A variable force of 6x−2 pounds moves an object along a straight line when it is x feet from the origin. Calculate the work done in moving the object from x = 1 ft to x = 18 ft. (Round your answer to two decimal places.) ft-lb

Answer:

ω(t) = 0.4 + 0.036 t²

Explanation:  

The angular displacement of the disk is given as the function of time:

β(t) = Ct + B t³

where,

C = 0.4 rad/s

B = 0.012 rad/s³

Therefore,

β(t) = 0.4 t + 0.012 t³

Now, for angular velocity ω(t), we must take derivative of angular displacement with respect to t:

ω(t) = dβ/dt = (d/dt)(0.4 t + 0.012 t³)

ω(t) = 0.4 + 0.036 t²

Answer:

Explanation:

Let the velocity of astronaut be u and the velocity of flat sheet of metal plate be v . They will move in opposite direction ,  so their relative velocity

= u + v = 2.3 m /s ( given )

We shall apply conservation of momentum law for the movement of astronaut and metal plate

mu  = M v where m is mass of astronaut , M is mass of metal plate

71 u = 230 x v

71 ( 2.3 - v ) = 230 v

163.3 = 301 v

v = .54 m / s

u = 1.76 m / s

honeycomb will be at rest  because honeycomb surface  is frictionless . Plate will slip over it . Over plate astronaut is walking .

a ) velocity of metal sheet relative to honeycomb will be - 1.76 m /s

b ) velocity of astronaut relative to honeycomb will be + .54 m /s

Here + ve direction is assumed to be the direction of astronaut .  

Explanation:

It is given that,

Angle of incidence from air to another medium, i = 26°

The angle of reflection, r = 32°

We need to find the refractive index of the medium. The ratio of sine of angle of incidence to the sine of angle of reflection is called refractive index. It can be given by :

[tex]n=\dfrac{\sin i}{\sin r}\\\\n=\dfrac{\sin (26)}{\sin (32)}\\\\n=0.82[/tex]

So, the index of refraction is 0.82. Hence, the correct option is C.

Answer:

The average velocity for this trip is 0 km/hr

Explanation:

We know that average velocity = total displacement/total time.

Now, its displacement is d = final position - initial position.

Since the  car starts and ends at its initial position at Hither, if we assume its initial position is 0 km, then its final position is also 0 km.

So, its displacement is d = 0 km - 0 km = 0 km.

Since the total time for the round trip is 2 hours, the average velocity is

total displacement/ total time = 0 km/2 hr = 0 km/hr.

So the average velocity for this trip is 0 km/hr  

Answer:

0.098 N

Explanation:

From the question,

Spring scale reading = W-U............... Equation 1

Where W = weight of the cube, U = upthrust.

W = mg

Where m =  mass of the cube, g = acceleration due to gravity.

Given: m = 11 g = 0.011 kg, g = 9.8 m/s².

W = 0.011(9.8)

W = 0.1078 N.

From Archimedes principle,

Upthrust = weight of water displaced.

U = (Density of water×volume of metal cube)×acceleration due to gravity.

U = (D×V)g

Given: D = 1000 kg/m², V = 1 cm³ = (1/1000000) = 1×10⁻⁶ m³, g - 9.8 m/s²

U = 1000(9.8)(10⁻⁶)

U = 0.0098 N.

Substitute the value of W and U into equation 1

Reading of the spring scale = 0.1078-0.0098

Reading of the spring scale = 0.098 N

Answer:

Sleep, behavior patterns, mental state, and job

Explanation:

Answer:

If I exercise right after school, I might be low on energy. I suppose that I could eat a snack and drink something before my workout. If I exercise before school, I might be tired. But, as long as I keep getting eight to nine hours of sleep, I think that my body will adjust to the new schedule after a while. The trick will be getting to bed on time and eating a little breakfast before I work out. I’m kind of worried that the gym won’t be open early in the morning. On the weekends, my friends might keep me from exercising. I suppose I can try to get them to do it with me. We can pick things that we all like to do. I know they like to play tennis sometimes.

