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Roman55 [17]
3 years ago
11

Hich would increase the speed of a sound wave? Check all that apply.

Physics
2 answers:
asambeis [7]3 years ago
6 0

The correct answer is:

B. The medium increases in temperature while remaining in the same phase.

D. A wave passes from a gas to a liquid while remaining the same temperature.

These would increase the speed of a sound wave.

|Huntrw6|

Serggg [28]3 years ago
4 0
<span>Hich would increase the speed of a sound wave? Check all that apply. where are the answer chices?</span>
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Can someone plz check if what I typed is good :)
pochemuha

Answer:

Explanation:

6 0
3 years ago
Leila is building an aluminum-roofed shed in her backyard to store her garden tools.The flat roof will measure 2.0 x 3.0m in are
IrinaK [193]
Answer: 0.0138 m^2  = 138 cm^2

Explanation:

The thermal expansion is the term use for the physical phenomena of dilation of the objects when they are exposed to changes in temperature.

The objects dilate when they are heated and contract when they are cooled.

The dilation is proportional to the change in temperatur.

For linear dilation, the proportionality constant is called linear dilation coefficient of the materials, it is named α and is measured in °C ^-1.

ΔL = α * Lo * ΔT, which means that the dilation (or contraction) is proportional to the product of the original length (Lo) and the change of temperature (ΔT).

There is also superficial dilation, for which the dilation is:

ΔA = β * Ao * ΔT, which means that the superficial dilation (or contraction) is proportional to the product of the original area (Ao) and the change of temperature (ΔT).

It is very interesting and important to solve problems that β = 2α, because regularly you will find the values of α for different materials and so, you just to multiply it times 2 to use β.

For this problem:

- Original area, Ao = area of the flat roof at - 10°C = 2.0m * 3.0m = 6.0 m^2.
- α for aluminum = 24 * 10^ -6 °C^-1.
- ΔT = 38°C - (-10°C) = 48°C

So, ΔA = 6.0m^2 * (2 * 24*10^-6 °C&-1) * 48°C = 0.0138 m^2

And that is the area that should stick out in summer to fit the structure during cold winter nights.

You can pass that number to cm^2 to grasp better the idea of this size:

0.0138 m^2 * (100 cm)^2 / m^2 = 138 cm^2




3 0
3 years ago
NEED ASAP PLEASE !!
Svetllana [295]
The balloon was 30.65 meters above ground.

ANSWER: A
3 0
3 years ago
Read 2 more answers
During a chemical reaction, some substances are completely consumed while others my not be. What is the substance that is comple
LUCKY_DIMON [66]

Answer:

It is Limiting Reactant

Explanation:

5 0
3 years ago
An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,
insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}

Where

t_{hf} = Temperature at the inside of the pipe = 300 °C

t_{f} = Temperature at the outside of the pipe = 20 °C

r₁ =internal  radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

k_A = 15 W/m·K

k_B = 0.038 W/m·K

h_{hf} = 75 W/m²·K

h_{cf} = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

Q = \frac{t_A-t_B}{R_T} Where R_T for the air film on the pipe outer surface is given by

R_T= \frac{1}{\alpha A}

where A =area of the outside of the pipe

= \frac{1}{10*2\pi*0.07*1 } = 0.227 K/W

Therefore

131.32 W = \frac{t_A-20}{0.227} which gives

t_A = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

8 0
3 years ago
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