Answer:
1) θ = 0.00118 rad, 2) θ = 0.00236 rad , 3) I / I₀ = 0.1738, 4) I / Io = 0.216
Explanation:
In the double-slit interference phenomenon it is explained for constructive interference by the equation
d sin θ = m λ
1) the first order maximum occurs for m = 1
sin θ = λ / d
θ = sin⁻¹ λ / d
let's reduce the magnitudes to the SI system
λ = 520 nm = 520 10⁻⁹ θ = 0.00118 radm
d = 0.440 mm = 0.440 10⁻³ m ³
let's calculate
θ = sin⁻¹ (520 10⁻⁹ / 0.44 10⁻³)
θ = sin⁻¹ (1.18 10⁻³)
θ = 0.00118 rad
2) the second order maximum occurs for m = 2
θ = sin⁻¹ (m λ / d)
θ = sin⁻¹ (2 5¹20 10⁻⁹ / 0.44 10⁻³)
θ = 0.00236 rad
3) To calculate the intensity of the interference spectrum, the diffraction phenomenon must be included, so the equation remains
I = I₀ cos² (π d sin θ /λ ) sinc² (pi b sin θ /λ )
where the function sinc = sin x / x
and b is the width of the slits
we caption the values
x = π 0.310 10⁻³ sin 0.00118 / 520 10⁻⁹)
x = 2.21
I / I₀ = cos² (π 0.44 10⁻³ sin 0.00118 / 520 10⁻⁹) (sin (2.21) /2.21)²
remember angles are in radians
I / I₀ = cos² (3.0945) [0.363] 2
I / I₀ = 0.9978 0.1318
I / I₀ = 0.1738
4) the maximum second intensity is
I / I₀ = cos² (π d sinθ / λ) sinc² (πb sin θ /λ)
x =π 0.310 10⁻³ sin 0.00236 / 520 10⁻⁹)
x = 4.41
I / Io = cos² (π 0.44 10⁻³ sin 0.00236 / 520 10⁻⁹) (sin 4.41 / 4.41)²
I / Io = cos² 6.273 0.216
I / Io = 0.216
.
Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find k (spring constant) and c (damping constant). Find a formula for the mass in terms of the frequency in Hz. Note that there may be more than one possible mass for a given frequency. b) For an unknown object you measured 0.2 Hz, what is the mass of the object? Suppose that you know that the mass of the unknown object is more than a kilogram.
Answer:
a) k = 95.54 N / m, c = 19.55 , b) m₃ = 0.9078 kg
Explanation:
In a simple harmonic movement with friction, we can assume that this is provided by the speed
fr = -c v
when solving the system the angular value remains
w² = w₀² + (c / 2m)²
They give two conditions
1) m₁ = 1 kg
f₁ = 1.1 Hz
the angular velocity is related to frequency
w = 2π f₁
Let's find the angular velocity without friction is
w₂ = k / m₁
we substitute
(2π f₁)² = k / m₁ + (c / 2m₁)²
2) m₂ = 2 kg
f₂ = 0.8 Hz
(2π f₂)² = k / m₂ + (c / 2m₂)²
we have a system of two equations with two unknowns, so we can solve it
we solve (c / 2m)² is we equalize the expression
(2π f₁)² - k / m₁ = (2π f₂²) 2 - k / m₁
k (1 / m₂ - 1 / m₁) = 4π² (f₂² - f₁²)
k = 4π² (f₂² -f₁²) / (1 / m₂ - 1 / m₁)
a) Let's calculate
k = 4 π² (0.8² -1.1²) / (½ -1/1)
k = 39.4784 (1.21) / (-0.5)
k = 95.54 N / m
now we can find the constant of friction
(2π f₁) 2 = k / m₁ + (c / 2m₁)²
c2 = ((2π f₁)² - k / m₁) 4m₁²
c2 = (4ππ² f₁² - k / m₁) 4 m₁²
let's calculate
c² = (4π² 1,1² - 95,54 / 1) 4 1²
c² = (47.768885 - 95.54) 8
c² = -382.1689
c = 19.55
b) f₃ = 0.2 Hz
m₃ =?