Explanation:

Answer:

The speed of m2 just before it hits the ground is 2.1 m/s

Explanation:

mass on the ground m1 = 30 kg

mass oat rest at the above the ground m2 = 35 kg

height of m2 above the ground =2.9 m

Let the tension on the string be taken as T

for the mass m2 to reach the ground, its force equation is given as

[tex]m_{2} g - T = m_{2}a[/tex]    ....equ 1

where g is acceleration due to gravity = 9.81 m/s^2

and a is the acceleration with which it moves down

For mass m1 to move up, its force equation is

[tex]T - m_{1} g = m_{1} a[/tex]

[tex]T = m_{1}a + m_{1}g[/tex]

[tex]T = m_{1}(a + g)[/tex]    ....equ 2

substituting T in equ 1, we have

[tex]m_{2} g - m_{1}(a+g) = m_{2}a[/tex]

imputing values, we have

 [tex](35*9.81) - 30(a+9.81) = 35a[/tex]

 [tex]343.35 - 30a-294.3 = 35a[/tex]

[tex]343.35 -294.3 = 35a+ 30a[/tex]

[tex]49.05 = 65a[/tex]

a = 49.05/65 = 0.755 m/s^2

The initial velocity of mass m2 = u = 0

acceleration of mass m2 = a = 0.755 m/s^2

distance to the ground = d = 2.9 m

final velocity = v = ?

using Newton's equation of motion

[tex]v^{2}= u^{2} + 2ad[/tex]

substituting values, we have

[tex]v^{2}= 0^{2} + 2*0.755*2.9[/tex]

[tex]v^{2}= 2*0.755*2.9 = 4.379\\v = \sqrt{4.379}[/tex]

v = 2.1 m/s

Answer:

Explanation:

a )

Moment of inertial of four masses about axis that coincides with one side :

Out of four masses . location of two masses will lie on the axis so their moment of inertia will be zero .

Moment of inertia of the two remaining masses

= m L² + m L²

= 2 mL²

b )

Axis that bisects two opposite sides

Each of the four masses will lie at a distance of L / 2 from this axis so moment of inertia of the four masses

= 4 x m x ( L/2 )²

= 4 x  mL² / 4

= m L² .

Answer:

The speed of the light signal as viewed from the observer is c.

Explanation:

Recall the basic postulate of the theory of relativity that the speed of light is the same in ALL inertial frames. Based on this, the speed of light is independent of the motion of the observer.

Answer:

a

    [tex]\alpha = 2327.7 \ rev/s^2[/tex]

b

   [tex]\theta = 9124.5 \ rev[/tex]

Explanation:

From the question we are told that

    The maximum  angular   speed is  [tex]w_{max} = 391000 \ rpm = \frac{2 \pi * 391000}{60} = 40950.73 \ rad/s[/tex]

     The  time  taken is  [tex]t = 2.8 \ s[/tex]

     The  minimum angular speed is  [tex]w_{min}= 0 \ rad/s[/tex] this is because it started from rest

Apply the first equation of motion to solve for acceleration we have that

       [tex]w_{max} = w_{mini} + \alpha * t[/tex]

=>     [tex]\alpha = \frac{ w_{max}}{t}[/tex]

substituting values

       [tex]\alpha = \frac{40950.73}{2.8}[/tex]

       [tex]\alpha = 14625 .3 \ rad/s^2[/tex]

converting to [tex]rev/s^2[/tex]

  We have

           [tex]\alpha = 14625 .3 * 0.159155 \ rev/s^2[/tex]

           [tex]\alpha = 2327.7 \ rev/s^2[/tex]

According to the first equation of motion the angular displacement is  mathematically represented as

       [tex]\theta = w_{min} * t + \frac{1}{2} * \alpha * t^2[/tex]

substituting values

      [tex]\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2[/tex]

      [tex]\theta = 57331.2 \ radian[/tex]

converting to revolutions  

        [tex]revolution = 57331.2 * 0.159155[/tex]

        [tex]\theta = 9124.5 \ rev[/tex]

Answer:

The maximum displacement of the mass m₂ [tex]= \frac{2(m_1-m_2)g}{k}[/tex]

Explanation:

Kinetic Energy (K) = 1/2mv²

Potential Energy (P) = mgh

Law of Conservation of energy states that total energy of the system remains constant.

i.e; Total energy before collision = Total energy after collision

This implies that: the gravitational potential energy lost by m₁ must be equal to sum of gravitational energy gained by m₂ and the elastic potential energy stored in the spring.

