A Van de Graaff generator produces a beam of 2.02-MeV deuterons, which are heavy hydrogen nuclei containing a proton and a neutron.
A) If the beam current is 10.0 μA, how far apart are the deuterons?
B) Is the electrical force of repulsion among them a significant factor in beam stability? Explain.
Answer:
A) The distance of the deuterons from one another = 2.224× 10⁻⁷ m
B) The electrical force of repulsion among them shows a small effect in beam stability.
Explanation:
Given that:
A Van de Graaff generator produces a beam of 2.02-MeV deuterons
If the beam current is 10.0 μA, the distance of the deuterons from one another can be determined by using the concept of kinetic energy of the generator.
[tex]\mathtt{K.E = \dfrac{1}{2}mv^2}[/tex]
2 K.E = mv²
[tex]\mathtt{v^2 = \dfrac{2 K.E }{m}}[/tex]
[tex]\mathtt{v =\sqrt{ \dfrac{2 K.E }{m}}}[/tex]
so, v is the velocity of the deuterons showing the distance of the deuterons apart from one another.
[tex]\mathtt{v =\sqrt{ \dfrac{2 (2.02 \ MeV) \times \dfrac{10^6 \ eV}{ 1 \ MeV} \times \dfrac{1.60 \times 10^{-19} \ J }{1 \ eV} }{ 3.34 \times 10^ {-27} \ kg}}}[/tex]
[tex]\mathtt{v =\sqrt{ \dfrac{6.464 \times 10^{-13} \ J }{ 3.34 \times 10^ {-27} \ kg}}}[/tex]
v = 13911611.49 m/s
v = 1.39 × 10⁷ m/s
So, If the beam current is 10.0 μA.
We all know that:
[tex]I = \dfrac{q}{t}[/tex]
[tex]t = \dfrac{q}{I}[/tex]
[tex]\mathtt{ t = \dfrac{1.6 * 10 ^{-19} \ C}{10.0 * 10^{-6} \ A}}[/tex]
t = 1.6 × 10⁻¹⁴ s
Finally, the distance of the deuterons from one another = v × t
the distance of the deuterons from one another = (1.39 × 10⁷ m/s × 1.6 × 10⁻¹⁴ s)
the distance of the deuterons from one another = 2.224× 10⁻⁷ m
B) Is the electrical force of repulsion among them a significant factor in beam stability? Explain.
The electrical force of repulsion among them shows a small effect in beam stability. This is because, one nucleus tends to put its nearest neighbor at potential V = (k.E × q) / r = 7.3e⁻⁰³ V. This is very small compared to the 2.02-MeV accelerating potential, Thus, repulsion within the beam is a small effect.
Give an example of hypothesis for an experiment and then identify its dependent and independent variables. Write all the steps of the scientific method. Explain why it is good to limit an experiment to test only one variable at a time whenever possible ?
Please somebody !!!!
The magnetic field at the center of a 1 cm diameter loop is 2.5 mT. If a long straight wire carries the same current as the loop of wire, at what distance from the wire is the magnetic field 2.5 mT? A. 0.10 m B. 1.6x10-3 m C. 0.01 m D. 20 m
Answer:
B. 1.6 x 10⁻³ m
Explanation:
The magnetic field at the center of the loop is given by;
[tex]B = \frac{\mu_o I }{2R}[/tex]
Where;
μ₀ is the permeability of free space
I is the current in the loop
R is the radius of the circular loop
B is the magnetic field
Given;
diameter of the loop = 1cm
radius of the loop, r = 0.5 cm = 0.005 m
magnetic field, B = 2.5mT = 2.5 x 10⁻³ T
The current in the loop is calculated as;
[tex]I = \frac{2BR}{\mu_o} \\\\I = \frac{2*2.5*10^{-3}*0.005}{4\pi*10^{-7}} \\\\I = 19.89 \ A[/tex]
The magnetic at a distance from the long straight wire is calculated as;
[tex]B = \frac{\mu_o I}{2\pi d}[/tex]
where;
d is the distance from the wire;
[tex]d = \frac{\mu_o I}{2\pi B} \\\\d = \frac{4\pi *10^{-7} * 19.89}{2\pi *2.5*10^{-3}} \\\\d = 1.6 *10^{-3} \ m[/tex]
Therefore, the distance from the wire where the magnetic field is 2.5 mT is 1.6 x 10⁻³ m.
