Answer:
It would cost approximately $925,455,484 million to cover continental US and Alaska with $1 bills.
Explanation:
the area of a one dollar bill = 6.5 cm x 15.5 cm = 100.75 sq cm
the approximate area of continental US + Alaska = (1,000 miles x 3,000 miles) x 1.2 = 3,600,000 sq miles
each sq mile is roughly 2.58999 sq km, so the total area in sq km = 9,323,964 sq km
1 sq km = 1,000,000 sq meters
each sq meter = 10,000 sq cm
1 sq m = 10,000 / 100.75 = 99.25558 bills
1 sq km = 99,255,583.13 bills
9,323,964 sq km = 925,455,483,900,000 bills
If the US includes the area of the Hawaii and the Alaska. It would be costing approx. $925,455,484 mn to cover continental with the $1 bills.
As the area of 1 dollar bill = 6.5 cm x 15.5 cm = 100.75 cm2 The area of US + Alaska = (1,000 miles x 3,000 miles) x 1.2 = 3,600,000 sq Thus the total area in km2 = 9,323,964 sq km 1 km2 equals to 1,000,000 sq meters each sq meter = 10,000 sq cm 1 sq m = 10,000 / 100.75 = 99.25558 bills 1 km2 equals to 99,255,583.13 bills 9,323,964 sq km = 925,455,483,900,000 bills .Learn more about the would it cost to cover the entire land area.
brainly.com/question/17333335.
A "swing" ride at a carnival consists of chairs that are swung in a circle by 19.8 m cables attached to a vertical rotating pole, as the drawing shows. Suppose the total mass of a chair and its occupant is 137 kg. (a) Determine the tension in the cable attached to the chair. (b) Find the speed of the chair.
Answer:
a) T = 1342.6 cos θ, b) v = 13.93 √(sin θ tan θ)
Explanation:
We can solve this problem using Newton's second law
Let's fix a reference system with a horizontal axis and the other vertical, therefore the only force to decompose is the tension, in these problems the most common is to measure the angle with respect to the vertical. Let's use trigonometry to find the components of the dispute
cos θ = [tex]T_{y}[/tex] / T
T_{y} = T cos tea
sin θ = Tₓ / T
Tₓ = T sin θ
let's write Newton's second law
axis and vertical
T cos θ - W = 0
T = mg / cos θ
let's calculate
T = 137 9.8 cos θ
T = 1342.6 cos θ
unfortunately there is no drawing or indication of the angle
Axis x Horizontal
T sin θ = m a
acceleration is centripetal
a = v² / R
T sin θ = m v² / R
v² = (g / cos θ) R sin θ
v = √ (gR tan θ)
let's use trigonometry to find radius of gyration
sin θ = R / L
R = L sin θ
v = √ (g L sin θ tan θ)
let's calculate
v = √(9.8 19.8 sin θ tant θ)
v = 13.93 √(sin θ tan θ)
they do not give the angle for which the calculation cannot be finished
Suppose the kicker launches the ball at 60∘ instead of 30∘. Assuming that the goal is 4.55 m high and 40 m away, what minimum initial speed v0 would the ball need to have in order to just clear the goal?
Answer:22 m/s
Explanation:
Given
launch angle [tex]\theta =60^{\circ}[/tex]
height of goal [tex]h=4.55\ m[/tex]
and horizontal distance [tex]x=40\ m[/tex]
Suppose initial speed is [tex]u[/tex]
Trajectory of a Projectile is given by
[tex]y=x\tan \theta -\frac{1}{2}\frac{gx^2}{u^2\cos ^2\theta }[/tex]
substituting the values we get
[tex]4.55=40\tan (60)-0.5\times \frac{9.8\times (40)^2}{u^2\cdot \cos ^260 }[/tex]
[tex]4.55=69.28-0.5\times \frac{15,680}{u^2\cdot 0.25}[/tex]
[tex]\frac{31,360}{u^2}=69.28-4.55[/tex]
[tex]\frac{31,360}{64.73}=u^2[/tex]
[tex]u^2=484.47[/tex]
[tex]u=22.01\ m/s[/tex]
So, initial launch speed is [tex]22\ m/s[/tex]
High speed stroboscopic photographs show that the head of a 244 g golf club is traveling at 57.6 m/s just before it strikes a 45.2 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 39.9 m/s. Find the speed of the golf ball just after impact.
