Answer:
221 lines per millimetre
Explanation:
We know that for a diffraction grating, dsinθ =mλ where d = spacing between grating, θ = angle to maximum, m = order of maximum and λ = wavelength of light.
Since the grating is 42.0 cm from the screen and its first order maximum (m = 1) is at 6.09 cm from the center of the pattern,
tanθ = 6.09 cm/42.0 cm = 0.145
From trig ratios, cot²θ + 1 = cosec²θ
cosecθ = √((1/tanθ)² + 1) = √((1/0.145)² + 1) = √48.562 = 6.969
sinθ = 1/cosecθ = 1/6.969 = 0.1435
Also, sinθ = mλ/d at the first-order maximum, m = 1. So
sinθ = (1)λ/d = λ/d
Equating both expressions we have
0.1435 = λ/d
d = λ/0.1435
Now, λ = 650 nm = 650 × 10⁻⁹ m
d = 650 × 10⁻⁹ m/0.1435
d = 4529.62 × 10⁻⁹ m per line
d = 4.52962 × 10⁻⁶ m per line
d = 0.00452962 × 10⁻³ m per line
d = 0.00452962 mm per line
Since d = width of grating/number of lines of grating
Then number of lines per millimetre = 1/grating spacing
= 1/0.00452962
= 220.77 lines per millimetre
≅ 221 lines per millimetre since we can only have a whole number of lines.
Kasek rides his bicycle down a 6.0° hill (incline is
6° with the horizontal) at a steady speed of 4.0
m/s. Assuming a total mass of 75 kg (bicycle and
Kasek), what must be Kasek's power output to
climb the same hill at the same speed?
Answer:
P = 2923.89 W
Explanation:
Power is
P = F v
for which we must calculate the force, let's use Newton's second law, let's set a coordinate system with a flat parallel axis and the other axis (y) perpendicular to the plane
X Axis
F - Wₓ = 0
F = Wₓ
Y Axis
N - [tex]W_{y}[/tex] = 0
let's use trigonometry for the components of the weight
sin 6 = Wₓ / W
cos 6 = W_{y} / W
Wₓ = W sin 6
W_{y} = W cos 6
F = mg cos 6
F = 75 9.8 cos 6
F = 730.97 N
let's calculate the power
P = F v
P = 730.97 4.0
P = 2923.89 W
2.) Is it possible to have negative velocity but positive acceleration? If so, what would
this mean?
Answer:
Yes, yes it would
Explanation:
Suppose that a sound source is emitting waves uniformly in all directions. If you move to a point twice as far away from the source, the frequency of the sound will be:________.
a. one-fourth as great.
b. half as great.
c. twice as great.
d. unchanged.
Answer:
d. unchanged.
Explanation:
The frequency of a wave is dependent on the speed of the wave and the wavelength of the wave. The frequency is characteristic for a wave, and does not change with distance. This is unlike the amplitude which determines the intensity, which decreases with distance.
In a wave, the velocity of propagation of a wave is the product of its wavelength and its frequency. The speed of sound does not change with distance, except when entering from one medium to another, and we can see from
v = fλ
that the frequency is tied to the wave, and does not change throughout the waveform.
where v is the speed of the sound wave
f is the frequency
λ is the wavelength of the sound wave.
Proposed Exercises: Strength and Acceleration in Circular Movement In the situation illustrated below, a 7kg sphere is connected to a rope so that it can rotate in a vertical plane around an O axis perpendicular to the plane of the figure. When the sphere is in position A, it has a speed of 3m/s. Determine for this position the modulus of tension on the string and the rate at which the tangential velocity is increased.
Answer:
81 N
7.1 m/s²
Explanation:
Draw a free body diagram of the sphere. There are two forces:
Weight force mg pulling straight down,
and tension force T pulling up along the rope.
Sum of forces in the centripetal direction:
∑F = ma
T − mg sin 45° = m v² / r
T = m (g sin 45° + v² / r)
T = (7 kg) (10 m/s² sin 45° + (3 m/s)² / 2 m)
T = 81 N
Sum of forces in the tangential direction:
mg cos 45° = ma
a = g cos 45°
a = (10 m/s²) cos 45°
a = 7.1 m/s²
A 10-cm-long thin glass rod uniformly charged to 6.00 nC and a 10-cm-long thin plastic rod uniformly charged to - 6.00 nC are placed side by side, 4.4 cm apart. What are the electric field strengths E1 to E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?