B. 1.6 x 10⁻³ m
This question involves the concepts of the magnetic field due to a loop and a current-carrying wire and current.
A long straight wire carrying the same current as the loop of wire has a magnetic field of 2.5 mT at a distance of b "B. 1.5 x 10⁻³ m".
The magnetic field at the center of a loop of wire is given by the following formula:
[tex]B=\frac{\mu_o I}{2r}[/tex]
where,
B = Magnetic Field = 2.5 mT = 2.5 x 10⁻³ T
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
I = current = ?
r = radius = diameter/2 = 1 cm/2 = 0.5 cm = 0.005 m
Therefore,
[tex]I = \frac{(2.5\ x\ 10^{-3}\ T)(2)(0.005\ m)}{4\pi\ x\ 10^{-7}\ N/A^2}[/tex]
I = 19.9 A
Now, the magnetic field at a distance from the straight wire is given by the following formula:
[tex]B=\frac{\mu_o I}{2\pi R}[/tex]
where,
R = distance from wire = ?
Therefore,
[tex]R = \frac{(4\pi \ x \ 10^{-7}\ N/A^2)(19.9\ A)}{2\pi(2.5\ x\ 10^{-3}\ T)}[/tex]
R = 1.6 x 10⁻³ m
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The switch on the electromagnet, initially open, is closed. What is the direction of the induced current in the wire loop (as seen from the left)?
Answer:
The induced current is clockwise
Bob and Lily are riding on a merry-go-round. Bob rides on a horse near the outer edge of the circular platform, and Lily rides on a horse near the center of the circular platform. When the merry-go-round is rotating at a constant angular speed, Bob's angular speed is:_____.
a. the same as Lily's.
b. larger than Lily's.
c. exactly half as much as Lily's.
d. exactly twice as much as Lily's.
e. smaller than Lily's.
Answer: the same as Lily's
Explanation:
Angular velocity has to do with the speed at which an object will be able to rotate. We are informed that Bob and Lily are riding on a merry-go-round.
Since we are further told that Bob rides on a horse near the outer edge of the circular platform, and Lily rides on a horse near the center of the circular platform and that he merry-go-round is rotating at a constant angular speed.
Based on the above analysis, Bob's angular speed will be thesame as that of Lily.
1. What does the acronym LASER stand for? What characteristic of a laser makes it suitable for today's experiment?
Answer:Light Amplification by Stimulated Emission of Radiation. It is able to convert light or electrical energy into focused high energy beam to treat some sickness and diseases.
Explanation:
Answer:
Light amplification by stimulated emission of radiation
A rectangular coil having N turns and measuring 15 cm by 25 cm is rotating in a uniform 1.6-T magnetic field with a frequency of 75 Hz. The rotation axis is perpendicular to the direction of the field. If the coil develops a sinusoidal emf of maximum value 56.9 V, what is the value of N?
A) 2
B) 4
C) 6
D) 8
E) 10
Answer:
A) 2
Explanation:
Given;
magnetic field of the coil, B = 1.6 T
frequency of the coil, f = 75 Hz
maximum emf developed in the coil, E = 56.9 V
area of the coil, A = 0.15 m x 0.25 m = 0.0375 m²
The maximum emf in the coil is given by;
E = NBAω
Where;
N is the number of turns
ω is the angular velocity = 2πf = 2 x 3.142 x 75 = 471.3 rad/s
N = E / BAω
N = 56.9 / (1.6 x 0.0375 x 471.3)
N = 2 turns
Therefore, the value of N is 2
A) 2
PLEASE HELP FAST WILL GIVE BRAINLIEST The sentence, "The popcorn kernels popped twice as fast as the last batch," is a(n) _____. 1.experiment 2.hypothesis 3.observation 4.control
The answer is 3. Observation
Explanation:
The sentence "The popcorn kernels popped twice as fast as the last batch" is the result of observing or measuring the time popcorn kernels require to pop. In this context, the sentence best matches the word "observation" which the term used in the Scientific method to refer to statements that are the result of studying a phenomenon, either through the senses such as sight or through precise instruments that allow scientists to understand numerically variables such as time, speed, temperature, etc.
