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3 years ago

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What is the difference in light that is refracted compared to light that is reflected? Think in terms of speed of light as well

as what happens to light waves when they interact with a medium.

1 answer:

Ber [7]3 years ago

4 0

Answer:

The refracted light wave is bent at an angle while the reflected light wave is bounced back either at 90° or at angle less than 180°.

The refracted light wave changes its speed when it moves from one medium to another based on the density of the medium.

The reflected light does not change its speed once it contacts another medium. It just bounces back with the same speed.

Explanation:

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Three metal fishing weights, each with a mass of 1.00x102 g and at a temperature of 100.0°C, are placed in 1.00x102 g of water a

worty [1.4K]

Answer:

Approximately $0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}$ assuming no heat exchange between the mixture and the surroundings.

Explanation:

Consider an object of specific heat capacity $c$ and mass $m$ . Increasing the temperature of this object by $\Delta T$ would require $Q = c\, m \, \Delta T$ .

Look up the specific heat of water: $c(\text{water}) = 4.182\; {\rm J \cdot g^{-1} \cdot K^{-1}}$ .

It is given that the mass of the water in this mixture is $m(\text{water}) = 1.00 \times 10^{2}\; {\rm g}$ .

Temperature change of the water: $\Delta T(\text{water}) = (45 - 35)\; {\rm K} = 10\; {\rm K}$ .

Thus, the water in this mixture would have absorbed :

$\begin{aligned}Q &= c\, m\, \Delta T \\ &= 4.182\; {\rm J \cdot g^{-1}\cdot K^{-1}} \\ &\quad \times 1.00 \times 10^{2}\; {\rm g} \times 10\; {\rm K} \\ &= 4.182 \times 10^{3}\; {\rm J}\end{aligned}$ .

Thus, the energy that water absorbed was: $Q(\text{water}) = 4.182 \times 10^{3}\; {\rm J}$ .

Assuming that there was no heat exchange between the mixture and its surroundings. The energy that the water in this mixture absorbed, $Q(\text{water})$ , would be the opposite of the energy that the metal in this mixture released.

Thus: $Q(\text{metal}) = -Q(\text{water}) = -4.182 \times 10^{3}\; {\rm J}$ (negative because the metal in this mixture released energy rather than absorbing energy.)

Mass of the metal in this mixture: $m(\text{metal}) = 3 \times 1.00 \times 10^{2}\; {\rm g} = 3.00 \times 10^{2}\; {\rm g}$ .

Temperature change of the metal in this mixture: $\Delta T(\text{metal}) = (100 - 45)\; {\rm K} = 55\; {\rm K}$ .

Rearrange the equation $Q = c\, m \, \Delta T$ to obtain an expression for the specific heat capacity: $c = Q / (m\, \Delta T)$ . The (average) specific heat capacity of the metal pieces in this mixture would be:

$\begin{aligned}c &= \frac{Q}{m\, \Delta T} \\ &= \frac{-4.182 \times 10^{3}\; {\rm J}}{3.00 \times 10^{2}\; {\rm g} \times (-55\; {\rm K})} \\ &\approx 0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}\end{aligned}$ .

6 0

2 years ago

More work was done. Which statement best describes the cause? There was more force applied; therefore, there was less movement.

yanalaym [24]

There was more force applied therefore there was more movement. Hope this helps.

8 0

4 years ago

Read 2 more answers

An experiment is performed aboard the International Space Station to verify that linear momentum is conserved during collisions

Anni [7]

Answer:

$v=15.9554\ m.s^{-1}$

Explanation:

Given:

mass of the honey drop 1, $m_1=35.5\times 10^{-3}\ kg$

velocity of the honey drop 1, $v_1=13.1\ m.s^{-1}$

mass of the honey drop 2, $m_1=52.3\times 10^{-3}\ kg$

velocity of the honey drop 2, $v_2=14.5\ m.s^{-1}$

mass of the honey drop 3, $m_1=75.7\times 10^{-3}\ kg$

velocity of the honey drop 3, $v_3=18.3\ m.s^{-1}$

<em>In ISS there is zero gravity an the collision is completely inelastic.</em>

<u>So, applying the law of conservation of momentum:</u>

$m_1.v_1+m_2.v_2+m_3.v_3=(m_1+m_2+m_3).v$

$35.5\times 10^{-3}\times 13.1+52.3\times 10^{-3}\times 14.5+75.7\times 10^{-3}\times 18.3=(35.5+52.3+75.7)\times 10^{-3}\times v$

$v=15.9554\ m.s^{-1}$

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3 years ago

Which of the following must an engineer take into account when designing a roller coaster?

saveliy_v [14]

Answer:

The correct answer would be A

8 0

3 years ago

What is the formula used to calculate the speed of an object

navik [9.2K]

The formula for speed is v=dt

Where
v=velocity (speed with direction)
d=distance
t=time of displacement

4 0

3 years ago