A. Specify the electric field strength E1
B. Specify the electric field strength E2
C. Specify the electric field strength E3
Answer:
A) E(r) = 1.3957 × 10^(5) N/C
B) E(r) = 9.8864 × 10⁴ N/C
C) E(r) = 1.13 × 10^(5) N/C
Explanation:
We are given;
q = 6 nc = 6 × 10^(-9) C
L = 10 cm = 0.1 m
d = 4.4 cm = 0.044 m
r1 = 1 cm = 0.01 m
r2 = 2 cm = 0.02 m
r3 = 3 cm = 0.03 m
Formula for the electric field strength in this question is given as;
E(r) = q/(2π(ε_o)rL) + q/(2π(ε_o)(d - r)L)
When factorized, we have;
E(r) = q/(2π(ε_o)L) × [(1/r) + (1/(d - r))]
Plugging in the relevant values for q/(2π(ε_o)L)
We know that (ε_o) has a constant value of 8.854 × 10^(−12) C²/N².m
Thus; q/(2π(ε_o)L) = (6 × 10^(-9))/(2π(8.854 × 10^(−12)0.1) = 1078.53
Thus;
E(r) = 1078.52 [1/r + 1/(d - r)]
A) E1 is at r = 1 cm = 0.01m
Thus;
E(r) = 1078.52 (1/0.01 + (1/(0.044 - 0.01))
E(r) = 1.3957 × 10^(5) N/C
B) E2 is at r = 2 cm = 0.02 m
Thus;
E(r) = 1078.52 (1/0.02 + (1/(0.044 - 0.02))
E(r) = 9.8864 × 10⁴ N/C
C) E2 is at r = 3 cm = 0.03 m
Thus;
E(r) = 1078.52 (1/0.03 + (1/(0.044 - 0.03))
E(r) = 1.13 × 10^(5) N/C
An electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelength of 880 nm. What is the potential difference though which this electron was accelerated
Answer:
3x10⁴v
Explanation:
Using
Wavelength= h/ √(2m.Ke)
880nm = 6.6E-34/√ 2.9.1E-31 x me
Ke= 6.6E-34/880nm x 18.2E -31.
5.6E-27/18.2E-31
= 3 x 10⁴ Volts
A long solenoid consists of 1700 turns and has a length of 0.75 m.The current in the wire is 0.48 A. What is the magnitude of the magnetic field inside the solenoid
Answer:
1.37 ×10^-3 T
Explanation:
From;
B= μnI
μ = 4π x 10-7 N/A2
n= number of turns /length of wire = 1700/0.75 = 2266.67
I= 0.48 A
Hence;
B= 4π x 10^-7 × 2266.67 ×0.48
B= 1.37 ×10^-3 T
Water is pumped with a 120 kPa compressor entering the lower pipe (1) and flows upward at a speed of 1 m/s. Acceleration due to gravity is 10 m/s and water density is1000 kg/m-3. What is the water pressure on the upper pipe (II).
Answer:
The water pressure on the upper pipe is 92.5 kPa.
Explanation:
Given that,
Pressure in lower pipe= 120 kPa
Speed of water in lower pipe= 1 m/s
Acceleration due to gravity = 10 m/s²
Density of water = 1000 kg/m³
Radius of lower pipe = 12 m
Radius of uppes pipe = 6 m
Height of upper pipe = 2 m
We need to calculate the velocity in upper pipe
Using continuity equation
[tex]A_{1}v_{1}=A_{2}v_{1}[/tex]
[tex]\pi r_{1}^2\times v_{1}=\pi r_{2}^2\times v_{2}[/tex]
[tex]v_{2}=\dfrac{r_{1}^2\times v_{1}}{r_{2}^2}[/tex]
Put the value into the formula
[tex]v_{2}=\dfrac{12^2\times1}{6^2}[/tex]
[tex]v_{2}=4\ m/s[/tex]
We need to calculate the water pressure on the upper pipe
Using bernoulli equation
[tex]P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}[/tex]
Put the value into the formula
[tex]120\times10^{3}+\dfrac{1}{2}\times1000\times1^2+1000\times10\times0=P_{2}+\dfrac{1}{2}\times1000\times(4)^2+1000\times10\times2[/tex]
[tex]120500=P_{2}+28000[/tex]
[tex]P_{2}=120500-28000[/tex]
[tex]P_{2}=92500\ Pa[/tex]
[tex]P_{2}=92.5\ kPa[/tex]
Hence, The water pressure on the upper pipe is 92.5 kPa.
