Answer:
0.7707
Explanation:
From Snell's law,
n(1) * sin θ1 = n(2) * sinθ2
Where n(1) = refractive index of air = 1.0003
θ1 = angle of incidence
n(2) = refractive index of second substance
θ2 = angle of refraction
The angle of reflection through the unknown substance is the same as the angle of incidence of air. Thus this means that θ1 = 29°
=> 1.0003 * sin29 = n(2) * sin39
n(2) = (1.0003 * sin29) / sin39
n(2) = 0.7707
Explanation:
The index of refraction n of the substance is 0.7707
Snell law:Here we know that
n(1) * sin θ1 = n(2) * sinθ2
here
n(1) = refractive index of air = 1.0003
θ1 = angle of incidence
n(2) = refractive index of second substance
θ2 = angle of refraction
The angle of reflection should be via the unknown substance that represent the same as the angle of incidence of air.
So,
θ1 = 29°
1.0003 * sin29 = n(2) * sin39
n(2) = (1.0003 * sin29) / sin39
n(2) = 0.7707
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In a double-slit experiment using light of wavelength 486 nm, the slit spacing is 0.600 mm and the screen is 2.00 m from the slits. Find the distance along the screen between adjacent bright fringes.
Answer:
The distance is [tex]y = 0.00162 \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 486 \ nm = 486 *10^{-9} \ m[/tex]
The slit spacing is [tex]d = 0.600 \ mm = 0.60 *10^{-3} \ m[/tex]
The distance of the screen is [tex]D = 2.0 \ m[/tex]
Generally the distance along the screen between adjacent bright fringes is mathematically represented as
[tex]y = \frac{\lambda * D}{d}[/tex]
substituting values
[tex]y = \frac{ 486 *10^{-9} * 2}{0.6*10^{-3}}[/tex]
[tex]y = 0.00162 \ m[/tex]
The entropy of any substance at any temperature above absolute zero is called the: Select the correct answer below:
a. absolute entropy
b. Third Law entropy
c. standard entropy
d. free entropy
e. none of the above
Answer:
b. Third Law entropy
Explanation:
Third law entropy: In physics, the term "third law entropy" or "the third law of thermodynamics" states that the specific entropy of a particular system at "absolute zero" is considered as a "well-defined constant". It occurs because any system at "zero temperature" tends to exists or persists in its "ground state" in order for the entropy to be determined or described only by the "degeneracy" of the given ground state.
In the question above, the correct answer is option b.
A soccer ball of mass 0.4 kg is moving horizontally with a speed of 20 m/s when it is kicked by a player. The kicking force is so large that the ball flies up at an angle of 30 degrees above the ground. The player however claims (s)he aimed her/his foot at a 40 degree angle above the ground. Calculate the average kicking force magnitude and the final speed of the ball, if you are given that the foot was in contact with the ball for one hundredth of a second.
Answer:
v_{f} = 74 m/s, F = 230 N
Explanation:
We can work on this exercise using the relationship between momentum and moment
I = ∫ F dt = Δp
bold indicates vectors
we can write this equations in its components
X axis
Fₓ t = m ( -v_{xo})
Y axis
t = m (v_{yf} - v_{yo})
in this case with the ball it travels horizontally v_{yo} = 0
Let's use trigonometry to write the final velocities and the force
sin 30 = v_{yf} / vf
cos 30 = v_{xf} / vf
v_{yf} = vf sin 30
v_{xf} = vf cos 30
sin40 = F_{y} / F
F_{y} = F sin 40
cos 40 = Fₓ / F
Fₓ = F cos 40
let's substitute
F cos 40 t = m ( cos 30 - vₓ₀)
F sin 40 t = m (v_{f} sin 30-0)
we have two equations and two unknowns, so the system can be solved
F cos 40 0.1 = 0.4 (v_{f} cos 30 - 20)
F sin 40 0.1 = 0.4 v_{f} sin 30
we clear fen the second equation and subtitles in the first
F = 4 sin30 /sin40 v_{f}
F = 3.111 v_{f}
(3,111 v_{f}) cos 40 = 4 v_{f} cos 30 - 80
v_{f} (3,111 cos 40 -4 cos30) = - 80
v_{f} (- 1.0812) = - 80
v_{f} = 73.99
v_{f} = 74 m/s
now we can calculate the force
F = 3.111 73.99
F = 230 N
Consider two isolated spherical conductors each having net charge Q. The spheres have radii a and b, where b > a.
