Answer:
a) 238U, 40K and 87Rb, b) 235U and to a lesser extent 40K , c) he 235U,
d) possibility is 14C , e)this period would be ideal for 14C , f) 14C should be used since it is the one with the least average life time, even though the measurements must be very careful
Explanation:
One of the applications of radioactive decay is the dating of different systems.
To do this, the quantity of radioactive material in a meter is determined and with the average life time, the time of the sample is found.
Let's write the half-life times of the given materials
87Rb T ½ = 4.75 1010 years
147Sm T ½ = 1.06 1011 years
235U = 7,038 108 years
238U = 4.47 109 years
40K = 1,248 109 years
14C = 5,568 103 years
we already have the half-life of the different elements given
a) meteors. As these decomposed in the formation of the solar system, their life time is around 3 109 to 5 109 years, so it is necessary to look for elements that have a life time of this order, among the candidates we have 238U, 40K and 87Rb if these elements were at the moment of the formation of these meteors, there must still be rations in them, instead elements 14C already completely adequate
b) rock. The formation period is 4.20-108 years, therefore one of the most promising elements is 235U and to a lesser extent 40K since it is more abundant in rocks. The other elements with higher life times have not decayed and therefore will not give a true value and the 14C is completely decayed
c) volcanic ash. Formation time 6107 years, the only element that has the possibility of having a count is the 235U, the others have a life time so long that they have not decayed and the 14C is complete, unbent
d) scarp of an earthquake formation time 5 101 years, The only one that has any possibility is 14C even when it has declined very little, all the others, you have time to long that has not decayed
e) INCA excavation. The time of this civilization is about 10000 to 500 years (104 to 5 102 years), we see that this period would be ideal for 14C since it has some period of cementation, the others have not decayed
f) Tree in Blepharitis. 14C should be used since it is the one with the least average life time, even though the measurements must be very careful because of a period of disintegration. We have such a long time that they have not decayed
Two objects attract each other with a gravitational force of magnitude 1.02 10-8 N when separated by 19.7 cm. If the total mass of the two objects is 5.14 kg, what is the mass of each
Answer:
The two masses are 3.39 Kg and 1.75 Kg
Explanation:
The gravitational force of attraction between two bodies is given by the formula;
F = Gm₁m₂/d²
where G is the gravitational force constant = 6.67 * 10⁻¹¹ Nm²Kg⁻²
m₁ = mass of first object; m₂ = mass of second object; d = distance of separation between the objects
Further calculations are provided in the attachment below
A wave with a frequency of 1200 Hz propagates along a wire that is under a tension of 800 N. Its wavelength is 39.1 cm. What will be the wavelength if the tension is decreased to 600 N and the frequency is kept constant
Answer:
The wavelength will be 33.9 cm
Explanation:
Given;
frequency of the wave, F = 1200 Hz
Tension on the wire, T = 800 N
wavelength, λ = 39.1 cm
[tex]F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}[/tex]
Where;
F is the frequency of the wave
T is tension on the string
μ is mass per unit length of the string
λ is wavelength
[tex]\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu = \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\[/tex]
when the tension is decreased to 600 N, that is T₂ = 600 N
[tex]T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2 = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2 = \sqrt{0.11466} \\\\\lambda _2 =0.339 \ m\\\\\lambda _2 =33.9 \ cm[/tex]
Therefore, the wavelength will be 33.9 cm
When a particular wire is vibrating with a frequency of 6.3 Hz, a transverse wave of wavelength 53.3 cm is produced. Determine the speed of wave pulses along the wire.
Answer:
335.79cm/s
Explanation:
When a transverse wave of wavelength λ is produced during the vibration of a wire, the frequency(f), and the speed(v) of the wave pulses are related to the wavelength as follows;
v = fλ ------------------(ii)
From the question;
f = 6.3Hz
λ = 53.3cm
Substitute these values into equation (i) as follows;
v = 6.3 x 53.3
v = 335.79cm/s
Therefore, the speed of the wave pulses along the wire is 335.79cm/s
A very long, solid cylinder with radius R has positive charge uniformly distributed throughout it, with charge per unit volume \rhorho.
(a) Derive the expression for the electric field inside the volume at a distance r from the axis of the cylinder in terms of the charge density \rhorho.