A ball travels with velocity given by [21] [ 2 1 ], with wind blowing in the direction given by [3−4] [ 3 −4 ] with respect to some co-ordinate axes. What is the size of the velocity of the ball in the direction of the wind?
Answer:
2/5 m/s
Explanation:
There are two vectors v and w . Let θ be angle b/w the two vector.
[tex]cos\theta =\frac{\overleftarrow{v}\cdot \overleftarrow{w}}{\left | v \right |\left | w \right |}\\=\frac{6-4}{\sqrt(2^2+1^2)\sqrt(3^2+4^2)} =\frac{2}{5\sqrt(5)}[/tex]
velocity of the ball in direction of the the wind
[tex]\left | vcos\theta \right |\\\left | v \right |cos\theta\\\sqrt(2^2+1^2)\frac{2}{5\sqrt(5)} = \frac{2}{5}[/tex]
The size of the velocity of the ball in the direction of the wind is 2/5 ms.
Calculation of the size of velocity:Since there are two vectors v and w
Also, here we assume θ be angle b/w the two vector.
So
Cos θ = 6-4 / √(2^2 + 1^2) √(3^2 + 4^2)
= 2/5√5
Now the velocity of the ball should be
= √(2^2 + 1^2) 2 ÷ 5√(5)
= 2 /5
hence, The size of the velocity of the ball in the direction of the wind is 2/5 ms.
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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
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You have three resistors: R1 = 1.00 Ω, R2 = 2.00 Ω, and R3 = 4.00 Ω in parallel. Find the equivalent resistance for the combination
Answer:
4 / 7
Explanation:
1/total resistance = 1/1 + 1/2 + 1/4
= 1¾
total resistance = 1 ÷ 1¾
= 4/7
What is the separation in meters between two slits for which 594 nm orange light has its first maximum at an angle of 32.8°?
Answer:
1.1micro meter
Explanation:
Given that
Constructive interference is
ma = alpha x sin theta
Alpha = 1 x 594 x10^ -9/ sin 32.8°
= 1.1 x 10^ -6m
Explanation:
Given a double slit apparatus with slit distance 2 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 500 nm on the slits
Answer:
The values is [tex]m_{max} = 8001 \ bright \ spots[/tex]
Explanation:
From the question we are told that
The slit distance is [tex]d = 2 \ mm = 2*10^{-3} \ m[/tex]
The wavelength is [tex]\lambda = 500 \ nm = 500 *10^{-9} \ m[/tex]
At the first half of the screen from the central maxima
The number of bright spot according to the condition for constructive interference is
[tex]n = \frac{d * sin (\theta )}{\lambda}[/tex]
For maximum number of spot [tex]\theta = 90^o[/tex]
So
[tex]n = \frac{2*10^{-3} * sin (90 )}{500 *10^{-9}}[/tex]
[tex]n =4000[/tex]
Now for the both sides plus the central maxima we have
[tex]m_{max} = 2 * n + 1[/tex]
substituting values
[tex]m_{max} = 2 * 4000 + 1[/tex]
[tex]m_{max} = 8001 \ bright \ spots[/tex]
Which one of the following actions would make the maxima in the interference pattern from a grating move closer together?
A. Increasing the number of lines per length.
B. Decreasing the number of lines per length.
C. Increasing the distance to the screen.
D. Increasing the wavelength of the laser.
Answer:
Answer:
A. Increasing the number of lines per length.
A 28.0 kg child plays on a swing having support ropes that are 2.30 m long. A friend pulls her back until the ropes are 45.0 ∘ from the vertical and releases her from rest.
A: What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing?
B: How fast will she be moving at the bottom of the swing?