Which sphere has the higher potential?
1. the sphere of radius a
2. the sphere of radius b
3. They have the same potential
Answer:
1. the sphere of the radius a
Explanation:
Because the charge distribution for each case is spherically symmetric, we can choose a spher- ical Gaussian surface of radius r , concentric with the sphere in question.
So E = k (Q /r 2) (for r ≥ R ) , where R is the radius of the sphere being considered, either a or b .
With the choice of potential at r = ∞ being zero, the electric potential at any distance r from the center of the sphere can be expressed as V = - integraldisplay r ∞ E dr = k /Q r
(for r ≥ R ) .
On the spheres of radii a and b , we have V a = k (Q/ a)and V b = k (Q/ b), respectively.
So Since b > a , the sphere of radius a will have the higher potential.
Also recall Because E = 0 inside a conductor, the potential
A 10kg block with an initial velocity of 10 m/s slides 1o m across a horizontal surface and comes to rest. it takes the block 2 seconds to stop. The stopping force acting on the block is about
Answer:
-50N
Explanation:
F=ma=m(Vf-Vi)/t
m=10kgVf=0m/sVi=10m/st=2sF=(10)(-10)/(2)=-50N
So the force acting on the block is -50N, where the negative sign simply tells us that the force is opposite to the direction of movement.
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
A man using a 70kg garden roller on a level surface, exerts a force of 200N at 45 degrees to the ground. find the vertical force of the roller on the ground if,
i.he pulls
ii.he pushes the roller
Answer:
i) 545.2 N upwards
ii) 828.2 N downwards
Explanation:
mass of the roller = 70 kg
force exerted = 200 N
angle the force makes with the ground ∅ = 45°
weight of the roller W = mg
where
m is the mass of the roller
g is the acceleration due to gravity = 9.81 m/s^2
weight of the roller = 70 x 9.81 = 686.7 N
The effective vertical force exerted by the man = F sin ∅ = 200 x sin 45°
==> F = 200 x 0.707 = 141.5 N
i) if the man pulls, then the exerted force will be in opposite direction to the weight of the roller vertically upwards
Resultant vertical force = 686.7 N - 141.5 N = 545.2 N upwards
ii) if he pushes, then the exerted force will be in the direction of the weight vertically downwards
Resultant vertical force = 686.7 N + 141.5 N = 828.2 N downwards
You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and hang a 1.8 g mass from it. This stretches the spring to a length of 5.1 cm . You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.3 cm .
Requried:
What is the magnitude of the charge (in nC) on each bead?
Answer:
2.2nC
Explanation:
Call the amount by which the spring’s unstretched length L,
the amount it stretches while hanging x1
and the amount it stretches while on the table x2.
Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,
we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,
applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,
where ke is the Coulomb constant. Combining these,
we get q = √(mgx2(L+x2)²/x1ke =2.2nC
A uniform bar has two small balls glued to its ends. The bar is 2.10 m long and with mass 3.70 kg , while the balls each have mass 0.700 kg and can be treated as point masses.
Required:
Find the moment of inertia of this combination about an axis
a. perpendicular to the bar through its center.
b. perpendicular to the bar through one of the balls.
c. parallel to the bar through both balls.
d. parallel to the bar and 0.500 m from it.
