Answer:
R = 36.885 km
Explanation:
In order to distinguish the two planes we must use the Rayleigh criterion that establishes two distinguishable objects if in their diffraction the central maximum of one coincides with the first minimum of the other
The diffraction equation for slits is
a sin θ = m λ
the first minimum occurs for m = 1
sin θ = λ a
as the diffraction experiments the angles are very small, we approximate
sin θ = θ
θ = λ / a
This expression is for a slit, in the case of circular objects, when solving the system in polar coordinates, a numerical constant appears, leaving the expression of the form
θ = 1.22 λ / a
In this problem they give us the frequency, let's find the wavelength with the relation
c = λ f
λ = c / f
θ = 1.22 c/ f a
since they ask us for the distance between the planes, we can use the definition of radians
θ = s / R
if we assume that the distance is large, we can approximate the arc to the horizontal distance
s = x
we substitute
x / R = 1.22 c / fa
R = x f a / 1.22c
Let's reduce the magnitudes to the SI system
f = 9000 MHz = 9 109 Hz
a = 15 m
x = 100 m
let's calculate
R = 100 10⁹ 15 / (1.22 3 108)
R = 3.6885 10⁴ m
let's reduce to km
R = 3.6885 10¹ km
R = 36.885 km
A car whose tire have radii 50cm travels at 20km/h. What is the angular velocity of the tires?
Convert to m/s
[tex]\\ \sf\longmapsto v=20\times 5/18=5.5m/s[/tex]
We know
[tex]\boxed{\sf \omega=\dfrac{rv}{|r|^2}}[/tex]
[tex]\\ \sf\longmapsto \omega=\dfrac{0.5(5.5)}{|0.5|^2}[/tex]
[tex]\\ \sf\longmapsto \omega=\dfrac{2.75}{0.25}[/tex]
[tex]\\ \sf\longmapsto \omega=11rad/s[/tex]
If a disk rolls on a rough surface without slipping, the acceleration of the center of gravity (G) will _ and the friction force will b
Answer:
Will be equal to alpha x r; less than UsN
What is 3/4 of 12 and 24
Answer:
3/4 of 12 = 16
3/4 of 24 = 32
Those are the answers based on how your question sounded
Explanation:
Answer:
27
Explanation:
3/4 of (12 and 24)
of means ×
and means +
therefore,3/4× (12+24)
3/4×(36)
3×9
27
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
A thin film of soap with n = 1.37 hanging in the air reflects dominantly red light with λ = 696 nm. What is the minimum thickness of the film?
Answer:
The thickness is [tex]t = 1.273 *10^{-7} \ m[/tex]
Explanation:
From the question we are told that
The refractive index of the film is [tex]n = 1.37[/tex]
The wavelength is [tex]\lambda = 696 \ nm = 696 *10^{-9 } \ m[/tex]
Generally the condition for constructive interference in a film is mathematically represented as
[tex]2 * t = [m + \frac{1}{2} ] \lambda_k[/tex]
Here t is the thickness of the film , m is the order number (0, 1, 2, 3 ... )
[tex]\lambda _k[/tex] is the wavelength of light that is inside the film , this is mathematically evaluated as
[tex]\lambda _k = \frac{ \lambda }{ n}[/tex]
[tex]\lambda _k = \frac{ 696 *10^{-9}}{ 1.37}[/tex]
[tex]\lambda _k = 5.095 *10^{-7 } \ m[/tex]
So for m = 0
[tex]t = [ 0 + \frac{1}{2} ] \lambda _k * \frac{1}{2}[/tex]
substituting values
[tex]t = [ 0 + \frac{1}{2} ] (5.095 *10^{-7}) * \frac{1}{2}[/tex]
[tex]t = 1.273 *10^{-7} \ m[/tex]
6. What is the bulk modulus of oxygen if 32.0 g of oxygen occupies 22.4 L and the speed of sound in the oxygen is 317 m/s?