(2πf₃)² = k / m₃ + (c / 2m₃) 2
we substitute the values
(4π² 0.2²) = 95.54 / m₃ + 382.1689 2/4 m₃²
1.579 = 95.54 / m₃ + 95.542225 / m₃²
let's call
x = 1 / m₃
x² = 1 / m₃²
- 1.579 + 95.54 x + 95.542225 x² = 0
60.5080 x² + 60.5080 x -1 = 0
x² + x - 1.65 10⁻² = 0
x = [1 ±√ (1- 4 (-1.65 10⁻²)] / 2
x = [1 ± 1.03] / 2
x₁ = 1.015 kg
x₂ = -0.015 kg
Since the mass must be positive we eliminate the second results
x₁ = 1 / m₃
m₃ = 1 / x₁
m₃ = 1 / 1.1015
A nozzle with a radius of 0.22 cm is attached to a garden hose with a radius of 0.89 cm that is pointed straight up. The flow rate through hose and nozzle is 0.55 L/s.
Randomized Variables
rn = 0.22 cm
rh = 0.94 cm
Q = 0.55
1. Calculate the maximum height to which water could be squirted with the hose if it emerges from the nozzle in m.
2. Calculate the maximum height (in cm) to which water could be squirted with the hose if it emerges with the nozzle removed assuming the same flow rate.
Answer:
1. 0.2m
1. 66m
Explanation:
See attached file
The expressions of fluid mechanics allows to find the result for the maximum height that the water leaves through the two points are;
1) The maximum height when the water leaves the hose is: Δy = 0.20 m
2) The maximum height of the water leaves the nozzle is: Δy = 68.6m
Given parameters
The flow rate Q = 0.55 L/s = 0.55 10⁻³ m³ / s Nozzle radius r₁ = 0.22 cm = 0.22 10⁻² m Hose radius r₂ = 0.94 cm = 0.94 10⁻² mTo find
1. Maximum height of water in hose
2. Maximum height of water at the nozzle
Fluid mechanics studies the movement of fluids, liquids and gases in different systems, for this it uses two expressions:
The continuity equation. It is an expression of the conservation of mass in fluids.
A₁v₁ = A₂.v₂
Bernoulli's equation. Establishes the relationship between work and the energy conservation in fluids.P₁ + ½ ρ g v₁² + ρ g y₁ = P₂ + ½ ρ g v₂² + ρ g y₂
Where the subscripts 1 and 2 represent two points of interest, P is the pressure, ρ the density, v the velocity, g the acceleration of gravity and y the height.
1, Let's find the exit velocity of the water in the hose.
Let's use subscript 1 for the nozzle and subscript 2 for the hose.
The continuity equation of the flow value that must be constant throughout the system.
Q = A₁ v₁
v₁ = [tex]\frac{Q}{A_1 }[/tex]
The area of a circle is:
A = π r²
Let's calculate the velocity in the hose.
A₁ = π (0.94 10⁻²) ²
A₁ = 2.78 10⁻⁴ m²
v₁ = [tex]\frac{0.55 \ 10^{-3}}{2.78 \ 10^{-4}}[/tex]
v₁ = 1.98 m / s
Let's use Bernoulli's equation.
When the water leaves the hose the pressure is atmospheric and when it reaches the highest point it has not changed P1 = P2
½ ρ v₁² + ρ g y₁ = ½ ρ v₂² + ρ g v₂
y₂-y₁ = ½ [tex]\frac{v_i^2 - v_2^2}{g}[/tex]
At the highest point of the trajectory the velocity must be zero.
y₂- y₁ = [tex]\frac{v_1^2}{2g}[/tex]
Let's calculate
y₂-y₁ = [tex]\frac{1.98^2}{2 \ 9.8}[/tex]
Δy = 0.2 m
2. Let's find the exit velocity of the water at the nozzle
A₁ = π r²
A₁ = π (0.22 10⁻²) ²
A₁ = 0.152 10⁻⁴ m / s
With the continuity and flow equation.