Answer:
Explanation:
a )
moment of inertia in the first case will be sum of moment of inertia of two balls + moment of inertia of bar
= 2 x .700 x (2.1 / 2 )² + 3.7 x 2.1² / 12
= 1.5435 + 1.35975
= 2.90325 kg m²
b )
moment of inertia required
= moment of inertia of bar + moment of inertia of the other ball
= 3.70 x (2.1² / 3 ) + .7 x 2.1²
= 5.439 + 3.087
= 8.526 kg m²
c )
In this case moment of inertia of the combination = 0 as distance of masses from given axis is zero .
d )
masses = 3.7 + .7 = 4.4 kg
distance from axis = .5 m
moment of inertia about given axis
= 4.4 x .5²
= 1.1 kg m².
An electron moves on a circular orbit in a uniform magnetic field of 7.83×10-4 T. The kinetic energy of the electron is 55.3 eV. What is the diameter of the orbit?
Answer:
3.9E-8
Explanation:
We know that
Mv²/r = Bqv
So
r= mv/Bq
But E is 1/2mv² which is 5.53eV
m²v² =2m x 5.53eV
mv = √( 2 x 9.1E-31)
So
r= √( 2 x9.1E-31 x 5.53)/ 7.83x10^-4 x1.6E-19
= 3.9x10-8cm
Explanation:
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
A 0.50-T magnetic field is directed perpendicular to the plane of a circular loop of radius 0.25 m. What is the magnitude of the magnetic flux through the loop
Answer:
The magnitude of the magnetic flux through the loop is 0.0982 T.m²
Explanation:
Given;
magnitude of magnetic field, B = 0.5 T
radius of the loop, r = 0.25 m
Area of the loop is given by;
A = πr²
A = 3.142 x (0.25)²
A = 0.1964 m²
The magnitude of the magnetic flux through the loop is given by;
Ф = BA
Where;
B is the magnitude of the magnetic field
A is area of the field
Ф = 0.5 x 0.1964
Ф = 0.0982 T.m²
Therefore, the magnitude of the magnetic flux through the loop is 0.0982 T.m²
A spherical balloon has a radius of 6.95m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of air is 1.29 kg/m3. The skin and structure of the balloon has a mass of 960kg . Neglect the buoyant force on the cargo volume itself. Determine the largest mass of cargo the balloon can lift.
Answer:
602.27 kg
Explanation:
The computation of the largest mass of cargo the balloon can lift is shown below:-
Volume of helium inside the ballon= (4 ÷ 3) × π × r^3
= (4 ÷ 3) × 3.14 × 6.953
= 1406.19 m3
Mass the balloon can carry = volume × (density of air-density of helium)
= 1406.19 × (1.29-0.179)
= 1562.27 kg
Mass of cargo it can carry = Mass it can carry - Mass of structure
= 1562.27 - 960
= 602.27 kg
In order to waken a sleeping child, the volume on an alarm clock is doubled. Under this new scenario, how much more energy will be striking the child's ear drums each second?
Answer:4 times more energy will be striking the childbearing
Explanation:
Because Volume is directly proportional to amplitude of sound. Energy is proportional to amplitude squared. If you triple the amplitude, you multiply the energy by 4
A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 6.0 cm and released. If the ball makes 30 oscillations in 20 s, what are its (a) mass and (b) maximum speed?
Answer:
a
[tex]m = 0.169 \ kg[/tex]
b
[tex]|v_{max} |= 0.5653 \ m/s[/tex]
Explanation:
From the question we are told that
The spring constant is [tex]k = 14 \ N/m[/tex]
The maximum extension of the spring is [tex]A = 6.0 \ cm = 0.06 \ m[/tex]
The number of oscillation is [tex]n = 30[/tex]
The time taken is [tex]t = 20 \ s[/tex]
Generally the the angular speed of this oscillations is mathematically represented as
[tex]w = \frac{2 \pi}{T}[/tex]
where T is the period which is mathematically represented as
[tex]T = \frac{t}{n}[/tex]
substituting values
[tex]T = \frac{20}{30 }[/tex]
[tex]T = 0.667 \ s[/tex]
Thus
[tex]w = \frac{2 * 3.142 }{ 0.667}[/tex]
[tex]w = 9.421 \ rad/s[/tex]
this angular speed can also be represented mathematically as
[tex]w = \sqrt{\frac{k}{m} }[/tex]
=> [tex]m =\frac{k }{w^2}[/tex]
substituting values
[tex]m =\frac{ 15 }{(9.421)^2}[/tex]
[tex]m = 0.169 \ kg[/tex]
In SHM (simple harmonic motion )the equation for velocity is mathematically represented as
[tex]v = - Awsin (wt)[/tex]
The velocity is maximum when [tex]wt = \(90^o) \ or \ 1.5708\ rad[/tex]
[tex]v_{max} = - A* w[/tex]
=> [tex]|v_{max} |= A* w[/tex]
=> [tex]|v_{max} |= 0.06 * 9.421[/tex]
=> [tex]|v_{max} |= 0.5653 \ m/s[/tex]
Categorize each ray tracing statement as relating to ray 1, ray 2, or ray 3.
A. Drawn from the top of the object so that it passes through the center of the lens at the optical axis.
B. Drawn from the top of the object so that it passes through the focal point on the same side of the lens as the object.
C. Drawn parallel to the optical axis from the top of the object.
D. Ray bends parallel to the optical axis.
E. Ray bends so that it passes through the focal point on the opposite side of the lens as the object.
F. Ray does not bend.
Answer:
statement 1 with answer C
statement 2 with answer F
statement 3 with answer B
Statement 1 with E
Statement 2 with A
Statement 3 with D
Explanation:
In this exercise you are asked to relate each with the answers
In general, in the optics diagram,
* Ray 1 is a horizontal ray that after stopping by the optical system goes to the focal point
* Ray 2 is a ray that passes through the intercept point between the optical axis and the system and does not deviate
* Ray 3 is a ray that passes through the focal length and after passing the optical system, it comes out horizontally.
With these statements, let's review the answers
statement 1 with answer C
statement 2 with answer F
statement 3 with answer B
Statement 1 with E
Statement 2 with A
Statement 3 with D
Which best identifies the requirements for work to be performed? an object that has a force acting on it an object that is moving and has no net force a force acting on a motionless object a force that moves an object
Answer:
a force that moves an object
Explanation:
the formula for work is force * distance
This question involves the concepts of work, force, and displacement.
The statement that best identifies the requirements for work to be performed is "a force that moves an object".
Work is defined as the product of force applied on an object and the distance moved by the object. Mathematically,
Work = (Force)(Displacement)
Hence, both the applied force and the displacement of the object as a result of the application of the force is necessary for the work to be done. If any one of these values becomes zero, the work automatically becomes zero, which means no work is performed.
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Three capacitors C1 = 10.7 µF, C2 = 23.0 µF, and C3 = 29.3 µF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Answer:
E = 1336.71875 J
Explanation:
We are given;. Capacitance of Capacitor 1; C1 = 10.7 µF
Capacitor 2; C2 = 23.0 µF
Capacitor 3; C3 = 29.3 µF
Supply voltage;V = 125 V
Formula for capacitance in series is;
Capacitors in series circuit: C(eq) = 1/C(1) +1/C(2) +1/C(3) .......
Thus, equivalent capacitance is;
C(eq) = (1/10.7) + (1/23) + (1/29.3) = 0.1711 µF = 0.1711 × 10^(6) F
Now, the formula for maximum energy stored is;
E = ½ × C(eq) × V²
E = ½ × 0.1711 × 10^(-6) × 125²
E = 1336.71875 J
If you were to come back to our solar system in 6 billion years, what might you expect to find?
A) a red giant star
B) a rapidly spinning pulsar
C) a white dwarf
D) a black hole
E) Everything will be essentially the same as it is now
Answer:
A)a red giant star
Currents in DC transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers.
A. For a line that has current 150 A and a height of 8.0 m above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas.
B. What magnetic field does the line produce at ground level as a percent of earth's magnetic field which is 0.50 G?
C. Is this value of magnetic field cause for worry? Choose your answer below.
i. Yes. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.
ii. No. Since this field is much lesser than the earth's magnetic field, it would be expected to have less effect than the earth's field.
iii. Yes. Since this field is much greater than the earth's magnetic field, it would be expected to have more effect than the earth's field.
iv. No. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.
Answer:
Explanation:
magnetic field due to an infinite current carrying conductor
B = k x 2I / r where k = 10⁻⁷ , I is current in conductor and r is distance from wire
putting the given data
B = 10⁻⁷ x 2 x 100 / 8
= 25 x 10⁻⁷ T .
B )
earth's magnetic field = .5 gauss
= .5 x 10⁻⁴ T
= 5 x 10⁻⁵ T
percent required = (25 x 10⁻⁷ / 5 x 10⁻⁵) x 100
= 5 %
C )
ii. No. Since this field is much lesser than the earth's magnetic field, it would be expected to have less effect than the earth's field.
An interference pattern is produced by light with a wavelength 520 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.440 mm.
1. If the slits are very narrow, what would be the angular position of the first-order, two-slit, interference maxima?
2. What would be the angular position of the second-order, two-slit, interference maxima in this case?
3. Let the slits have a width 0.310 mm . In terms of the intensity I0 at the center of the central maximum, what is the intensity at the angular position of θ1?
4. What is the intensity at the angular position of θ2?
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
.
An inductor is hooked up to an AC voltage source. The voltage source has EMF V0 and frequency f. The current amplitude in the inductor is I0.
Part A
What is the reactance XL of the inductor?
Express your answer in terms of V0 and I0.
Part B
What is the inductance L of the inductor?
Express your answer in terms of V0, f, and I0.
Answer:
a. The reactance of the inductor is XL = V₀/I₀
b. The inductance of the inductor is L = V₀/2πfI₀
Explanation:
PART A
Since the voltage across the inductor V₀ = I₀XL where V₀ = e.m.f of voltage source, I₀ = current amplitude and XL = reactance of the inductor,
XL = V₀/I₀
So, the reactance of the inductor is XL = V₀/I₀
PART B
The inductance of the inductor is gotten from XL = 2πfL where f = frequency of voltage source and L = inductance of inductor
Since XL = V₀/I₀ = 2πfL
V₀/I₀ = 2πfL
L = V₀/2πfI₀
So the inductance of the inductor is L = V₀/2πfI₀
A) The reactance XL of the inductor : [tex]\frac{V_{0} }{I_{0} }[/tex]
B) The Inductance L of the inductor : [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex]
A) Expressing the Reactance of the inductor
Voltage across the Inductor = V₀ = I₀XL ---- ( 1 )
Where : V₀ = emf voltage , I₀ = current
from equation ( 1 )
∴ XL ( reactance ) = [tex]\frac{V_{0} }{I_{0} }[/tex]
B ) Expressing the Inductance of the Inductor
Inductance of an inductor is expressed as : XL = 2πfL
from part A
XL = [tex]\frac{V_{0} }{I_{0} }[/tex] = 2πfL
∴ The inductance L of the Inductor expressed in terms of V₀, F and I₀
L = [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex]
Hence we can conclude that The reactance XL of the inductor : [tex]\frac{V_{0} }{I_{0} }[/tex] and The Inductance L of the inductor : [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex] .
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Experiments are performed with ultracold neutrons having velocities of 7.54 m/s. (a) What is the wavelength (in nm) of such a neutron
Answer:
λ = 52.5 nm
Explanation:
De Broglie's duality principle states that all matter has wave and particle characteristics, being related by the expression
p = h / λ
where the moment
p = mv
λ = h / mv
let's calculate
λ = 6.63 10⁻³⁴ / 1.675 10⁻²⁷ 7.54
λ = 5.25 10⁻⁸ m
Let's reduce anm
λ = 5.25 10⁻⁸ m (10⁹ nm / 1m)
λ = 52.5 nm
If 1.7 kg of 238Pu is required to power the spacecraft's data transmitter, for how long after launch would scientists be able to receive data? Round to the nearest year. Do not round intermediate calculations.
The question is incomplete. Here is the complete question.
The isotope of Plutonium 238Pu is used to make thermoeletric power sources for spacecraft. Suppose that a space probe was launched in 2012 with 3.5 kg of 238Pu.