Answer:
[tex] \boxed{\sf Bulk \ modulus \ of \ oxygen \approx 143.5 \ kPa} [/tex]
Given:
Mass of oxygen (m) = 32.0 g = 0.032 kg
Volume occupied by oxygen (V) = 22.4 L = 0.0224 m³
Speed of sound in oxygen (v) = 317 m/s
To Find:
Bulk modulus of oxygen
Explanation:
[tex]\sf Density \ of \ oxygen \ (\rho) = \frac{m}{V}[/tex]
[tex]\sf \implies Bulk \ modulus \ of \ oxygen \ (B) = v^{2} \rho[/tex]
[tex]\sf \implies B = v^{2} \times\frac{m}{V}[/tex]
[tex]\sf \implies B = {(317)}^{2} \times \frac{0.032}{0.0224} [/tex]
[tex]\sf \implies B = {(317)}^{2} \times 1.428[/tex]
[tex]\sf \implies B = 100489 \times 1.428[/tex]
[tex]\sf \implies B = 143498.292 \: Pa[/tex]
[tex]\sf \implies B \approx 143.5 \: kPa[/tex]
The capacitor is originally charged. How does the current I in the ammeter behave as a function of time after the switch is closed?
1. I = 0 always.
2. I = constant, not equal to 0.
3. I increases, then is constant.
4. I instantly jumps up, then slowly decreases.
5. None of the above.
Answer:
The current in the ammeter is zero.
(1) is correct option.
Explanation:
Given that,
The capacitor is charged.
We need find the current after closed switched
We know that,
When switch is closed then the capacitor behave as a short circuit, and the all current flows through it. the current is zero.
Then, the ammeter reads zero.
Hence, The current in the ammeter is zero.
(1) is correct option.
The advantage of a hydraulic lever is A : it transforms a small force acting over a large distance into a large force acting over a small distance. B : it transforms a small force acting over a small distance into a large force acting over a large distance. C : it allows you to exert a larger force with less work. D : it transforms a large force acting over a large distance into a small force acting over a small distance. E : it transforms a large force acting over a small distance into a small force acting over a large distance.
Answer:
A) it transforms a small force acting over a large distance into a large force acting over a small distance.
Explanation:
The hydraulic lever works based on Pascal's law of transmission of pressure through a fluid. In the hydraulic lever, the pressure transmitted is the same.
Pressure transmitted P = F/A
where F is the force applied
and A is the area over which the force is applied.
This pressure can be manipulated on the input end as a small force applied over a small area, and then be transmitted to the output end as a large force over a large area.
F/A = f/a
where the left side of the equation is for the output, and the right side is for the input.
The volume of the displaced fluid will be the same on both ends of the hydraulic lever. Since we know that
volume V = (area A) x (distance d)
this means that the the piston on the input smaller area of the hydraulic lever will travel a greater distance, while the piston on the larger output area of the lever will travel a small distance.
From all these, we can see that the advantage of a hydraulic lever is that it transforms a small force acting over a large distance into a large force acting over a small distance.
g A projectile is fired from the ground at an angle of θ = π 4 toward a tower located 600 m away. If the projectile has an initial speed of 120 m/s, find the height at which it strikes the tower
Answer:
The projectile strikes the tower at a height of 354.824 meters.
Explanation:
The projectile experiments a parabolic motion, which consist of a horizontal motion at constant speed and a vertical uniformly accelerated motion due to gravity. The equations of motion are, respectively:
Horizontal motion
[tex]x = x_{o}+v_{o}\cdot t \cdot \cos \theta[/tex]
Vertical motion
[tex]y = y_{o} + v_{o}\cdot t \cdot \sin \theta +\frac{1}{2} \cdot g \cdot t^{2}[/tex]
Where:
[tex]x_{o}[/tex], [tex]x[/tex] - Initial and current horizontal position, measured in meters.
[tex]y_{o}[/tex], [tex]y[/tex] - Initial and current vertical position, measured in meters.
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]t[/tex] - Time, measured in seconds.
