash more uranium, and the cycle continues forever.The neutrons are absorbed into a nucleus in such a nuclear reactor. It transforms into uranium-236, which is extremely unstable. Its nucleus is split into two enormous shards known as 'daughter nuclei'.Therefore, the final answer is "Option D".
Find out more about the nuclear fission here:
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A flat circular coil of wire having 400 turns and diameter 6.0 cm carries a current of 7.0 A. It is placed in a magnetic field of with the plane of the coil making an angle of 30° with the magnetic field. What is the magnitude of the magnetic torque on the coil?
Answer:
6.8 N.m
Explanation:
The computation of the magnitude of the magnetic torque on the coil is given below:
Given that
n = 400
d = 6.0 cm
Current is I = 7.0 A
Angle is [tex]\theta[/tex] = 30 degree
Now
We know that
the magnitude of the magnetic torque is
= nIABsin[tex]\theta[/tex]
= (400) (7.0) π ÷ 4 (0.06m)^2 sin(90° - 30°)
As
[tex]\theta[/tex] = (90° - Ф)
= (400) (7.0) π ÷ 4 (0.06m)^2 sin 60°
= 6.8 N.m
What is the centripetal force for a roller coaster if the mass is 10 kg and the normal force is 25 N?
Answer:
Fc = 123 Newton
Explanation:
Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.
Mathematically, net force is given by the formula;
[tex] Fnet = Fapp + Fg[/tex]
Where;
Fnet is the net force.
Fapp is the applied force.
Fg is the force due to gravitation.
Given the following data;
Normal force = 25N
Mass = 10kg
To find the centripetal force;
From the net force, we have the following formula;
Fc = N + mg
Where;
Fc is the centripetal force.
N is the normal force.
mg is the the weight of the object.
Substituting into the formula, we have;
Fc = 25 + 10(9.8)
Fc = 25 + 98
Fc = 123 Newton
A boy is pulling a sled with a net force of 10 N. If the mass of the sled is 20 kg, what is the acceleration of the sled?
Answer:
0.5 m/s
Explanation:
acceleration= force times mass
Given the amount of force applied on the sled as well as its mass, the acceleration of the sled is 0.5m/s².
What is force?A force is simply referred to as either a push or pull of an object resulting from the object's interaction with another object.
From Newton's Second Law, force is expressed as;
F = m × a
Where is mass of object and a is the acceleration
Given the data in the question;
Force applied F = 10N = 10kgm/s²Mass of the sled = 20kgAcceleration a = ?F = m × a
10kgm/s² = 20kg × a
a = 10kgm/s² ÷ 20kg
a = 0.5m/s²
Given the amount of force applied on the sled as well as its mass, the acceleration of the sled is 0.5m/s².
Learn more about force here: brainly.com/question/27196358
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When a narrow laser beam passes through a fine wire mesh before arriving at the wall, it forms a complicated pattern of bright spots on the wall. This pattern of spots would not occur if you sent a flashlight beam through the mesh because light from the flashlight is not a single electromagnetic wave. cannot be sent through a single opening of the mesh. is horizontally polarized, while laser light is vertically polarized. is vertically polarized, while laser light is horizontally polarized.
Answer:
this pattern to occur there must be coherence in the light beams.
you use a flashlight, the rays are incoherent so diffraction patterns cannot occur.
Explanation:
The point pattern that appears in the wall is the result of the interference and diffraction processes through each space of the mesh, for this pattern to occur there must be coherence in the light beams.
The coherence process is that all the rays have the same constant and phase, before the appearance of the lasers, the light is stopped by a small opening and this ray is the one that passes through the slits, with the appearance of the laser this it is consistent from its production process, so opening is not necessary, with this there is much greater intensity and the measurement process is simplified.
When you use a flashlight, the rays are incoherent so diffraction patterns cannot occur.
Polarization has no effect on diffraction patterns so it does not matter if it is vertical or horizontal.
This is 100 points. When i find out how i will put the first person to answer as brainiest.
i have no question. God bless you all
Answer:
Thank you so much! Have a great day!