(a) If the half-life of 238Pu is 87.7 yr, write a function of the form [tex]Q(t)=Q_{0}e^{-kt}[/tex] to model the quantity Q(t) of 238Pu left after t years. Round ythe value of k to 3 decimal places. Do not round intermediate calculations.
(b) If 1.7kg of 238Pu is required to power the spacecraft's data transmitter, for low long after launch would scientists be able to receive data? Round to the nearest year. Do not round intermediate calculations.
Answer: (a) [tex]Q(t)=3.5e^{-0.0079t}[/tex]
(b) 91 years.
Explanation:
(a) Half-life is time it takes a substance to decrease to half of itself, i.e.:
Q(t) = [tex]0.5Q_{0}[/tex]
[tex]0.5Q_{0}=Q_{0}e^{-87.7k}[/tex]
[tex]0.5=e^{-87.7k}[/tex]
[tex]ln(0.5)=ln(e^{-87.7k})[/tex]
[tex]ln(0.5)=-87.7k[/tex]
[tex]k = \frac{ln(0.5)}{-87.7}[/tex]
k = 0.0079
Knowing k and [tex]Q_{0}[/tex]=3.5kg, function is [tex]Q(t)=3.5e^{-0.0079t}[/tex]
(b) Using function:
[tex]Q(t)=3.5e^{-0.0079t}[/tex]
[tex]1.7=3.5e^{-0.0079t}[/tex]
[tex]e^{-0.0079t}=\frac{1.7}{3.5}[/tex]
[tex]e^{-0.0079t}=0.4857[/tex]
[tex]ln(e^{-0.0079t})=ln(0.4857)[/tex]
[tex]-0.0079t=-0.7221[/tex]
[tex]t = \frac{-0.7221}{-0.0079}[/tex]
t = 91.41
t ≈ 91 years
Scientists will be able to receive data for approximately 91 years.
A woman was told in 2020 that she had exactly 15 years to live. If she travels away from the Earth at 0.8 c and then returns at the same speed, the last New Year's Day the doctors expect her to celebrate is:
Answer:
2035
Explanation:
The doctor does not travel with the woman, and therefore, he won't experience any relativistic effect on his time. The doctor will judge time by the time here on earth. Technically, the last new year's day the doctor, who is here on earth, would expect the woman to celebrate will be in 2020 + 15 years = 2035
What happens when two polarizers are placed in a straight line, one behind the other? A. They allow light to pass only if they are polarized in exactly the same direction. B. They block all light if they are polarized in exactly the same direction. C. They allow light to pass only if their directions of polarizations are exactly 90° apart. D. They block all light if their directions of polarizations are exactly 90° apart. E. They block all light if their directions of polarizations are either exactly the same or exactly 90° apart.
Answer:
C
They allow light to pass only if their directions of polarizations are exactly 90° apart.
Unpolarized light is passed through three successive Polaroid filters, each with its transmission axis at 45.0° to the preceding filter. What percentage of light gets through?