(b) What is the electric field at a point outside the volume in terms of the charge per unit length \lambdaλ in the cylinder?
(c) Compare the answers to parts (a) and (b) for r = R.
(d) Graph the electric-field magnitude as a function of r from r = 0 to r = 3R.
Answer:
the answers are provided in the attachments below
Explanation:
Gauss law state that the net electric field coming out of a closed surface is directly proportional to the charge enclosed inside the closed surface
Applying Gauss law to the long solid cylinder
A) E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]
B) E = 2K λ / r
C) Answers from parts a and b are the same
D) attached below
Applying Gauss's law which states that the net electric field in an enclosed surface is directly ∝ to the charge found in the enclosed surface.
A ) The expression for the electric field inside the volume at a distance r
Gauss law : E. A = [tex]\frac{q}{e_{0} }[/tex] ----- ( 1 )
where : A = surface area = 2πrL , q = p(πr²L)
back to equation ( 1 )
E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]
B) Electric field at point Outside the volume in terms of charge per unit length λ
Given that: linear charge density = area * volume charge density
λ = πR²P
from Gauss's law : E ( 2πrL) = [tex]\frac{q}{e_{0} }[/tex]
∴ E = [tex]\frac{\pi R^{2}P }{2e_{0}r\pi }[/tex] ----- ( 2 )
where : πR²P = λ
Back to equation ( 2 )
E = λ / 2e₀π*r where : k = 1 / 4πe₀
∴ The electric field ( E ) at point outside the volume in terms of charge per unit Length λ
E = 2K λ / r
C) Comparing answers A and B
Answers to part A and B are similar
Hence we can conclude that Applying Gauss law to the long solid cylinder
E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex], E = 2K λ / r also Answers from parts a and b are the same.
Learn more about Gauss's Law : https://brainly.com/question/15175106
The magnitude of the magnetic flux through the surface of a circular plate is 6.80 10-5 T · m2 when it is placed in a region of uniform magnetic field that is oriented at 43.0° to the vertical. The radius of the plate is 8.50 cm. Determine the strength of the magnetic field. mT A circular plate of radius r is lying flat. A field of arrows labeled vector B rising up and to the right pass through the plate.
Answer:
B = 4.1*10^-3 T = 4.1mT
Explanation:
In order to calculate the strength of the magnetic field, you use the following formula for the magnetic flux trough a surface:
[tex]\Phi_B=S\cdot B=SBcos\alpha[/tex] (1)
ФB: magnetic flux trough the circular surface = 6.80*10^-5 T.m^2
S: surface area of the circular plate = π.r^2
r: radius of the circular plate = 8.50cm = 0.085m
B: magnitude of the magnetic field = ?
α: angle between the direction of the magnetic field and the normal to the surface area of the circular plate = 43.0°
You solve the equation (1) for B, and replace the values of the other parameters:
[tex]B=\frac{\Phi_B}{Scos\alpha}=\frac{6.80*10^{-5}T.m^2}{(\pi (0.085m)^2)cos(43.0\°)}\\\\B=4.1*10^{-3}T=4.1mT[/tex]
The strength of the magntetic field is 4.1mT
An ice skater spinning with outstretched arms has an angular speed of 5.0 rad/s . She tucks in her arms, decreasing her moment of inertia by 11 % . By what factor does the skater's kinetic energy change? (Neglect any frictional effects.)
Answer:
K_{f} / K₀ =1.12
Explanation:
This problem must work using the conservation of angular momentum (L), so that the moment is conserved in the system all the forces must be internal and therefore the torque is internal and the moment is conserved.