[tex]m_1gd = m_2gd+\frac{1}{2}kd^2\\\\m_1g = m_2g+\frac{1}{2}kd\\\\d = \frac{2(m_1-m_2)g}{k}[/tex]

d = maximum displacement of the mass m₂

Answer:

A) W = 103.55 N

B) mass of displaced water = 4186 g

C) W_displaced water = 41.06 N

D) Buoyant force = 41.06 N.

E) ZERO

F) 62.54 N

Explanation:

We are given;

mass of statuette;m = 10,566 g = 10.566 kg

volume = 4,064 cm³

Density of seawater;ρ = 1.03 g/mL = 1.03 g/cm³

A) The dry weight of the statuette can be calculated as;

W = mg

So;

W = 10.556 × 9.81

W = 103.55 N

B) Mass of displaced water is calculated from;

Density = mass/volume

So, mass = Density × Volume

m = 1.03 × 4,064 = 4186 g

C) Weight of displaced water is given by;

W_displaced water = (m_displaced water) × g

W_displaced water = 4.186 kg × 9.81 m/s^2 = 41.06 N

D) The buoyant force is the same as the weight of the displaced water.

Thus, Buoyant force = 41.06 N.

E) The apparent weight of the statuette is calculated from;

Apparent weight = Dry weight - Weight of displaced water

Apparent weight = 103.6 N - 41.06 N = 62.54 N. It is sitting on the bottom of the sea, so the sea floor is providing an opposite force that is equal but opposite the weight so that the net force on the statuette is zero. Since It has zero acceleration, in any direction, hence the net force on it is zero.

F. From E above, The Force required to lift the statuette = 62.54 N

Answer:

[tex]1.357\times 10^{-12}[/tex]

Explanation:

Relevant Data provided

Area which indicates A = 2.3 cm^2 = 2.3 x 10^-4 m^2

Distance which indicates d = 1.50 x 10^-3 m

Voltage which indicates V = 12 V

According to the requirement, the computation of value of its capacitance is shown below:-

[tex]Capacitance, C = \frac{\epsilon oA}{D}[/tex]

[tex]= \frac{= 8.854\times 10^{-12}\times 2.3\times 10^{-4}}{(1.5 \times 10^{-3})}[/tex]

= [tex]1.357\times 10^{-12}[/tex]

Therefore for computing the capacitance we simply applied the above formula.

Answer:

(A)

Explanation:

We know , electric potential energy between two charge particles of charges "q" and "Q" respectively is given by kqQ/r where r is the distance between them.

Since the two charged particles are moving apart, the distance between them (r) increases and thus electrical potential energy decreases.

A) The free-body diagram of the forces acting on the brick is attached.

B) The free-body diagram of the forces acting on the rubber cushion is attached.

C) The action and reaction forces of the entire brick–cushion–planet system has been enumerated below.

A) The brick has a Mass M placed on top of a rubber cushion of mass m.

This means that there will be a normal force acting acting upwards on the brick and also a gravitational force acting downward. These forces are denoted as;

Normal force of rubber cushion acting on brick = [tex]n_{cb}[/tex]

Gravitational force acting on brick = Mg

Find attached the free body diagram.

B) The forces acting on the cushion will be;

Normal force of parking lot pavement on rubber cushion  = [tex]n_{pc}[/tex]

Gravitational force of earth acting on cushion = mg

Force of brick acting on the rubber cushion = [tex]F_{bc}[/tex]

C)  The action pairs of forces are;

i) Force; Normal force of rubber cushion acting on brick  = [tex]n_{cb}[/tex]

Reaction Force; Force of brick acting on the rubber cushion = [tex]F_{bc}[/tex]

ii) Action Force; Gravitational force acting on brick = Mg

Reaction; Gravitational force of brick acting on the earth

iii) Action Force; Normal force of parking lot pavement on rubber cushion = [tex]n_{pc}[/tex]

Reaction; Force of rubber cushion on parking lot pavement

iv) Action Force; Gravitational force of earth acting on rubber cushion = mg

Reaction Force; Gravitational force of rubber cushion on the earth.

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Answer:

Explanation:

correct options

a ) Their electrical potential energy keeps decreasing

Actually as they move apart , their electrical potential energy decreases due to increase of distance between them and kinetic energy increases

so a ) option is correct

b ) Their acceleration keeps decreasing

As they move apart , their mutual force of repulsion decreases due to increase of distance between them so the acceleration decreases .

c ) c. Their kinetic energy keeps increasing

Their kinetic energy increases because their electrical potential energy decreases . Conservation of energy law will apply .

