Example of centrifugal force <br>(wrong answer will get reported)(right answer will get brainliest)

A parallel beam of light in air makes an angle of 43.5 ∘ with the surface of a glass plate having a refractive index of 1.68. Yo

aniked [119]

Answer:

a) 46.5º b) 64.4º

Explanation:

To solve this problem we will use the laws of geometric optics

a) For this part we will use the law of reflection that states that the reflected and incident angle are equal

θ = 43.5º

This angle measured from the surface is

θ_r = 90 -43.5

θ_s = 46.5º

b) In this part the law of refraction must be used

n₁ sin θ₁ = n₂. Sin θ₂

sin θ₂ = n₁ / n₂ sin θ₁

The index of air refraction is n₁ = 1

The angle is this equation is measured between the vertical line called normal, if the angles are measured with respect to the surface

θ_s = 90 - θ

θ_s = 90- 43.5

θ_s = 46.5º

sin θ₂ = 1 / 1.68 sin 46.5

sin θ₂ = 0.4318

θ₂ = 25.6º

The angle with respect to the surface is

θ₂_s = 90 - 25.6

θ₂_s = 64.4º

measured in the fourth quadrant

3 0

3 years ago

Yes this nees helppppppppppppppp

kozerog [31]

Answer:

lol

Explanation:

it was funny

6 0

3 years ago

The luxury liner Queen Elizabeth 2 has a diesel-electric powerplant with a maximum power of 90 MW at a cruising speed of 31.5 kn

AlexFokin [52]

Answer:

5558643.69 N

Explanation:

F = Force

v = Velocity = 31.5 knots

Converting to m/s

$1\ knot=0.514\ m/s$

$31.5\ knot=31.5\times 0.514\ m/s=16.191\ m/s$

Power is given by

$P=Fv\\\Rightarrow F=\frac{P}{v}\\\Rightarrow F=\frac{90\times 10^6}{16.191}\\\Rightarrow F=5558643.69\ N$

The forward force is exerted on the ship at this highest attainable speed is 5558643.69 N

5 0

3 years ago

The distance between adjacent nodes in a standing wave pattern in a length of string is 25.0 cm:A. What is the wavelength of wav

mina [271]

A) 50 cm

B) 10000 cm/s

Explanation

Step 1

A)

If you know the distance between nodes and antinodes then use this equation:

$\begin{gathered} \frac{\lambda}{2}=D \\ \text{where}\lambda\text{ is the wavelength} \\ D\text{ is the distance betw}een\text{ nodes} \end{gathered}$

then, let

$D=\text{ 25 cm }$

now, replace to find the wavelength

$\begin{gathered} \frac{\lambda}{2}=25 \\ \text{Multiply both sides by 2} \\ \frac{\lambda}{2}\cdot2=25\cdot2 \\ \lambda=50\text{ Cm} \end{gathered}$

so, the wavelength is

A) 50 cm

Step 2

The speed of a wave can be found using the equation

$v=\lambda f$

or velocity = wavelength x frequency,

then,let

$\begin{gathered} \lambda=50\text{ cm} \\ f=200\text{ Hz} \end{gathered}$

replace and evaluate

$\begin{gathered} v=\lambda f \\ v=50\text{ cm }\cdot200\text{ HZ} \\ v=10000\text{ }\frac{\text{cm}}{s} \end{gathered}$

so

B) 10000 cm/s

I hope this helps you

6 0

1 year ago

To determine the muzzle velocity of a bullet fired from a rifle, you shoot the 2.47-g bullet into a 2.43-kg wooden block. The bl

Elza [17]

Answer:

The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 2.47 g = 0.00247 kg

mass of the wooden block, m₂ = 2.43 kg

initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

Apply the principle of conservation of linear momentum;

Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

0.00247v₁ + 2.43 x 0 = v₂(2.43 + 0.00247)

0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

K.E = ¹/₂ (2.4325)v₂²

The potential energy of the bullet-block system after collision;

P.E = mgh

P.E = (2.4325)(9.8)(0.00295)

P.E = 0.07032

Apply the principle of conservation of mechanical energy;

K.E = P.E

¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂² = 0.07032

v₂² = 0.07032 / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

5 0

3 years ago

Example of centrifugal force (wrong answer will get reported)(right answer will get brainliest)​

Example of centrifugal force (wrong answer will get reported)(right answer will get brainliest)