The time spent for the projectile to strike the tower is obtained from first equation:
[tex]t = \frac{x-x_{o}}{v_{o}\cdot \cos \theta}[/tex]
If [tex]x = 600\,m[/tex], [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]\theta = \frac{\pi}{4}[/tex], then:
[tex]t = \frac{600\,m-0\,m}{\left(120\,\frac{m}{s} \right)\cdot \cos \frac{\pi}{4} }[/tex]
[tex]t \approx 7.071\,s[/tex]
Now, the height at which the projectile strikes the tower is: ([tex]y_{o} = 0\,m[/tex], [tex]t \approx 7.071\,s[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex])
[tex]y = 0\,m + \left(120\,\frac{m}{s} \right)\cdot (7.071\,s)\cdot \sin \frac{\pi}{4}+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (7.071\,s)^{2}[/tex]
[tex]y \approx 354.824\,m[/tex]
The projectile strikes the tower at a height of 354.824 meters.
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.
A small wave pulse and a large wave pulse approach each other on a string; the large pulse is moving to the right.
Sometime after the pulses have met and passed each other, which of the following statements is correct? (More than one answer may be correct)
- the large pulse continues moving to the right
- the large pulse continues unchanged, moving to the right
- the small pulse is reflected and moves off to the right with a smaller amplitude
- the small pulse is reflected and moves off to the right with its original amplitude
- the two pulses combine into a single pulse moving to the right
Answer:
the large amplitude wave keeps moving to the right
the small amplitude wave continues to move to the left.
When checking the answers, the correct ones are 1, 2
Explanation:
The waves fulfill the principle of superposition, which states that the value of the function at a point is the algebraic sum of the waves at a given instant.
The two waves in this exercise travel in the opposite direction, so when they are close, the resulting wave is the sum of the two waves, having a complicated shape. But when the waves follow their movement, they give in the same way as the initial a,
the large amplitude wave keeps moving to the right
the small amplitude wave continues to move to the left.
When checking the answers, the correct ones are 1, 2
An organ pipe open at both ends is 1.5 m long. A second organ pipe that is closed at one end and open at the other is 0.75 m long. The speed of sound in the room is 330 m/s. Which of the following sets of frequencies consists of frequencies which can be produced by both pipes?
a. 110Hz,220Hz, 330 Hz
b. 220Hz 440Hz 66 Hz
c. 110Hz, 330Hz, 550Hz
d. 330 Hz, 550Hz, 440Hz
e. 660Hz, 1100Hz, 220Hz
Answer:
A. 110Hz,220Hz, 330 Hz
Explanation:
for organ open at open both ends;
the length of the organ for the fundamental frequency, L = A---->N + N----->A
A---->N = λ /4 and N----->A = λ /4
L = λ /4 + λ /4 = λ /2
[tex]L = \frac{\lambda}{2} \\\\\lambda = 2L[/tex]
λ = 2 x 1.5m = 3.0 m
Wave equation is given by;
V = Fλ
Where;
V is the speed of sound
F is the frequency of the wave
F = V/ λ
F₀ = V / 2L
Where;
F₀ is the fundamental frequency
F₀ = 330 / 2(1.5)
F₀ = 330 / 3
F₀ = 110 Hz
the length of the organ for the first overtone, L = A---->N + N----->A + A----->N + N----->A
L = 4λ /4
L = λ
λ = 1.5 m
F₁ = 330 / 1.5
F₁ = 220 Hz
Thus, F₁ = 2F₀
For open organ at one end
the length of the organ for the fundamental frequency, L = N------A
L = λ /4
λ = 4L
F₀ = V/4L
F₀ = 330 / (4 x 0.75)
F₀ = 110 Hz
the length of the organ for the first overtone, L = N-----N + N-----A
L = λ/2 + λ / 4
L = 3λ /4
F₁ = 3F₀
F₁ = 3 x 110
F₁ = 330 Hz
Thus the fundamental frequency for both organs is 110 Hz,
The first overtone for the organ open at both ends is 220 Hz
The first overtone for the organ open at one end is 330 Hz
The correct option is "A. 110Hz,220Hz, 330 Hz"
The correct option is option (A)
the frequencies produced by the pipes are (A) 110Hz,220Hz, 330 Hz
Frequencies and overtones:(I) For an organ pipe open at open both ends the frequency of different modes is given by:
F = nv/2L
where
F is the frequency
L is the length of the organ pipe
v is the speed of the wave
and, n is the mode of frequency
the fundamental frequency corresponds to n = 1, given by:
F₀ = v/2L
F₀ = 330 / 2(1.5)
F₀ = 330 / 3
F₀ = 110 Hz
The first overtone corresponds to n = 2, the second overtone corresponds to n = 3, and so on...