A main-sequence star at a distance of 20 pc is barely visible through a certain telescope. The star subsequently ascends the giant branch, during which time its temperature drops by a factor of three and its radius increases a hundredfold. What is the new maximum distance at which the star would still be visible in the same telescope?
Answer:
Explanation:
The surface area of a star estimated by the energy emitted per sq meter yields the overall luminosity, which can be represented mathematically as:
[tex]L= 4 \pi R^2 \sigma T^4 --- (1)[/tex]
where;
L ∝ R²T⁴
and;
R = radius of the sphere
σ = Stefans constant
T = temperature
Also; The following showcase the relationship between flux density as well as illuminated surface area as:
[tex]F = \dfrac{L}{A}[/tex]
where
A = 4πd² and L ∝ R²T⁴
[tex]F = \dfrac{R^2T^4}{4 \pi d^2} \\ \\ F \alpha \dfrac{R^2T^4}{ d^2} --- (2)[/tex]
Given that:
distance d₁ = 20 pc
Then, using equation (2)
[tex]F_1 \ \alpha \ \dfrac{R^2_1T^4_1}{ d^2_1}[/tex]
However, we are also being told that there is a temp. drop by a factor of 3;
So, the final temp. [tex]T_2 = \dfrac{T_1}{3}[/tex]; and the final radius is [tex]R_2 = 100R_1[/tex] since there is increment by 100 folds.
Now;
[tex]F_2 \ \alpha \ \dfrac{R^2_2T^4_2}{ d^2_2}[/tex]
SInce;
[tex]F_1 = F_2[/tex]
It implies that:
[tex]\dfrac{R^2_1T^4_1}{ d^2_1 } = \dfrac{R^2_2T^4_2}{ d^2_2} \\ \\ d_2 = \sqrt{\dfrac{R_2^2T_2^4}{R_1^2T_1^4}}(d_1)[/tex]
Replacing all our values, we have:
[tex]d_2 = \sqrt{\dfrac{(100R_1)^2 \times (\dfrac{T_1}{3})^4}{R_1^2T_1^4}}(20 ) \\ \\ d_2 = \sqrt{\dfrac{(100)^2 }{3^4}}(20 ) \\ \\ d_2 = \sqrt{\dfrac{(100)^2 }{3^4}}(20 ) \\ \\ d_2 =222 \ pc[/tex]
Your _________ is simply your weight-to-height ratio which can be calculated using certain body measurements.
O health
O size
O weight loss
O BMI
Answer:
SiZe
Explanation:
bad bunny is the best!!!
A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing through the cross-sectional area in 10s is
Answer:
n = 1.56 x 10¹⁷ electrons
Explanation:
First of all, we will calculate the current passing through wire:
[tex]I = \frac{q}{t}[/tex]
where,
I = current = ?
q = charge = 9 mC = 0.009 C
t = time = 3.6 s
Therefore,
[tex]I = \frac{0.009\ C}{3.6\ s}\\\\I = 0.0025\ A = 2.5\ mA[/tex]
Now, for the same current in 10 s time the charge will be:
q = It = (0.0025 A)(10 s)
q = 0.025 C
Now, the number of electrons can be given as:
[tex]q = ne\\\\n = \frac{q}{e}\\\\[/tex]
where,
n = no. of electrons = ?
q = charge = 0.025 C
e = charge on single electron = 1.6 x 10⁻¹⁹ C
Therefore,
[tex]n = \frac{0.025\ C}{1.6\ x\ 10^{-19}\ C}[/tex]
n = 1.56 x 10¹⁷ electrons
What is this answer? A blank is a combination of two or more substances, elements, compounds, or both, in the same place but not chemically combined
On a distance-time graph, the faster the object, the _______ the slope of the line. *
A.curvier
B.flatter
C.lower
D.steeper
The following words are used to describe the direction of a moving object EXCEPT... *
A. East
B. North
C.South
D. Today
Answer:
..In a distance-time graph, the slope or gradient of the line is equal to the speed of the object. The steeper the line (and the greater the gradient) the faster the object is moving. Calculate the speed of the object represented by the green line in the graph, from 0 to 3 s.
d) steeper
D)Today
b
Jasmine plays a note of wavelength 22 cm on her clarinet. The speed of
sound in air is 340 m/s. What is the frequency of this note? Give your
answer to 3 significant figures.