C: How much work does the tension in the ropes do as the child swings from the initial position to the bottom?
Answer
A)184.9J
B)=3.63m/s
C) Zero
Explanation:
A)potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is
U=Mgh
Where m=28kg
g= 9.8m/s
h= difference in height between the initial position and the bottom position
We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical
h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)
=0.674m
Her Potential Energy will now
= 28× 9.8×0.674
=184.9J
B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:
E= 0.5mv^2
where
m = 28.0 kg is the mass of the child
v is the speed of the child at the bottom position
Solving the equation for v, we find
V=√2k/m
V=√(2×184.9/28
=3.63m/s
C)we can find work done by the tension in the rope is given using expresion below
W= Tdcosx
where W= work done
T is the tension
d = displacement of the child
x= angle between the directions of T and d
In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. x= 90∘ and cos90∘=0 hence, the work done is zero.
The frequency of light emitted from hydrogen present in the Andromeda galaxy has been found to be 0.10% higher than that from hydrogen measured on Earth.
Is this galaxy approaching or receding from the Earth, and at what speed?
Answer:
3x10^5m/s
Explanation:
See attached file
Explanation:
The speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].
Doppler's Effect
According to the Doppler effect, the difference between the frequency at which light wave leave a source and reaches an observer is caused by the relative motion of the observer and the wave source.
Given that the difference in the frequency is 0.10 %. The speed of light emitted from the galaxy can be calculated by the Doppler effect.
[tex]\dfrac {\Delta f}{f} = \dfrac {v}{c}[/tex]
Where f is the frequency of the light, v is the speed of light emitted from the galaxy and c is the speed of light emitted from the earth.
[tex]\dfrac {0.10 f}{100 f} = \dfrac {v}{3\times 10^8}[/tex]
[tex]v = 3\times 10^5\;\rm m/s[/tex]
Hence we can conclude that the speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].
To know more about the doppler effect, follow the link given below.
https://brainly.com/question/1330077.
What is the density of the unknown fluid in Figure below? ρwater = 1000 kgm−3
Answer:
2500 kg/m³
Explanation:
P = P
ρgh = ρgh
ρh = ρh
(1000 kg/m³) (8.9 cm) = ρ (3.5 cm)
ρ ≈ 2500 kg/m³
Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consider points along the line connecting the two sources.Required:a. At what distance from source A is there constructive interference between points A and B?b. At what distances from source A is there destructive interference between points A and B?
Answer:
a
[tex]z= 2.5 \ m[/tex]
b
[tex]z = (1 \ m , 4 \ m )[/tex]
Explanation:
From the question we are told that
Their distance apart is [tex]d = 5.00 \ m[/tex]
The wavelength of each source wave [tex]\lambda = 6.0 \ m[/tex]
Let the distance from source A where the construct interference occurred be z
Generally the path difference for constructive interference is
[tex]z - (d-z) = m \lambda[/tex]
Now given that we are considering just the straight line (i.e points along the line connecting the two sources ) then the order of the maxima m = 0
so
[tex]z - (5-z) = 0[/tex]
=> [tex]2 z - 5 = 0[/tex]
=> [tex]z= 2.5 \ m[/tex]
Generally the path difference for destructive interference is
[tex]|z-(d-z)| = (2m + 1)\frac{\lambda}{2}[/tex]
=> [tex]|2z - d |= (0 + 1)\frac{\lambda}{2}[/tex]
=> [tex]|2z - d| =\frac{\lambda}{2}[/tex]
substituting values
[tex]|2z - 5| =\frac{6}{2}[/tex]
=> [tex]z = \frac{5 \pm 3}{2}[/tex]
So
[tex]z = \frac{5 + 3}{2}[/tex]
[tex]z = 4\ m[/tex]
and
[tex]z = \frac{ 5 -3 }{2}[/tex]
=> [tex]z = 1 \ m[/tex]
=> [tex]z = (1 \ m , 4 \ m )[/tex]
How wide is the central diffraction peak on a screen 2.20 mm behind a 0.0328-mmmm-wide slit illuminated by 588-nmnm light?
