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4 years ago

11

What physical property of light changes and causes the light to refract, or bend?

1 answer:

Tcecarenko [31]4 years ago

3 0

I think it would be by the change in speed
For example if light travels from air to water it will slow down causing it to travel at a different angle/direction

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Protons have electrical charge; electrons have. electrical charge

tekilochka [14]

Answer:

<u>true</u>

Explanation:

they both have an electrical charge

protons have a positive charge while electrons have a negative charge but both still have an electrical charge

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3 years ago

Elan Coil [88]

Answer:

Explanation:

$a_{x}=0$ $v_{xo}= v_{o}cos(delta)t$ $a_{y}=-g$ $v_{yo}=sin$ θ

X-direction | Y-direction

$x=x_o+v_{xo}t$ ⇒ $x=v_{xo}cos(delta)t$ |

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3 years ago

Because light travels in a straight line and casts a shadow, Isaac Newton hypothesized that light is

KonstantinChe [14]

A stream of particles

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4 years ago

Marc attaches a falling 500-kg object with a rope through a pulley to a paddle wheel shaft. He places the system in a well-insul

Klio2033 [76]

Answer:

4.68227 °C

Explanation:

$m_o$ = Mass of object = 500 kg

$m_w$ = Mass of water = 25 kg

c = Specific heat of water at 20°C = 4186 J/kg°C

h = Height from which the object falls = 100 m

g = Acceleration due to gravity = 9.8 m/s²

The potential energy and heat will balance each other

$PE=Q\\\Rightarrowmc m_ogh=m_oc\Delta T\\\Rightarrow \Delta T=\frac{m_ogh}{m_oc}\\\Rightarrow \Delta T=\frac{500\times 9.8\times 100}{25\times 4186}\\\Rightarrow \Delta=4.68227\ ^{\circ}C$

The temperature change in the water is 4.68227 °C

5 0

4 years ago

An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper

dimulka [17.4K]

Answer:

The power output of this engine is $P = 17.5 W$

The the maximum (Carnot) efficiency is $\eta_c = 0.7424$

The actual efficiency of this engine is $\eta _a = 0.46$

Explanation:

From the question we are told that

The temperature of the hot reservoir is $T_h = 1250 \ K$

The temperature of the cold reservoir is $T_c = 322 \ K$

The energy absorbed from the hot reservoir is $E_h = 1.37 *10^{5} \ J$

The energy exhausts into cold reservoir is $E_c = 7.4 *10^{4} J$

The power output is mathematically represented as

$P = \frac{W}{t}$

Where t is the time taken which we will assume to be 1 hour = 3600 s

W is the workdone which is mathematically represented as

$W = E_h -E_c$

substituting values

$W = 63000 J$

So

$P = \frac{63000}{3600}$

$P = 17.5 W$

The Carnot efficiency is mathematically represented as

$\eta_c = 1 - \frac{T_c}{T_h}$

$\eta_c = 1 - \frac{322}{1250}$

$\eta_c = 0.7424$

The actual efficiency is mathematically represented as

$\eta _a = \frac{W}{E_h}$

substituting values

$\eta _a = \frac{63000}{1.37*10^{5}}$

$\eta _a = 0.46$

7 0

4 years ago