Q = A v
v₁ = [tex]\frac{Q}{A}[/tex]
v₁ = [tex]\frac{0.55 \ 10{-3} }{0.152 \ 10^{-4} }[/tex]
v₁ = 36.67 m / s
Using Bernoulli's equation, where the speed of the water at the highest point is zero.
y₂- y₁ = [tex]\frac{v^1^2}{g}[/tex]
Let's calculate.
Δy = [tex]\frac{36.67^2 }{2 \ 9.8 }[/tex]
Δy = 68.6m
In conclusion using the expressions of fluid mechanics we can find the results the maximum height that the water leaves through the two cases are:
1) The maximum height when the water leaves the hose is:
Δy = 0.20 m
2) The maximum height of the water when it leaves the nozzle is:
Δy = 68.6 m
Learn more here: https://brainly.com/question/4629227
?a wire is stretched 30% what is the percentage change in resistance
Answer:
The percentage change in resistance of the wire is 69%.
Explanation:
Resistance of a wire can be determined by,
R = (ρl) ÷ A
Where R is its resistance, l is the length of the wire, A its cross sectional area and ρ its resistivity.
When the wire is stretched, its length and area changes but its volume and resistivity remains constant.
[tex]l_{o}[/tex] = 1.3l, and [tex]A_{o}[/tex] = [tex]\frac{A}{1.3}[/tex]
So that;
[tex]R_{o}[/tex] = (ρ[tex]l_{o}[/tex]) ÷ [tex]A_{o}[/tex] = (ρ × 1.3l) ÷ ([tex]\frac{A}{1.3}[/tex])
= (1.3lρ) ÷ ([tex]\frac{A}{1.3}[/tex])
= [tex](1.3)^{2}[/tex] × [(ρl) ÷ A]
= 1.69R (∵ R = (ρl) ÷ A)
[tex]R_{o}[/tex] = 1.69R
Where [tex]R_{o}[/tex] is the new resistance, [tex]l_{o}[/tex] is the new length, and [tex]A_{o}[/tex] is the new area after stretching the wire.
The change in resistance of the wire = [tex]R_{o}[/tex] - R
= 1.69R - 1R
= 0.69R
The percentage change in resistance = [tex]\frac{0.69R}{R}[/tex] × 100
= 0.69 × 100
= 69%
The percentage change in resistance of the wire is 69%.
Each proton-proton cycle generates 26.7 MeV of energy. If 9.9 Watts are generated via by the proton-proton cycle, how many billion neutrinos are produced
Answer:
4.635 *10^12 Neutrinos
Explanation:
Here in this question, we are to determine the number of neutrinos in billions produced, given the power generated by the proton-proton cycle.
We proceed as follows;
In proton-proton cycle generates 26.7 MeV of energy and in this cycle two neutrinos are produced.
From the question, we are given that
Power P = 9.9 watts = 9.9 J/s
Watts is same as J/s
The number of proton-proton cycles required to generate E energy is N = E / E '
Where E ' = Energy generated in proton-proton cycle which is given as 26.7 Mev in the question
Converting Mev to J, we have
= 26.7 x1.6 x10 -13 J
To get the number N which is the number of proton-proton cycle required, we have;
N = 9.9 /(26.7 x1.6 x10^-13) = 2.32 * 10^12
Since we have two proton cycles( proton-proton), it automatically means 2 neutrinos will be produced.
Therefore number of neutrions produced = 2 x Number of proton-proton cycles = 2 * 2.32 * 10^12 = 4.635 * 10^12 neutrinos
Two waves are traveling in the same direction along a stretched string. The waves are 45.0° out of phase. Each wave has an amplitude of 7.00 cm. Find the amplitude of the resultant wave.
Answer:
The amplitude of the resultant wave is 12.93 cm.