The moving apart should be true statements:

a. The electrical potential energy should be reduced.

b. The acceleration should be reduced.

c. The kinetic energy should be increased.

At the time when the moving part, there is the reduction of the electric potential energy because there is a rise in the distance due to which the increment of the kinetic energy.  The reduction of the mutual force of repulsion because of increment in the distance due to this the acceleration should be reduced. There is the increase in the kinetic energy due to the reduction of the electrical potential energy. here the law of conversation of energy should be applied.

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Answer:

Explanation:

induced emf = ?

induced emf = induced current x resistance of the circuit

= .54 x 3.6

= 1.944 V

induced emf required = 1.944 V .

Answer:

`1. charge Q, on the capacitor increases, while the current will decrease

2. τ = t = secs

Explanation:

1. consider RC  of a circuit to be am external source

voltage across the circuit is given as

v =v₀(1 - [tex]e^{\frac{t}{τ} }[/tex])

where v = voltage

v₀ = peak voltage

t = time taken

τ= time constant

as the charge across the capacitor increases, current decreases

the charge across the circuit is given as

Q= Q₀(1 - [tex]e^{\frac{t}{τ} }[/tex])

charge Q is inversely proportional to the current I

hence the charge across the circuit increases

2. τ = RC

unit of time constant, τ,

= Ω × F

=[tex]\frac{V}{I}[/tex] ˣ [tex]\frac{C}{V}[/tex]

=[tex]\frac{C}{A}[/tex]

=[tex]\frac{C}{C/t}[/tex]

τ = t = secs

Answer:

μ = 0.03

Explanation:

In order for the trunk not to slide the frictional force between the turntable and the trunk must be equal to the unbalanced force applied on the trunk by the motion of the turntable. Therefore,

Unbalanced Force = Frictional Force

but,

Unbalanced Force = ma (Newton's second law of motion)

Frictional Force = μN = μW = μmg

Therefore,

ma = μmg

a = μg

μ = a/g

where,

μ = coefficient of static friction between the trunk and the turntable = ?

a = tangential acceleration of trunk = 0.3 m/s²

g = 9.8 m/s²

Therefore,

μ = (0.3 m/s²)/(9.8 m/s²)

μ = 0.03

Answer:

A)   V = -136.36 V , B)  V = 4.85 10³ V , C)  V = 1.62 10⁴ V

Explanation:

To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is

          V = k ∑ [tex]q_{i} / r_{i}[/tex]

where [tex]q_{i}[/tex] and [tex]r_{i}[/tex] are the loads and the point distances.

A) We apply this equation to our case

          V = k (q₁ / r₁ + q₂ / r₂)

They ask us for the potential at the midpoint of separation

         r = 3.30 / 2 = 1.65 m

this distance is much greater than the radius of the spheres

let's calculate

         V = 9 10⁹ (1.1 10⁻⁸ / 1.65  + (-3.6 10⁻⁸) / 1.65)

         V = 9 10¹ / 1.65 (1.10 - 3.60)

         V = -136.36 V

B) The potential at the surface sphere A

r₂ is the distance of sphere B above the surface of sphere A

              r₂ = 3.30 -0.02 = 3.28 m

              r₁ = 0.02 m

we calculate

             V = 9 10⁹ (1.1 10⁻⁸ / 0.02  - 3.6 10⁻⁸ / 3.28)

             V = 9 10¹ (55 - 1,098)

             V = 4.85 10³ V

C) The potential on the surface of sphere B

      r₂ = 0.02 m

      r₁ = 3.3 -0.02 = 3.28 m

      V = 9 10⁹ (1.10 10⁻⁸ / 3.28  - 3.6 10⁻⁸ / 0.02)

       V = 9 10¹ (0.335 - 180)

       V = 1.62 10⁴ V

Answer:

F = 44.22 N

Explanation:

Let force 1, [tex]F_1=19\ N[/tex]

Force 2, [tex]F_2=32\ N[/tex]

The angle between forces, [tex]\theta=118^{\circ}[/tex]

We need to find the magnitude of the resultant force. It is based on the law of cosines. The formula is given by :

[tex]F^2=F_1^2+F_2^2-2AB\cos\theta\\\\F^2=(19)^2+(32)^2-2\times 19\times 32\times \cos(118)\\\\F=\sqrt{(19)^{2}+(32)^{2}-2\times19\times32\times\cos(118)}\\\\F=44.22\ N[/tex]

So, the magnitude of resultant force is 44.22 N.

Correct question: Two forces act on a point on the plane. The angle between the two forces is given. Find the magnitude of the resultant force. forces of 19 and 32 newtons, forming an angle of 118 degrees.

the magnitude of the resultant force is 28.53 N.

The resultant of the two vectors can be calculated using parallelogram theorem.

parallelogram theorem states that if two vectors are represented by the adjacent side of a parallelogram, the resultant of the vectors is the diagonal of the parallelogram drawn from the point of intersection of the vectors.

This can be expressed mathematically as

R² = P²+Q²-2PQcos(180-∅).............. Equation 1

Where R = resultant of the vectors, P and Q = the two vectors respectively, ∅ = angle between the vectors.

From the question,

Given: P = 19 N, Q = 32 N, ∅ = 118°

Substitute these values into equation 2

R² = 19²+32²-2×19×32cos(180-118)

R² = 361+1024-1216cos62°

R² = 1385-1216(0.4695)

R² = 1385-570.878

R² = 814.122

R = √(814.122)

R = 28.53 N

Hence, the magnitude of the resultant force is 28.53 N

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Answer:

(a) aₓ = 1.33 m/s² and aᵧ = -0.770 m/s²

(b) x = 0.665 m and y = 4.62 m

(c) 3.61 s

Explanation:

(a) There are two ways we can solve this.  The first way is to sum the forces in the x and y direction, then use the relation tan 30° = -aᵧ/aₓ, where aᵧ is the acceleration in the +y direction (up) and aₓ is the acceleration in the +x direction (right).

The second way is to sum the forces in the parallel and perpendicular directions to find the acceleration parallel to the incline, a.  Then, use the relations aᵧ = -a sin 30° and aₓ = a cos 30°.

Let's try the first method.  Sum of forces in the +y direction:

∑F = ma

N cos 30° + Nμ sin 30° − mg = maᵧ

N cos 30° + Nμ sin 30° − mg = -maₓ tan 30°

Sum of forces in the +x direction:

∑F = ma

N sin 30° − Nμ cos 30° = maₓ

Substituting:

N cos 30° + Nμ sin 30° − mg = -(N sin 30° − Nμ cos 30°) tan 30°

N cos 30° + Nμ sin 30° − mg = -N sin 30° tan 30° + Nμ sin 30°

N cos 30° − mg = -N sin 30° tan 30°

N (cos 30° + sin 30° tan 30°) = mg

N = mg / (cos 30° + sin 30° tan 30°)

N = (10 kg) (10 m/s²) / (cos 30° + sin 30° tan 30°)

N = 86.6 N

Now, solving for the accelerations:

N sin 30° − Nμ cos 30° = maₓ

aₓ = N (sin 30° − μ cos 30°) / m

aₓ = (86.6 N) (sin 30° − 0.4 cos 30°) / 10 kg

aₓ = 1.33 m/s²

N cos 30° + Nμ sin 30° − mg = maᵧ

aᵧ = N (cos 30° + μ sin 30°) / m − g

aᵧ = (86.6 N) (cos 30° + 0.4 sin 30°) / 10 kg − 10 m/s²

aᵧ = -0.770 m/s²

Now let's try the second method.

Sum of forces in the perpendicular direction:

∑F = ma

N − mg cos 30° = 0

N = mg cos 30°

Sum of forces in the parallel direction:

∑F = ma

mg sin 30° − Nμ = ma

mg sin 30° − mgμ cos 30° = ma

a = g (sin 30° − μ cos 30°)

a = (10 m/s²) (sin 30° − 0.4 cos 30°)

a = 1.536 m/s²

Solving for the accelerations:

aₓ = a cos 30°

aₓ = 1.33 m/s²

aᵧ = -a sin 30°

aᵧ = -0.770 m/s²

As you can see, the second method is faster and easier, but both methods will give you the same answer.

(b) In the x direction:

Given:

x₀ = 0 m

v₀ = 0 m/s

aₓ = 1.33 m/s²

t = 1 s

Find: x

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (1 s) + ½ (1.33 m/s²) (1 s)²

x = 0.665 m

In the y direction:

Given:

y₀ = 5 m

v₀ = 0 m/s

aᵧ = -0.770 m/s²

t = 1 s

Find: y

y = y₀ + v₀ t + ½ at²

y = 5 m + (0 m/s) (1 s) + ½ (-0.770 m/s²) (1 s)²

y = 4.62 m

(c) In the y direction:

Given:

y₀ = 5 m

y = 0 m

v₀ = 0 m/s

aᵧ = -0.770 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 m = 5 m + (0 m/s) t + ½ (-0.770 m/s²) t²

t = 3.61 s

Answer:

F= 175.5N

Explanation:

Given:

Torque which can also be called moment is defined as rotational equivalent of linear force. It is the product of the external force and perpendicular distance

torque of 38N⋅m

angle of 120∘

Torque(τ): 38Nm

position r relative to its axis of rotation: 25cm , if we convert to metre for consistency we have 0.25m

Angle: 120°

To find the Force, the torque equation will be required which is expressed below

τ = Frsinθ

We need to solve for F, if we rearrange the equation, we have the expression below

F= τ/rsinθ

Note: the torque is maximum when the angle is 90 degrees

But θ= 180-120=60

F= 38/0.25( sin(60) )

F= 175.5N

Answer:

2.9tons

Explanation:

Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:

parallel ("tangential"): F_t = F sin a

perpendicular ("normal"): F_n = F cos a

At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:

F sin a ~~ (46 tons)*0.064 ~~ 2.9tons

Also see attached file

The required force parallel to the incline to hold the monolith on this​ causeway will be "2.9 tons".

According to the question,

Angle, a = 3.7 degrees or,

Sin a = 0.064

Force, F = 46 tons

We know the relation,

Parallel (tangential), [tex]F_t[/tex] = F Sin a

By substituting the values,

                                       = 46 × 0.064

                                       = 2.9 tons

Thus the response above is appropriate answer.

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Answer:

its supposed to be (a) 1W

Answer:

f = 500 x 10^12Hz

Explanation:

E=hc/wavelength

E=hf

hc/wavelength =hf

c/wavelength =f

f = 3 x 10^8 / 600 x 10^-9 = 500 x 10^12Hz

Answer:

D &B

Explanation:

Using Fleming right hand rule that States that if the fore-finger, middle finger and the thumb of left hand are stretched mutually perpendicular to each other, such that fore-finger points in the direction of magnetic field, the middle finger points in the direction of the motion of positive charge, then the thumb points to the direction of the force

Answer:

The  critical angle is  [tex]\theta_c = \ 63.68^o[/tex]

Explanation:

From the question we are told that

   The refractive index of the core is  [tex]n_c = 1.35[/tex]

   The refractive index of the cladding  is [tex]n_s = 1.21[/tex]

Generally according to Snell's law

      [tex]\frac{sin i }{sin r } = \frac{n_s}{n_c }[/tex]

Here for total internal reflection the refractive angle is  [tex]r = 90^o[/tex] and  the critical angle is equal to the critical angle so  [tex]i = \theta_c[/tex]

      [tex]\frac{sin \theta_c }{sin (90) } = \frac{n_s}{n_c }[/tex]

substituting values

       [tex]\frac{sin \theta_c }{sin (90) } = \frac{1.21}{1.35 }[/tex]

       [tex]\theta_c = sin^{-1} [\frac{1.21}{1.35} ][/tex]

      [tex]\theta_c = \ 63.68^o[/tex]

Answer:

Explanation:

Given data

angle of incidence, i=40°

angle of refraction,r = ?

index of refraction u=  1.33

applying the formula

[tex]u=\frac{ sin i}{ sin r}[/tex]

According to Snell's law, the incident, normal and refracted rays all act on the same point.

Substituting and solving for r we have.

[tex]1.33= \frac{sin (40)}{sin (r)} \\\\sin(40)= 1.33* sin(r)\\\0.642= 1.33* sin(r)[/tex]

divide both sides by 1.33 we have

[tex]sin(r)= \frac{0.642}{1.33} \\\sin(r)= 0.48\\\r= sin^-^10.48\\\r= 28.68[/tex]

r= 28.68°