F₁ =2v/2L
F₁ = 330 / 1.5
F₁ = 220 Hz
Thus, F₁ = 2F₀
The difference between successive overtones is F₀
(II) For an organ pipe open at one end the frequency of different modes is given by:
F = nv/4L
where
F is the frequency
L is the length of the organ pipe
v is the speed of the wave
and, n is the mode of frequency
the fundamental frequency corresponds to n = 1, given by:
F₀ = V/4L
F₀ = 330 / (4 x 0.75)
F₀ = 110 Hz
For an organ pipe open at one end, only those overtones are present which correspond to odd n, that is n = 3,5,...so:
F₁ = 3F₀
F₁ = 3 x 110
F₁ = 330 Hz
Learn more about overtones:
https://brainly.com/question/1515875?referrer=searchResults
A projectile is shot from the edge of a cliff 80 m above ground level with an initial speed of 60 m/sec at an angle of 30° with the horizontal. Determine the time taken by the projectile to hit the ground below.
Answer:
8 seconds
Explanation:
Answer:
Explanation:
Going up
Time taken to reach maximum height= usin∅/g
=3 secs
Maximum height= H+[(usin∅)²/2g]
=80+[(60sin30)²/20]
=125 meters
Coming Down
Maximum height= ½gt²
125= ½(10)(t²)
t=5 secs
A plastic dowel has a Young's Modulus of 1.50 ✕ 1010 N/m2. Assume the dowel will break if more than 1.50 ✕ 108 N/m2 is exerted.
(a) What is the maximum force (in kN) that can be applied to the dowel assuming a diameter of 2.40 cm?
______Kn
(b) If a force of this magnitude is applied compressively, by how much (in mm) does the 26.0 cm long dowel shorten? (Enter the magnitude.)
mm
Answer:
a
[tex]F = 67867.2 \ N[/tex]
b
[tex]\Delta L = 2.6 \ mm[/tex]
Explanation:
From the question we are told that
The Young modulus is [tex]Y = 1.50 *10^{10} \ N/m^2[/tex]
The stress is [tex]\sigma = 1.50 *10^{8} \ N/m^2[/tex]
The diameter is [tex]d = 2.40 \ cm = 0.024 \ m[/tex]
The radius is mathematically represented as
[tex]r =\frac{d}{2} = \frac{0.024}{2} = 0.012 \ m[/tex]
The cross-sectional area is mathematically evaluated as
[tex]A = \pi r^2[/tex]
[tex]A = 3.142 * (0.012)^2[/tex]
[tex]A = 0.000452\ m^2[/tex]
Generally the stress is mathematically represented as
[tex]\sigma = \frac{F}{A}[/tex]
=> [tex]F = \sigma * A[/tex]
=> [tex]F = 1.50 *10^{8} * 0.000452[/tex]
=> [tex]F = 67867.2 \ N[/tex]
Considering part b
The length is given as [tex]L = 26.0 \ cm = 0.26 \ m[/tex]
Generally Young modulus is mathematically represented as
[tex]E = \frac{ \sigma}{ strain }[/tex]
Here strain is mathematically represented as
[tex]strain = \frac{ \Delta L }{L}[/tex]
So
[tex]E = \frac{ \sigma}{\frac{\Delta L }{L} }[/tex]
[tex]E = \frac{\sigma }{1} * \frac{ L}{\Delta L }[/tex]
=> [tex]\Delta L = \frac{\sigma * L }{E}[/tex]
substituting values
[tex]\Delta L = \frac{ 1.50*10^{8} * 0.26 }{ 1.50 *10^{10 }}[/tex]
[tex]\Delta L = 0.0026[/tex]
Converting to mm
[tex]\Delta L = 0.0026 *1000[/tex]
[tex]\Delta L = 2.6 \ mm[/tex]
A thermos bottle works well because:
a. its glass walls are thin
b. silvering reduces convection
c. vacuum reduces heat radiation
d. silver coating is a poor heat conductor
e. none of the above
Answer:
A thermos bottle works well because:
A) Its glass walls are thin
Answer:
A thermos bottle works well because:
C
Vacuum reduces heat radiation
A swimmer is treading water with their head above the surface of a pool and sees a penny at the bottom of the pool 5.0 mm below. How deep does the coin appear to be? (Index of refraction of water = 1.33) [Conceptual note: Does the coin appear to be shallower or deeper?]