Explanation:
The amplitude of resultant of two waves, y₁ and y₂, is given as;
Y = y₁ + y₂
Let y₁ = A sin(kx - ωt)
Since the wave is out phase by φ, y₂ is given as;
y₂ = A sin(kx - ωt + φ)
Y = y₁ + y₂ = 2A Cos (φ / 2)sin(kx - ωt + φ/2 )
Given;
phase difference, φ = 45°
Amplitude, A = 7.00 cm
Y = 2(7) Cos (45 /2) sin(kx - ωt + 22.5° )
Y = 12.93 cm
Therefore, the amplitude of the resultant wave is 12.93 cm.
which is example of radiation
Answer:
Ultraviolet light from the sun.
Explanation:
This is an example of radiation.
Answer:
X-Ray
Explanation:
x-Ray is an example of radiation.
At what frequency should a 200-turn, flat coil of cross sectional area of 300 cm2 be rotated in a uniform 30-mT magnetic field to have a maximum value of the induced emf equal to 8.0 V
Answer:
The frequency of the coil is 7.07 Hz
Explanation:
Given;
number of turns of the coil, 200 turn
cross sectional area of the coil, A = 300 cm² = 0.03 m²
magnitude of the magnetic field, B = 30 mT = 0.03 T
Maximum value of the induced emf, E = 8 V
The maximum induced emf in the coil is given by;
E = NBAω
Where;
ω is angular frequency = 2πf
E = NBA(2πf)
f = E / 2πNBA
f = (8) / (2π x 200 x 0.03 x 0.03)
f = 7.07 Hz
Therefore, the frequency of the coil is 7.07 Hz
If one could transport a simple pendulum of constant length from the Earth's surface to the Moon's, where acceleration due to gravity is one-sixth (1/6) that on the Earth, by what factor would be the pendulum frequency be changed
Answer:
The frequency will change by a factor of 0.4
Explanation:
T = 2(pi)*sqrt(L/g)
Since g(moon) = (1/6)g(earth), the period would change by sqrt[1/(1/6)] = sqrt(6) ~ 2.5 times longer on the moon. Since the period & frequency are inverses, the frequency would be 1/2.5 or 0.4 times shorter on the moon.
A circular loop of wire of area 25 cm2 lies in the plane of the paper. A decreasing magnetic field B is coming out of the paper. What is the direction of the induced current in the loop?
Answer:
counterclockwise
Explanation:
given data
area = 25 cm²
solution
We know that a changing magnetic field induces the current and induced emf is express as
[tex]\epsilon = -N \frac{d \phi }{dt}[/tex] ..................................1
and we will get here direction of the induced current in the loop that is express by the Lens law that state that the direction of induces current is such that the magnetic flux due to the induced current opposes the change in magnetic flux due to the change in magnetic field
so when magnetic field decrease and point coming out of the paper.
so induced current in the loop will be counterclockwise
5. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 ????. To what value must the tension be changed to raise the wave speed to 180 m/s?
Answer:
The tension on string when the speed was raised is 134.53 N
Explanation:
Given;
Tension on the string, T = 120 N
initial speed of the transverse wave, v₁ = 170 m/s
final speed of the transverse wave, v₂ = 180 m/s
The speed of the wave is given as;
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
where;
μ is mass per unit length
[tex]v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}[/tex]
The final tension T₂ will be calculated as;
[tex]T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N[/tex]
Therefore, the tension on string when the speed was raised is 134.53 N
Three resistors, each having a resistance, R, are connected in parallel to a 1.50 V battery. If the resistors dissipate a total power of 3.00 W, what is the value of R
Answer:
The value of resistance of each resistor, R is 2.25 Ω
Explanation:
Given;
voltage across the three resistor, V = 1.5 V
power dissipated by the resistors, P = 3.00 W
the resistance of each resistor, = R
The effective resistance of the three resistors is given by;
R(effective) = R/3
Apply ohms law to determine the current delivered by the source;
V = IR
I = V/R
I = 3V/R
Also, power is calculated as;
P = IV
P = (3V/R) x V
P = 3V²/R
R = 3V² / P
R = (3 x 1.5²) / 3
R = 2.25 Ω
Therefore, the value of resistance of each resistor, R is 2.25 Ω
Air flows through a converging-diverging nozzle/diffuser. A normal shock stands in the diverging section of the nozzle. Assuming isentropic flow, air as an ideal gas, and constant specific heats determine the state at several locations in the system. Solve using equations rather than with the tables.