I WILL GIVE BRAINLIEST Identify two types of motion where an object's speed remains the same while it continues to change direction
Answer:
velocity and acceleration
Answer:
Hey there!
Centripetal (Circular Motion) and Oscillating Motion.
Let me know if this helps :)
Rank the following types of electromagnetic waves by the wavelength of the wave.
a. Microwaves
b. X-rays
c. Radio waves
d. Visible light
Explanation:
In order of Increasing Wavelength of the Electromagnetic Spectrum :
B) X rays
D) Visible light
A) Microwave
C) Radio Waves
Electromagnetic waves in order of decreasing wavelength is X-rays,visible light,microwaves and radio waves.
What are electromagnetic waves?The electromagnetic radiation consists of waves made up of electromagnetic field which are capable of propogating through space and carry the radiant electromagnetic energy.
The radiation are composed of electromagnetic waves which are synchronized oscillations of electric and magnetic fields . They are created due to change which is periodic in electric as well as magnetic fields.
In vacuum ,all the electromagnetic waves travel at the same speed that is with the speed of air.The position of an electromagnetic wave in an electromagnetic spectrum is characterized by it's frequency or wavelength.They are emitted by electrically charged particles which undergo acceleration and subsequently interact with other charged particles.
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An L-R-C series circuit has L = 0.450 H, C=2.50×10^−5F, and resistance R.
Required:
a. What is the angular frequency of the circuit when R = 0?
b. What value must R have to give a decrease in angular frequency of 10.0 % compared to the value calculated in Part a.
Answer:
298rad/s and 116.96 ohms
Explanation:
Given an L-R-C series circuit where
L = 0.450 H,
C=2.50×10^−5F, and resistance R= 0
In this situation we have a simple LC circuit with angular frequency
Wo = 1√LC
= 1/√(0.450)(2.50×10^-5)
= 1/√0.00001125
= 1/0.003354
= 298rad/s
B) Now we need to find the value of R such that it gives a 10% decrease in angular frequency.
Wi/W° = (100-10)/100
Wi/W° = 90/100
Wi/W° = 0.90 ............... 1
Angular frequency of oscillation
The complete aspect of the solution is attached, please check.
a. The angular frequency of the circuit when R = 0 Ohms is 294.12 rad/s.
b. The value R must have to give a decrease in angular frequency of 10.0 % compared to the initial value is equal to 116.96 Ohms.
Given the following data:
Inductance, L = 0.450 HenryCapacitance, C = [tex]2.50\times 10^{-5}[/tex] Faradsa. To determine the angular frequency of the circuit when R = 0 Ohms:
Mathematically, the angular frequency of a LC circuit is given by the formula:
[tex]\omega = \frac{1}{\sqrt{LC} } \\\\\omega =\frac{1}{\sqrt{0.450 \times 2.50\times 10^{-5}}} \\\\\omega =\frac{1}{\sqrt{1.125 \times 10^{-5}}} \\\\\omega = \frac{1}{0.0034} \\\\\omega = 294.12\;rad/s[/tex]
b. To find the value R must have to give a decrease in angular frequency of 10.0 % compared to the value calculated above:
The mathematical expression is given as follows:
[tex]\frac{\omega_f}{\omega_i} = \frac{100-10}{100} \\\\\frac{\omega_f}{\omega_i} =\frac{90}{100} \\\\\frac{\omega_f}{\omega_i} =0.9[/tex]
[tex](\frac{\omega_f}{\omega_i})^2 = 1 - \frac{R^2C}{4L} \\\\0.90^2=1 - \frac{R^2C}{4L}\\\\R=\sqrt{\frac{4L(1-0.81)}{C}} \\\\R=\sqrt{\frac{4\times 0.450 \times (0.19)}{2.50\times 10^{-5}}}\\\\R = \sqrt{\frac{0.342}{2.50\times 10^{-5}} }\\\\R =\sqrt{13680}[/tex]
R = 116.96 Ohms.
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A resistor and an inductor are connected in series to an ideal battery of constant terminal voltage. At the moment contact is made with the battery, the voltage across the inductor is
Answer:
The voltage is equal to the batteries terminal voltage
Explanation:
Explanation:
A rigid container holds 4.00 mol of a monatomic ideal gas that has temperature 300 K. The initial pressure of the gas is 6.00 * 104 Pa. What is the pressure after 6000 J of heat energy is added to the gas?