Initial moment. With arms outstretched
L₀ = I₀ w₀
the wo value is 5.0 rad / s
final moment. After he shrugs his arms
[tex]L_{f}[/tex] = I_{f} w_{f}
indicate that the moment of inertia decreases by 11%
I_{f} = I₀ - 0.11 I₀ = 0.89 I₀
L_{f} = L₀
I_{f} w_{f} = I₀ w₀
w_{f} = I₀ /I_{f} w₀
let's calculate
w_{f} = I₀ / 0.89 I₀ 5.0
w_{f} = 5.62 rad / s
Having these values we can calculate the change in kinetic energy
[tex]K_{f}[/tex] / K₀ = ½ I_{f} w_{f}² (½ I₀ w₀²)
K_{f} / K₀ = 0.89 I₀ / I₀ (5.62 / 5)²
K_{f} / K₀ =1.12
g At some point the road makes a right turn with a radius of 117 m. If the posted speed limit along this part of the highway is 25.1 m/s, how much should Raquel bank the turn so that a vehicle traveling at the posted speed limit can make the turn without relying on the frictional force between the tires and the road
Answer:
Ф = 28.9°
Explanation:
given:
radius (r) = 117m
velocity (v) = 25.1 m/s
required: angle Ф
Ф = inv tan (v² / (r * g)) we know that g = 9.8
Ф = inv tan (25.1² / (117 * 9.8))
Ф = 28.9°
Monochromatic coherent light shines through a pair of slits. If the wavelength of the light is decreased, which of the following statements are true of the resulting interference pattern? (There could be more than one correct choice.)
a. The distance between the maxima decreases.
b. The distance between the minima decreases.
c. The distance between the maxima stays the same.
d. The distance between the minima increases.
e. The distance between the minima stays the same.
Answer:
he correct answers are a, b
Explanation:
In the two-slit interference phenomenon, the expression for interference is
d sin θ= m λ constructive interference
d sin θ = (m + ½) λ destructive interference
in general this phenomenon occurs for small angles, for which we can write
tanθ = y / L
tan te = sin tea / cos tea = sin tea
sin θ = y / La
un
derestimate the first two equations.
Let's do the calculation for constructive interference
d y / L = m λ
the distance between maximum clos is and
y = (me / d) λ
this is the position of each maximum, the distance between two consecutive maximums
y₂-y₁ = (L 2/d) λ - (L 1 / d) λ₁ y₂ -y₁ = L / d λ
examining this equation if the wavelength decreases the value of y also decreases
the same calculation for destructive interference
d y / L = (m + ½) κ
y = [(m + ½) L / d] λ
again when it decreases the decrease the distance
the correct answers are a, b
Use Coulomb’s law to derive the dimension for the permittivity of free space.
Answer:
Coulomb's law is:
[tex]F = \frac{1}{4*pi*e0} *(q1*q2)/r^2[/tex]
First, force has units of Newtons, the charges have units of Coulombs, and r, the distance, has units of meters, then, working only with the units we have:
N = (1/{e0})*C^2/m^2
then we have:
{e0} = C^2/(m^2*N)
And we know that N = kg*m/s^2
then the dimensions of e0 are:
{e0} = C^2*s^2/(m^3)
(current square per time square over cubed distance)
And knowing that a Faraday is:
F = C^2*S^2/m^2
The units of e0 are:
{e0} = F/m.
An asteroid that has an orbit with a semi-major axis of 4 AU will have an orbital period of about ______ years.
Answer:
16 years.
Explanation:
Using Kepler's third Law.
P2=D^3
P=√d^3
Where P is the orbital period and d is the distance from the sun.
From the question the semi major axis of the asteroid is 4 AU= distance. The distance is always express in astronomical units.
P=?
P= √4^3
P= √256
P= 16 years.
Orbital period is 16 years.
A force of 44 N will stretch a rubber band 88 cm (0.080.08 m). Assuming that Hooke's law applies, how far will aa 11-N force stretch the rubber band? How much work does it take to stretch the rubber band this far?
Answer:
The rubber band will be stretched 0.02 m.
The work done in stretching is 0.11 J.
Explanation:
Force 1 = 44 N
extension of rubber band = 0.080 m
Force 2 = 11 N
extension = ?
According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.
F = ke
where k = constant of elasticity
e = extension of the material
F = force applied.
For the first case,
44 = 0.080K
K = 44/0.080 = 550 N/m
For the second situation involving the same rubber band
Force = 11 N
e = 550 N/m
11 = 550e
extension e = 11/550 = 0.02 m
The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.
potential energy stored = [tex]\frac{1}{2}ke^{2}[/tex]
==> [tex]\frac{1}{2}* 550* 0.02^{2}[/tex] = 0.11 J
A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to the object to change the period to 2.00 s
Answer:
389 kg
Explanation:
The computation of mass is shown below:-
[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]
Where K indicates spring constant
m indicates mass
For the new time period
[tex]T^' = 2\pi \sqrt{\frac{m'}{k} }[/tex]
Now, we will take 2 ratios of the time period
[tex]\frac{T}{T'} = \sqrt{\frac{m}{m'} }[/tex]
[tex]\frac{1.50}{2.00} = \sqrt{\frac{0.500}{m'} }[/tex]
[tex]0.5625 = \sqrt{\frac{0.500}{m'} }[/tex]
[tex]m' = \frac{0.500}{0.5625}[/tex]
= 0.889 kg
Since mass to be sum that is
= 0.889 - 0.500
0.389 kg
or
= 389 kg
Therefore for computing the mass we simply applied the above formula.
