Answer:
A) 60 Hz
B) 0.04 T
Explanation:
Given that.
Number of turns, N = 200
Length of the side, l = 20 cm = 0.2 m
Speed if rotation, w = 3600 rpm
Voltage, V = 120 V
First, we try to convert the speed from rpm to rad/s
3600 * (2π/60)
3600 * 0.10473
3600 rpm = 377 rad/s
Now, we use that as our w, speed of rotation
Frequency of output, f =
w/2π
f = 377 / 6.284
f = 59.99 Hz or approximately, 60 Hz.
B
Strength of the magnetic field in which the coil is turning
E• = NABw
Where, A = l² = 0.2² = 0.04, on substituting the values to the equation, we have
120 = 200 * 0.04 * 377 * B
120
Making B subject of formula,
B = 120/ 3016
B = 0.04 T..
The frequency of the output voltage is 60 Hz and the strength of the magnetic field is 0.04 T
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
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]
on which principle does water pump work ?
Answer:
The working principle of a water pump mainly depends upon the positive displacement principle as well as kinetic energy to push the water.
Explanation:
it mainly depends upon the positive displacement principle and also kinetic energy to push water. hope this hepls!
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.
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:
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A cook preparing a meal for a group of people is an example of
O kinetic energy because he has the ability to make a meal
O potential energy because he has the ability to make a meal
O kinetic energy because he is making the meal
o potential energy because he is making the meal
The concentration of sodium and potassium ions in the blood and body fluid is regulated by :
Answer:
Kidney
Explanation:
One of the main function of the kidney is to maintain the homeostasis of sodium and potassium ions in the blood and body.
Aldosterone is a key steroid hormone that balances sodium and potassium ions in the blood and body fluid. Potassium and sodium ions generate electric impulse in the body which helps to perform different activities such as muscles flexing.
Kidney function for reabsorption and secretion, in which reabsorption of Na is done nd balances the sodium and potassium in the blood and body.
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low mass sail and the energy and momentum of sunlight for propulsion.
(a) Should the sail be absorbing or reflective? Why?
(b) How large a sail is necessary to propel a 10000kg
spacecraft against the gravitational force of the sun? Express your result in square kilometers.
(c) Explain why your answer to part (b) is independent of the distance from the sun.
The gravitational constant is G=6.67×10−11m3⋅s−2⋅kg−1.
The mass of the sun is Ms=1.99×1030kg.
Answer:
The complete question is
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective? Why? (b) The total power output of the sun is 3.9 x 10^26 W. How large a sail is necessary to propel a 10,000-kg spacecraft against the gravitational force of the sun? Express your result in square kilometers. (c) Explain why your answer to part (b) is independent of the distance from the sun.
a) The sail should be reflective because, an incident electromagnetic wave, in this case, light wave, impacts twice the energy density on a reflective sail, and hence twice the force on a totally reflective sail as would be impacted on a sail that is totally absorbing.
For totally reflective, F = (2I/c)A ....1
for totally reflective, F = (I/c)A ....2
where I is the intensity of the light
c is the speed of light = 3 x 10^8 m/s
A is the area the sail
b) The intensity of the light from the sun = power/area
==> I = [tex]\frac{3.9*10^{26}}{4\pi r^{2} }[/tex]
where r is the distance from the sun and the sail
The Force from the sail from equation 1 is therefore
[tex]F[/tex] = [tex]\frac{2*3.9*10^{26}*A}{4\pi r^{2} *3*10^{8}}[/tex] = [tex]2.069*10^{17}\frac{A}{r^{2}}[/tex]
gravitational force between the sail and the sun [tex]F_{g}[/tex] = [tex]\frac{GMm}{r^{2}}[/tex]
where
G is the gravitational constant = 6.67 x 10^−11 m^3⋅s−2⋅kg−1.
m is the mass of the sail = 10000 kg
M is the mass of the sun = 1.99 x 10^30 kg.
==> [tex]F_{g}[/tex] = [tex]\frac{6.67*10^{-11}*1.99*10^{30}*10000}{r^{2}}[/tex] = [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]
Equating the forces, we have
[tex]2.069*10^{17}\frac{A}{r^{2}}[/tex] = [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]
the distance cancels out
A = (1.33 x 10^24)/(2.069 x 10^17) = 6428226.196 m^2
==> 6428.2 km^2
c) The force of the solar radiation is proportional to the intensity of the sun from the light, and the intensity is inversely proportional to the square of the distance from the source. Also, the force of gravitation is inversely proportional to the square of the distance, so they both cancel out.
When a monochromatic light of wavelength 433 nm incident on a double slit of slit separation 6 µm, there are 5 interference fringes in its central maximum. How many interference fringes will be in the central maximum of a light of wavelength 632.9 nm for the same double slit?
Answer:
The number of interference fringes is [tex]n = 3[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 433 \ nm = 433 *10^{-9} \ m[/tex]
The distance of separation is [tex]d = 6 \mu m = 6 *10^{-6} \ m[/tex]
The order of maxima is m = 5
The condition for constructive interference is
[tex]d sin \theta = n \lambda[/tex]
=> [tex]\theta = sin^{-1} [\frac{5 * 433 *10^{-9}}{ 6 *10^{-6}} ][/tex]
=> [tex]\theta = 21.16^o[/tex]
So at
[tex]\lambda_1 = 632.9 nm = 632.9*10^{-9} \ m[/tex]
[tex]6 * 10^{-6} * sin (21.16) = n * 632.9 *10^{-9}[/tex]
=> [tex]n = 3[/tex]
What would happen in a State if its citizens lack relevant knowledge, skills
and positive attitude?
Answer:
If the older generation is lacking, the younger generation would likely have knowledge, skill, or a positive attitude in some combination, but it is relative to the culture.