Answer:
The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.
Explanation:
When a container is rigid, the process is supposed to be isochoric, that is, at constant volume. Then, the equation of state for ideal gases can be simplified into the following expression:
[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}[/tex]
Where:
[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressures, measured in pascals.
[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Initial and final temperatures, measured in Kelvins.
In addtion, the specific heat at constant volume for monoatomic ideal gases, measured in joules per mole-Kelvin is given by:
[tex]\bar c_{v} = \frac{3}{2}\cdot R_{u}[/tex]
Where:
[tex]R_{u}[/tex] - Ideal gas constant, measured by pascal-cubic meters per mole-Kelvin.
If [tex]R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}[/tex], then:
[tex]\bar c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{2}}{mol\cdot K} \right)[/tex]
[tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex]
And change in heat energy ([tex]Q[/tex]), measured by joules, by:
[tex]Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})[/tex]
Where:
[tex]n[/tex] - Molar quantity, measured in moles.
The final temperature of the monoatomic ideal gas is now cleared:
[tex]T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}[/tex]
Given that [tex]T_{1} = 300\,K[/tex], [tex]Q = 6000\,J[/tex], [tex]n = 4\,mol[/tex] and [tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex], the final temperature is:
[tex]T_{2} = 300\,K + \frac{6000\,J}{(4\,mol)\cdot \left(12.471\,\frac{J}{mol\cdot K} \right)}[/tex]
[tex]T_{2} = 420.279\,K[/tex]
The final pressure of the system is calculated by the following relationship:
[tex]P_{2} = \left(\frac{T_{2}}{T_{1}}\right) \cdot P_{1}[/tex]
If [tex]T_{1} = 300\,K[/tex], [tex]T_{2} = 420.279\,K[/tex] and [tex]P_{1} = 6.00\times 10^{4}\,Pa[/tex], the final pressure is:
[tex]P_{2} = \left(\frac{420.279\,K}{300\,K} \right)\cdot (6.00\times 10^{4}\,Pa)[/tex]
[tex]P_{2} = 8.406\times 10^{4}\,Pa[/tex]
The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.
Can someone log into my acc FOR ME I will pay you to complete my physics assignments for money or points?!!
Answer: no sorry../
Explanation:
A small helium-neon laser emits red visible light with a power of 5.40 mW in a beam of diameter 2.30 mm.
Required:
a. What is the amplitude of the electric field of the light? Express your answer with the appropriate units.
b. What is the amplitude of the magnetic field of the light?
c. What is the average energy density associated with the electric field? Express your answer with the appropriate units.
d. What is the average energy density associated with the magnetic field? Express your answer with the appropriate units.
E) What is the total energy contained in a 1.00-m length of the beam? Express your answer with the appropriate units.
Answer:
A. 990v/m
B.330x10^-8T
C.2.19x10^-6J/m³
D.1.45x10^-11J
Explanation:
See attached file
A 300 MWe (electrical power output) Power Plant having a thermal efficiency of 40% is cooled by sea water. Due to environmental regulations the seawater can only increase temperature by 5 C during the process. How much sea water (minimum) must be used in kg/s for cooling if the plant operates at it's rated capacity?
Answer:
m = 22,877 kg / s
Explanation:
Let's solve this exercise in parts, first look for the amount of heat generated by the plant and then the amount of water to dissipate this heat
The plant generates a power of 300 MW at a rate of 40%, let's use a direct ratio rule to find the heat. If the power is 400 MW it corresponds to 40%, what heat (Q) corresponds to the other 60%
Q = 300 60% / 40%
Q = 450 MW
having the amount of heat generated we can use the calorimeter equation,
Q = m [tex]c_{e}[/tex] [tex](T_{f} - T_{o})[/tex]
m = Q / c_{e} (T_{f} - T_{o})
let's use the maximum temperature change allowed
(T_{f} - T_{o}) = 5
the specific heat of sea water is 3934 J / kg ºC, note that it is less than that of pure water, due to the salts dissolved in sea water
power and energy are related
W = Q / t
Q = W t
let's calculate
m = 450 10⁶ / (3934 5)
m = 22,877 kg / s
Problem 25.40 What is the energy (in eV) of a photon of visible light that has a wavelength of 500 nm
Answer:
E = 2.48 eV
Explanation:
The energy of a photon is given by the following formula:
E = hυ
where,
E = Energy of Photon = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
υ = frequency of photon = c/λ
Therefore,
E = hc/λ
where,
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of light = 500 nm = 5 x 10⁻⁷ m
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5 x 10⁻⁷ m)
E = (3.97 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)
E = 2.48 eV
A photon of visible light that has a wavelength of 500 nm, has an energy of 2.48 eV.
