Answer:
9.43*10^3 year
Explanation:
For this question, we ought to remember, or know that the half life of carbon 14 is 5730, and that would be vital in completing the calculation
To start with, we use the formula
t(half) = In 2/k,
if we make k the subject of formula, we have
k = in 2/t(half), now we substitute for the values
k = in 2 / 5730
k = 1.21*10^-4 yr^-1
In(A/A•) = -kt, on rearranging, we find out that
t = -1/k * In(A/A•)
The next step is to substitite the values for each into the equation, giving us
t = -1/1.21*10^-4 * In(5.4/15.3)
t = -1/1.21*10^-4 * -1.1041
t = 0.943*10^4 year
The mass (M) of a piece of metal is directly proportional to its volume (V), where the proportionality constant is the density (D) of the metal. (1) Write an equation that represents this direct proportion, in which D is the proportionality constant. The density of lead metal is 11.3 g/cm3. (2) What is the mass of a piece of lead metal that has a volume of 17.3 cm3
Answer:
1) M = 11.3V2) 195.49 gramsExplanation:
1) If the mass (M) of a piece of metal is directly proportional to its volume (V), where the proportionality constant is the density (D) of the metal, this is expressed mathematically as shown;
M ∝ V
M = kV
For every proportionality sign, there will always be a proportionality constant 'k'
Since the proportionality constant is the density (D) of the metal, the equation will become;
M = DV
Given the density to be 11.3 g/cm3, the equation will become;
M = 11.3V
Hence, the equation that represents this direct proportion, in which D is the proportionality constant with metal density of 11.3g/cm³ is M = 11.3V
2) If the volume of the metal is 17.3cm³, on substituting this values into the equation in (1) to get the mass of the metal, we will have;
M = 11.3V
M = 11.3 * 17.3
M = 195.49 grams
Hence, the mass of a piece of lead metal that has a volume of 17.3 cm³ is 195.49 grams.
Rod cells in the retina of the eye detect light using a photopigment called rhodopsin. 1.8 eV is the lowest photon energy that can trigger a response in rhodopsin. Part A What is the maximum wavelength of electromagnetic radiation that can cause a transition
Answer:
The maximum wavelength of the e-m wave is 6.9 x 10^-7 m
Explanation:
Energy required to trigger a response = 1.8 eV
we convert to energy in Joules.
1 eV = 1.602 x 10^-19 J
1.8 eV = [tex]x[/tex] J
[tex]x[/tex] = 1.8 x 1.602 x 10^-19 = 2.88 x 10^-19 J
The energy of an electromagnetic wave is gotten as
E = hf
where
h is the Planck's constant = 6.63 x 10^-34 J-s
and f is the frequency of the wave.
substituting values, we have
2.88 x 10^-19 = 6.63 x 10^-34 x f
f = (2.88 x 10^-19)/(6.63 x 10^-34)
f = 4.34 x 10^14 Hz
We know that the frequency of an e-m wave is given as
f = c/λ
where
c is the speed of light = 3 x 10^8 m/s
λ is the wavelength of the e-m wave
From this we can say that
λ = c/f
λ = (3 x 10^8)/(4.34 x 10^14)
λ = 6.9 x 10^-7 m
In the lab, you shoot an electron towards the south. As it moves through a magnetic field, you observe the electron curving upward toward the roof of the lab. You deduce that the magnetic field must be pointing:_______.
a. to the west.
b. upward.
c. to the north.
d. to the east.
e. downward.
Answer:
a. to the west.
Explanation:
An electron in a magnetic field always experience a force that tends to change its direction of motion through the magnetic field. According to Lorentz left hand rule (which is the opposite of Lorentz right hand rule for a positive charge), the left hand is used to represent the motion of an electron in a magnetic field. Hold out the left hand with the fingers held out parallel to the palm, and the thumb held at right angle to the other fingers. If the thumb represents the motion of the electron though the field, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the particle.
In this case, if we point the thumb (which shows the direction we shot the electron) to the south (towards your body), with the palm (shows the direction of the force) facing up to the roof, then the fingers (the direction of the field) will point west.
Question 2.
In the US, lengths are often measured in inches, feet, yards and miles. Let's do
some conversions. The definition of the inch is: 1 inch = 25.4 mm, exactly. A foot is
12 inches and a mile is 5280 ft, exactly. A centimetre is exactly 0.01 m or 10 mm.
Sammy is 5 feet and 5.3 inches tall.
a). What is Sammy's height in Inches? (answer to 3 significant figures)
(3)
b). What is Sammy's height in Feet? (answer to 3 significant figures)
what is Sammy's hight in feet according to this statement
Explanation:
1 inch = 25.4 mm
1 foot = 12 inches
1 mile = 5260 feet
1 cm = 0.01 m or 10 mm
Now Sammy's height is 5 feet and 5.3 inches.
(a) We need to find Sammy's height in inches.
Since, 1 foot = 12 inches
5 feet = 5 × 12 inches = 60 inches
Now, 5 feet and 5.3 inches = 60 inches + 5.3 inches = 65.3 inches
Sammy's height is 65.3 inches.
(b) We need to find Sammy's height in feet.
Since, 1 foot = 12 inches
[tex]1\ \text{inch}=\dfrac{1}{12}\ \text{feet}[/tex]
So,
[tex]5.3\ \text{inch}=\dfrac{5.3}{12}\ \text{feet}=0.4416\ \text{feet}[/tex]
5 feet and 5.3 inches = 5 feet + 0.4416 feet = 5.44 feet
Sammy's height is 5.44 feet.
NASA is doing research on the concept of solar sailing. A solar sailing craft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion.
A) Should the sail be absorptive or reflective? Why?
B)The total power output of the sun is 3.90 × 1026 W . How large a sail is necessary to propel a 1.06 × 104 kg spacecraft against the gravitational force of the sun?
Answer:
A = 6.8 km²
Explanation:
A) The sail should be reflective. This is so that, it can produce the maximum radiation pressure.
B) let's begin with the formula used to calculate the average solar sail in orbit around the sun. Thus;
F_rad = 2IA/c
I is given by the formula;
I = P/(4πr²)
Thus;
F_rad = (2A/c) × (P/(4πr²)) = PA/2cπr²
Where;
A is the area of the sail
r is the distance of the sail from the sun
c is the speed of light = 3 × 10^(8) m/s
P is total power output of the sun = 3.90 × 10^(26) W
Now,F_rad = F_g
Where F_g is gravitational force.
Thus;
PA/2cπr² = G•m•M_sun/r²
r² will cancel out to givw;
PA/2cπ = G•m•M_sun
Making A the subject, we have;
A = (2•c•π•G•m•M_sun)/P
Now, m = 1.06 × 10⁴ kg and M_sun has a standard value of 1.99 × 10^(30) kg
G is gravitational constant and has a value of 6.67 × 10^(-11) Nm²/kg²
Thus;
A = (2 × 3 × 10^(8) × π × 6.67 × 10^(-11) × 1.06 × 10^(4) × 1.99 × 10^(30))/(3.90 × 10^(26))
A = 6.8 × 10^(6) m² = 6.8 km²
The charger for your electronic devices is a transformer. Suppose a 60 Hz outlet voltage of 120 V needs to be reduced to a device voltage of 3.0 V. The side of the transformer attached to the electronic device has 45 turns of wire.
How many turns are on the side that plugs into the outlet?
Answer:
N₁ = 1800 turns
So, the side of the transformer that plugs into the outlet has 1800 turns.
Explanation:
The transformer turns ratio is given by the following equation:
V₁/V₂ = N₁/N₂
where,
V₁ = Voltage of outlet = 120 V
V₂ = Device Voltage = 3 V
N₁ = No. of turns on outlet side = ?
N₂ = No. of turns on side of device = 45
Therefore,
120 V/3 V = N₁/45
N₁ = (40)(45)
N₁ = 1800 turns
So, the side of the transformer that plugs into the outlet has 1800 turns.
hi guys!!! i have no more points, can someone nice guess all of these for me? :)
1.What happens to the ocean water before the precipitation part of the water cycle
2.During which stage of the water cycle does water from the ocean form clouds?
3.what is a runoff??
4.Which statement about oceans is incorrect? A.Evaporation occurs when water is warmed by the sun. B.Most evaporation and precipitation occur over the ocean. C.97 percent of Earth's water is fresh water from the ocean. D.Water leaves the ocean by the process of evaporation
5.How does most ocean water return to the ocean in the water cycle
tysm to u who answers :)
1. The ocean water collects back in the ocean.
2. Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds.
3. an excessive amount of water flowing from downslope along earths surface
4. A.Evaporation occurs when water is warmed by the sun.
5. The water returns into the ocean by the water cycle . It evaporates , then it condensates , then it participates ( Rains ) and then goes back into the ocean.
Hope this answer correct ✌️
If a convex lens were made out of very thin clear plastic filled with air, and were then placed underwater where n = 1.33 and where the lens would have an effective index of refraction n = 1, the lens would act in the same way
a. as a flat refracting surface between water and air as seen from the water side.
b. as a concave mirror in air.
c. as a concave lens in air.
d. as the glasses worn by a farsighted person.
e. as a convex lens in air.
Answer:
D. A convex lens in air
Explanation:
This is because the air tight plastic under water will reflect light rays in the same manner as a convex lens
A 137 kg horizontal platform is a uniform disk of radius 1.53 m and can rotate about the vertical axis through its center. A 68.7 kg person stands on the platform at a distance of 1.19 m from the center, and a 25.9 kg dog sits on the platform near the person 1.45 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.
Answer:
The moment of inertia is [tex]I= 312.09 \ kg \cdot m^2[/tex]
Explanation:
From the question we are told that
The mass of the platform is m = 137 kg
The radius is r = 1.53 m
The mass of the person is [tex]m_p = 68.7 \ kg[/tex]
The distance of the person from the center is [tex]d_c =1.19 \ m[/tex]
The mass of the dog is [tex]m_d = 25.9 \ kg[/tex]
The distance of the dog from the person [tex]d_d = 1.45 \ m[/tex]
Generally the moment of inertia of the system is mathematically represented as
[tex]I = I_1 + I_2 + I_3[/tex]
Where [tex]I_1[/tex] is the moment of inertia of the platform which mathematically represented as
[tex]I_1 = \frac{m * r^2}{2}[/tex]
substituting values
[tex]I_1 = \frac{ 137 * (1.53)^2}{2}[/tex]
[tex]I_1 = 160.35 \ kg\cdot m^2[/tex]
Also [tex]I_2[/tex] is the moment of inertia of the person about the axis which is mathematically represented as
[tex]I_2 = m_p * d_c^2[/tex]
substituting values
[tex]I_2 = 68.7 * 1.19^2[/tex]
[tex]I_2 = 97.29 \ kg \cdot m^2[/tex]
Also [tex]I_3[/tex] is the moment of inertia of the dog about the axis which is mathematically represented as
[tex]I_3 = m_d * d_d^2[/tex]
substituting values
[tex]I_3 = 25.9 * 1.45^2[/tex]
[tex]I_3 = 54.45 \ kg \cdot m^2[/tex]
Thus
[tex]I= 160.35 + 97.29 + 54.45[/tex]
[tex]I= 312.09 \ kg \cdot m^2[/tex]
In the lab , you have an electric field with a strength of 1,860 N/C. If the force on a particle with an unknown charge is 0.02796 N, what is the value of the charge on this particle.
Answer:
The charge is [tex]q = 1.50 *10^{-5} \ C[/tex]
Explanation:
From the question we are told that
The electric field strength is [tex]E = 1860 \ N/C[/tex]
The force is [tex]F = 0.02796 \ N[/tex]
Generally the charge on this particle is mathematically represented as
[tex]q = \frac{F}{E}[/tex]
=> [tex]q = \frac{0.02796}{ 1860}[/tex]
=> [tex]q = 1.50 *10^{-5} \ C[/tex]
An electric heater draws 13 amperes of current when connected to 120 volts. If the price of electricity is $0.10/kWh, what would be the approximate cost of running the heater for 8 hours?
(A) $0.19
(B) $0.29
(C) $0.75
(D) $1.25
(E) $1.55
Answer:
C $0.75 my friend I wish it is right answer
Two 1.0 nF capacitors are connected in series to a 1.5 V battery. Calculate the total energy stored by the capacitors.
Answer:
1.125×10⁻⁹ J
Explanation:
Applying,
E = 1/2CV²................... Equation 1
Where E = Energy stored in the capacitor, C = capacitance of the capacitor, V = Voltage of the battery.
Given; C = 1.0 nF, = 1.0×10⁻⁹ F, V = 1.5 V
Substitute into equation 1
E = 1/2(1.0×10⁻⁹×1.5²)
E = 1.125×10⁻⁹ J
Hence the energy stored by the capacitor is 1.125×10⁻⁹ J
A radiation worker is subject to a dose of 200 mrad/h of maximum QF neutrons for one 40 h work week. How many times the yearly allowable effective dose did she receive?
Answer:
16 times.
Explanation:
The rate of the radiation dose is , R = 200 ×10^{-3} rad/hr
Time consumed, t = 40 hr
The magnitude of Q.F for the neutrons, Q.F = 2
Thus the effective radiation dose is:
[tex]R_{Eff} = Rt(Q.F) \\= 200 \times 10^{-3} \frac{rad}{hr} (40hr)(2) \\= 16 \ rad[/tex]
Thus, the effective dose allowable yearly = 16 times
Without actually calculating any logarithms, determine which of the following intervals the sound intensity level of a sound with intensity 3.66×10^−4W/m^2 falls within?
a. 30 and 40
b. 40 and 50
c. 50 and 60
d. 60 and 70
e. 70 and 80
f. 80 and 90
g. 90 and 100
Answer:
f. 80 and 90
Explanation:
1 x 10⁻¹² W/m² sound intensity falls within 0 sound level
1 x 10⁻¹¹ W/m² sound intensity falls within 10 sound level
1 x 10⁻¹⁰ W/m² sound intensity falls within 20 sound level
1 x 10⁻⁹ W/m² sound intensity falls within 30 sound level
1 x 10⁻⁸ W/m² sound intensity falls within 40 sound level
1 x 10⁻⁷ W/m² sound intensity falls within 50 sound level
1 x 10⁻⁶ W/m² sound intensity falls within 60 sound level
1 x 10⁻⁵ W/m² sound intensity falls within 70 sound level
1 x 10⁻⁴ W/m² sound intensity falls within 80 sound level
1 x 10⁻³ W/m² sound intensity falls within 90 sound level
Given sound intensity (3.66 x 10⁻⁴ W/m²) falls with 1 x 10⁻⁴ W/m² of intensity which is within 80 and 90 sound level.
f. 80 and 90
The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the ground. To maximize the potential difference between one end of the fluorescent tube and the other, how should the tube be held?a. The tube should be held horizontally, parallel to the ground b. The potential difference between the ends of the tube does not depend on the tube's orientation. c. The tube should be held vertically perpendicular to the ground
Answer:
b) True. potencial diferencie does not depend on orientation
Explanation:
In this exercise we are asked to show which statements are true.
The expression the potential with respect to earth or the electric field with respect to earth refers to the potential or electric charge of the planet that is assumed to be very large and does not change in value during work.
It does not refer to the height of the system.
We can now review the claims
a) False. Potential not to be refers to height
b) True. Does not depend on orientation
c) False The potential does not refer to the altitude but to the Earth's charge
an electron travels at 0.3037 times the speed of light through a magnetic field and feels a force of 1.2498 pN. What is the magnetic field in teslas
Answer:
Explanation:
Charge on an electron (q) = 1.6 * 10 ^ -19 C
Velocity of electron (v) = 0.3037 * 300,000,000 = 91,110,000 m/sec
We know that, Force exerted on moving particle moving through a magnetic field :
[tex]F= q * v * B ( q,v\ and\ B\ are\ mutually\ perpendicular)[/tex]
1.2498 * 10 ^ -12 = 1.6 * 10^ -19 * 91110000 * B
B = 0.08573 T
I MIND TRICK PLZ HELP LOL
Troy and Abed are running in a race. Troy finishes the race in 12 minutes. Abed finishes the race in 7 minutes and 30 seconds. If Troy is running at an average speed of 3 miles per hour and speed varies inversely with time, what is Abed’s average speed for the race?
Answer:
Explanation:
Let the race be of a fixed distance x
[tex]Average Speed = \frac{Total Distance}{Total Time}[/tex]
Troy's Average speed = 3 miles/hr = x / 0.2 hr
x = 0.6 miles
Abed's Average speed = 0.6 / 0.125 = 4.8 miles/hr
Intelligent beings in a distant galaxy send a signal to earth in the form of an electromagnetic wave. The frequency of the signal observed on earth is 2.2% greater than the frequency emitted by the source in the distant galaxy. What is the speed vrel of the galaxy relative to the earth
Answer:
Vrel= 0.75c
Explanation:
See attached file
A ball is thrown upward from a height of 432 feet above the ground, with an initial velocity of 96 feet per second. From physics it is known that the velocity at time t is v (t )equals 96 minus 32 t feet per second. a) Find s(t), the function giving the height of the ball at time t. b) How long will the ball take to reach the ground? c) How high will the ball go?
Answer;
A)S(t)=96t-16t² +432
B)it will take 9 seconds for the ball to reach the ground.
C)864feet
Explanation:
We were given an initial height of 432 feet.
And v(t)= 96-32t
A) we are to Find s(t), the function giving the height of the ball at time t
The position, or heigth, is the integrative of the velocity. So
S(t)= ∫(96-32)dt
S(t)=96t-16t² +K
S(t)=96t-16t² +432
In which the constant of integration K is the initial height, so K= 432
b) we need to know how long will the ball take to reach the ground
This is t when S(t)= 0
S(t)=96t-16t² +432
-16t² +96t +432=0
This is quadratic equation, if you solve using factorization method we have
t= -3 or t= 9
Therefore, , t is the instant of time and it must be a positive value.
So it will take 9 seconds for the ball to reach the ground.
C)V=s/t
Velocity= distance/ time
=96=s/9sec
S=96×9
=864feet
By applying the integrations,
(a) [tex]S = 96t-16t^2+432[/tex]
(b) Time will be "t = 9".
(c) Height will be "576"
Given:
Height,
423 feetInitial velocity,
96 feet/secAccording to the question,
(a)
Integrate v:
[tex]S = 96t-16t^2+C[/tex]Initial Condition,
→ [tex]S = 96t-16t^2+432[/tex]
(b)
Hits the ground when,
S = 0→ [tex]0=96t-16t^2+432[/tex]
→ [tex]t =9[/tex]
(c)
Maximum height when,
v = 0→ [tex]0 = 96-32 t[/tex]
→ [tex]t = 3[/tex]
Now,
→ [tex]S = 96\times 3-16\times 3^2+432[/tex]
[tex]= 576[/tex]
Thus the answer above is correct.
Learn more:
https://brainly.com/question/16105731
"A satellite requires 88.5 min to orbit Earth once. Assume a circular orbit. 1) What is the circumference of the satellites orbit
Answer:
circumference of the satellite orbit = 4.13 × 10⁷ m
Explanation:
Given that:
the time period T = 88.5 min = 88.5 × 60 = 5310 sec
The mass of the earth [tex]M_e[/tex] = 5.98 × 10²⁴ kg
if the radius of orbit is r,
Then,
[tex]\dfrac{V^2}{r} = \dfrac{GM_e}{r^2}[/tex]
[tex]{V^2} = \dfrac{GM_e r}{r^2}[/tex]
[tex]{V^2} = \dfrac{GM_e }{r}[/tex]
[tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]
Similarly :
[tex]T = \sqrt{\dfrac{ 2 \pi r} {V} }[/tex]
where; [tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]
Then:
[tex]T = {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {GM_e }} }[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {6.674\times 10^{-11} \times 5.98 \times 10^{24} }} }[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ 3.991052 \times 10^{14} }}[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {19977617.48}[/tex]
[tex]5310 \times 19977617.48= 2 \pi r^{3/2}}[/tex]
[tex]1.06081149 \times 10^{11}= 2 \pi r^{3/2}}[/tex]
[tex]\dfrac{1.06081149 \times 10^{11}}{2 \pi}= r^{3/2}}[/tex]
[tex]r^{3/2}} = \dfrac{1.06081149 \times 10^{11}}{2 \pi}[/tex]
[tex]r^{3/2}} = 1.68833392 \times 10^{10}[/tex]
[tex]r= (1.68833392 \times 10^{10})^{2/3}}[/tex]
[tex]r= 2565.38^2[/tex]
r = 6579225 m
The circumference of the satellites orbit can now be determined by using the formula:
circumference = 2π r
circumference = 2π × 6579225 m
circumference = 41338489.85 m
circumference of the satellite orbit = 4.13 × 10⁷ m
If you wish to observe features that are around the size of atoms, say 5.5 × 10^-10 m, with electromagnetic radiation, the radiation must have a wavelength of about the size of the atom itself.
Required:
a. What is its frequency?
b. What type of electromagnetic radiation might this be?
Answer:
a) 5.5×10^17 Hz
b) visible light
Explanation:
Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;
λ= 5.5 × 10^-10 m
Since;
c= λ f and c= 3×10^8 ms-1
f= c/λ
f= 3×10^8/5.5 × 10^-10
f= 5.5×10^17 Hz
The electromagnetic wave is visible light
light of wavelength 550 nm is incident on a diffraction grating that is 1 cm wide and has 1000 slits. What is the dispersion of the m = 2 line?
Answer:
The dispersion is [tex]D = 2.01220 *10^{5} \ rad/m[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 550 \ = 550 *10^{-9} \ n[/tex]
The width of the grating is[tex]k = 1\ cm = 0.01 \ m[/tex]
The number of slit is N = 1000 slits
The order of the maxima is m = 2
Generally the spacing between the slit is mathematically represented as
[tex]d = \frac{k}{N}[/tex]
substituting values
[tex]d = \frac{ 0.01}{1000}[/tex]
[tex]d = 1.0 *10^{-5} \ m[/tex]
Generally the condition for constructive interference is
[tex]d\ sin(\theta ) = m * \lambda[/tex]
substituting values
[tex]1.0 *10^{-5} sin (\theta) = 2 * 550 *10^{-9}[/tex]
[tex]\theta = sin^{-1} [\frac{ 2 * 550 *10^{-9}}{ 1.0 *10^{-5}} ][/tex]
[tex]\theta = 6.315^o[/tex]
Generally the dispersion is mathematically represented as
[tex]D = \frac{ m }{d cos(\theta )}[/tex]
substituting values
[tex]D = \frac{ 2 }{ 1.0 *10^{-5} cos(6.315 )}[/tex]
[tex]D = 2.01220 *10^{5} \ rad/m[/tex]
Determine the next possible thickness of the film (in nm) that will provide the proper destructive interference. The index of refraction of the glass is 1.58 and the index of refraction of the film material is 1.48.
Answer:
I know the answer
Explanation:
We want to choose the film thickness such that destructive interference occurs between the light reflected from the air-film interface (call it wave 1) and from the film-lens interface (call it wave 2). For destructive interference to occur, the phase difference between the two waves must be an odd multiple of half-wavelengths.
You can think of the phases of the two waves as second hands on a clock; as the light travels, the hands tick-tock around the clock. Consider the clocks on the two waves in question. As both waves travel to the air-film interface, their clocks both tick-tock the same time-no phase difference. When wave 1 is reflected from the air-film boundary, its clock is set forward 30 seconds; i.e., if the hand was pointing toward 12, it's now pointing toward 6. It's set forward because the index of refraction of air is smaller than that of the film.
Now wave 1 pauses while wave two goes into and out of the film. The clock on wave 2 continues to tick as it travels in the film-tick, tock, tick, tock.... Clock 2 is set forward 30 seconds when it hits the film-lens interface because the index of refraction of the film is smaller than that of the lens. Then as it travels back through the film, its clock still continues ticking. When wave 2 gets back to the air-film interface, the two waves continue side by side, both their clocks ticking; there is no change in phase as they continue on their merry way.
So, to recap, since both clocks were shifted forward at the two different interfaces, there was no net phase shift due to reflection. There was also no phase shift as the waves travelled into and out from the air-film interface. The only phase shift occured as clock 2 ticked inside the film.
Call the thickness of the film t. Then the total distance travelled by wave 2 inside the film is 2t, if we assume the light entered pretty much normal to the interface. This total distance should equal to half the wavelength of the light in the film (for the minimum condition; it could also be 3/2, 5/2, etc., but that wouldn't be the minimum thickness) since the hand of the clock makes one revolution for each distance of one wavelength the wave travels (right?).
Lamar has been running sprints to prepare for his next football game.He has found that he can maintain his maximum speed for 45 yards.He’s thinking of running in a 5km race in a few months,but doesn’t know if he can maintain his maximum speed for the entire 5 km.Can you help him determine how far he can?
Answer:
Kindly check explanation
Explanation:
Length of race = 5km
Maximum speed = 45 yards
Converting from yards to kilometer :
1km = 1093.613 yards
x = 45 yards
(1093.613 * x) = 45
x = 45 / 1093.613
x = 0.0411480 km
Where x = maximum length for which he can maintain his maximum speed expressed in kilometers.
Therefore, with the available information, it can be concluded that Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.
Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.
The calculation is as follows;
Length of race = 5km
Maximum speed = 45 yards
Converting from yards to kilometer :
1km = 1093.613 yards
x = 45 yards
[tex](1093.613 \times x) = 45[/tex]
[tex]x = 45 \div 1093.613[/tex]
x = 0.0411480 km
here x represent maximum length for which he can maintain his maximum speed expressed in kilometers.
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Which is one criterion that materials of a technological design should meet? They must be imported. They must be affordable. They must be naturally made. They must be locally produced.
Answer:
they must be affordable because they have to pay for it or they wont get the stuff they are bying.
Explanation:
need a brainliest please.
Answer: B, they must be affordable.
Explanation:
The Bohr model pictures a hydrogen atom in its ground state as a proton and an electron separated by the distance a0 = 0.529 × 10−10 m. The electric potential created by the electron at the position of the proton is
Answer:
E = -8.23 10⁻¹⁷ N / C
Explanation:
In the Bohr model, the electric potential for the ground state corresponding to the Bohr orbit is
E = k q₁ q₂ / r²
in this case
q₁ is the charge of the proton and q₂ the charge of the electron
E = - k e² / a₀²
let's calculate
E = - 9 10⁹ (1.6 10⁻¹⁹)² / (0.529 10⁻¹⁰)²
E = -8.23 10⁻¹⁷ N / C
The block moves up an incline with constant speed. What is the total work WtotalWtotalW_total done on the block by all forces as the block moves a distance LLL
Answer:
External force W₁ = F L
Friction force W₂ = - fr L
weight component W₃ = - mg sin θ L
Y Axis Force W=0
Explanation:
When the block rises up the plane with constant velocity, it implies that the sum of the forces is zero.
For these exercises it is indicated to create a reference system with the x axis parallel to the plane and the y axis perpendicular
let's write the equations of translational equilibrium in given exercise
X axis
F - fr -Wₓ = 0
F = fr + Wₓ
the components of the weight can be found using trigonometry
Wₓ = W sin θ
[tex]W_{y}[/tex] = W cos θ
let's look for the work of these three forces
W = F x cos θ
External force
W₁ = F L
since the displacement and the force have the same direction
Friction force
W₂ = - fr L
since the friction force is in the opposite direction to the displacement
For the weight component
W₃ = - mg sin θ L
because the weight component is contrary to displacement
Y Axis
N- Wy = 0
in this case the forces are perpendicular to the displacement, the angle is 90º and the cosine 90 = 0
therefore work is worth zero
6. What is the bulk modulus of oxygen if 32.0 g of oxygen occupies 22.4 L and the speed of sound in the oxygen is 317 m/s?
Answer:
[tex] \boxed{\sf Bulk \ modulus \ of \ oxygen \approx 143.5 \ kPa} [/tex]
Given:
Mass of oxygen (m) = 32.0 g = 0.032 kg
Volume occupied by oxygen (V) = 22.4 L = 0.0224 m³
Speed of sound in oxygen (v) = 317 m/s
To Find:
Bulk modulus of oxygen
Explanation:
[tex]\sf Density \ of \ oxygen \ (\rho) = \frac{m}{V}[/tex]
[tex]\sf \implies Bulk \ modulus \ of \ oxygen \ (B) = v^{2} \rho[/tex]
[tex]\sf \implies B = v^{2} \times\frac{m}{V}[/tex]
[tex]\sf \implies B = {(317)}^{2} \times \frac{0.032}{0.0224} [/tex]
[tex]\sf \implies B = {(317)}^{2} \times 1.428[/tex]
[tex]\sf \implies B = 100489 \times 1.428[/tex]
[tex]\sf \implies B = 143498.292 \: Pa[/tex]
[tex]\sf \implies B \approx 143.5 \: kPa[/tex]
A circular conducting loop of radius 31.0 cm is located in a region of homogeneous magnetic field of magnitude 0.700 T pointing perpendicular to the plane of the loop. the loop is connected in series with a resistor of 265 ohms. The magnetic field is now increased at a constant rate by a factor of 2.30 in 29.0 s.