Answer:
HELLO your question has some missing parts below are the missing parts
note: The specific heat ratio and gas constant for air are given as k=1.4 and R=0.287 kJ/kg-K respectively.
--Given Values--
Inlet Temperature: T1 (K) = 325
Inlet pressure: P1 (kPa) = 560
Inlet Velocity: V1 (m/s) = 97
Throat Area: A (cm^2) = 5.3
Pressure upstream of (before) shock: Px (kPa) = 207.2
Mach number at exit: M = 0.1
Answer: A) match number at inlet = 0.2683
B) stagnation temperature at inlet = 329.68 k
C) stagnation pressure = 588.73 kPa
D) ) Throat temperature = 274.73 k
Explanation:
Determining states at several locations in the system
A) match number at inlet
= V1 / C1 = 97/ 261.427 = 0.2683
C1 = sound velocity at inlet = [tex]\sqrt{K*R*T}[/tex] = [tex]\sqrt{1.4 *0.287*10^3}[/tex] = 361.427 m/s
v1 = inlet velocity = 97
B) stagnation temperature at inlet
= T1 + [tex]\frac{V1 ^2}{2Cp}[/tex] = 325 + [tex]\frac{97^2}{2 * 1.005*10^{-3} }[/tex]
stagnation temperature = 329.68 k
C) stagnation pressure
= [tex]p1 ( 1 + 0.2Ma^2 )^{3.5}[/tex]
Ma = match number at inlet = 0.2683
p1 = inlet pressure = 560
hence stagnation pressure = 588.73 kPa
D) Throat temperature
= [tex]\frac{Th}{T} = \frac{2}{k+1}[/tex]
Th = throat temperature
T = stagnation temp at inlet = 329.68 k
k = 1.4
make Th subject of the relation
Th = 329.68 * (2 / 2.4 ) = 274.73 k
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm^2, separated by a distance of 1.70 mm. A 25.0-V potential difference is applied to these plates. Calculate: a. the electric field between the plates b. the surface charge density c. the capacitance d. the charge on each plate.
Answer:
(a) 1.47 x 10⁴ V/m
(b) 1.28 x 10⁻⁷C/m²
(c) 3.9 x 10⁻¹²F
(d) 9.75 x 10⁻¹¹C
Explanation:
(a) For a parallel plate capacitor, the electric field E between the plates is given by;
E = V / d -----------(i)
Where;
V = potential difference applied to the plates
d = distance between these plates
From the question;
V = 25.0V
d = 1.70mm = 0.0017m
Substitute these values into equation (i) as follows;
E = 25.0 / 0.0017
E = 1.47 x 10⁴ V/m
(c) The capacitance of the capacitor is given by
C = Aε₀ / d
Where
C = capacitance
A = Area of the plates = 7.60cm² = 0.00076m²
ε₀ = permittivity of free space = 8.85 x 10⁻¹²F/m
d = 1.70mm = 0.0017m
C = 0.00076 x 8.85 x 10⁻¹² / 0.0017
C = 3.9 x 10⁻¹²F
(d) The charge, Q, on each plate can be found as follows;
Q = C V
Q = 3.9 x 10⁻¹² x 25.0
Q = 9.75 x 10⁻¹¹C
Now since we have found other quantities, it is way easier to find the surface charge density.
(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e
σ = Q / A
σ = 9.75 x 10⁻¹¹ / 0.00076
σ = 1.28 x 10⁻⁷C/m²
Can abnormality exist outside of a cultural context
Scientists today learn about the world by _____. 1. using untested hypotheses to revise theories 2. observing, measuring, testing, and explaining their ideas 3. formulating conclusions without testing them 4. changing scientific laws
Answer:
Option 2 (observing, measuring, testing, and explaining their ideas) is the correct choice.