Answer:
Frequency = 1,550Hz
Explanation:
To solve this we can use the equation: [tex]f=\frac{v}{\lambda}[/tex]
(frequency = velocity/wavelength).
We are given the information that the wavelength is 22cm and the speed is 340m/s. The first step is to make sure everything is in the correct units (SI units), and to convert them if needed. The SI Units for velocity and wavelength are m/s and m respectively. This means we need to convert 22cm into meters, which we can do by dividing by 100, (as there are 100cm in a meter). 22/100 = 0.22m
Now we can substitute these values into the formula and calculate to solve:
[tex]f=\frac{340}{0.22} \\\\f=1545.454...[/tex]
Simplify to 3 significant figures:
f = 1,550Hz
(Which I believe is just below a G6 if you were interested)
Hope this helped!
use a trigonometric equation to determine the leg of this triangle
C=90°
A=30°
c=10m
What is a?
Answer: 5
Explanation: B is for sure 60°, c* cosB = 10*1/2 =5
Please help I’ll mark you brainliest
Answer:
Percentage:
Rr = 50% because it's 2/4 (for both or 25% each since you have them separate)
rr = also 50%, because it's also 2/4.
Phenotype:
Rr = heterozygous
rr = "hozygous" recessive
In addition, RR is "hozygous" dominant
Explanation:
They said the hozygous is a swearword LOL.
A column of argon is open at one end and closed at the other. The shortest length of such a column that will resonate with a 200 Hz tuning fork is 42.5 cm. The speed of sound in argon must be: Group of answer choices
Answer:
The speed of sound in the argon is 340 m/s.
Explanation:
Given;
fundamental frequency, f₀ = 200 Hz
length of the pipe, L = 42.5 cm = 0.425 m
A pipe that is open at one end and closed at another end is known as a closed pipe.
The wavelength for the first harmonic is calculated as;
L = Node -------> Antinode
L = λ/4
λ = 4L
The speed of the sound is calculated as;
v = fλ
where;
v is the speed of the sound
v = f x 4L
v = 200 x (4 x 0.425)
v = 340 m/s
Therefore, the speed of sound in the argon is 340 m/s.
A weight lifter lifts a 1470N barbell above his head from the floor the a height of 2m. he holds the barbell there for 5 sec. how much work does he do?
Answer:
Workdone = 2940 Nm
Explanation:
Given the following data;
Force = 1470 N
Distance = 2m
Time = 5 seconds
To find the work done;
Mathematically, workdone is given by the formula;
Workdone = force * distance
Substituting into the formula, we have;
Workdone = 1470 * 2
Workdone = 2940 Nm
When an object is fully converted into energy the amount of energy liberated is
3.6×1017 J, what is the mass of the substance?
Answer:
Mass, m = 4 kg
Explanation:
Given the following data;
Energy = 3.6 * 10^17 Joules
We know that the speed of light is equal to 3 * 10⁸ m/s.
To find the mass of the substance;
The theory of special relativity by Albert Einstein gave birth to one of the most famous equation in science.
The equation illustrates, energy equals mass multiplied by the square of the speed of light.
Mathematically, the theory of special relativity is given by the formula;
[tex] E = mc^{2} [/tex]
Where;
E is the energy possessed by a substance.m is the mass.c is the speed of light.Substituting into the formula, we have;
[tex] 3.6 * 10^{17} = m * 300000000^{2} [/tex]
[tex] 3.6 * 10^{17} = m * 9*10^{16} [/tex]
[tex] m = \frac {3.6 * 10^{17}}{9*10^{16}} [/tex]
Mass, m = 4 kg
In addition to producing images, ultrasound can be used to heat tissues of the body for therapeutic purposes. An emitter is placed against the surface of the skin; the amplitude of the ultrasound wave at this point is quite large. When a sound wave hits the boundary between soft tissue and bone, most of the energy is reflected. The boundary acts like the closed end of a tube which can lead to standing waves. Suppose 0.7 MHz ultrasound is directed through a layer of tissue at a bone 0.55 cm below the surface. Remember, sound waves in the body travel at 1540 m/s. Explain.