The simple reason is the desirability for genetic variation using recessive genes.
In other words, if the older generation lacks something, it tends to be something they don’t need, but something that will look good on young people. But mostly relative to the culture and education system.
Hope this helps
We can reasonably model a 75 W incandescent light bulb as a sphere 6.0 cm in diameter. Typically only about 5% of the energy goes to visible light; the rest goes largely to non-visible infrared radiation. (a) What is the visible light intensity at the surface of the bulb
Answer:
Visible light intensity at the surface of the bulb (I) = 331 W/m²
Explanation:
Given:
Energy = 75 W
Radius = 6 /2 = 3 cm = 3 × 10⁻² m
Energy goes to visible light = 5% = 0.05
Find:
Visible light intensity at the surface of the bulb (I)
Computation:
Visible light intensity at the surface of the bulb (I) = P / 4A
Visible light intensity at the surface of the bulb (I) = (0.05)(75) / 4π(3 × 10⁻²)²
Visible light intensity at the surface of the bulb (I) = 3.75 / 4π(9 × 10⁻⁴)
Visible light intensity at the surface of the bulb (I) = 331 W/m²
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.
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.
Learn more about velocity here: https://brainly.com/question/1303810
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
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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The tibia is a lower leg bone (shin bone) in a human. The maximum strain that the tibia can experience before fracturing corresponds to a 1 % change in length.
A. Young's modulus for bone is about Y = 1.4 x 10 N/m². The tibia (shin bone) of a human is 0.35 m long and has an average cross-sectional area of 2.9 cm. What is the effective spring constant of the tibia?
B. If a man weighs 750 N, how much is the tibia compressed if it supports half his weight?
C. What is the maximum force that can be applied to a tibia with a cross-sectional area, A = 2.90 cm?
Answer:
a
[tex]k = 11600000 N/m[/tex]
b
[tex]\Delta L = 3.2323 *10^{-5} \ m[/tex]
c
[tex]F = 3750.28 \ N[/tex]
Explanation:
From the question we are told that
The Young modulus is [tex]E = 1.4 *10^{10} \ N/m^2[/tex]
The length is [tex]L = 0.35 \ m[/tex]
The area is [tex]2.9 \ cm^2 = 2.9 *10^{-4} \ m ^2[/tex]
Generally the force acting on the tibia is mathematically represented as
[tex]F = \frac{E * A * \Delta L }{L}[/tex] derived from young modulus equation
Now this force can also be mathematically represented as
[tex]F = k * \Delta L[/tex]
So
[tex]k = \frac{E * A }{L}[/tex]
substituting values
[tex]k = \frac{1.4 *10^{10} * 2.9 *10^{-4} }{ 0.35}[/tex]
[tex]k = 11600000 N/m[/tex]
Since the tibia support half the weight then the force experienced by the tibia is
[tex]F_k = \frac{750 }{2} = 375 \ N[/tex]
From the above equation the extension (compression) is mathematically represented as
[tex]\Delta L = \frac{ F_k * L }{ A * E }[/tex]
substituting values
[tex]\Delta L = \frac{ 375 * 0.35 }{ (2.9 *10^{-4}) * 1.4*10^{10} }[/tex]
[tex]\Delta L = 3.2323 *10^{-5} \ m[/tex]
From the above equation the maximum force is
[tex]F = \frac{1.4*10^{10} * (2.9*10^{-4}) * 3.233*10^{-5} }{ 0.35}[/tex]
[tex]F = 3750.28 \ N[/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]
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°
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.
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
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]
A 1300-turn coil of wire 2.40 cm in diameter is in a magnetic field that increases from 0 T to 0.120 T in 9.00 ms . The axis of the coil is parallel to the field. What is the emf of the coil?
Answer:
The induced emf in the coil is 7.843 V
Explanation:
Given;
number of turns of the coil, N = 1300 turn
diameter of the coil, d = 2.4 cm = 0.024 m
initial magnetic field, B₁ = 0 T
final magnetic field, B₂ = 0.12 T
change in time, dt = 9.0 ms = 9 x 10⁻³ s
Area of the coil is given by;
A = πr²
radius of the coil, r = 0.024 / 2
radius of the coil, r = 0.012 m
A = π(0.012)²
A = 4.525 x 10⁻⁴ m²
The induced emf in the coil is given by;
E = NA(dB/dt)
E = NA [(B₂ - B₁) /dt]
E = 1300 x 4.525 x 10⁻⁴ (0.12 - 0) / (9 x 10⁻³)
E = 7.843 V
Therefore, the induced emf in the coil is 7.843 V
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
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:
Experiment to find ways to make rainbows.
a) Insert at least one setup where light passing through a prism gives a rainbow and describe why a rainbow is formed.
b) Explain why only some types of light will yield rainbows.
Answer:
Explanation:
a) To get a rainbow from a prism arrangement, we will need
A triangular prismA black cardboard boxA source of white light (light from the window will suffice)A pocket knifeFirst, you cut a slit in one end of the cardboard with the pen knife.
Next you open up a space on top of the cardboard through which you can observe the experiment and its result.
Next, you place the triangular prism with its slant face facing the the cut slit.
Finally, position the slit to face the light from the open window, and adjust the prism till the projected bands of colored light (rainbow) is very much obvious on the other end of the box, opposite the slit.
b) For a light to yield rainbow, it most be composed of different component colors of light. The colors of light is due to the difference in wavelength, and dispersion is due to the different in the wavelengths of the component light. So to get rainbow from a light source, the light must not be monochromatic. This means that only light composed of component light of different colors can produce rainbow. Light from the sun for example is composed of 7 distinct colors of light, and white light can be created with just three colors; blue, green, and red light.
"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
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:
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.
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
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.
To know more about friction:
https://brainly.com/question/28356847
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