Answer:
The speed will be "1.06 m/s".
Explanation:
The given values are:
Momentum,
m1 = 244 g
m2 = 45.2 g
On applying momentum conservation ,
Let v2 become the final golf's speed.
From Momentum Conservation
⇒ [tex]Total \ initial \ momentum = Total \ final \ momentum[/tex]
⇒ [tex]m1\times u1 + m2\times u2 = m1\times v1 + m2\times v2[/tex]
On putting the estimated values, we get
⇒ [tex]0.244\times 57.6+0=0.244\times 39.9+45.2\times v2[/tex]
⇒ [tex]57.844+0=9.7356+45.2\times v2[/tex]
⇒ [tex]48.1084=45.2\times v2[/tex]
⇒ [tex]v2=\frac{48.1084}{45.2}[/tex]
⇒ [tex]v2=1.06 \ m/s[/tex]
A long, East-West-oriented power cable carrying an
unknown current I is at a height of 8 m above the Earth's
surface. If the magnetic flux density recorded by a magnetic-
field meter placed at the surface is 15 ut when the current is
flowing through the cable and 20 ut when the current is zero,
what is the magnitude of 1?
Answer:
200A
Explanation:
Given that
the distance between earth surface and power cable d = 8m
when the current is flowing through cable , the magnitude flux density at the surface is 15μT
when the current flow throught is zero the magnitude flux density at the surface is 20μT
The change in flux density due to the current flowing in the power cable is
B = 20μT - 15μT
B =5μT -----(1)
The expression of magnitude flux density produced by the current carrying cable is
[tex]B=\frac{\mu_0I}{2\pi d}[/tex]-----(2)
Substitute the value of flux density
B from eqn 1 and eqn 2
[tex]\frac{\mu_0I}{2\pi d}=5\times 10^-^6\\\\\frac{(4\pi \times 10^-^7)I}{2 \pi (8)} =5\times 10^-^6\\\\I=200A[/tex]
Therefore, the magnitude of current I is 200A
The frames for a pair of eyeglasses have a radius of 2.14 cm at 20.0°C. Lenses with radius of 2.16 cm have to be inserted into these frames. To what temperature must the technician heat the frames to accommodate the lenses? The frames are made of a material whose thermal expansion coefficient is 1.30 10-4/°C.
Answer:
The technician must heat the frame to a temperature of T₂ = 0.0072°C
Explanation:
Change in one-dimension of an object upon heating is given by the following equation:
ΔL = α L ΔT
For the radius this equation can be re-written as follows:
ΔR = α R ΔT
where,
ΔR = Change in Radius = 2.16 cm - 2.14 cm = 0.02 cm
α = Thermal Expansion Coefficient = 1.3 x 10⁻⁴ /°C
R = Original Radius = 2.14 cm
ΔT = Change in Temperature = T₂ - 20°C
Therefore,
0.02 cm = (1.3 x 10⁻⁴ /°C)(2.14 cm)(T₂ - 20°C)
T₂ - 20°C = (0.02 cm)/(1.3 x 10⁻⁴ /°C)(2.14 cm)
T₂ = 0.0072°C + 20°C
T₂ = 20.0072°C
A long solenoid that has 1 080 turns uniformly distributed over a length of 0.420 m produces a magnetic field of magnitude 1.00 10-4 T at its center. What current is required in the windings for that to occur?
Answer:
Current, I = 0.073 A
Explanation:
It is given that,
Number of turns in a long solenoid is 1080
Length of the solenoid is 0.420 m
It produces a magnetic field of [tex]10^{-4}\ T[/tex] at its center.