We can calculate the energy (E) of a photon with a wavelength (λ) of 500 nm using the Planck's-Einstein relation.
[tex]E = \frac{h \times c}{\lambda } = \frac{(6.63 \times 10^{-34}J.s ) \times (3.00 \times 10^{8}m/s )}{500 \times 10^{-9}m } = 3.98 \times 10^{-19} J[/tex]
where,
h: Planck's constantc: speed of lightWe can convert 3.98 × 10⁻¹⁹ J to eV using the conversion factor 1 J = 6.24 × 10¹⁸ eV.
[tex]3.98 \times 10^{-19} J \times \frac{6.24 \times 10^{18} eV }{1J} = 2.48 eV[/tex]
A photon of visible light that has a wavelength of 500 nm, has an energy of 2.48 eV.
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A 70 kg human body typically contains 140 g of potassium. Potassium has a chemical atomic mass of 39.1 u and has three naturally occurring isotopes. One of those isotopes, 40K,is radioactive with a half-life of 1.3 billion years and a natural abundance of 0.012%. Each 40K decay deposits, on average, 1.0 MeV of energy into the body. What yearly dose in Gy does the typical person receive from the decay of 40K in the body?
Answer:
0.03143 Gy
Explanation:
Mass of the human body = 70 kg
Mass of potassium in the human body = 140 g
chemical atomic mass of potassium = 39.1
From avogadros number, we know that 1 atomic mass of an element contains 6.023 × 10^(23) atoms
Thus,
140g of potassium will contain;
(140 × 6.023 × 10^(23))/(39.1) = 2.1566 × 10^(24) atoms
We are told that the natural abundance of one of the 40K isotopes is 0.012%.
Thus;
Number of atoms of this isotope = 0.012% × 6.023 × 10^(23) = 7.2276 × 10^(19) K-40 atoms
Formula for activity of K-40 is given as;
Activity = (0.693 × number of K-40 atoms)/half life
Activity = (0.693 × 7.2276 × 10^(19))/1300000000
Activity = 3.85 × 10^(10)
We are told that each decay deposits 1.0 MeV of energy into the body.
Thus;
Total energy absorbed by the body in a year = 3.85 × 10^(10) × 1 × 365 = 1405.25 × 10^(10) MeV
Now, 1 MeV = 1.602 × 10^(-13) joules
Thus;
Total energy absorbed by the body in a year = 1405.25 × 10^(10) × 1.602 × 10^(-13) = 2.25 J
1 Gy = 1 J/kg
Thus;
Yearly dose = 2.25/70 = 0.03143 Gy
If the
refractive index of benzere is 2.419,
what is the speed of light in benzene?
Answer:
[tex]v=1.24\times 10^8\ m/s[/tex]
Explanation:
Given that,
The refractive index of benzene is 2.419
We need to find the speed of light in benzene. The ratio of speed of light in vacuum to the speed of light in the medium equals the refractive index. So,
[tex]n=\dfrac{c}{v}\\\\v=\dfrac{c}{n}\\\\v=\dfrac{3\times 10^8}{2.419}\\\\v=1.24\times 10^8\ m/s[/tex]
So, the speed of light in bezene is [tex]1.24\times 10^8\ m/s[/tex].
Define wave length as applied to wave motion
Answer: Wavelength can be defined as the distance between two successive crests or troughs of a wave. It is measured in the direction of the wave.
Explanation:
Wavelength refers to the length or distance between two identical points of neighboring cycles of a wave signal traveling in space or in any physical medium. ... The wavelength of a signal is inversely proportional to its frequency, that is, the higher the frequency, the shorter the wavelength.