Answer:
[tex]y = 0.0394 \ m[/tex]
Explanation:
From the question we are told that
The distance of the screen is [tex]D = 2.20 \ m[/tex]
The distance of separation of the slit is [tex]d = 0.0328 \ mm = 0.0328*10^{-3} \ m[/tex]
The wavelength of light is [tex]\lambda = 588 \ nm = 588 *10^{-9} \ m[/tex]
Generally the condition for constructive interference is
[tex]dsin\theta = n * \lambda[/tex]
=> [tex]\theta = sin^{-1} [ \frac{ n * \lambda }{d } ][/tex]
here n = 1 because we are considering the central diffraction peak
=> [tex]\theta = sin^{-1} [ \frac{ 1 * 588*10^{-9} }{0.0328*10^{-3} } ][/tex]
=> [tex]\theta = 1.0274 ^o[/tex]
Generally the width of central diffraction peak on a screen is mathematically evaluated as
[tex]y = D tan (\theta )[/tex]
substituting values
[tex]y = 2.20 * tan (1.0274)[/tex]
[tex]y = 0.0394 \ m[/tex]
The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume that the emissivity eee is equal to 1 for these surfaces.
Required:
a. Find the radius RRigel of the star Rigel, the bright blue star in the constellation Orion that radiates energy at a rate of 2.7 x 10^31 W and has a surface temperature of 11,000 K.
b. Find the radius RProcyonB of the star Procyon B, which radiates energy at a rate of 2.1 x 10^23 W and has a surface temperature of 10,000 K. Assume both stars are spherical. Use σ=5.67 x 10−8^ W/m^2*K^4 for the Stefan-Boltzmann constant.
Given that,
Energy [tex]H=2.7\times10^{31}\ W[/tex]
Surface temperature = 11000 K
Emissivity e =1
(a). We need to calculate the radius of the star
Using formula of energy
[tex]H=Ae\sigma T^4[/tex]
[tex]A=\dfrac{H}{e\sigma T^4}[/tex]
[tex]4\pi R^2=\dfrac{H}{e\sigma T^4}[/tex]
[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]
Put the value into the formula
[tex]R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}[/tex]
[tex]R=5.0\times10^{10}\ m[/tex]
(b). Given that,
Radiates energy [tex] H=2.1\times10^{23}\ W[/tex]
Temperature T = 10000 K
We need to calculate the radius of the star
Using formula of radius
[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]
Put the value into the formula
[tex]R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}[/tex]
[tex]R=5.42\times10^{6}\ m[/tex]
Hence, (a). The radius of the star is [tex]5.0\times10^{10}\ m[/tex]
(b). The radius of the star is [tex]5.42\times10^{6}\ m[/tex]
Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plates in Capacitor B are separated by twice the separation of the plates of Capacitor A. If Capacitor A has a capacitance of CA-17.8nF, what is the capacitance of Capacitor? .
Answer:
CB = 4.45 x 10⁻⁹ F = 4.45 nF
Explanation:
The capacitance of a parallel plate capacitor is given by the following formula:
C = ε₀A/d
where,
C = Capacitance
ε₀ = Permeability of free space
A = Area of plates
d = Distance between plates
FOR CAPACITOR A:
C = CA = 17.8 nF = 17.8 x 10⁻⁹ F
A = A₁
d = d₁
Therefore,
CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F ----------------- equation 1
FOR CAPACITOR B:
C = CB = ?
A = A₁/2
d = 2 d₁
Therefore,
CB = ε₀(A₁/2)/2d₁
CB = (1/4)(ε₀A₁/d₁)
using equation 1:
CB = (1/4)(17.8 X 10⁻⁹ F)
CB = 4.45 x 10⁻⁹ F = 4.45 nF
Consider two parallel wires where the magnitude of the left currentis 2 I0(io) and that of the right current is I0(io). Point A is midway between the wires,and B is an equal distance on the other side of the wires.