Answer:
ΔT = 4.058 10²⁰ [tex]\frac{S_o^2}{r^2 \ c_e}[/tex]
Explanation:
In this experiment the system can be approximated as a tube with one end open and the other closed.
The open end is where the ultrasound emitter is and the closed end where the limit between the tissue and the bone is, the length of the tube is L = 0.55 cm = 5.5 10⁻³ m
a node is formed at the closed end and a belly at the open end, so the resonance has the form
λ = 4L 1st harmonic
λ = 4/3 L third harmonic
λ = 4/5 L fifth harmonic
λ = 4L / (2n + 1) n = 0, 1, 2, (2n + 1)
This wave is a standing wave therefore energy density remains in place
P = 1/2 ρ v (w S₀)²
angular velocity is related to frequency
w = 2π f
we substitute
E = P = 2π² ρ v f² S₀²
if this energy per unit area is transformed into heat
E = m c_e DT
let's use the concept of density
ρ = m / V
m = ρ V
if there are no losses in the system
½ π² ρ v f² S₀² = ρ V c_e ΔT
ΔT = [tex]\frac{\pi ^2 \ v \f^2 S_o^2}{2V \ c_e}[/tex]
When analyzing this expression the temperature increase is
* quadratic at the frequency and maximum amplitude of the wave
* proportional to the speed of the wave in the tissue
* inversely proportional to tissue volume
we can approximate the volume of the tissue to the volume of a cylinder tube
V = π r² L
ΔT = [tex]\frac{\pi \ v \ f^2 S_o^2 }{r^2 \ L \ c_e}[/tex]
we calculate
ΔT = π 1450 (0.7 10⁶)² S₀² /( r² 5.5 10-3 c_e)
ΔT = 4.058 10²⁰ [tex]\frac{S_o^2}{r^2 \ c_e}[/tex]
A grindstone in the shape of a solid disk has a shaft attached to allow a force to be exerted on. The grindstone has a diameter of 0.650m and a mass of 55.0 kg. The shaft is 0.300 m from the center of the stone and has a mass of 4.00 kg. The grindstone has a motor attached and it is rotating at 450rev/min at a run when the motor is shut off. The grindstone comes to rest in 9.50 s.
a. What is the moment of inertia of the grindstone including the shaft?
b. What is the angular acceleration of the grindstone?
c. What average torque is done by friction to bring the grindstone to rest?
d. How many revolutions has it taken the grindstone to come to rest?
e. What is the linear acceleration of the shaft in this process?
f. What is the linear distance that the shaft traveled?
Answer:
a) 3.265 kg-m^2
b) - 4.96 rad/s^2
c) 16.1944 N-m
d) 35.625
e) - 1.488 m/s^2
f) 67.1175 m.
Explanation:
Given data:
Diameter of grindstone ( D ) = 0.650 m , Radius ( R ) = 0.325 m
mass of grindstone ( M ) = 55 kg
Radius of shaft ( r ) = 0.300 m
mass of shaft ( m ) = 4 kg
Initial Angular velocity = 450 rev/min = f = 7.5 rev/s = w =15π rad/s
time ( t ) = 9.50 secs
a) Determine the moment of inertia of the grindstone including the shaft
moment of inertia of grindstone = MR^2 / 2 = 55* (0.325)^2 / 2 = 2.905
moment of inertia of shaft = mr^2 = 4 *0.3^2 = 0.36
∴ moment of inertia including shaft = 2.905 + 0.36 = 3.265 kg-m^2
b) Determine the angular acceleration of the grindstone
∝ = - 15π / 9.5 ( i.e. angular velocity / time )
= - 4.96 rad/s^2 ( deceleration value )
c) Determine average torque done by friction to bring the grindstone to rest
Torque ( I * ∝ ) = 3.265 x 4.96 = 16.1944 N-m (magnitude)
d) Determine the number of revolutions before grindstone comes to rest
Total revolutions N before grindstone comes to rest
= ( f1 + f2)* t /2 = 7.5 * 9.5 / 2 = 35.625
Note : f2 = 0 as it comes to rest
f1 = 7.5 rev/s
e) Determine the Linear acceleration of the shaft in this process
This can be calculated using this relation
r * ∝ = 0.3 x (- 4.96 ) = - 1.488 m/s^2
f) Determine the linear distance travelled by the shaft
This can be calculated with the relation below
r * 2 * π * N = 0.3 * 2π * 35.625 = 67.1175 m.