We need to find the current is required in the winding for that to occur. The magnetic field at the center of the solenoid is given by :
[tex]B=\mu_0 NI[/tex]
I is current
[tex]I=\dfrac{B}{\mu_o N}\\\\I=\dfrac{10^{-4}}{4\pi \times 10^{-7}\times 1080}\\\\I=0.073\ A[/tex]
There has long been an interest in using the vast quantities of thermal energy in the oceans to run heat engines. A heat engine needs a temperature difference, a hot side and a cold side. Conveniently, the ocean surface waters are warmer than the deep ocean waters. Suppose you build a floating power plant in the tropics where the surface water temperature is ~ 35.0 C. This would be the hot reservoir of the engine. For the cold reservoir, water would be pumped up from the ocean bottom where it is always ~ 5.00 C.
What is the maximum possible efficiency of such a power plant?
Answer:
Explanation:
The maximum efficient power plant will be the plant based on carnot cycle whose efficiency is given by the following formula
Efficiency = (T₁ - T₂) / T₁
T₁ is temperature of hot reservoir and T₂ is temperature of cold reservoir.
Putting the given values
efficiency of power plant = (35 - 5) / (273 + 35 )
= 30 / 308
= .097
= 9.7 %
The density of gasoline is 730 kg/m3 at 00C. Its volume expansion coefficient is 0.000960C-1. If one gallon of gasoline occupies 0.0038 m3, how many extra kilograms of gasoline are obtained when 10 gallons of gasoline are bought at 00C rather than at 200C
Answer:
Explanation:
volume expansion coefficient = .000960 C⁻¹
d₀ = d ( 1 + .000960 x t ) , d is density at t and d₀ is density at 0 degree Celsius .
if t = 200
d₀= d ( 1 + .00096 x 200 )
= d ( 1 + .00096 x 200 )
= 1.192d
10 gallons of gasolene
= 10 x .0038 m³ of gasolene
= .038 m³
difference of mass = volume x difference of density
= .038 x ( d₀ - d )
.038 x ( d₀ - d₀ / 1.192 )
= .038 x ( 1 - 1 / 1.192 ) d₀
= .00612 x 730
= 4.468 kg .
A 1500 kg car travelling at 25 m/s collides with a 2500 kg van which had
stopped at a traffic light. As a result of the collision the two vehicles become
entangled. Which of the following pairs given in the table below shows the
initial speed the entangled mass will move and the type of collision
A car starting from rest, travels 0.40 km in 11.0 s. What is the
magnitude of its constant acceleration?
Answer:
6.61 m/s²
Explanation:
Given:
Δx = 0.40 km = 400 m
v₀ = 0 m/s
t = 11.0 s
Find: a
Δx = v₀ t + ½ at²
400 m = (0 m/s) (11.0 s) + ½ a (11.0 s)²
a = 6.61 m/s²
The initial mass of a radioisotope is 10.0 g. If the radioisotope has a half life of 2.75 years, how much remains after four half lives?
After four half-lives, the amount that remains is (1/2)⁴ of the original sample.
That's (1/16) of 10 g, or 0.625 g.
It doesn't matter how long the half-life is.
A radioisotope has a starting mass of 10.0 g. Half lives of the isotopes are 0-10, 1-5, 2-2.25, 3-1.25, and 4-0.625 if the radioisotope has a half life of 2.75 years.
Who or what are radioisotopes?A chemical element in an unstable state that emits radiation as it decomposes and becomes more stable. Both the natural environment and a laboratory can produce radioisotopes. They are utilized in imaging studies and therapy in medicine.
Why do you use the term "half-life"?In radioactivity, the half-life is the amount of time needed for half of a radioactive sample's atomic nuclei to spontaneously decay (convert into different nuclear species by emitting particles and energy), or, more precisely, the amount of time needed for the number of disintegrations per second.
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What are the 2 types of electricity
What is the gravitational force between mars and Phobos
Answer:
[tex]F=5.16\times 10^{15}\ N[/tex]
Explanation:
We have,
Mass of Mars is, [tex]m_M=6.42\times 10^{23}\ kg[/tex]
Mass of its moon Phobos, [tex]m_P=1.06\times 10^{16}\ kg[/tex]
Distance between Mars and Phobos, d = 9378 km
It is required to find the gravitational force between Mars and Phobos. The force between two masses is given by
[tex]F=G\dfrac{m_Mm_P}{d^2}[/tex]
Plugging all values, we get :
[tex]F=6.67\times 10^{-11}\times \dfrac{6.42\times 10^{23}\times 1.06\times 10^{16}}{(9378\times 10^3)^2}\\\\F=5.16\times 10^{15}\ N[/tex]
So, the gravitational force is [tex]5.16\times 10^{15}\ N[/tex].
Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of these seeds can be determined with a high-speed camera. In an experiment on one type of plant, seeds are projected at 20 cm above ground level with initial speeds between 2.3 m/s and 4.6 m/s. The launch angle is measured from the horizontal, with +90∘ corresponding to an initial velocity straight up and -90∘ straight down. The experiment is designed so that the seeds move no more than 0.20 mm between photographic frames. What minimum frame rate for the high-speed camera is needed to achieve this?
a. 250 frames/s
b. 2500 frames/s
c. 25,000 frames/s
d. 250,000 frames/s.
Answer:
c. 25,000 frames/s
Explanation:
For computing the minimum frame rate for high speed first we have to determine the time by applying the following equation
[tex]t = \frac{d}{s}[/tex]
[tex]= \frac{0.2\ mm}{4.6\ m/s }[/tex]
[tex]= \frac{0.2 \times 10 ^{-3}}{4.6\ m/s }[/tex]
[tex]= 4.347 \times 10^{-5} sec[/tex]
Now the frame rate is
[tex]Frame\ rate = \frac{1}{t}[/tex]
[tex]= \frac{1}{4.347 \times 10^{-5} sec}[/tex]
= 23,000 frame per sec
≈ 25,000 frame per sec
First we have find the time then after finding out the time we calculate the frame time by applying the above formula so that the minimum frame rate could come
standing on a hill that is 3.5 m high above level ground. A wall that is on level ground is 15 m away from the student. The student decides to throw a ball horizontally at a combined height of 4.5 m, at the wall. The student measures the time it takes to make an impact and finds it to be 0.65 s. At what height, from the ground, did the student hit the side of the wal
Answer:
The height will be "2.42 m".
Explanation:
The given values are:
time, t = 0.65 seconds
∴ g = 9.8
As we know,
⇒ [tex]x=vt+\frac{1}{2}gt^2[/tex]
∴ v = 0
On putting the estimated values, we get
⇒ [tex]=0+\frac{1}{2}\times 9.8\times (0.65)^2[/tex]
⇒ [tex]=\frac{1}{2}\times 9.8\times 0.4225[/tex]
⇒ [tex]=\frac{1}{2}\times 4.1405[/tex]
⇒ [tex]=2.07025 \ m[/tex]
Now,
Height, [tex]h=4.5-2.07025[/tex]
⇒ [tex]=2.42 \ m[/tex]
A particle moves with a velocity v in a circle of radius r, then its angular velocity is equal
to………….. and acts along the…………..
Answer:
Given that,
Speed = v
Radius = r
We have to ascertain the precise speed
Utilizing equation of speed
Where, v = velocity/speed
r = radius
= precise speed/ angular velocity
Angular Velocity :
The precise speed is characterized by the speed of turn.
The precise speed is straightforwardly corresponding to the direct speed and contrarily relative to the range of the molecule.
Subsequently, The precise speed is v/r
Which of the following statements about this system of lumps must be true? A. The momentum of the system is conserved during the collision. B. The kinetic energy of the system is conserved during the collision. C. The two masses lose all their kinetic energy during the collision. D. The velocity of the center of mass of the system is the same after the collision as it was before the collisio
Answer:
option A.
Explanation:
Since the two lumps collide together, it is an inelastic collision. An inelastic collision, in contrast to an elastic collision, is a collision in which kinetic energy is not conserved due to the action of internal friction. Part of the kinetic energy is changed to some other form of energy in the collision
Momentum is conserved in inelastic collisions. This means the correct option is option A.
A girl walks South at 2.7 m/s. What is the y component of her velocity
Answer:
-2.7m/s
Explanation:
...............
In order for an external force to do work on a system, Question 9 options: a) there must be a component of the force perpendicular to the motion. b) there must be a component of force parallel to the motion of the object. c) the force can be at any angle relative to the motion of the object.
Answer:
b) there must be a component of force parallel to the motion of the object.