The ratio ofthe magnitude of the magnetic field at point A to that at point Bis________
Answer:
Explanation:
At the point midway between wires
magnetic field due to wire having current 2I₀
= 10⁻⁷ x 2 x2I₀ / r where 2r is the distance between wires .
magnetic field due to wire having current I₀
= 10⁻⁷ x 4 I₀ / r
magnetic field due to wire having current I₀
= 10⁻⁷ x 2I₀ / r
= 10⁻⁷ x 2 I₀ / r where 2r is the distance between wires .
these fields are in opposite direction as direction of current is same in both .
net magnetic field = (4 - 2 )x 10⁻⁷ x I₀ / r
= 2 x 10⁻⁷ x I₀ / r
At point A net magnetic field = 2 x 10⁻⁷ x I₀ / r
At point B , we shall calculate magnetic field
magnetic field due to nearer wire having current 2 I₀ = 10⁻⁷ x 4 I₀ / r
magnetic field due to wire far away = 10⁻⁷ x 2 I₀ / 3r
These magnetic fields act in the same direction so they will add up
net magnetic field = [ (4 I₀ / r) + (2 I₀ / 3r) ] x 10⁻⁷
= (14 I₀ / 3r ) x 10⁻⁷
Magnetic field at point B = (14 I₀ / 3r ) x 10⁻⁷
Ratio of field at A and B
= 3 / 7 . Ans
The ratio of the magnitude of the magnetic field at point A to point B is :
3 / 7
Given data :
Magnitude of the left current is 2I₀
Magnitude of the right current is I₀
First step : Determine the magnetic field at point A
The magnetic field due to the left current ( 2I₀ )
10⁻⁷ * 2 * 2I₀ / r ( 2r = distance between wires )
The magnetic field due to the right current ( I₀ )
10⁻⁷ * 2 I₀ / r
From the expressions above the magnetic fields are in opposite direction
∴ Net magnetic field = (4 - 2 )* 10⁻⁷ * I₀ / r = 2 * 10⁻⁷ * I₀ / r
Hence The magnetic field at point A = 2 * 10⁻⁷ * I₀ / r
Next step : determine the magnetic field at point B
Magnetic field due to the closest wire to point B ( i.e.2I₀ ) = 10⁻⁷ * 4 I₀ / r
Magnetic field due to the wire away from point A = 10⁻⁷ * 2 I₀ / 3r
Since the fields acts in the same directions
The net magnetic field = (4 I₀ / r) + (2 I₀ / 3r) ] * 10⁻⁷ = ( 14 I₀ / 3r ) * 10⁻⁷
Hence The magnetic field at point A = ( 14 I₀ / 3r ) * 10⁻⁷
Therefore the ratio of the magnitude of the magnetic field at point A to point B = 3/ 7
Hence we can conclude that the ratio of the magnitude of the magnetic field at point A to point B = 3 / 7
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A radar installation operates at 9000 MHz with an antenna (dish) that is 15 meters across. Determine the maximum distance (in kilometers) for which this system can distinguish two aircraft 100 meters apart.
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
1. Why do you see colors when you look at reflected light from a CD or DVD disk, or when you look at a soap bubble or oil film on water?
2. What do you think causes the colors on the artwork panels on the side of HLS2 (Health Sciences building) which change with time of day and the angle from which you view them?
Explanation:
1.The light reflected from the CD/DVD or soap bubble or oil film forms an interference with the surrounding light. The inference both constructive and destructive making some color appear and some disappear.
2.As light behaves as wave it will interfere differently at different angles. At certain angle it will interfere constructively and at certain angle it will interfere destructively making some color brighter and some disappear. So, at different angles the color are different.
Interference pattern is responsible for the formation of different colour when a light reflected from CD or soap bubble.
We can see colors when we look at reflected light from a CD or DVD disk, or a soap bubble or oil film on water because of the interference pattern. The colors that we see on the CD are created due to the reflection of white light from ridges in the metal. When light passes through something with many small ridges or scratches, we often see rainbow colors and interesting patterns.
These patterns are called interference patterns. White light is made up of 7 colors i.e. red, orange, yellow, green, blue, indigo, violet. The CD converts or separates the white light into 7 colors so we can conclude that interference pattern is responsible for the formation of different colour when a light reflected from CD OR soap bubble.
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48. A patient presents with a thrombosis in
the popliteal vein. This thrombosis most likely
causes reduction of blood flow in which of the
following veins?
Answer:
the interation blood veins
Explanation:
A 17.0 g bullet traveling horizontally at 785 m/s passes through a tank containing 13.5 kg of water and emerges with a speed of 534 m/s.