Explanation:
A traditional perception of such a scientist is those of an individual who performs experiments in some kind of a white coat. The reality of the situation is, a researcher can indeed be described as an individual interested in the comprehensive as well as a recorded review of the occurrences occurring in nature but perhaps not severely constrained to physics, chemistry as well as biology alone.The other three choices have no relation to a particular task. So the option given here is just the right one.
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]
PLEASE HELP FAST Five-gram samples of brick and glass are at room temperature. Both samples receive equal amounts of energy due to heat flow. The specific heat capacity of brick is 0.22 cal/g°C and the specific heat capacity of glass is 0.22 cal/g°C. Which of the following statements is true? 1.The temperature of each sample will increase by the same amount. 2.The temperature of each sample will decrease by the same amount. 3.The brick will get hotter than the glass. 4.The glass will get hotter than the brick.
Answer:
1.The temperature of each sample will increase by the same amount
Explanation:
This is because, since their specific heat capacities are the same and we have the same mass of each substance, and the same amount of energy due to heat flow is supplied to both the glass and brick at room temperature, their temperatures would thereby increase by the same amount.
This is shown by the calculation below
Q = mcΔT
ΔT = Q/mc where ΔT = temperature change, Q = amount of heat, m = mass of substance and c = specific heat capacity of substance.
Since Q, m and c are the same for both substances, thus ΔT will be the same.
So, the temperature of each sample will increase by the same amount
2. The nuclear model of the atom held that
a. electrons were randomly spread through "a sphere of uniform positive
electrification."
b. matter was made of tiny electrically charged particles that were smaller than the
atom
C. matter was made of tiny, indivisible particles.
d. the atom had a dense, positively charged nucleus.
Answer:
the atom had a dense, positively charged nucleus.
Explanation:
Ernest Rutherford, based on the experiment carried out by two of his graduate students, established the authenticity of the nuclear model of the atom.
According to the nuclear model, an atom is made up of a dense positive core called the nucleus. Electrons are found to move round this nucleus in orbits. This is akin to the movement of the planets round the sun in the solar system.
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:
Q9 A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance. Find (a) the height of the tabletop above the floor; (b) the horizontal distance from the edge of the table to the point where the book strikes the floor; (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor.
Answer:
(a) 0.613 m
(b) 0.385 m
(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s
v = 3.68 m/s², θ = 72.6° below the horizontal
Explanation:
(a) Take down to be positive.
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 0.350 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²
Δy = 0.613 m
(b) Given in the x direction:
v₀ = 1.10 m/s
a = 0 m/s²
t = 0.350 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²
Δx = 0.385 m
(c) Find: vₓ and vᵧ
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (0.350 s) + 1.10 m/s
vₓ = 1.10 m/s
vᵧ = aᵧt + v₀ᵧ
vᵧ = (10 m/s²) (0.350 s) + 0 m/s
vᵧ = 3.50 m/s
The magnitude is:
v² = vₓ² + vᵧ²
v = 3.68 m/s²
The direction is:
θ = atan(vᵧ / vₓ)
θ = 72.6° below the horizontal
If a negatively charged rod is held near a neutral metal ball, the ball is attracted to the rod. This happens:_______
a. because of magnetic effects
b. because the ball tries to pull the rod's electrons over to it
c. because the rod polarizes the metal
d. because the rod and the ball have opposite charges
Answer:
c. because the rod polarizes the metal.
Explanation:
Bringing the negatively charged rod close to the neutral metal ball causes the neutral metal ball to be polarized with induced positive charge on it. The polarizing of the formally neutral metal ball is due to the negative charge on the metal rod (bodies induce a charge opposite of their own charge on a nearby neutral body). The ball and rod then attract themselves because bodies with opposite charges attract each other, unlike bodies with same charges that repel each other.
3. What color of laser light shines through a diffraction grating with a line density of 500 lines/mm if the third maxima from the central maxima (m=3) is at an angle of 45°?
Answer:
Wavelength is 471 nm
Explanation:
Given that,
Lines per unit length of diffraction grating is 500 lines/mm.
The third maxima from the central maxima (m=3) is at an angle of 45°
We need to find the color of laser light shines through a diffraction grating.