Answer:
The apparent depth is [tex]D' = 0.00376 \ m[/tex]
Explanation:
From the question we are told that
The depth of the water is [tex]D = 5.0 \ mm = 5.0 *10^{-3} \ m[/tex]
The refractive index of water is [tex]n = 1.33[/tex]
Generally the apparent depth of the coin is mathematically represented as
[tex]D' = D * [\frac{ n_a}{n} ][/tex]
Here [tex]n_a[/tex] is the refractive index of air the value is [tex]n_a = 1[/tex]
So
[tex]D' = 5.0 *10^{-3} * [\frac{1}{1.33} ][/tex]
[tex]D' = 0.00376 \ m[/tex]
The apparent depth will be 0.00376 m.
What is an index of refraction?
The index of refraction of a substance also known as the refraction index is a dimensionless quantity that specifies how quickly light passes through it in optics.
d is the depth of the water =5.0 mm =5.0 ×10⁻³
n is the refractive index of water =1.33
[tex]\rm n_a[/tex] is the refractive index of wire=1
The apparent depth of the coin is given as;
[tex]\rm D'=D \times \frac{n_a}{n} \\\\ \rm D'=5.0 \times 10^{-3} \times \frac{1}{1.33} \\\\ \rm D'=0.00376 \ m[/tex]
Hence the apparent depth will be 0.00376 m.
To learn more about the index of refraction refer to the link;
https://brainly.com/question/23750645
Assume that Mars and Earth are in the same plane and that their orbits around the Sun are circles (Mars is ≈230×10^6km from the Sun and Earth is ≈150×10^6km from the Sun).
a) How long would it take a message sent as radio waves from Earth to reach Mars when Mars is nearest Earth?
b) How long would it take a message sent as radio waves from Earth to reach Mars when Mars is farthest from Earth?
Answer:
Explanation:
speed of light ≈ 300 000 000 m/s
a) How long would it take a message sent as radio waves from Earth to reach Mars when Mars is nearest Earth?
(230e9 - 150e9) / 3e8 = 277 s or 4 minutes 27 seconds.
b) How long would it take a message sent as radio waves from Earth to reach Mars when Mars is farthest from Earth?
(230e9 + 150e9) / 3e8 = 1267 s or 21 minutes 6 seconds.
If the solenoid is 45.0 cm long and each winding has a radius of 8.0 cm , how many windings are in the solenoid
Answer:
The number of windings is 1.
Explanation:
The radius of the solenoid = 8.0 cm = 0.08 m
Length of the solenoid = 45.0 cm = 0.45 m
number of turn = ?
circumference of each winding = 2πr = 2 x 3.142 x 0.08 = 0.503 m
The number of windings = (Length of the solenoid)/(circumference of each winding)
==> 0.45/0.503 = 0.89 ≅ 1
In order to waken a sleeping child, the volume on an alarm clock is doubled. Under this new scenario, how much more energy will be striking the child's ear drums each second?
Answer:4 times more energy will be striking the childbearing
Explanation:
Because Volume is directly proportional to amplitude of sound. Energy is proportional to amplitude squared. If you triple the amplitude, you multiply the energy by 4
When a battery is connected to a lightbulb properly, current flows through the lightbulb and makes it glow. How much current flows through the battery compared with the lightbulb
Answer:
The same amount of current flows through the battery and light bulb
Explanation:
Because for a single loop, the current is the same at every point in the loop. Thus, the amount of current that flows through the lightbulb is the same as the amount that flows through the battery
Answer:
The same amount of current flows through the battery and light bulb
Explanation:
A 50.0 ohm and a 30.0 ohm resistor are connected in parallel. What is their equivalent resistance? Unit=Ohms
R(parallel) = product/ sum
50×30/50+30
1500/80
18,75 ohms
Answer: 18.75
above is right but you need to put a dot after the number 18
A load of 1 kW takes a current of 5 A from a 230 V supply. Calculate the power factor.