What’s the answer to 19 please please help me
Answer:
A.
Explanation:
A po sa tingin ko
sana po makatulong
When electrons are accelerated by 2450 V in an electron microscope, they will have
wavelengths of:
a) 8.113 nm
b) 0.622 nm
c) 0.811 nm
The wavelength of the electron microscope is 2.472*10⁻²⁰ m.
Calculation:
Provided, accelerating voltage = 2450 V
According to de Broglie's equation:
λ = h/p
where,λ = the wavelength of the particle that moves
P= momentum of moving particle
h= plank's constant = 6.63 x 10⁻³ Js
We know that,
P = √(2mk)
P = √(2mqV)
where P= momentum of moving particle
M= mass of the moving object
K= kinetic energy of the object
V= Electric potential of the object(accelerating voltage)
q= charge of the object(electron) = e=1.6 x 10⁻¹⁹C
∴λ = h/√(2meV) = [tex]\frac{6,63 *10^{-34} }{2*9.1*10^{-31}*1.6*10^{-19}*2450 }[/tex]
=2.472*10⁻²⁰ m
S0, the wavelength of an electron microscope is 2.472*10⁻²⁰ m. Think it matches option C.
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Light is traveling in glass, and hits a glass/unknown surface. In the glass the light beam is making an angle of 45.0 o with the normal to the surface. The glass has an index of refraction of 1.52. (A) If the refracted light ray leaves the glass at an 55o from the normal, what is the index of refraction for the unknown surface
Answer:
n = 1.31
Explanation:
When a ray of light crosses the separation surface between two transparent media, there exists a fixed relationship between the indexes of refraction of both media, related with the angles of incidence and refraction, which is known as Snell's Law.The Snell's Law can be written as follows:[tex]n_{i} * sin( \theta_{i}) = n_{r} * sin( \theta_{r}) (1)[/tex]
In our case the ray is incident from the glass, so ni = n glass = 1.52The angle of incidence is the angle that the ray makes with the normal to the separation surface, so θi=45º.The angle of refraction is the angle that the refracted ray makes with the normal, so θr= 55ºReplacing by the values in (1), and solving for nr, we have:[tex]n_{r} =\frac{n_{i} * sin \theta_{i} }{sin \theta_{r} } = \frac{sin (45)*1.52}{sin (55)} = 1.31 (2)[/tex]
Five lamp, each labbled "6V,3W" are operated at normal brightness. What is the total energy supplied to the lamps in five seconds.
Answer:
E = 75 J
Explanation:
First, we will calculate the total power consumed by the five lamps:
[tex]Total\ Power = P = (5)(Power\ of\ one\ lamp)\\P = (5)(3\ W)\\P = 15\ W[/tex]
Now, the energy supply can be calculated as follows:
[tex]E = Pt[/tex]
where,
E = Energy = ?
t = time = 5 s
Therefore,
E = (15 W)(5 s)
E = 75 J
Monochromatic light with lambda equals 630 space n m is incident on a single slit. The slit width is 0.8 mm. If the distance between the screen and slit is 1.8 m, what is the width of the central bright fringe on the screen
Answer:
The angular width is "[tex]1.575\times 10^{-3} \ rad[/tex]" and the linear width is "[tex]2.835\times 10^{-3} \ m[/tex]".
Explanation:
Given:
Wavelength,
[tex]\lambda = 630 \ nm[/tex]
or,
[tex]=630\times 10^{-9} \ m[/tex]
Slit width,
[tex]a = 0.8 \ mm[/tex]
or,
[tex]=0.8\times 10^{-3} \ m[/tex]
Distance between slit and screen,
[tex]D = 1.8 \ m[/tex]
As we know,
The central bright fringe's angular height,
⇒ [tex]\theta = \frac{2 \lambda}{a}[/tex]
[tex]= \frac{2\times 630\times 10^{-9}}{0.8\times 10^{-3}}[/tex]
[tex]=1.575\times 10^{-3} \ rad[/tex]
and,
The linear width will be:
⇒ [tex]x_o=\frac{2D \lambda}{a}[/tex]
By substituting the values, we get
[tex]=\frac{3\times 1.8\times 630\times 10^{-9}}{0.8\times 10^{-3}}[/tex]
[tex]=2.835\times 10^{-3} \ m[/tex]
A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis of 18kg⋅m2. She then tucks into a small ball, decreasing this moment of inertia to 3.6kg⋅m2. While tucked, she makes two complete revolutions in 1.2s.