The mass added to the object to change the period to 2.00 s is 0.389 kg and this can be determined by using the formula of the time period.
Given :
A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s.
The formula of the time period is given by:
[tex]\rm T = 2\pi\sqrt{\dfrac{m}{K}}[/tex] ---- (1)
where m is the mass and K is the spring constant.
The new time period is given by:
[tex]\rm T'=2\pi\sqrt{\dfrac{m'}{K}}[/tex] ---- (2)
where m' is the total mass after the addition and K is the spring constant.
Now, divide equation (1) by equation (2).
[tex]\rm \dfrac{T}{T'}=\sqrt{\dfrac{m}{m'}}[/tex]
Now, substitute the known terms in the above expression.
[tex]\rm \dfrac{1.50}{2}=\sqrt{\dfrac{0.5}{m'}}[/tex]
Simplify the above expression in order to determine the value of m'.
[tex]\rm m'=\dfrac{0.5}{0.5625}[/tex]
m' = 0.889 Kg
Now, the mass added to the object to change the period to 2.00 s is given by:
m" = 0.889 - 0.500
m" = 0.389 Kg
For more information, refer to the link given below:
https://brainly.com/question/2144584
the density of gold is 19 300kg/m^3. what is the mass of gold cube with the length 0.2015m?
Answer:
The mass is [tex]157.87m^3[/tex]Explanation:
Given data
length of cube= 0.2015 m
density = 19300 kg/m^3.
But the volume of cube is given as [tex]l*l*l= l^3[/tex]
[tex]volume -of- cube= 0.2015*0.2015*0.2015= 0.00818 m^3[/tex]
The density is expressed as = mass/volume
[tex]mass=19300*0.00818= 157.87m^3[/tex]
how do a proton and neutron compare?
Answer:
c.they have opposite charges.
Explanation:
because the protons have a positive charge and the neutrons have no charge.
Two beams of coherent light start out at the same point in phase and travel different paths to arrive at point P. If the maximum destructive interference is to occur at point P, the two beams must travel paths that differ by
Answer:
the two beams must travel paths that differ by one-half of a wavelength.
When looking at the chemical symbol, the charge of the ion is displayed as the
-superscript
-subscript
-coefficient
-product
Answer:
superscript
Explanation:
When looking at the chemical symbol, the charge of the ion is displayed as the Superscript. This is because the charge of ions is usually written up on the chemical symbol while the atom/molecule is usually written down the chemical symbol. The superscript refers to what is written up on the formula while the subscript is written down on the formula.
An example is H2O . The 2 present represents two molecule of oxygen and its written as the subscript while Fe2+ in which the 2+ is written up is known as the superscript.
Answer:
superscript
Explanation:
An ice skater is in a fast spin with her arms held tightly to her body. When she extends her arms, which of the following statements in NOT true?
A. Het total angular momentum has decreased
B. She increases her moment of inertia
C. She decreases her angular speed
D. Her moment of inertia changes
Answer:
A. Her total angular momentum has decreased
Explanation:
Total angular momentum is the product of her moment of inertia and angular velocity. In this scenario it doesn’t decrease but rather remains constant as the movement of the arms doesn’t have any effect on the total angular momentum.
The movement of the arm under certain conditions however has varying effects and changes on parameters such as the moment of inertia and the angular speed.
