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My name is Ann [436]
3 years ago
5

A ray diagram shows the ray passing from medium 1 to medium 2. After passing into medium 2, the ray bends away from the normal.

Physics
1 answer:
nalin [4]3 years ago
7 0

If a ray passing from one medium to other medium then it will refract either away from normal or towards the normal

This all we can find out using snell's law

\mu_1sin\theta_1 = \mu_2sin\theta_2

now if we wish to find the angle inside other medium then we will have

sin\theta_2 = \frac{\mu_1}{\mu_2} sin\theta_1

now if light ray bend towards the normal then

\mu_2 > \mu_1

so second medium must be optical denser

otherwise if light ray is bending away from the normal then

\mu_2 < \mu_1

so second medium is rarer than first medium

so for the given condition of the question correct answer will be

<em>refraction with medium 1 denser than medium 2</em>

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pentagon [3]
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3 years ago
Problem:
pav-90 [236]

Answer:

Total energy is constant

Explanation:

The laws of thermodynamics state that thermal energy (heat) is always transferred from a hot body (higher temperature) to a cold body (lower temperature).

This is because in a hot body, the molecules on average have more kinetic energy (they move faster), so by colliding with the molecules of the cold body, they transfer part of their energy to them. So, the temperature of the hot body decreases, while the temperature of the cold body increases.

This process ends when the two bodies reach the same temperature: we talk about thermal equilibrium.

In this problem therefore, this means that the thermal energy is transferred from the hot water to the cold water.

However, the law of conservation of energy states that the total energy of an isolated system is constant: therefore here, if we consider the hot water + cold water as an isolated system (no exchange of energy with the surroundings), this means that their total energy remains constant.

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3 years ago
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 29.4 m/s2 m /
Sidana [21]

Answer:

The maximum height is 2881.2 m.

Explanation:

Given that,

Acceleration = 29.4 m/s²

Time = 7.00 s

We need to calculate the distance

Using equation of motion

s=ut+\dfrac{1}{2}at^2

Put the value into the formula

s=0+\dfrac{1}{2}\times29.4\times7^2

s=720.3\ m

We need to calculate the velocity

Using formula of velocity

v=a\times t

Put the value into the formula

v=29.4\times7

v=205.8\ m/s

We need to calculate the height

Using formula of height

H=\dfrac{v^2}{2g}

Put the value into the formula

H=\dfrac{(205.8)^2}{2\times9.8}

H=2160.9\ m

We need to calculate the maximum height

Using formula for maximum height

H'=H+s

Put the value into the formula

H'=2160.9+720.3

H'=2881.2\ m

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4 0
3 years ago
A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the
lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

l^2 = x^2 + y^2-----(1)

Differentiating with respect to t ( time ),

0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}  ( l = 26 feet = constant )

\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}

\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}

We have,

y = 10, \frac{dx}{dt}= -3\text{ feet per min}

\frac{dy}{dt}=\frac{3x}{10}-----(X)

(i) From equation (1),

26^2 = x^2 + 10^2

676=x^2 + 100

576 = x^2

\implies x = 24\text{ feet}

From equation (X),

\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}

(ii) From equation (X),

\frac{dy}{dt}\propto x

Thus, for different value of x the value of \frac{dy}{dt} would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

\frac{dy}{dt}=0\text{ feet per min}

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3 years ago
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puteri [66]

When you set a heavy bag down on the ground, you are doing negative work on it.

4 0
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