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ASHA 777 [7]
4 years ago
11

Suppose the glass paperweight has index of refraction n=1.38. a) find the value of θ for which the reflection on the vertical su

rface of the paperweight exactly satisfies the condition for total internal reflection. b) If θ is increased, is the reflection at the vertical surface still total?
Physics
1 answer:
zimovet [89]4 years ago
8 0

Answer:

a)θ=71.89°

b)NO

Explanation:

Given that

For glass n= 1.38

We know that for air n'=1

The angle  for total internal reflection θc given as

sin θc=n'/n

By putting the values

sin θc=n'/n

sin θc=1/1.38

θc=46.43°

n'sinθ = n sinθref

sinθref = cosθc

n'sinθ =  n cosθc

1 x sinθ =1.38 x cos 46.43°

θ=71.89°

b)

NO

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A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.
Nesterboy [21]

Answer:

<em>The velocity after the collision is 2.82 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}

\displaystyle v'=\frac{22.287}{7.91}

\boxed{v' = 2.82\ m/s}

The velocity after the collision is 2.82 m/s

6 0
2 years ago
How is the temperature of a gas related to the kinetic energy of its particles?
devlian [24]

Answer:

As the temperature increases, the kinetic energy of the particles increases.

Explanation:

When the temperature of the substance increases, the velocity increases which makes the movement of the particles to speed up. This causes the particles to increase. Therefore, as the temperature increases, the kinetic energy of the particles also increases.

4 0
3 years ago
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We now have two tappers that are tapping the surface of the water together. They are tapping in sync (both are hitting the surfa
ollegr [7]

Answer:

The cork moves up and down.

Explanation:

The cork moves up and down because of the waves produce due to tapping the surface of water. The cork experience movement due to placed on the water surface so when the wave passes through the medium by tapping of water, movement of cork also occur. Cork has high volume and lower mass so it floats on the water surface and experience movement due to waves.

5 0
3 years ago
An 800-g block of ice at 0.00°C is resting in a large bath of water at 0.00°C insulated from the environment. After an entropy c
Allisa [31]

Answer:

Unmeltedd ice = 308.109 g

Explanation:

Gibbs Free energy:

A systems Gibbs Free Energy is defined as the free energy of the product of the absolute temperature and the entropy change less than the enthalpy change.

Therefore, G = ΔH-TΔS

where G is Gibbs Free Energy

          ΔH is enthalpy change

          T is absolute temperature

          ΔS is entropy change

Here since there is a phase change, therefore G will be 0.

∴ΔH = TΔS

Given: Temperature, T = 0°C = 273 K

           Entropy change,ΔS = 600 J/K

           Latent heat of fusion of water = 333 J/g

∴ΔH = TΔS

  ∴ΔH = 273 x 600

           = 163800 J

So this is the amount of enthalpy that will be used into melting of ice.

  ∴ΔH = mass of ice melted x latent heat of fusion of water

    Mass of ice melted = ΔH / latent heat of fusion of water

                                     = 163800 / 333

                                     = 491.891 g

This is the mass of ice melted.

And initial amount of ice is 800 g

Amount of ice left after melting = Initial amount of ice - amount of ice melted

                                                   = 800-491.891

                                                  = 308.109 g

Amount of ice remained after melting = 308.109 g

8 0
3 years ago
A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m
MissTica

Answer:

  • The magnitude of the vector \vec{C} is 107.76 m

Explanation:

To find the components of the vectors we can use:

\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | is the magnitude of the vector, and θ is the angle over the positive x axis.

The negative x axis is displaced 180 ° over the positive x axis, so, we can take:

\vec{A} = 56.0 \ m \ ( \ cos( 180 \° + 30 \°) \ , \ sin (180 \° + 30 \°) \ )

\vec{A} = 56.0 \ m \ ( \ cos( 210 \°) \ , \ sin (210 \°) \ )

\vec{A} = ( \ -48.497 \ m \ , \ - 28 \ m \ )

\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )

\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )

\vec{B} = ( \ -53.797 \ m \ , \ 61.886\ m \ )

Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)

So, for our vectors:

\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ ,  ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )

\vec{C} = ( \ - 102.294 \ m , \ 33.886 m \ )

To find the magnitude of this vector, we can use the Pythagorean Theorem

|\vec{C}| = \sqrt{C_x^2 + C_y^2}

|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}

|\vec{C}| =107.76 m

And this is the magnitude we are looking for.

5 0
3 years ago
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