What is the maximum temperature increase that the water could have as a result of this event? (in degrees)
Answer:
The maximum temperature increase is [tex]\Delta T = 0.0497 \ ^oC[/tex]
Explanation:
From the question we are told that
The mass of the bullet is [tex]m = 17.0 \ g =0.017 \ kg[/tex]
The speed is [tex]v_1 = 785 \ m/s[/tex]
The mass of the water is [tex]m_w = 13.5 \ kg[/tex]
The velocity it emerged with is [tex]v_2 = 534 \ m/s[/tex]
Generally due to the fact that energy can nether be created nor destroyed but transferred from one form to another then
the change in kinetic energy of the bullet = the heat gained by the water
So
The change in kinetic energy of the water is
[tex]\Delta KE = \frac{1}{2} m (v_1^2 - v_2 ^2 )[/tex]
substituting values
[tex]\Delta KE =0.5 * 0.017 * (( 785)^2 - (534) ^2 )[/tex]
[tex]\Delta KE = 2814.1 \ J[/tex]
Now the heat gained by the water is
[tex]Q = m_w* c_w * \Delta T[/tex]
Here [tex]c_w[/tex] is the specific heat of water which has a value [tex]c_w = 4190 J/kg \cdot K[/tex]
So since [tex]\Delta KE = Q[/tex]
we have that
[tex]2814.1 = 13.5 * 4190 * \Delta T[/tex]
[tex]\Delta T = 0.0497 \ ^oC[/tex]
At what temperature (degrees Fahrenheit) is the Fahrenheit scale reading equal to:_____
(a) 3 times that of the Celsius and
(b) 1/5 times that of the Celsius
Answer:
C = 26.67° and F = 80°C = -20° and F = -4°Explanation:
Find:
3 times that of the Celsius and 1/5 times that of the CelsiusComputation:
F = (9/5)C + 32
3 times that of the Celsius
If C = x
So F = 3x
So,
3x = (9/5)x + 32
15x = 9x +160
6x = 160
x = 26.67
So, C = 26.67° and F = 80°
1/5 times that of the Celsius
If C = x
So F = x/5
So,
x/5 = (9/5)x + 32
x = 9x + 160
x = -20
So, C = -20° and F = -4°
You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the
most likely final temperature of the mixture?
O A. 80°C
OB. 10-C
OC. 20°C
O D. 60°C
Answer:
Option (c) : 20°C
Explanation:
[tex]t(final) = \frac{w1 \times t1 + w2 \times t2}{w1 + w2} [/tex]
T(final) = 500* 10 + 100*70/600 = 20°C
Suppose you drop paperclips into an open cart rolling along a straight horizontal track with negligible friction. As a result of the accumulating paper clips, explain whether the momentum and kinetic energy increase, decrease, or stay the same.
Answer:
Stay the same
Explanation:
Since, friction is negligible:
Initial Momentum = Final Momentum
Initial KE = Final KE
m1 * v1 = m2 * v2
When m increases v decreases.
The momentum and kinetic energy remain the same if you drop paper clips into an open cart rolling along a straight horizontal track with negligible friction.
What is friction?Between two surfaces that are sliding or attempting to slide over one another, there is a force called friction. For instance, friction makes it challenging to push a book down the floor. Friction always moves an object in a direction that is counter to the direction that it is traveling or attempting to move.
Given:
The paperclips into an open cart rolling along a straight horizontal track with negligible friction,
Calculate the momentum, Since friction is negligible,
Initial Momentum = Final Momentum
Initial Kinetic Energy = Final Kinetic Energy
m₁ × v₁ = m₁ × v₂
When m increases, v decreases,
Thus, momentum will remain the same.
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A train on one track moves in the same direction as a second train on the adjacent track. The first train, which is ahead of the second train and moves with a speed of 36.4 m/s , blows a horn whose frequency is 123 Hz .what is its speed?
Answer:
51. 7m/s
Explanation:
Take speed of sound in air = 340 m/s
fp = fs (V + Vp)/(V + Vs)
128 = 123 (340 + Vp)/(340 + 36.4)
Vp = 51.7m/s
Explanation:
