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3 years ago

What does a refractive index from medium 1 to medium 2 mean? Isn't one of the medium must be vacuum?

Physics

1 answer:

Musya8 [376]3 years ago

6 0

Refractive index is the measure of bending of a ray of light when passing from one medium into another. It is not necessary that one medium must be vaccum.

For example, the image below shows refraction, where first medium is air and second medium is water.

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Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The mass of the

IRISSAK [1]

Answer:

a₁ = 0.63 m/s² (East)

a₂ = -1.18 m/s² (West)

Explanation:

m₁ = 95 Kg

m₂ = 51 Kg

F = 60 N

a₁ = ?

a₂ = ?

To get the acceleration (magnitude and direction) of the man we apply

∑Fx = m*a (⇒)

F = m₁*a₁ ⇒ 60 N = 95 Kg*a₁

⇒ a₁ = (60N / 95Kg) = 0.63 m/s² (⇒) East

To get the acceleration (magnitude and direction) of the woman we apply

∑Fx = m*a (⇒)

F = -m₂*a₂ ⇒ 60 N = -51 Kg*a₂

⇒ a₂ = (60N / 51Kg) = -1.18 m/s² (West)

For every case we apply Newton’s 3 d Law

8 0

3 years ago

The phenomenon of vehicle "tripping" is investigated here. The sport-utility vehicle is sliding sideways with speed v1 and no an

Mrrafil [7]

Answer:

$v_1 = 3.5 \ m/s$

Explanation:

Given that :

mass of the SUV is = 2140 kg

moment of inertia about G , i.e $I_G$ = 875 kg.m²

We know from the conservation of angular momentum that:

$H_1= H_2$

$mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2$

$2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2$

$1637.1 v_1$ $= 3841.575 \omega_2$

$\omega_2 = \frac{1637.1 v_1}{3841.575}$

$\omega _2 = 0.4626 \ v_1$

From the conservation of energy as well;we have :

$T_2 +V_{2 \to 3} = T_3 \\ \\ \\ \frac{1}{2} I_A \omega_2^2 - mgh =0$

$[\frac{1}{2} [875+2140(0.765^2+0.895^2)](0.4262 \ v_1)^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0$

$706.93 \ v_1^2 - 8657.49 =0$

$706.93 \ v_1^2 = 8657.49$

$v_1^2 = \frac{8657.49}{706.93 }$

$v_1 ^2 = 12.25$

$v_1 = \sqrt{ 12.25$

$v_1 = 3.5 \ m/s$

6 0

3 years ago

Una patinadora de 50 kg parte del reposo y después de recorrer 3k alcanza una velocidad de 15 m/s. ¿Qué fuerza neta experimenta

e-lub [12.9K]

Answer:

$F_{net} = 1.875\,N$

Explanation:

Asúmase que la patinadora experimenta una aceleración constante. La fuerza neta experimentada por la patinadora:

$F_{net} = (50\,kg)\cdot \left[\frac{\left(15\,\frac{m}{s}\right)^{2}-\left(0\,\frac{m}{s}\right)^{2} }{2\cdot (3000\,m)} \right]$

$F_{net} = 1.875\,N$

6 0

3 years ago

A 60 kg man jumps down from a 0.8 m table. What is the speed when he

Rom4ik [11]

Answer: Speed = 4 m/s

Explanation:

The parameters given are

Mass M = 60 kg

Height h = 0.8 m

Acceleration due to gravity g= 10 m/s2

Before the man jumps, he will be experiencing potential energy at the top of the table.

P.E = mgh

Substitute all the parameters into the formula

P.E = 60 × 9.8 × 0.8

P.E = 470.4 J

As he jumped from the table and hit the ground, the whole P.E will be converted to kinetic energy according to conservative of energy.

When hitting the ground,

K.E = P.E

Where K.E = 1/2mv^2

Substitute m and 470.4 into the formula

470.4 = 1/2 × 60 × V^2

V^2 = 470.4/30

V^2 = 15.68

V = square root (15.68)

V = 3.959 m/s

Therefore, the speed of the man when hitting the ground is approximately 4 m/s

4 0

3 years ago

Helppppp pleaseee :(

Elza [17]

1) push down on the end of the lever, and 2) 3/4 of the way from the fulcrum

7 0

3 years ago