A 3-liter container has a pressure of 4 atmospheres. The container is sent underground, with resulting compression into 2 L. Applying Boyle's Law, what will the new pressure be? choices: 2.3 atm 8 atm 6 atm 1.5 atm
Answer:
6 atm
Explanation:
PV = PV
(4 atm) (3 L) = P (2 L)
P = 6 atm
How far apart (in mm) must two point charges of 90.0 nC (typical of static electricity) be to have a force of 3.80 N between them
Answer:
The distance between the two charges is =4.4mm
What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3
Answer:
3) Transmitted intensity of light if unpolarized light passes through a single polarizing filter = 40 W/m²
- Transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described = 7.5 W/m²
Explanation:
Complete Question
3) What is the transmitted intensity of light if unpolarized light passes through a single polarizing filter and the initial intensity is 80 W/m²?
- What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3 (the setup)? Show all work in your answer.
The image of this setup attached to this question as obtained from online is attached to this solution.
Solution
3) When unpolarized light passes through a single polarizer, the intensity of the light is cut in half.
Hence, if the initial intensity of unpolarized light is I₀ = 80 W/m²
The intensity of the light rays thay pass through the first single polarizer = I₁ = (I₀/2) = (80/2) = 40 W/m²
- According to Malus' law, the intensity of transmitted light through a polarizer is related to the intensity of the incident light and the angle at which the polarizer is placed with respect to the major axis of the polarizer before the current polarizer of concern.
I₂ = I₁ cos² θ
where
I₂ = intensity of light that passes through the second polarizer = ?
I₁ = Intensity of light from the first polarizer that is incident upon the second polarizer = 40 W/m²
θ = angle between the major axis of the first and second polarizer = 30°
I₂ = 40 (cos² 30°) = 40 (0.8660)² = 30 W/m²
In the same vein, the intensity of light that passes through the third/additional polarizer is related to the intensity of light that passes through the second polarizer and is incident upon this third/additional polarizer through
I₃ = I₂ cos² θ
I₃ = intensity of light that passes through the third/additional polarizer = ?
I₂ = Intensity of light from the second polarizer that is incident upon the third/additional polarizer = 30 W/m²
θ = angle between the major axis of the second and third/additional polarizer = 60° (although, it is 90° with respect to the first polarizer, it is the angle it makes with the major axis of the second polarizer, 60°, that matters)
I₃ = 30 (cos² 60°) = 30 (0.5)² = 7.5 W/m²
Hope this Helps!!!
An 88.0 kg spacewalking astronaut pushes off a 645 kg satellite, exerting a 110 N force for the 0.450 s it takes him to straighten his arms. How far apart are the astronaut and the satellite after 1.40 min?
Answer:
The astronaut and the satellite are 53.718 m apart.
Explanation:
Given;
mass of spacewalking astronaut, = 88 kg
mass of satellite, = 645 kg
force exerts by the satellite, F = 110N
time for this action, t = 0.45 s
Determine the acceleration of the satellite after the push
F = ma
a = F / m
a = 110 / 645
a = 0.171 m/s²
Determine the final velocity of the satellite;
v = u + at
where;
u is the initial velocity of the satellite = 0
v = 0 + 0.171 x 0.45
v = 0.077 m/s
Determine the displacement of the satellite after 1.4 m
d₁ = vt
d₁ = 0.077 x (1.4 x 60)
d₁ = 6.468 m
According to Newton's third law of motion, action and reaction are equal and opposite;
Determine the backward acceleration of the astronaut after the push;
F = ma
a = F / m
a = 110 / 88
a = 1.25 m/s²
Determine the final velocity of the astronaut
v = u + at
The initial velocity of the astronaut = 0
v = 1.25 x 0.45
v = 0.5625 m/s
Determine the displacement of the astronaut after 1.4 min
d₂ = vt
d₂ = 0.5625 x (1.4 x 60)
d₂ = 47.25 m
Finally, determine the total separation between the astronaut and the satellite;
total separation = d₁ + d₂
total separation = 6.468 m + 47.25 m
total separation = 53.718 m
Therefore, the astronaut and the satellite are 53.718 m apart.
mention two similarities of citizen and aliens
Answer:
The main points of difference between a citizen and alien are: (a) A citizen is a permanent resident of a state, while an alien is a temporary resident, who comes for a specific duration of time as a tourist or on diplomatic assignment. ... Aliens do not possess such rights in the state where they reside temporarily
Explanation:
An isolated capacitor with capacitance C = 1 µF has a charge Q = 45 µC on its plates.a) What is the energy stored in the capacitor?Now a conductor is inserted into the capacitor. The thickness of the conductor is 1/3 the distance between the plates of the capacitor and is centered inbetween the plates of the capacitor.b) What is the charge on the plates of the capacitor?c) What is the capacitance of the capacitor with the conductor in place?d) What is the energy stored in the capacitor with the conductor in place?