The condition for maxima is :
[tex]d\sin\theta=m\lambda[/tex]
d = 1/N, N = number of lines per mm
[tex]\lambda=\dfrac{1}{Nm}\sin\theta\\\\\lambda=\dfrac{10^{-3}}{500\times 3}\sin(45)\\\\\lambda=4.31\times 10^{-7}\\\\\text{or}\\\\\lambda=471\ nm[/tex]
The target variable is the speed of light v in the glass, which you can determine from the index of refraction n of the glass. Which equations will you use to find n and v?
Answer:
n= speed of light in vacuum/ speed of light in the other medium.
Explanation:
If light is moving from medium 1 into medium 2 where medium 1 is vacuum (approximated to mean air) and we are required to find the velocity of light; then we can confidently write;
n= speed of light in vacuum/ speed of light in the other medium.
Hence;
n= c/v
Where;
n= refractive index of the material
c= speed of light in vacuum
v = speed of light in another medium.
Note that the refractive index is the amount by which a transparent medium decreases the speed of light.
The ability of a water strider to move along the surface of water without breaking the surface is due to:
Answer:
The ability of a water strider to move along the surface of water without breaking the surface is due to:
SURFACE TENSION
Explanation:
this is because Water molecules are more attracted to each other than they are to other materials, so they generate a force to stay together called surface tension. Which allows the strider to move without breaking the surface
An electron moving at 3.94 103 m/s in a 1.23 T magnetic field experiences a magnetic force of 1.40 10-16 N. What angle does the velocity of the electron make with the magnetic field? There are two answers between 0° and 180°. (Enter your answers from smallest to largest.)
Answer:
10.4⁰ and 169.6⁰Explanation:
The force experienced by the moving electron in the magnetic field is expressed as F = qvBsinθ where;
q is the charge on the electron
v is the velocity of the electron
B is the magnetic field strength
θ is the angle that the velocity of the electron make with the magnetic field.
Given parameters
F = 1.40*10⁻¹⁶ N
q = 1.6*10⁻¹⁹C
v = 3.94*10³m/s
B = 1.23T
Required
Angle that the velocity of the electron make with the magnetic field
Substituting the given parameters into the formula:
1.40*10⁻¹⁶ = 1.6*10⁻¹⁹ * 3.94*10³ * 1.23 * sinθ
1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁹⁺³sinθ
1.40*10⁻¹⁶ = 7.75392 * 10⁻¹⁶sinθ
sinθ = 1.40*10⁻¹⁶/7.75392 * 10⁻¹⁶
sinθ = 1.40/7.75392
sinθ = 0.1806
θ = sin⁻¹0.1806
θ₁ = 10.4⁰
Since sinθ is positive in the 1st and 2nd quadrant, θ₂ = 180-θ₁
θ₂ = 180-10.4
θ₂ = 169.6⁰
Hence, the angle that the velocity of the electron make with the magnetic field are 10.4⁰ and 169.6⁰
Find the focal length of contact lenses that would allow a nearsighted person with a 130 cmcm far point to focus on the stars at night.
Answer:
130cmExplanation:
The lens equation is expressed as;
1/f = 1/u+1/v where;
f is the focal length of the lens
u is the object distance
v is the image distance
Since the near sighted person wants focus the starts at nigt, the stars at night are the images located that infinity. Hence the image distance v = ∞.
