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VMariaS [17]
3 years ago
14

You adjust the temperature so that a sound wave travels more quickly through the air. You increase the temperature from 30 degre

es c to 36 degrees c. what is the velocity if the sound wave at this new temperature?
Physics
2 answers:
katen-ka-za [31]3 years ago
7 0
To calculate the velocity of the sound wave, we use this formula:
V = 331 + [0.6*T], 
Where V is the velocity and T represents temperature.
When the temperature is 36 degree Celsius, we have 
V = 331 + [0.6 * 36]
V = 331 + 21.6 = 352.6 
Therefore, V = 352.6 m/s.
777dan777 [17]3 years ago
3 0

352.6 m/s just took the test

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A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS
Flura [38]

Answer:

(a) the electrical power generated for still summer day is 1013.032 W

(b)the electrical power generated for a breezy winter day is 1270.763 W

Explanation:

Given;

Area of panel = 2 m × 4 m, = 8m²

solar flux  GS = 700 W/m²

absorptivity of the panel, αS = 0.83

efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp

panel emissivity , ε = 0.90

Apply energy balance equation to determine he electrical power generated;  

transferred energy + generated energy = 0

(radiation + convection) +  generated energy = 0

[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0

[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s

(a) the electrical power generated for still summer day

T_s = T_{\infty} = 35 ^oC = 308 \ k

[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \  \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k

P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W

(b)the electrical power generated for a breezy winter day

T_s = T_{\infty} = -15 ^oC = 258 \ k

[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \  \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k

P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W

3 0
3 years ago
An airplane flying at an altitude of 6 miles passes directly over a radar antenna. When the airplane is 10 miles away (s = 10),
Novosadov [1.4K]

Answer:

Explanation:

Given

altitude of the Plane h=6\ miles

When Airplane is s=10\ miles away

Distance is changing at the rate of \frac{\mathrm{d} s}{\mathrm{d} t}=290\ mph

From diagram we can write as

h^2+x^2=s^2

differentiate above equation w.r.t time

2h\frac{\mathrm{d} h}{\mathrm{d} t}+2x\frac{\mathrm{d} x}{\mathrm{d} t}=2s\frac{\mathrm{d} s}{\mathrm{d} t}

as altitude is not changing therefore \frac{\mathrm{d} h}{\mathrm{d} t}=0

0+x\frac{\mathrm{d} x}{\mathrm{d} t}=s\frac{\mathrm{d} s}{\mathrm{d} t}

at s=10\ miles\ and\ h=6\ miles

substitute the value we get x=\sqrt{10^2-6^2}=8\ miles

8\times \frac{\mathrm{d} x}{\mathrm{d} t}=10\times 290

\frac{\mathrm{d} x}{\mathrm{d} t}=362.5\ mph

5 0
4 years ago
When two license plates are issued they need to be attached to the _________ of the vehicle they were issued for.
Sergio039 [100]

Answer:

Front and the rear bumpers.

Explanation:

When two license plates are issued they need to be attached to the Front and the rear bumpers of the vehicle they were issued for.

7 0
3 years ago
Water has a specific heat of 4.184
LuckyWell [14K]

Answer:

Water.

Explanation:

This means:

1) For the temperature of water to raise at any point to the next degree by 1°C, will require a specific heat capacity of  4.184  J/Kg°C

2) For the temperature of wood to raise at any point to the next degree by 1°C, will require a specific heat capacity of  1.760  J/Kg°C

Note that: specific heat is directly proportional to energy, therefore the higher the heat capacity, the higher the energy.

4.184  J/Kg°C is higher than 1.760  J/Kg°C, hence WATER needs more energy.

8 0
3 years ago
A small sculpture made of brass (rho brass = 8470 kg/m 3 ) is believed to have a secret central cavity. The weight of the sculpt
GaryK [48]

Answer:

Volume of secret cavity = 4 x 10⁻⁶m³

Explanation:

The weight of the sculpture in air is 15.76 N

Mass of the sculpture = 1.61 kg

Mass = Volume x Density

1.61 = V x 8470

Volume of brass =1.90 x 10⁻⁴ m³

When it is submerged in water, the weight is 13.86 N.

That is

  Weight of sculpture - Weight of water displaced = 13.86 N

  15.76 - Weight of water displaced = 13.86

   Weight of water displaced = 1.9 N

   Mass of water displaced = 0.194 kg

 Mass = Volume x Density

  0.194 = V x 1000

  Volume of water displaced =1.94 x 10⁻⁴ m³

Volume of secret cavity =  Volume of water displaced - Volume of brass material

Volume of secret cavity = 1.94 x 10⁻⁴-1.94 x 10⁻⁴ = 0.04x 10⁻⁴ = 4 x 10⁻⁶m³

7 0
4 years ago
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