Required:
If she hadn't tucked at all, how many revolutions would she have made in the 1.5 s from board to water?
Answer:
θ₁ = 0.5 revolution
Explanation:
We will use the conservation of angular momentum as follows:
[tex]L_1=L_2\\I_1\omega_1=I_2\omega_2[/tex]
where,
I₁ = initial moment of inertia = 18 kg.m²
I₂ = Final moment of inertia = 3.6 kg.m²
ω₁ = initial angular velocity = ?
ω₂ = Final Angular velocity = [tex]\frac{\theta_2}{t_2} = \frac{2\ rev}{1.2\ s}[/tex] = 1.67 rev/s
Therefore,
[tex](18\ kg.m^2)\omega_1 = (3.6\ kg.m^2)(1.67\ rev/s)\\\\\omega_1 = \frac{(3.6\ kg.m^2)(1.67\ rev/s)}{(18\ kg.m^2)}\\\\\omega_1 = \frac{\theta_1}{t_1} = 0.333\ rev/s\\\\\theta_1 = (0.333\ rev/s)t_1[/tex]
where,
θ₁ = revolutions if she had not tucked at all = ?
t₁ = time = 1.5 s
Therefore,
[tex]\theta_1 = (0.333\ rev/s)(1.5\ s)\\[/tex]
θ₁ = 0.5 revolution
8. Consider a capacitor that is made of two large conducting plates that are rectangular in shape (1 cm by 6 cm), aligned parallel to each other, and separated by an air-filled gap of 0.001 cm. This capacitor is included in a circuit where a battery provides 15,000 V of potential difference. When the capacitor is fully charged in this circuit, what is the c
Answer: [tex]7.96\ \mu C[/tex]
Explanation:
Given
The dimension of the plate is [tex]1\ cm\times 6\ cm[/tex]
The gap between the plate is [tex]0.001\ cm[/tex]
Voltage applied [tex]V=15,000\ V[/tex]
The capacitance of the capacitor is
[tex]C=\dfrac{\epsilon_o A}{d}\\\\C=\dfrac{8.85\times 10^{-12}\times 1\times 6\times 10^{-4}}{10^{-5}}\\\\C=53.1\times 10^{-11}\ F[/tex]
Charge acquired by the capacitor
[tex]\Rightarrow Q=CV\\\Rightarrow Q=53.1\times 10^{-11}\times 15,000\\\Rightarrow Q=796.5\times 10^{-8}\\\Rightarrow Q=7.96\times 10^{-6}\ C[/tex]
_PCi5 + _H20 + _H3PO4 _HCI
A.) 1,3,1,5
B.) 2,4,1,5
C.) 1,4,1,5
D.) 2,1,2,5
Answer:
C
Explanation:
because of the kids are doing well and staying in the middle of the day early to get a new phone number for home now deep
when a light beam emerges from water into air, the average light speed ____.
a. increases
b. decreases
c. remains the same
d. not provided
Answer:
Increases
Explanation:
The light is emerging from a denser medium to rarer medium so the average light speed increases.
When a light beam emerges from water into air, the average light speed increases.
What is refraction?The bending of a ray passes at an angle from one medium to another in which the speed will be different, as when light passes from air into water and also from water into air can be known as refraction.
When light travels from water to air, the angle from water to air the light gets speeds up and then its changes direction. The light turns away from the normal line. Because the light moves from a denser medium to the rarer medium. But from air to water the speed will be decreased because in that case the light travels from the rarer medium to the denser medium.
Hence, the light beam when passes, the average speed gets increased.
So, Option A is the correct answer.
Learn more about refraction,
https://brainly.com/question/2660868
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