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Katarina [22]
3 years ago
10

2. Jenny picks up a guitar and strums. If the sound from the string emanates out at 440 Hz,

Physics
1 answer:
erik [133]3 years ago
7 0

Answer:

6.8×10^6 m

Explanation:

f = velocity of light/ wavelength

440= 3×10^8 / wavelength

so

wavelength = 3×10^7/ 44

= 6.8×10^6 m

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A submersible pump is put under the water at the bottom of a well and is used to push water up through a pipe. What minimum outp
Maslowich

Answer:

695800 N/m^2 or Pa

Explanation:

Height of the water from the ground H  =  71 m

Acceleration due to gravity g =9.8 m/s^2

density of water ρ= 1000 kg/m^3

The minimum output gauge pressure to make water reach height H

P= ρgH

= 1000×9.8×71= 695800 N/m^2 or Pa

5 0
3 years ago
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)

ṁ = 23445.7/(4182 × 11.111)

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
The X-ray source Cygnus X-1 has a mass of at least 11 solar masses and a diameter of only about one-quarter the diameter of the
SOVA2 [1]

Answer:

Answer is It was deduced from the rate at which it glimmers.

Refer below.

Explanation:

The X-ray source Cygnus X-1 has a mass of at least 11 solar masses and a diameter of only about one-quarter the diameter of the Earth. With such a small diameter it must be a compact object, and with such a large mass it can't be a white dwarf or a neutron star, so a black hole is the only possibility remaining. The diameter of Cygnus X-1 found:

It was deduced from the rate at which it glimmers.

7 0
3 years ago
A wall (of thermal conductivity 0.98 W/m · ◦ C) of a building has dimensions of 3.7 m by 15 m. The average inside and outside
disa [49]

Answer:

Answer:

the amount of energy flowing is 1.008x10⁹J

Explanation:

To calculate how much heat flows, the expression is the following:

Where

K=thermal conductivity=0.81W/m°C

A=area=6.2*12=74.4m²

ΔT=30-8=22°C

L=thickness=8cm=0.08m

t=time=16.9h=60840s

Replacing:

Explanation:

4 0
3 years ago
A sinusoidal voltage is given by the expression ????(????)=20cos(5π×103 ????+60°) V. Determine its (a) frequency in hertz, (b) p
MA_775_DIABLO [31]

<em>There are some placeholders in the expression, but they can be safely assumed</em>

Answer:

(a) f=1617.9\ Hz

(b) T=0.618\ ms

(c) A=20 \ Volts

(d) \varphi=60^o

Explanation:

<u>Sinusoidal Waves </u>

An oscillating wave can be expressed as a sinusoidal function as follows

V(t)&=A\cdot \sin(2\pi ft+\varphi )

Where

A=Amplitude

f=frequency

\varphi=Phase\  angle

The voltage of the question is the sinusoid expression  

V(t)=20cos(5\pi\times 103t+60^o)

(a) By comparing with the general formula we have

f=5\pi\times 103=1617.9\ Hz

\boxed{f=1617.9\ Hz}

(b) The period is the reciprocal of the frequency:

\displaystyle T=\frac{1}{f}

\displaystyle T=\frac{1}{1617.9\ Hz}=0.000618\ sec

Converting to milliseconds

\boxed{T=0.618\ ms}

(c) The amplitude is

\boxed{A=20 \ Volts}

(d) Phase angle:

\boxed{\varphi=60^o}

4 0
3 years ago
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