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3 years ago

Contrasting the Doppler effect with shock waves, the one that requires the faster source is?

Physics

2 answers:

MAXImum [283]3 years ago

6 0

I think shock waves require more speed they travel at the speed of sound

WITCHER [35]3 years ago

5 0

Answer:

Shock waves

Explanation:

The Doppler effect is well noticed at moments where the speed of the source is going slower than the speed of the waves. E.g. the change in apparent frequency of the sound of a train horn. As the train nears the observer, the blast of its horn is observed at a high pitch and as the train accelerates further, the blast of its horn is observed at a low pitch.

A shock wave is observed if the source is going at an equal speed as or faster than the wave can go. When the origin of sound moves at an equal speed as sound, it results in the source being at the front edge of the waves that it forms at all times.

The attached image shows a snapshots in time of diverse wavefronts forned by an aircraft that is accelerating at an equal speed as sound. The circular lines represent compressional wavefronts of the sound waves. Notice that these circles are bunched up at the front of the aircraft. This phenomenon is known as a shock wave. Shock waves are also produced if the aircraft moves faster than the speed of sound. If a moving source of sound moves faster than sound, the source will always be ahead of the waves that it produces. The diagram at the right depicts snapshots in time of a variety of wavefronts produced by an aircraft that is moving faster than sound. Note that the circular compressional wavefronts fall behind the faster moving aircraft (in actuality, these circles would be spheres).

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What is the science behind the making of pop-up books?

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Answer

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Explanation

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3 years ago

All else equal, the payback period for a project will decrease whenever the_______.

telo118 [61]

Answer:

(B) cash inflows are moved earlier in time.

Explanation:

The payback period stated time-frame during which the initial amount of investment should be recovered. It is expressed in the year form

The formula to compute the payback period is shown below:

Payback period = Initial investment ÷ Net cash flow

where,

The net cash flow = annual net operating income + depreciation expenses

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3 years ago

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3 years ago

Moist air initially at 1258C, 4 bar, and 50% relative humidity is contained in a 2.5-m3 closed, rigid tank. The tank contents ar

brilliants [131]

Here is the missing part of the question

To Determine the heat transfer, in kJ if the final temperature in the tank is 110 deg C

Answer:

Explanation:

The image attached below shows the process on T - v diagram

At State 1:

The first step is to find the vapor pressure

$P_{v1} = \rho_1 P_g_1$

$= \phi_1 P_{x \ at \ 125^0C}$

= 0.5 × 232 kPa

= 116 kPa

The initial specific volume of the vapor is:

$P_{v_1} v_{v_1} = \dfrac{\overline R}{M_v}T_1$

$116 \times 10^3 \times v_{v_1} = \dfrac{8314}{18} \times (125 + 273)$

$116 \times 10^3 \times v_{v_1} = 183831.7778$

$v_{v_1} = 1.584 \ m^3/kg$

At State 1:

The next step is to determine the mass of water vapor pressure.

$m_{v1} = \dfrac{V}{v_{v1}}$

$= \dfrac{2.5}{1.584}$

= 1.578 kg

Using the ideal gas equation to estimate the mass of the dry air $m_a$ $P_{a1} V = m_a \dfrac{\overline R}{M_a}T_1$

$(P_1-P_{v1}) V = m_a \dfrac{\overline R}{M_a}T_1$

$(4-1.16) \times 10^5 \times 2.5 = m_a \dfrac{8314}{28.97}\times ( 125 + 273)$

$710000= m_a \times 114220.642$

$m_a = \dfrac{710000}{114220.642}$

$m_a = 6.216 \ kg$

For the specific volume $v_{v_1} = 1.584 \ m^3/kg$ , we get the identical value of saturation temperature

$T_{sat} = 100 + (110 -100) \bigg(\dfrac{1.584-1.673}{1.210 - 1.673}\bigg)$

$T_{sat} =101.92 ^0\ C$

Thus, at $T_{sat} =101.92 ^0\ C$ , condensation needs to begin.

However, since the exit temperature tends to be higher than the saturation temperature, then there will be an absence of condensation during the process.

Heat can now be determined by using the formula

Q = ΔU + W

Recall that: For a rigid tank, W = 0

Q = ΔU + 0

Q = ΔU

Q = U₂ - U₁

Also, the mass will remain constant given that there will not be any condensation during the process from state 1 and state 2.

At State 1;

The internal energy is calculated as:

$U_1 = (m_a u_a \ _{ at \ 125^0 C})+ ( m_{v1} u_v \ _{ at \ 125^0 C} )$