An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-28 kg, and that of the other is 1.86 10-27 kg. If the lighter fragment has a speed of 0.844c after the breakup, what is the speed of the heavier fragment
Answer: Speed = [tex]3.10^{-31}[/tex] m/s
Explanation: Like in classical physics, when external net force is zero, relativistic momentum is conserved, i.e.:
[tex]p_{f} = p_{i}[/tex]
Relativistic momentum is calculated as:
p = [tex]\frac{mu}{\sqrt{1-\frac{u^{2}}{c^{2}} } }[/tex]
where:
m is rest mass
u is velocity relative to an observer
c is light speed, which is constant (c=[tex]3.10^{8}[/tex]m/s)
Initial momentum is zero, then:
[tex]p_{f}[/tex] = 0
[tex]p_{1}-p_{2}[/tex] = 0
[tex]p_{1} = p_{2}[/tex]
To find speed of the heavier fragment:
[tex]\frac{mu_{1}}{\sqrt{1-\frac{u^{2}_{1}}{c^{2}} } }=\frac{mu_{2}}{\sqrt{1-\frac{u^{2}_{2}}{c^{2}} } }[/tex]
[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=\frac{3.10^{-28}.0.844.3.10^{8}}{\sqrt{1-\frac{(0.844c)^{2}}{c^{2}} } }[/tex]
[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=1.42.10^{-19}[/tex]
[tex]1.86.10^{-27}u_{1} = 1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }[/tex]
[tex](1.86.10^{-27}u_{1})^{2} = (1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } })^{2}[/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38}.(1-\frac{u_{1}^{2}}{9.10^{16}} )[/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -[2.02.10^{-38}(\frac{u_{1}^{2}}{9.10^{16}} )][/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -2.24.10^{-23}.u^{2}_{1}[/tex]
[tex]3.46.10^{-54}.u_{1}^{2}+2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]
[tex]2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]
[tex]u^{2}_{1} = \frac{2.02.10^{-38}}{2.24.10^{-23}}[/tex]
[tex]u_{1} = \sqrt{9.02.10^{-62}}[/tex]
[tex]u_{1} = 3.10^{-31}[/tex]
The speed of the heavier fragment is [tex]u_{1} = 3.10^{-31}[/tex]m/s.
A solenoid with 3,000.0 turns is 70.0 cm long. If its self-inductance is 25.0 mH, what is its radius? (The value of μ0 is 4π x 10-7 N/A2 .) A. 0.02219 m B. 327 m C. 52 m D. 0.00199 m
Answer:
A. 2.2*10^-2m
Explanation:
Using
Area = length x L/ uo xN²
So A = 0.7m * 25 x 10^-3H /( 4π x10^-7*
3000²)
A = 17.5*10^-3/ 1.13*10^-5
= 15.5*10^-2m²
Area= π r ²
15.5E-2/3.142 = r²
2.2*10^2m
Explanation:
A rectangular conducting loop of wire is approximately half-way into a magnetic field B (out of the page) and is free to move. Suppose the magnetic field B begins to decrease rapidly in strength
Requried:
What happens to the loop?
1. The loop is pushed to the left, toward the magnetic field.
2. The loop doesn’t move.
3. The loop is pushed downward, towards the bottom of the page.
4. The loop will rotate.
5. The loop is pushed upward, towards the top of the page.
6. The loop is pushed to the right, away from the magnetic field
Answer:
. The loop is pushed to the right, away from the magnetic field
Explanation
This decrease in magnetic strength causes an opposing force that pushes the loop away from the field
The law of reflection is quite useful for mirrors and other flat, shiny surfaces. (This sort of reflection is called specular reflection). However, you've likely been told that when you look at something, you are seeing light reflected from the object that you are looking at. This is reflection of a different sort of diffuse reflection.
Suppose that the spotlight shines so that different parts of the beam reflect off of different two surfaces, one inclined at an angle alpha (from the horizontal) and one inclined at an angle beta. What would the angular separation between the rays reflected from the two surfaces?
Answer:
Explanation:
Suppose initially the plane was horizontal and light was reflected back at some angle θ from the normal .
Now the reflecting surface is twisted so that is becomes inclined at angle alpha .
The reflected light will be deviated from its original direction by angle
2 x alpha .
