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Y_Kistochka [10]
2 years ago
13

what will happen when two wave pulses that are the same size approach one another from opposite directions?

Physics
1 answer:
xenn [34]2 years ago
5 0
When two or more waves meet, they interact with each other. The interaction of waves with other waves is called wave interference. Wave interference may occur when two waves that are traveling in opposite directions meet. The two waves pass through each other, and this affects their amplitude.
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A sample of gold has a mass of 30.94 grams and density of 19.32g/cm^3. What volume of space will this sample of gold occupy?
sdas [7]
Data:
mass, m = 30.94 g
density, d = 19.32 g/cm^3

Formula: d = m / v => v = m / d = 30.94 g / 19.32 g/cm^3 = 1.60 cm^3

Then, the answer is the option C. 
6 0
3 years ago
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During the storm a tree fell over into a river what might happen to the tree
mart [117]
The tree might get swept away by the current and it will disappear when it catches on something
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Adding heat to a liquid causes which of the following physical changes?
Nadya [2.5K]
Decreases density because gases have less density than liquids
4 0
3 years ago
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A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. ten second later, he is moving at 15 m/s. what is his
pantera1 [17]

\Large {{ \sf {Question :}}}

<h3>A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. Ten second later, he is moving at 15 m/s. What is his acceleration?</h3>

\Large {{ \sf {Given :}}}

<h3>Initial Velocity (<em>u</em>) - 5 m/s</h3><h3>Final Velocity (<em>v</em>) - 15 m/s</h3><h3>Time (<em>t</em>) - 10 sec</h3>

\Large {{ \sf {Formulae  :}}}

<h3>If the velocity of an object changes from an initial value <em>u </em>to the final value <em>v </em>in time <em>t,</em><em> </em>the acceleration <em>a</em> is, </h3><h3>a \:  =  \frac{v - u}{t}</h3><h3>\Large {{ \sf {Step-by-step explanation :}}}</h3>

a \:  =  \frac{v - u}{t}  \\ or \:  \: a =  \frac{(15 - 5)}{10} m \: s^{ - 2}  \\ or \:  \: a \:  =  \frac{10}{10}m \: s^{ - 2} \\ or \:  \: a = 1m \: s^{ - 2}

\Large {{ \sf {Answer :}}}

<h3>His acceleration is </h3><h3>1m \: s^{ - 2}</h3><h3 /><h3 />
5 0
3 years ago
An air bubble of volume 20 cm³ is at the bottom of a lake 40 m deep, where the temperature is 4.0°C. The bubble rises to the sur
soldi70 [24.7K]

Answer:

100 cm³

Explanation:

Use ideal gas law:

PV = nRT

where P is absolute pressure, V is volume, n is number of moles, R is ideal gas constant, and T is absolute temperature.

n and R are constant, so:

P₁V₁/T₁ = P₂V₂/T₂

If we say point 1 is at 40m depth and point 2 is at the surface:

P₂ = 1.013×10⁵ Pa

T₂ = 20°C + 273.15 = 293.15 K

P₁ = ρgh + P₂

P₁ = (1000 kg/m³ × 9.8 m/s² × 40 m) + 1.013×10⁵ Pa

P₁ = 4.933×10⁵ Pa

T₁ = 4.0°C + 273.15 = 277.15 K

V₁ = 20 cm³

Plugging in:

(4.933×10⁵ Pa) (20 cm³) / (277.15 K) = (1.013×10⁵ Pa) V₂ / (293.15 K)

V₂ = 103 cm³

Rounding to 1 sig-fig, the bubble's volume at the surface is 100 cm³.

6 0
3 years ago
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