Explanation:
We know that work done on a body by an external force is calculated by the formula given below:
W = F.d = Fd Cos θ
where,
W = Work Done by the force on the body
F = Magnitude of force
d = displacement of the body
θ = The angle between the direction of motion of the body and the force applied
It is clear from the formula of the work done, that "F Cosθ" represents the component of the force, that is acting in the direction of motion of the object or parallel to the direction of motion of the object. So, if there is no component of force parallel to motion of object, then this factor will become zero. As a result, the work done will also be zero.
Therefore, the correct option will be:
b) there must be a component of force parallel to the motion of object.
You are at the controls of a particle accelerator, sending a beam of 2.10×107 m/s protons (mass m) at a gas target of an unknown element. Your detector tells you that some protons bounce straight back after a collision with one of the nuclei of the unknown element. All such protons rebound with a speed of 1.80×107 m/s. Assume that the initial speed of the target nucleus is negligible and the collision is elastic.
A) Find the mass of one nucleus of the unknown element.
B) What is the speed of the unknown nucleus immediately after such a collision?
Answer:
a
The mass is [tex]m_2 =21.75*10^{-27} \ kg[/tex]
b
The velocity is [tex]v_2 = 3.0*10^{6} m/s[/tex]
Explanation:
From the question we are told that
The speed of the protons is [tex]u_1 = 2.10*10^{7} m/s[/tex]
The mass of the protons is [tex]m[/tex]
The speed of the rebounding protons are [tex]v_1 = -1.80 * 10^{7} \ m/s[/tex]
The negative sign shows that it is moving in the opposite direction
Now according to the law of energy conservation mass of one nucleus of the unknown element. is mathematically represented as
[tex]m_2 = [\frac{u_1 -v_1}{u_1 + v_1} ] m_1[/tex]
Where [tex]m_1[/tex] is the mass of a single proton
So substituting values
[tex]m_2 = \frac{2.10 *10^{7} - (-1.80 *10^{7})} {(2.10 *10^7) + (-1.80 *10^{7})} m_1[/tex]
[tex]m_2 =13 m_1[/tex]
The mass of on proton is [tex]m_1 = 1.673 * 10^{-27} \ kg[/tex]
So [tex]m_2 =13 ( 1.673 * 10^{-27} )[/tex]
[tex]m_2 =21.75*10^{-27} \ kg[/tex]
Now according to the law of linear momentum conservation the speed of the unknown nucleus immediately after such a collision is mathematically evaluated as
[tex]m_1 u_1 + m_2u_2 = m_1 v_1 + m_2v_2[/tex]
Now [tex]u_2[/tex] because before collision the the nucleus was at rest
So
[tex]m_1 u_1 = m_1 v_1 + m_2v_2[/tex]
=> [tex]v_2 = \frac{m_1(u_1 -v_1)}{m_2}[/tex]
Recall that [tex]m_2 =13 m_1[/tex]
So
[tex]v_2 = \frac{m_1(u_1 -v_1)}{13m_1}[/tex]
=> [tex]v_2 = \frac{(u_1 -v_1)}{13}[/tex]
substituting values
[tex]v_2 = \frac{( 2.10*10^{7} -(-1.80 *10^{7}))}{13}[/tex]
[tex]v_2 = 3.0*10^{6} m/s[/tex]
There is a known potential difference between two charged plates of 12000 Volts. An object with a charge of 6.5 x 10-6 C charge and a mass of 0.02 kg is placed next to the positive plate. How fast will it be traveling when it gets to the negative plate
Answer:
1.97 m/s.
Explanation:
From the question,
Using the law of conservation of energy,
The energy stored in the charged plate = Kinetic energy of the mass
1/2(qV) = 1/2mv².......................... Equation 1
Where q = charge, V = voltage, m = mass, v = velocity.
make v the subject of the equation
v = √(qV/m)......................... Equation 2
Given: q = 6.5×10⁻⁶ C, V = 12000 Volts, m = 0.02 kg
Substitute these values into equation 2
v = √(6.5×10⁻⁶×12000 /0.02)
v = √3.9
v = 1.97 m/s.