Similarly when the reflecting surface is further twisted so that it becomes inclined at angle beta then again the reflected beam will deviated by angle
2 x beta
Hence angle between these two reflected beam
= 2 beta - 2 alpha
= 2 ( β - α )
So, angular separation between the rays reflected from the two surfaces
= 2 ( β - α ) .
A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the slit width is doubled
Answer:
y ’= y / 2
thus when the slit width is doubled the pattern width is halved
Explanation:
The diffraction of a slit is given by the expressions
a sin θ = m λ
where a is the width of the slit, λ is the wavelength and m is an integer that determines the order of diffraction.
sin θ = m λ / a
If this equation
a ’= 2 a
we substitute
2 a sin θ'= m λ
sin θ'= (m λ / a) 1/2
sin θ ’= sin θ / 2
We can use trigonometry to find the width
tan θ = y / L
as the angle is small
tan θ = sin θ / cos θ = sin θ
sin θ = y / L
we substitute
y ’/ L = y/L 1/2
y ’= y / 2
thus when the slit width is doubled the pattern width is halved
A particle with charge q and momentum p, initially moving along the x-axis, enters a region where a uniform magnetic field* B=(B0)(k) extends over a width x=L. The particle is deflected a distance d in the +y direction as it traverses the field. Determine the magnitude of the momentum (p).
Answer:
Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.
Explanation:
So, from the question, we are given that the charge = q, the momentum = p.
=> From the question We are also given that, "initially, there is movement along the x-axis which then enters a region where a uniform magnetic field* B = (B0)(k) which then extends over a width x = L, the distance = d in the +y direction as it traverses the field."
Momentum,P = mass × Velocity, v -----(1).
We know that for a free particle the magnetic field is equal to the centrepetal force. Thus, we have the magnetic field = mass,.m × (velocity,v)^2 / radius, r.
Radius,r = P × v / B0 -----------------------------(2).
Centrepetal force = q × B0 × v. ----------(3).
(If X = L and distance = d)Therefore, the radius after solving binomially, radius = (d^2 + 2 L^2) / 2d.
Equating Equation (2) and (3) gives;
P = B0 × q × r.
Hence, the Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.
The AB rope is fixed to the ground at its A end, and forms 30º with the vertical. Its other end is connected to two ropes by means of the B-ring of negligible weight. The vertical rope supports the E block and the other rope passes through the grounded articulated pulley C to join at its end to the 80 N weight block D. The inclined section of the BD rope forms 60º with the vertical one; determine the weight of the E block necessary for the balance of the system and calculate the tension in the AB rope.
Answer:
T = 80√3 N ≈ 139 N
W = 160 N
Explanation:
Sum of forces on B in the x direction:
∑F = ma
80 N sin 60° − T sin 30° = 0
T = 80 N sin 60° / sin 30°
T = 80√3 N
T ≈ 139 N
Sum of forces on B in the y direction:
∑F = ma
80 N cos 60° + T cos 30° − W = 0
W = 80 N cos 60° + T cos 30°
W = 40 N + 120 N
W = 160 N
Calculate the focal length (in m) of the mirror formed by the shiny bottom of a spoon that has a 3.40 cm radius of curvature. m (b) What is its power in diopters? D
Answer:
The power of the mirror in diopters is 58.8 D
Explanation:
Given;
radius of curvature of the spoon, R = 3.4 cm = 0.034 m
The focal length of a mirror is given by;
[tex]f = \frac{R}{2} \\\\f = \frac{0.034}{2} \\\\f = 0.017 \ m[/tex]
The focal length of the mirror is 0.017 m
(b) The power of the mirror is given by;
[tex]P = \frac{1}{f}[/tex]
where;
P is the power of the mirror
f is the focal length
[tex]P = \frac{1}{f}\\\\P= \frac{1}{0.017}\\\\P = 58.8 \ D[/tex]
Thus, the power of the mirror in diopters is 58.8 D
A stereo speaker produces a pure "G" tone, with a frequency of 392 Hz. What is the period T of the sound wave produced by the speaker?
Answer:
The period is [tex]T = 0.00255 \ s[/tex]
Explanation:
From the question we are told that
The frequency is [tex]f = 392 \ Hz[/tex]
Generally the period is mathematically represented as
[tex]T = \frac{1}{f}[/tex]
=> [tex]T = \frac{1}{ 392}[/tex]
=> [tex]T = 0.00255 \ s[/tex]
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