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miss Akunina [59]
4 years ago
7

A laser beam is incident at an angle of 30.0° from the vertical onto a solution of corn syrup in water. The beam is refracted to

19.24° from the vertical. (a) What is the index of refraction of the corn syrup solution? Assume that the light is red, with vacuum wavelength 632.8 nm. Find its (b) wavelength, (c) frequency, and (d) speed in the solution.
Physics
1 answer:
dimaraw [331]4 years ago
5 0

Answer with Explanation:

We are given that

Angle of incidence,i=30^{\circ}

Angle of refraction,r=19.24^{\circ}

a.Refractive index of air,n_1=1

We know that

n_2sinr=n_1sini

n_2=\frac{n_1sin i}{sin r}=\frac{sin30}{sin19.24}=1.517

b.Wavelength of red light in vacuum,\lambda=632.8nm=632.8\times 10^{-9} m

1nm=10^{-9} m

Wavelength in the solution,\lambda'=\frac{\lambda}{n_2}

\lambda'=\frac{632.8}{1.517}=417nm

c.Frequency does not change .It remains same in vacuum and solution.

Frequency,\nu=\frac{c}{\lamda}=\frac{3\times 10^8}{632.8\times 10^{-9}}

Where c=3\times 10^8 m/s

Frequency,\nu=4.74\times 10^{14}Hz

d.Speed in the solution,v=\frac{c}{n_2}

v=\frac{3\times 10^8}{1.517}=1.98\times 10^8m/s

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What is the acceleration of a boat if the resultant force on it is 4000 N and its mass is 1240 kg
Marrrta [24]

Answer:

The acceleration of a boat if the resultant force on it is 4000 N and its mass is 1240 kg is 3.2 m/s².

Explanation:

Given

  • Force F = 4000 N
  • Mass m = 1240 kg

To determine

Acceleration a = ?

Important Tip:

  • The acceleration of a boat can be found using the formula F = ma

Using the formula

F =  ma

where

  • F is the force (N)
  • m is the mass (kg)
  • a is the acceleration (m/s²)

now substituting F = 4000, m = 1240 in the formula

F =  ma

4000\:=\:1240\:\times a

Switch sides

1240a=4000

Divide both sides by 1240

\frac{1240a}{1240}=\frac{4000}{1240}

Simplify

a=\frac{100}{31}

a=3.2 m/s²

Therefore, the acceleration of a boat if the resultant force on it is 4000 N and its mass is 1240 kg is 3.2 m/s².

5 0
3 years ago
A 3 Volt battery is connected in series to three resistors: 4, 6, and 2
DanielleElmas [232]

Answer:

Explanation:

   

7 0
3 years ago
When you mix a solution, the mass of the substances before and after you mix them should be the same.true or false
Leya [2.2K]

Answer:

True

Explanation:

The mixing substances can be taken to be the reactants and the resulting mixture as the product. When the substances change state during mixing, the mass of the substances does not change meaning that the mass of the product will equal the mass of the reactants .This is true due to the law of conservation of mass.

3 0
3 years ago
Read 2 more answers
A 2578-kg van runs into the back of a 825-kg compact car at rest. They move off together at 8.5 m/s. Assuming the friction with
allochka39001 [22]

Answer:

<em>The initial speed of the car = 11.22 m/s</em>

Explanation:

Law of conservation of energy: It states that when two bodied collide in a closed system, the sum of momentum before collision is equal to the sum of momentum after collision.

<em>Note:</em> A close system is one that is free from from external forces. E.g Frictional force.

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

m₁u₁ + m₂u₂ = (m₁ + m₂)V..................... Equation 1

Where m₁ = mass of the van, m₂ = mass of the car, u₁ = initial velocity of the van, u₂ = initial velocity of the car, V = Common velocity.

<em>Given: m₁ = 2578 kg, m₂ = 825 kg, u₂ = 0 ( the car was at rest), V= 8.5 m/s</em>

<em>Substituting these values into equation 1, and solving for u₁</em>

<em>2578(u₁) + (825 × 0) = (2578 + 825)8.5</em>

<em>2578u₁ = 28925.5</em>

<em>Dividing both side of the equation by the coefficient of u₁</em>

<em>2578u₁/2578 = 28925.5/2578</em>

<em>u₁ = 11.22 m/s</em>

<em>The initial speed of the car = 11.22 m/s</em>

3 0
3 years ago
7) A long straight wire of radius 1.00x10-3 m (1 mm) carries uniform current density J such that the magnetic field at the edge
Vedmedyk [2.9K]

Answer:A)0.5 and 2 mm

Explanation:

Given

radius R of wire is 1 mm

magnetic Field at edge(surface) =10^{-5} T

Magnetic Field at a distance r' from Center is B'=5\times 10^{-6} T

and we know

B=\frac{\mu _0I}{2\pi r}

where I= current

\mu _o=Permeability of free space

r=distance from center

For r=R

10^{-5}=\frac{\mu _0I}{2\pi R}---1

For r=r'

5\times 10^{-6}=\frac{\mu _0I}{2\pi r'}---2

Divide 1 and 2

\frac{10^{-5}}{0.5\times 10^{-5}}=\frac{r'}{R}

r'=2 R=2 mm

If r is inside the wire then

B=\frac{\mu _0rI}{2\pi R^2}

for r=R

10^{-5}=\frac{\mu _0R\cdot I}{2\pi R^2}---3

for r=r"

5\times 10^{-6}=\frac{\mu _0r"I}{2\pi R^2}----4

Divide 3 and 4

2=\frac{R}{r"}

r"=0.5 mm

3 0
3 years ago
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