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marta [7]
3 years ago
14

What total distance will a sound wave travel in air in 3.00 seconds at stp?

Physics
1 answer:
dimaraw [331]3 years ago
4 0
At stp conditions (T=0^{\circ}C), the speed of sound is
v=331.2 m/s
The sound wave moves by uniform motion, so we can use the basic relationship between space, time and velocity:
S=vt
where S is the distance covered by the sound wave in a time t. In our problem, t=3.00 s, therefore the distance covered by the sound wave is
S=vt=(331.2 m/s)(3.00 s)=993.6 m
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Explanation:

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g_X=GM/R^2

Here,g_X = 2g

Escape velocity is given by:

v =\sqrt{ \frac{2GM} {R}} = \sqrt{2aR}

Here, R=R_earth/2

and g_X = 2g

Therefore,v=\sqrt(2(2g)(R/2))=v_0

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3 years ago
Which statement best describes the characteristics of white light? A. White light is composed of a spectrum of many different co
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A white light is, by definition, apparently colorless light, for example ordinary daylight. It contains all the wavelengths of the visible spectrum at equal intensity.

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3 years ago
gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the
igomit [66]

Answer:

d = 68.5 x 10⁻⁶ m = 68.5 μm

Explanation:

The complete question is as follows:

An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is  1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?

The answer can be given by using the formula derived from Young's Double Slit Experiment:

y = \frac{\lambda L}{d}\\\\d  =\frac{\lambda L}{y}\\\\

where,

d = slit separation = ?

λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = distance from screen (detector) = 1.7 m

y = distance between bright fringes = 15.7 mm = 0.0157 m

Therefore,

d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\

<u>d = 68.5 x 10⁻⁶ m = 68.5 μm</u>

7 0
3 years ago
The temperature of a 700.96 gram piece of metal falls 120⁰C and in the process releases 2001 Joules of energy. What is the speci
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Answer:

The specific heat for the metal is 0.466 J/g°C.

Explanation:

Given,

Q = 1120 Joules

mass = 12 grams

T₁ = 100°C

T₂ = 300°C

The specific heat for the metal can be calculated by using the formula

Q = (mass) (ΔT) (Cp)

ΔT = T₂ - T₁ = 300°C  - 100°C   = 200°C

Substituting values,

1120 = (12)(200)(Cp)

Cp = 0.466 J/g°C.

Therefore, specific heat of the metal is 0.466 J/g°C.

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A metallic circular plate with radius r is fixed to a tabletop. An identical circular plate supported from above by a cable is f
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