Calculate the magnitude of induced emf in the loop while the magnetic field is increasing.
With the magnetic field held constant a ts its new value of 1.61 T, calculate the magnitude of its induced voltage in the loop while it is pulled horizontally out of the magnetic field region during a time interval of 3.90s.
Answer:
(a) The magnitude of induced emf in the loop while the magnetic field is increasing is 9.5 mV
(b) The magnitude of the induced voltage at a constant magnetic field is 124.7 mV
Explanation:
Given;
radius of the circular loop, r = 31.0 cm = 0.31 m
initial magnetic field, B₁ = 0.7 T
final magnetic field, B₂ = 2.3B₁ = 2.3 X 0.7 T = 1.61 T
duration of change in the field, t = 29
(a) The magnitude of induced emf in the loop while the magnetic field is increasing.
[tex]E = A*\frac{\delta B}{\delta t} \\\\[/tex]
[tex]E = A*\frac{B_2 -B_1}{\delta t}[/tex]
Where;
A is the area of the circular loop
A = πr²
A = π(0.31)² = 0.302 m²
[tex]E = A*\frac{B_2 -B_1}{\delta t} \\\\E = 0.302*\frac{1.61-0.7}{29} \\\\E = 0.0095 \ V\\\\E = 9.5 \ mV[/tex]
(b) the magnitude of the induced voltage at a constant magnetic field
E = A x B/t
E = (0.302 x 1.61) / 3.9
E = 0.1247 V
E = 124.7 mV
Therefore, the magnitude of the induced voltage at a constant magnetic field is 124.7 mV
(a) Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
pF
(b) Determine the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
kV
Explanation:
(a) Given that,
Area of a parallel plate capacitor, [tex]A=1.8\ cm^2=1.8\times 10^{-4}\ m^2[/tex]
The separation between the plates of a capacitor, [tex]d=0.01\ mm = 10^{-5}\ m[/tex]
The dielectric constant of, k = 2.1
When a dielectric constant is inserted between parallel plate capacitor, the capacitance is given by :
[tex]C=\dfrac{k\epsilon_o A}{d}[/tex]
Putting all the values we get :
[tex]C=\dfrac{2.1\times 8.85\times 10^{-12}\times 1.8\times 10^{-4}}{0.01\times 10^{-3}}\\\\C=3.345\times 10^{-10}\ F\\\\C=334.5\ pF[/tex]
(b) We know that the Teflon has dielectric strength of 60 MV/m, [tex]E=60\times 10^6\ V/m[/tex]
The voltage difference between the plates at this critical voltage is given by :
[tex]V=Ed\\\\V=60\times 10^6\times 0.01\times 10^{-3} \\\\V=600\ V[/tex]
or
V = 0.6 kV
We have that the Capacitance and potential difference is mathematically given as
[tex]Vmax=\frac{Q}{334.68pF}[/tex]C=334.68pF
Capacitance &potential differenceQuestion Parameters:
having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm
having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.
a)
Generally the equation for the Capacitance is mathematically given as
[tex]C=\frac{ke_0A}{d}\\\\Therefore\\\\C=\frac{2.1*1.80e-4*8.85e12}{0.01e-3}\\\\[/tex]
C=334.68pF
b)
Generally the equation for the Capacitance is mathematically given as
[tex]Vmax=\frac{Q}{C}[/tex]
Where
Q is the charge on the plates, and hence not given
Therefore, maximum potential difference is
[tex]Vmax=\frac{Q}{334.68pF}[/tex]
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