The object distance u = 130cm
Substituting the given parameters in the formula to get the focal length f
[tex]\frac{1}{f} = \frac{1}{\infty} + \frac{1}{130} \\\\As \ x \ tends \ to \ \infty, \, \frac{a}{x} \ tends \ to \ 0 \ where\ 'a' \ is \ a\ constant \\\\} \\\\[/tex]
[tex]\frac{1}{f} = 0+ \frac{1}{130}\\\\[/tex]
[tex]\frac{1}{f} =\frac{1}{130}\\cross\ multiply\\\\f = 130*1\\\\f = 130cm[/tex]
Hence the focal length of contact lenses that would allow a nearsighted person with a 130 cm far point to focus on the stars at night is 130cm
A steel ball attached to a spring moves in simple harmonic motion. The amplitude of the ball's motion is 11.0 cm, and the spring constant is 6.00 N/m. When the ball is halfway between its equilibrium position and its maximum displacement from equilibrium, its speed is 26.1 cm/s. (a) What is the mass of the ball (in kg)? kg (b) What is the period of oscillation (in s)? s (c) What is the maximum acceleration of the ball? (Enter the magnitude in m/s2.) m/s2
Answer:
a) m = 0.626 kg , b) T = 2.09 s , c) a = 1.0544 m / s²
Explanation:
In a spring mass system the equation of motion is
x = A cos (wt + Ф)
with w = √(k / m)
a) velocity is defined by
v = dx / dt
v = - A w sin (wt + Ф) (1)
give us that the speed is
v = 26.1 m / s
for the point
x = a / 2
the range of motion is a = 11.0 cm
x = 11.0 / 2
x = 5.5 cm
Let's find the time it takes to get to this distance
wt + Ф = cos⁻¹ (x / A)
wt + Ф = cos 0.5
wt + Ф = 0.877
In the exercise they do not indicate that the body started its movement with any speed, therefore we assume that for the maximum elongation the body was released, therefore the phase is zero f
Ф = 0
wt = 0.877
t = 0.877 / w
we substitute in equation 1
26.1 = -11.0 w sin (w 0.877 / w)
w = 26.1 / (11 sin 0.877))
w = 3.096 rad / s
from the angular velocity equation
w² = k / m
m = k / w²
m = 6 / 3,096²
m = 0.626 kg
b) angular velocity and frequency are related
w = 2π f
frequency and period are related
f = 1 / T
we substitute
w = 2π / T
T = 2π / w
T = 2π / 3,096
T = 2.09 s
c) maximum acceleration
the acceleration of defined by
a = dv / dt
a = - Aw² cos (wt)
the acceleration is maximum when the cosine is ±1
a = A w²
a = 11 3,096²
a = 105.44 cm / s²
we reduce to m / s
a = 1.0544 m / s²
What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 580 nm
Answer:
Explanation:
In case of soap film , light gets reflected from denser medium , hence interference takes place between two waves , one reflected from upper and second from lower surface . For destructive interference the condition is
2μt = nλ where μ is refractive index of water , t is thickness , λ is wavelength of light and n is an integer .
2 x 1.34 x t = 1 x 580
t = 216.42 nm .
Thickness must be 216.42 nm .
A metal sample of mass M requires a power input P to just remain molten. When the heater is turned off, the metal solidifies in a time T. The heat of fusion of this metal is
Answer:
L = Pt/M
Explanation:
Power, P= Q/t = mL/t
we know that, (Q=m×l)
Now ⇒l= Pt/M
Thus l= Pt/M
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that
Answer:
A) the moment of inertia of the system decreases and the angular speed increases.
Explanation:
The complete question is
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, It is true to say that
A) the moment of inertia of the system decreases and the angular speed increases.
B) the moment of inertia of the system decreases and the angular speed decreases.
C) the moment of inertia of the system decreases and the angular speed remains the same.
D) the moment of inertia of the system increases and the angular speed increases.
E) the moment of inertia of the system increases and the angular speed decreases
In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as
[tex]I_{1} w_{1} = I_{2} w_{2}[/tex] ....1
where [tex]I_{1}[/tex] and [tex]I_{2}[/tex] are the initial and final moment of inertia respectively.
and [tex]w_{1}[/tex] and [tex]w_{2}[/tex] are the initial and final angular speed respectively.
Also, we know that the moment of inertia of a rotating body is given as
[tex]I = mr^{2}[/tex] ....2
where [tex]m[/tex] is the mass of the rotating body,
and [tex]r[/tex] is the radius of the rotating body from its center.
We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.
From equation 1, we see that in order for the angular momentum to be conserved, the decrease from [tex]I_{1}[/tex] to [tex]I_{2}[/tex] will cause the angular speed of the system to increase from [tex]w_{1}[/tex] to [tex]w_{2}[/tex] .
From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.