Answer:
Power factor = 0.87 (Approx)
Explanation:
Given:
Load = 1 Kw = 1000 watt
Current (I) = 5 A
Supply (V) = 230 V
Find:
Power factor.
Computation:
Power factor = watts / (V)(I)
Power factor = 1,000 / (230)(5)
Power factor = 1,000 / (1,150)
Power factor = 0.8695
Power factor = 0.87 (Approx)
"A soap film is illuminated by white light normal to its surface. The index of refraction of the film is 1.33. Wavelengths of 479 nm and 798 nm and no wavelengths between are intensified in the reflected beam. The thickness of the film is"
Answer:
t = 8.98 10⁻⁷ m
Explanation:
This is an exercise in interference by reflection, let's analyze what happens on each surface of the film.
* When the light ray shifts from a medium with a lower refractive index to a medium with a higher refractive index, the reflected ray has a reflection of 180
* The beam when passing to the middle its wavelength changes
λ = λ₀ / n
if we take this into account, the constructive interference equation for normal incidence is
2t = (m + ½) λ₀ / n
let's apply this equation to our case
for λ₀ = 479 nm = 479 10⁻⁹ m
t = (m + ½) 479 10⁻⁹ / 1.33
(m + ½) = 1.33 t / 479 10⁻⁹
for λ₀ = 798 nm = 798 10⁻⁹ m
t = (m' + ½) 798 10⁻⁹ /1.33
(m' + ½) = 1.33 t / 798 10⁻⁹
as they tell us that no other constructive interference occurs between the two wavelengths, the order of interference must be consecutive, let's write the two equat⁻ions
(m + ½) = 1.33 t / 479 10⁻⁹
((m-1) + ½) = 1.33 t / 798 10⁻⁹
(m + ½) = 1.33 t / 798 10⁻⁹ +1
resolve
1.33 t / 479 10⁻⁹ = 1.33 t / 798 10⁻⁹ +1
1.33 t / 479 10⁻⁹ = (1.33t + 798 10⁻⁹) / 798 10⁻⁹
1.33t = (1 .33t + 798 10⁻⁹) 479/798
1.33t = (1 .33t + 798 10⁻⁹) 0.6
1.33 t = 0.7983 t + 477.6 10⁻⁹
t (1.33 - 0.7983) = 477.6 10⁻⁹
t = 477.6 10⁻⁹ /0.5315
t = 8.98 10⁻⁷ m
A patient prescribed a metered dose inhaler will find it
A. works for lower (not upper) respiratory diseases only.
B. should be filled with medication in aerosol form only.
C. is filled with medication used to administer a fixed amount of medication per inhalation through the mouth.
D. should be filled with medication in powder form only.
Answer:The correct answer should be C. is filled with medication used to administer a fixed amount of medication per inhalation through the mouth.
Explanation:
Answer:
c. is filled with medication used to administer a fixed amount of medication per inhalation through the mouth.
A charged capacitor and an inductor are connected in series. At time t = 0, the current is zero, but the capacitor is charged. If T is the period of the resulting oscillations, the next time, after t = 0 that the energy stored in the magnetic field of the inductor is a maximum is
Answer:
t = T / 2 all energy is stored in the inductor
Explanation:
The circuit described is an oscillating circuit where the charge of the condensation stops the inductor and vice versa, in this system the angular velocity of the oscillation is
w = √1/LC
2π / T =√1 / LC
T = 2π √LC
The energy is constant and for the initial instant it is completely stored in the capacitor
Uc = Q₀² / 2C
In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor
U = L I² / 2
in the intermediate instant the energy is stored in the two elements.
Since the period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor
After t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation (t = T/2).
The given problem is based on the charging and discharging concepts of capacitor. An oscillating circuit is a circuit where the charge of the capacitor stops the inductor and vice versa, in this system the angular frequency of the oscillation is given as,
[tex]\omega =\dfrac{1}{\sqrt{LC}}\\\\\\\dfrac{2 \pi}{T} =\dfrac{1}{\sqrt{LC}}\\\\\\T = 2\pi \times \sqrt{LC}[/tex]
here, T is the period of oscillation.