Answer:
a) Energy stored in the capacitor, [tex]E = 1.0125 *10^{-3} J[/tex]
b) Q = 45 µC
c) C' = 1.5 μF
d) [tex]E = 6.75 *10^{-4} J[/tex]
Explanation:
Capacitance, C = 1 µF
Charge on the plates, Q = 45 µC
a) Energy stored in the capacitor is given by the formula:
[tex]E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{2}\\\\E = 1012.5 *10^{-6}\\\\E = 1.0125 *10^{-3} J[/tex]
b) The charge on the plates of the capacitor will not change
It will still remains, Q = 45 µC
c) Electric field is non zero over (1-1/3) = 2/3 of d
From the relation V = Ed,
The voltage has changed by a factor of 2/3
Since the capacitance is given as C = Q/V
The new capacitance with the conductor in place, C' = (3/2) C
C' = (3/2) * 1μF
C' = 1.5 μF
d) Energy stored in the capacitor with the conductor in place
[tex]E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1.5* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{3}\\\\E = 675 *10^{-6}\\\\E = 6.75 *10^{-4} J[/tex]
Please Help!!!! I WILL GIVE BRAINLIEST!!!!!!!!!!!!!
Upon using Thomas Young’s double-slit experiment to obtain measurements, the following data were obtained. Use these data to determine the wavelength of light being used to create the interference pattern. Do this using three different methods.
The angle to the eighth maximum is 1.12°.
The distance from the slits to the screen is 302.0 cm.
The distance from the central maximum to the fifth minimum is 3.33 cm.
The distance between the slits is 0.000250 m.
The 3 equations I used were 1). d sin θ_m =(m)λ 2). delta x =λL/d and 3.) d(x_n)/L=(n-1/2)λ
but all my answers are different.
DID I DO SOMETHING WRONG!!!!!!!
Given info
d = 0.000250 meters = distance between slits
L = 302 cm = 0.302 meters = distance from slits to screen
[tex]\theta_8 = 1.12^{\circ}[/tex] = angle to 8th max (note how m = 8 since we're comparing this to the form [tex]\theta_m[/tex])
[tex]x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}[/tex] (n = 5 as we're dealing with the 5th minimum )
---------------
Method 1
[tex]d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}[/tex]
Make sure your calculator is in degree mode.
-----------------
Method 2
[tex]\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\[/tex]
-----------------
Method 3
[tex]\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\[/tex]
There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.
A 30 L electrical radiator containing heating oil is placed in a 50 m3room. Both the roomand the oil in the radiator are initially at 10◦C. The radiator with a rating of 1.8 kW is nowturned on. At the same time, heat is lost from the room at an average rate of 0.35 kJ/s.After some time, the average temperature is measured to be 20◦C for the air in the room,and 50◦C for the oil in the radiator. Taking the density and the specific heat of the oil to be950 kg/m3and 2.2 kJ/kg◦C, respectively, determine how long the heater is kept on. Assumethe room is well sealed so that there are no air leaks.
Answer:
Explanation:
Heat absorbed by oil
= mass x specific heat x rise in temperature
= 30 x 10⁻³ x 950 x 2.2 x 10³ x ( 50-10 )
= 25.08 x 10⁵ J
Heat absorbed by air
= 50 x 1.2 x 1.0054 x 10³ x ( 20-10 )
= 6.03 x 10⁵ J
Total heat absorbed = 31.11 x 10⁵ J
If time required = t
heat lost from room
= .35 x 10³ t
Total heat generated in time t
= 1.8 x 10³ t
Heat generated = heat used
1.8 x 10³ t = .35 x 10³ t + 31.11 x 10⁵
1.45 x 10³ t = 31.11 x 10⁵
t = 31.11 x 10⁵ / 1.45 x 10³
t = 2145.5 s
The interference of two sound waves of similar amplitude but slightly different frequencies produces a loud-soft-loud oscillation we call __________.
a. the Doppler effect
b. vibrato
c. constructive and destructive interference
d. beats
Answer:
the correct answer is d Beats
Explanation:
when two sound waves interfere time has different frequencies, the result is the sum of the waves is
y = 2A cos 2π (f₁-f₂)/2 cos 2π (f₁ + f₂)/2
where in this expression the first part represents the envelope and the second part represents the pulse or beatings of the wave.
