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Ray Of Light [21]

3 years ago

14

A ray of light strikes a smooth surface and is reflected. The angle of incidence is 35°. What can be predicted about the angle o

f reflection?

2 answers:

True [87]3 years ago

6 0

Answer:

C) Rays will follow the law of reflection, so the angle of reflection will be 35°.

Explanation:

andreev551 [17]3 years ago

5 0

By the law of reflection which says that the angle of reflection is equal to the angle of incidence, which in other words says, sin x = sin y, it can be said, based on the given that the angle of reflection is also equal to the angle of incidence of 35 degrees

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If you enjoyed teaching small or large classes and being active every day which career would be a good fit for you

Vlad [161]

You will probably be fit to be a elementary school teacher

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3 years ago

A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s aft

Anton [14]

Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

$m_1 v_1 = m_2 v_2$

$m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s$

Using the equation to find $v_1$ ,

$v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475$

Using the conservation of energy equation, we have,

$KE= \frac{1}{2}m*v^2$

$KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J$

$KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J$

$Total KE= 92077.75+13425530=92708.9J$

Now this energy over the cannonball

$KE=\frac{1}{2}m*v_2^2$

$92708.9=\frac{1}{2}15.5v_2^2$

$V_2 = 109.37m/s$

The gain in velocity is 0.37m/s

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3 years ago

suppose 384g of steam originally at 100C is quickly cooled to produce liquid water at 31C. How much heat must be removed from th

dlinn [17]

Answer:

$Q=977216.256\ J=977.216\ kJ$

Explanation:

Given:

mass of steam, $m=384\ g$

temperature of steam, $T_{is}=100^{\circ}C$

temperature of resultant water, $T_{fw}=31^{\circ}C$

We have,

latent heat of vapourization of water, $L=2256\ J.g^{-1}$

specific heat capacity of water, $c=4.186\ J.g^{-1}$

<em>When we cool the steam of 100°C then firstly it loses its latent heat to convert into water of 100°C and the further cools the water.</em>

<u>Now the heat removed from steam to achieve the final state of water:</u>

$\rm Q=latent\ heat\ of\ vapourization+sensible\ heat\ of\ water$

$Q=m(L+c.\Delta T)$

$Q=384(2256+4.186\times (100-31))$

$Q=977216.256\ J=977.216\ kJ$

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lozanna [386]

Answer:

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Explanation:

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