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ch4aika [34]
3 years ago
5

A student is investigating the differences between light waves and sound waves. The student does this by using a capsule filled

with solid glass at one end and a vacuum at the other end. The student will transmit waves into the capsule at a 30° angle to the (normal) centerline.
alert your teacher if the image is missing

Which >>TWO<< questions should the student ask, and which predictions are MOST LIKELY correct based on this investigation?

Group of answer choices

question: Can electromagnetic waves and mechanical waves travel from a solid glass medium into a liquid medium?

prediction: The electromagnetic waves will continue through the liquid medium, while the mechanical waves will go no farther.


question: How are electromagnetic waves and mechanical waves affected when traveling from a solid glass medium to a vacuum?

prediction: The electromagnetic waves will continue through the vacuum, while the mechanical waves will go no farther.


question: Can electromagnetic waves and mechanical waves travel from a solid glass medium into a liquid medium?

prediction: Both electromagnetic waves and mechanical waves will bend, showing that they have passed through each medium.


question: How are electromagnetic waves and mechanical waves affected when traveling from a solid glass medium to a vacuum?

prediction: The electromagnetic waves and mechanical waves will continue through the vacuum at a lower speed.


question: How is the speed of electromagnetic waves affected when traveling from a solid glass medium to a vacuum at a 30° angle?

prediction: The electromagnetic waves will travel in a straight line, showing that they have maintained a constant speed.


question: How is the speed of electromagnetic waves affected when traveling from a solid glass medium to a vacuum at a 30° angle?

prediction: The electromagnetic waves will bend downward, showing that they have sped up slightly
Physics
2 answers:
denis-greek [22]3 years ago
5 0

Answer:

Explanation:

The investigation is on differences between light waves and sound waves. Because the light and sound are transmitted at a 30° angle at a capsule with glass and vacuum, questions asked should show difference answers for light and sound.

First and third question talk about liquid which is not related here.

Fouth and sixth prediction are wrong as electromagnetic wave will not slow down or speed up in vacuum.

That leaves the second and fifth set as correct answers:

question: How are electromagnetic waves and mechanical waves affected when traveling from a solid glass medium to a vacuum?

prediction: The electromagnetic waves will continue through the vacuum, while the mechanical waves will go no farther.

question: How is the speed of electromagnetic waves affected when traveling from a solid glass medium to a vacuum at a 30° angle?

prediction: The electromagnetic waves will travel in a straight line, showing that they have maintained a constant speed.

Shalnov [3]3 years ago
3 0

Answer:

The two questions and predictions are:

question: How are electromagnetic waves and mechanical waves affected when traveling from a solid glass medium to a vacuum?

prediction: The electromagnetic waves will continue through the vacuum, while the mechanical waves will go no farther.

question: How is the speed of electromagnetic waves affected when traveling from a solid glass medium to a vacuum at a 30° angle?

prediction: The electromagnetic waves will travel in a straight line, showing that they have maintained a constant speed.

Explanation:

light is EM wave which is transverse while sound is longitudinal. speed changes for long wave but not EM.

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A puck of mass 0.70 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the movi
natali 33 [55]

Answer:

  • v_1  =  \ 5.196 \frac{m}{s}
  • v_2 =  3 \frac{m}{s}

Explanation:

For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

\vec{p}_i = \vec{p}_f

where the suffix i  means initial, and the suffix f means final.

The initial momentum will be:

\vec{p}_i = m_1 \ \vec{v}_{1_i} + m_2 \ \vec{v}_{2_i}

as the second puck is initially at rest:

\vec{v}_{2_i} = 0

Using the unit vector \vec{i} pointing in the original line of motion:

\vec{v}_{1_i} = 6.0 \frac{m}{s} \hat{i}

\vec{p}_i = 0.70 \ kg  \ 6.0 \frac{m}{s} \ \hat{i} + 0.70 \ kg \ 0

\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i}

So:

\vec{p}_i =  4.2 \ \frac{kg \ m}{s} \ \hat{i} = \vec{p}_f

\vec{p}_f =  4.2 \ \frac{kg \ m}{s} \ \hat{i}

Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

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

So, our velocity vectors will be:

\vec{v}_{1_f} = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )

\vec{v}_{2_f} = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )

We got

\vec{p}_f = 0.7 \ kg \ \vec{v}_{1_f} + 0.7 \ kg \ \vec{v}_{2_f}

4.2 \ \frac{kg \ m}{s} \ \hat{i} = 0.7 \ kg \   v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )  + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )

So, we got the equations:

4.2 \ \frac{kg \ m}{s}  = 0.7 \ kg \   v_1 \  cos(30 \°) + 0.7 \ kg \ v_2 \  cos(-60 \°)

and

0  = 0.7 \ kg \   v_1 \  sin(30 \°) + 0.7 \ kg \ v_2 \  sin(-60 \°).

From the last one, we get:

0  = 0.7 \ kg \  ( v_1 \  sin(30 \°) +  \ v_2 \  sin(-60 \°) )

0  =  v_1 \  sin(30 \°) +  \ v_2 \  sin(-60 \°)

v_1 \  sin(30 \°) = -  \ v_2 \  sin(-60 \°)

v_1  =  \ v_2 \  \frac{sin(60 \°)}{ sin(30 \°) }

and, for the first one:

4.2 \ \frac{kg \ m}{s}  = 0.7 \ kg  \ (  v_1 \  cos(30 \°) + v_2 \  cos(60 \°) )

\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} =    v_1 \  cos(30 \°) + v_2 \  cos(60 \°)

\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} =    v_1 \  cos(30 \°) + v_2 \  cos(60 \°)

6 \ \frac{m}{s} =    (\ v_2 \  \frac{sin(60 \°)}{ sin(30 \°) } ) \  cos(30 \°) + v_2 \  cos(60 \°)

6 \ \frac{m}{s} = v_2     (\   \frac{sin(60 \°)}{ sin(30 \°) } ) \  cos(30 \°) +   cos(60 \°)

6 \ \frac{m}{s} = v_2  * 2

so:

v_2 = 6 \ \frac{m}{s} / 2 = 3 \frac{m}{s}

and

v_1  =  \ 3 \frac{m}{s}  \  \frac{sin(60 \°)}{ sin(30 \°) }

v_1  =  \ 5.196 \frac{m}{s}

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3 years ago
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12 1/2 i would assume
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Assume you need to design a hydronic system that can deliver 80,000 Btu/hr. What flow rate of water is required if the temperatu
PolarNik [594]

Answer:

At 10°F change in temperature

Mass flowrate = 1.01 kg/s = 2.227 lbm/s

Volumetric flowrate = 1010 m³/s = 35667.8 ft³/s

At 20°F change in temperature

Mass flowrate = 0.505 kg/s = 1.113 lbm/s

Volumetric flowrate = 505 m³/s = 17833.9 ft³/s

Explanation:

80000 btu/hr = 23445.7 W

P = ṁc(ΔT)

ṁ = MASS flowrate

c = specific heat capacity of water = 4182 J/kg.K,

ΔT = change in temperature = 10°F

To convert, a change of 18°F is equal to a change of 10°C

A change of 10°F = 10×10/18 = 5.556°C = 5.556K

P = ṁc(ΔT)

23445.7 = ṁ(4182 × 5.556)

ṁ = 23445.7/(4182 × 5.556)

ṁ = 1.01 kg/s = 2.227 lbm/s

In volumetric flow rate, Q = density × mass flowrate = 1000 × 1.01 = 1010 m³/s = 35667.8 ft³/s

For a change of 20°F,

ΔT = change in temperature = 20°F

To convert, a change of 18°F is equal to a change of 10°C

A change of 20°F = 20×10/18 = 11.1111°C = 11.111K

P = ṁc(ΔT)

23445.7 = ṁ(4182 × 11.111)

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ṁ = 0.505 kg/s = 1.113 lbm/s

In volumetric flow rate, Q = density × mass flowrate = 1000 × 0.505 = 505 m³/s = 17833.9 ft³/s

Hope this Helps!!!

4 0
3 years ago
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Where they slide over each other.

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In a transform boundary, neither plate is added to at the boundary nor destroyed.  They are marked in some places by features like  stream beds that have been split in half and the two halves moved in opposite directions.



7 0
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