Electromagnetic waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, if it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current. In this case, the wave equation for an electric field:________
Answer:
Del(ρ/ε₀) - (Del)²E = -dμ₀J/dt
Explanation:
From Maxwell's fourth equation
Curl B = μ₀J + μ₀ε₀dE/dt (1) where the second term is the displacement current.
If the oscillation conduction current in the conductor is much larger than the displacement current then, the displacement current goes to zero. So we have
Curl B = μ₀J (2)(since μ₀ε₀dE/dt = 0)
From maxwell's third equation
Curl E = -dB/dt (3)
taking curl of the above from the left
Curl(Curl E) = Curl(-dB/dt)
Curl(Curl E) = (-d(CurlB)/dt) (4)
Substituting for Curl B into (4), we have
Curl(Curl E) = -dμ₀J/dt
Del(DivE) - (Del)²E = -dμ₀J/dt (5)
From Maxwell's first equation,
DivE = ρ/ε₀
Substituting this into (5), we have
Del(ρ/ε₀) - (Del)²E = -dμ₀J/dt
Learning Goal: To understand the behavior ofthe electric field at the surface of a conductor, and itsrelationship to surface charge on the conductor.
A conductor is placed in an external electrostatic field. Theexternal field is uniform before the conductor is placed within it.The conductor is completely isolated from any source of current orcharge.
PART A)
Which of the following describes the electricfield inside this conductor?
It is in thesame direction as the original external field.
It is in theopposite direction from that of the original externalfield.
It has adirection determined entirely by the charge on itssurface.
It is alwayszero.
PART B)
The charge density inside theconductor is:
0
non-zero;but uniform
non-zero;non-uniform
infinite
PART C)
Assume that at some point just outside thesurface of the conductor, the electric field has magnitudeE and is directed toward thesurface of the conductor. What is the charge density eta on the surface of the conductor at thatpoint?
Express your answer in terms ofE and epsilon_0.
eta =
Complete Question
The complete question is shown on the first uploaded image
Answer:
a it is always zero
b 0
c [tex]\eta = -\epsilon _o E[/tex]
Explanation:ss
Here the net charge is on the outer surface of the conductor thus this means that the net charge inside the conductor is zero
Generally the charge density of a conductor is dependent on the charge per unit area which implies that the charge density is dependent on the net charge so this means that the charge density inside the conductor is zero
Generally the direction of electric field this from the positive charge to the negative charge so from the question we can deduce that the negative charge is located on the surface of the conductor
So We can mathematically define the charge density on the surface of the electric field as
∮[tex]E \cdot dA = \frac{-Q}{\epsilon _o}[/tex]
Where E is the electric field
[tex]dA[/tex] change in unit area
[tex]-Q[/tex] is the negative charge
[tex]\epsilon _o[/tex] is the permittivity of free space
So
[tex]EA = \frac{-Q}{\epsilon _o }[/tex]
[tex]\frac{Q}{A} = -\epsilon _o E[/tex]
[tex]\eta = -\epsilon _o E[/tex]
Where [tex]\eta[/tex] is the charge density
Your friend says your body is made up of more than 99.9999% empty space. What do you think
Answer:false
Explanation:
A 1 900-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.00 m before coming into contact with the top of the beam, and it drives the beam 15.8 cm farther into the ground before coming to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.
Answer:
471392.4 N
Explanation:
From the question,
Just before contact with the beam,
mgh = Fd.................... Equation 1
Where m = mass of the beam, g = acceleration due to gravity, h = height. F = average Force on the beam, d = distance.
make f the subject of the equation
F = mgh/d................ Equation 2
Given: m = 1900 kg, h = 4 m, d = 15.8 = 0.158 m
Constant: g = 9.8 m/s²
Substitute into equation 2
F = 1900(4)(9.8)/0.158
F = 471392.4 N
A laser beam is incident from the air at an angle of 30.0° to the vertical onto a solution of Karo syrup in water. If the beam is refracted to 19.24° to the vertical, what is the index of refraction of the syrup solution?
Answer:
Index of Refraction = 1.52
Explanation:
The index of refraction or the refractive index is given as:
Index of Refraction = Sin i/Sin r
where,
i = angle of incidence
r = angle of refraction
In this case of karo syrup, we have the following data:
i = angle of incidence = 30°
r = angle of refraction = 19.24°
Therefore, substituting these values in the equation, we get:
Index of Refraction = Sin 30°/Sin 19.24°
Index of Refraction = 0.5/0.3295
Index of Refraction = 1.52
Hence, the refractive index or the index of refraction of the Karo Syrup is found to be 1.52.