Answer:
The percentage is [tex]k = 12.5 \%[/tex]
Explanation:
From the question we are told that
The axis is is at [tex]\theta = 45 ^o[/tex]
Generally the of intensity light emerging from the first polarizer is mathematically represented as
[tex]I_{1} = \frac{I_o}{ 2}[/tex]
Where [tex]I_o[/tex] is the intensity of unpolarized light
Now the light emerging from the second polarizer is mathematically represented as
[tex]I_2 = I_ 1 * cos ^2(\theta )[/tex]
[tex]I_2 = \frac{I_o}{2} * cos ^2(45 )[/tex]
[tex]I_2 = \frac{I_o}{2} * \frac{1}{2} = \frac{I_o}{4}[/tex]
Now the light emerging from the third polarizer is mathematically represented as
[tex]I_3 = I_ 2 * cos ^2(\theta )[/tex]
[tex]I_3 = \frac{I_o}{4} * cos ^2(45 )[/tex]
[tex]I_3 = \frac{I_o}{8}[/tex]
Now the percentage of the intensity of light that emerged with respect to the intensity of the unpolarized light is
[tex]k = \frac{\frac{I_o}{8} }{I_o } * 100[/tex]
[tex]k = 12.5 \%[/tex]
The percentage of light that gets through the three successive Polaroid filters is; 12.5%
We are given;
Angle of transmission axis; θ = 45°
Formula for intensity of light from first polarizer is;
I₁ = ¹/₂I₀
Formula for intensity of light from second polarizer is;
I₂ = I₁cos²θ
Formula for intensity of light from third polarizer is;
I₃ = I₂cos²(90 - θ)
Combining the 3 equations;
Put ¹/₂I₀ for I₁ in second formula to get;
I₂ = ¹/₂I₀cos²θ
Put ¹/₂I₀cos²θ for I₂ in third formula to get;
I₃ = ¹/₂I₀cos²θ*cos²(90 - θ)
Plugging in 45° for θ gives;
I₃ = ¹/₂I₀cos²45*cos²(90 - 45)
⇒ I₃ = ¹/₂I₀cos²45*cos²45
⇒ I₃ = ¹/₂I₀cos⁴45
Now, cos 45 in surd form is 1/√2. Thus;
I₃ = ¹/₂I₀(1/√2)⁴
I₃ = ¹/₂I₀(¹/₄)
I₃ = ¹/₈I₀
I₃/I₀ = ¹/₈
I₃/I₀ = 0.125
In percentage form, we have;
I₃/I₀ = 12.5%
Read more about unpolarized light at; https://brainly.com/question/1444040
Find an analytic expression for p(V)p(V)p(V), the pressure as a function of volume, during the adiabatic expansion.
Answer:
In an adiabatic process we have
pV γ = const..
This explains that the pressure is a function of volume, p ( V ) ,
So can be written as:
p ( V ) × V γ = p 0 V γ 0 ,
or p ( V ) = p 0 V 0 / V γ
= p 0 V 0 / V ^(7 / 5)
Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long. How long is the other pipe?
Answer:
The length of the longer pipe is L = 2.30 m
Explanation:
Given that:
Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long.
How long is the other pipe?
From above;
The formula for the frequency of open ended pipes can be expressed as:
[tex]f = \dfrac{nv}{2L}[/tex]
where n = 1 ( since half wavelength exist between those two pipes)
v = 343 m/s and L = 2.08 m
Thus, the shorter pipe produces a frequency of :
[tex]f = \dfrac{1*343}{2*2.08}[/tex]
[tex]f = \dfrac{343}{4.16}[/tex]
[tex]f =82.45 \ Hz[/tex]
Also; we know that the beat frequency was given as 8.0 Hz
Then,
The lower frequency of the longer pipe = ( 82.45 - 8.0 )Hz
The lower frequency of the longer pipe = 74.45 Hz
Finally;
From the above equation; make Length L the subject of the formula. Then,
The length of the longer pipe is L = [tex]\dfrac{nv}{2f}[/tex]
The length of the longer pipe is L = [tex]\dfrac{1*343}{2*74.45}[/tex]
The length of the longer pipe is L = [tex]\dfrac{343}{148.9}[/tex]
The length of the longer pipe is L = 2.30 m
Which is produced around a wire when an electrical current is in the wire? magnetic field solenoid electron flow electromagnet
Answer:
A. magnetic field
Explanation:
The magnetic field is produced around a wire when an electrical current is in the wire because of the magnetic effect of the electric current therefore the correct answer is option A .
What is a magnetic field ?A magnetic field could be understood as an area around a magnet, magnetic material, or an electric charge in which magnetic force is exerted.
As given in the problem statement we have to find out what is produced around a wire when an electrical current is in the wire.
The magnetic field is produced as a result when an electrical current is passed through the conducting wire .
Option A is the appropriate response because a wire's magnetic field is created when an electrical current flows through it due to the magnetic influence of the electric current .
Learn more about the magnetic fields here, refer to the link given below;
brainly.com/question/23096032
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