Also, the energy stored in the capacitor is constant and for the initial instant it is completely stored in the capacitor. So, the energy stored is given as,
[tex]U =\dfrac{Q^{2}}{2C}[/tex]
here, C is the capacitance.
In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor. So, the expression for the energy stored in the inductor is,
[tex]U'=\dfrac{L I^{2}}{2}[/tex]
here, L is the inductance and I is the current.
Note :- The period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor.
Thus, we conclude that after t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation (t = T/2).
Learn more about the capacitance here:
https://brainly.com/question/12644355
Two ice skaters push off against one another starting from a stationary position. The 45.0-kg skater acquires a speed of 0.375 m/s. What speed does the 60.0-kg skater acquire in m/s
Answer:
0.2812
Explanation:
Given that
mass of skater 1, m1 = 45 kg
mass of skater 2, m2 = 60 kg
speed of skater 1, v1 = 0.375 m/s
To attempt this question, we would be using the Law of conservation of momentum That says the momentum is constant, before and after the movement.
Thus, momentum p = mv
Law of conservation of momentum infers that,
m1v1 = m2v2
Now we proceed to substitute our values into the formula.
45 * 0.375 = 60 * v2
v2 = 16.875 / 60
v2 = 0.2812 m/s
Therefore the speed of the second skater has to be 0.2812 m/s
calculate horizontal distance travelled by a ball travelling with a speed of 20root2 mper sec without hitting ceiling of height 20 m per sec
Calculate the horizontal distance travelled by a ball throw with a velcoity 20 sqrt 2 ms^(-1) without hitting the ceiling of an anditorium of heith 20 m. Use g= 10 ms^(-2). =(20√2)2sin2×45∘10=9800×1)10=80m .
HOPE SO IT HELPS YOU
B. CO
A wave has frequency of 2 Hz and a wave length of 30 cm. the velocity of the wave is
A. 60.0 ms
B. 6.0 ms
D. 0.6 ms
Answer:
0.6 m/s
Explanation:
2Hz = 2^-1 = 2 /s
30cm = .3m
Velocity is in the units m/s, so multiplying wavelength in meters by the frequency will give you the velocity.
(.3m)*(2 /s) = 0.6 m/s
A small insect viewed through a convex lens is 1.5 cmcm from the lens and appears 2.5 times larger than its actual size. Part A What is the focal length of the lens
Answer:
The focal length of the lens is 2.5 cm
Explanation:
Use the two equations for thin lenses combined: the one for magnification (m), and the one that relates distances of object [tex]d_o[/tex], of image [tex]d_i[/tex], and focal length;
[tex]m=\frac{h_i}{h_o} =-\frac{d_i}{d_o} \\ \\\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}[/tex]
Since we know the value of the magnification (m), we can write the image distance in terms of the object distance, and then use it to replace the image distance in the second equation:
[tex]m=-\frac{d_i}{d_o} \\2.5=-\frac{d_i}{d_o}\\d_i=-2.5\,d_o[/tex]
then, solving for the focal distance knowing that the object distance is 1.5 cm:
[tex]\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}\\-\frac{1}{2.5\,d_o} +\frac{1}{d_o} =\frac{1}{f}\\(2.5\,d_o\,f)\,(-\frac{1}{2.5\,d_o} +\frac{1}{d_o}) =\frac{1}{f}\,(2.5\,d_o\,f)\\-f+2.5\,f=2.5\,d_o\\1.5\,f=2.5\,d_o\\f=\frac{2.5\,d_o}{1.5} \\f=\frac{2.5\,(1.5\,\,cm)}{1.5}\\f=2.5\,\,cm[/tex]
3. Which of the following accurately describes circuits?
O A. In a parallel circuit, the same amount of current flows through each part of the circuit
O B. In a series circuit, the amount of current passing through each part of the circuit may vary
O C. In a series circuit, the current can flow through only one path from start to finish
O D. In a parallel circuit, there's only one path for the current to travel.
Answer:
Option (c)
Explanation:
In a Series circuit, as the components are connected end-to-end ,the current can flow through only one path from start to finish.
(C.) is the only correct statement in the list of choices.
In a series circuit, the current can flow through only one path from start to finish.