The Michelson interferometer can be used to measure the index of refraction of a gas by placing an evacuated transparent tube in the light path along one arm of the device. Fringe shifts occur as the gas is slowly added to the tube. Assume 560-nm light is used, the tube is 5.10 cm long, and 168 bright fringes pass on the screen as the pressure of the gas in the tube increases to atmospheric pressure. What is the index of refraction of the gas
Answer:
The index of refraction of the gas is [tex]n_g = 1.000922[/tex]
Explanation:
From the question we are told that
The wavelength of light is [tex]\lambda = 560 \ nm = 560 *10^{-9} \ m[/tex]
The length of tube is [tex]L = 5.10 \ cm = \frac{5.10 }{100} =0.0510 \ m[/tex]
The number of fringes is [tex]N = 168[/tex]
Generally the index of refraction of the gas is mathematically repreented as
[tex]n_g = 1 + \frac{N \lambda }{2 L }[/tex]
substituting values
[tex]n_g = 1 + \frac{168 *[ 560* 10^{-9}] }{2 * 0.0510 }[/tex]
[tex]n_g = 1.000922[/tex]
A moving walkway has a speed of 0.5 m/s to the east. A stationary observer
sees a man walking on the walkway with a velocity of 0.8 m/s to the east.
What is the man's velocity relative to the moving walkway?
Answer: 0.3 m/sec
Explanation:
Vrel = (Vs-Vm) = (0.8-0.5) = 0.3 m/sec
Answer:
0.3 m/s east
Explanation:
Write the formula for the Newton’s Law of Gravitation.
please answer this question!!!
Answer:
F = G(m1m2)/R2.
Explanation:
Newton's law of universal gravitation:
Gravitational Force, in Newtons, between two objects =
(a constant)·(one mass)·(the other mass)/(distance between them)²
acting on EACH object, in the direction of the other object.
If the masses are in kilograms and the distance is in meters, then the constant is 6.67 x 10⁻¹¹ m³/kg-sec² .
I may be wrong, but I don't think Newton had any number to use for the constant. It had not been measured yet, the kilogram and the meter had not been invented yet, and there certainly was no unit called "a Newton" during his lifetime.
He might have been able to calculate the value of the constant by applying his law of gravity to the motion of one or two planets. But he would have needed to know the mass of the sun and the planets he used, and I don't think those were known yet in Newton's time.
A venturi meter used to measure flow speed in the pipe. Derive an expression for the flow speed "H1" interns of the crossectional areas "A1" and "A2" and the difference in height "h" of the liquid levels in the two vertical tubes ?
Answer:
v₁ = √[ 2gh / ((A₁ / A₂)² − 1) ]
Explanation:
Use Bernoulli's equation:
P₁ + ½ ρ v₁² + ρgz₁ = P₂ + ½ ρ v₂² + ρgz₂
Since there's no elevation change between points 1 and 2, z₁ = z₂.
P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²
Assuming incompressible fluid, the volumetric flow rate is the same at points 1 and 2.
Q₁ = Q₂
v₁ A₁ = v₂ A₂
v₂ = v₁ A₁ / A₂
Substituting:
P₁ + ½ ρ v₁² = P₂ + ½ ρ (v₁ A₁ / A₂)²
P₁ + ½ ρ v₁² = P₂ + ½ ρ v₁² (A₁ / A₂)²
P₁ − P₂ = ½ ρ v₁² (A₁ / A₂)² − ½ ρ v₁²
P₁ − P₂ = ½ ρ v₁² ((A₁ / A₂)² − 1)
v₁² = 2 (P₁ − P₂) / (ρ ((A₁ / A₂)² − 1))
v₁² = 2 (ρgh) / (ρ ((A₁ / A₂)² − 1))
v₁² = 2gh / ((A₁ / A₂)² − 1)
v₁ = √[ 2gh / ((A₁ / A₂)² − 1) ]