When examining the correct answer is d Beats
Two space ships collide in deep space. Spaceship P, the projectile, has a mass of 4M,
while the target spaceship T has a mass of M. Spaceship T is initially at rest and the
collision is elastic. If the final velocity of Tis 8.1 m/s, what was the initial velocity of
P?
Answer:
The initial velocity of spaceship P was u₁ = 5.06 m/s
Explanation:
In an elastic collision between two bodies the expression for the final velocity of the second body is given as follows:
[tex]V_{2} = \frac{(m_{2}-m_{1}) }{(m_{1}+m_{2})}u_{2} + \frac{2m_{1} }{(m_{1}+m_{2})}u_{1}[/tex]
Here, subscript 1 is used for spaceship P and subscript 2 is used for spaceship T. In this equation:
V₂ = Final Speed of Spaceship T = 8.1 m/s
m₁ = mass of spaceship P = 4 M
m₂ = mass of spaceship T = M
u₁ = Initial Speed of Spaceship P = ?
u₂ = Initial Speed of Spaceship T = 0 m/s
Using these values in the given equation, we get:
[tex]8.1 m/s = \frac{M-4M }{4M+M}(0 m/s) + \frac{2(4M) }{4M+M}u_{1}[/tex]
8.1 m/s = (8 M/5 M)u₁
u₁ = (5/8)(8.1 m/s)
u₁ = 5.06 m/s
A dipole moment is placed in a uniform electric field oriented along an unknown direction. The maximum torque applied to the dipole is equal to 0.1 N.m. When the dipole reaches equilibrium its potential energy is equal to -0.2 J. What was the initial angle between the direction of the dipole moment and the direction of the electric field?
Answer:
θ = 180
Explanation:
When an electric dipole is placed in an electric field, there is a torque due to the electric force
τ = p x E
by rotating the dipole there is a change in potential energy
ΔU = ∫ τ dθ
ΔU = p E (cos θ₂ - cos θ₁)
when the dipole starts from an angle to the equilibrium position for θ = 0
ΔU = pE (cos θ - cos 0)
cos θ = 1 + DU / pE)
let's apply this expression to our case, the change in potential energy is ΔU = -0.2J
let's calculate
cos θ = 1 -0.2 / 0.1
cos θ = -1
θ = 180
1. The uniform purely axial magnetic induction required by the experiment in a volume large enough to accommodate the Lorentz Tube is produced by the Helmholtz Coils. What is the magnetic induction due to a coil current 1.5 Ampere
Complete Question
The uniform purely axial magnetic induction required by the experiment in a volume large enough to accommodate the Lorentz Tube is produced by the Helmholtz Coils. What is the magnetic induction due to a coil current 1.5 Ampere? Convert the result in the still popular non-SI unit Gauss (1 Tesla = 10^4 Gauss).
B = N*mue*I/(2*r)
# of loops = 140
radius of the coil = 0.14m
Answer:
The magnetic induction is [tex]B = 2.639 \ Gauss[/tex]
Explanation:
From the question we are told that
The coil current is [tex]I = 1.5 \ A[/tex]
The number of loops is [tex]N = 140[/tex]
The magnetic field due to the current is mathematically represented as
[tex]B = \mu_o * N * I[/tex]
[tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting value
[tex]B = 4\pi * 10^{-7} * 140 * 1.5[/tex]
[tex]B = 2.639*19^{-4} \ T[/tex]
From question
(1 Tesla = [tex]10^4 \ Gauss[/tex]).
=> [tex]B = 2.693 *10^{-4} *10^4 = 2.63 \ Gauss[/tex]
=> [tex]B = 2.639 \ Gauss[/tex]
A 150m race is run on a 300m circular track of circumference. Runners start running from the north and turn west until reaching the south. What is the magnitude of the displacement made by the runners?
Answer:
95.5 m
Explanation:
The displacement is the position of the ending point relative to the starting point.
In this case, the magnitude of the displacement is the diameter of the circular track.
d = 300 m / π
d ≈ 95.5 m