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Wittaler [7]
3 years ago
7

What two factors affect the speed of a sound wave as it travels through a medium ?​

Physics
1 answer:
iragen [17]3 years ago
6 0
The material that makes up the medium such as air or water and the temperature.
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Which projectiles will be visibly affected by air resistance when they fall
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Balloon
this could be it
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A student performs a lab measuring the velocities of toy cars of different masses
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The answer is Car 1 and Car 2.
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4 years ago
Read 2 more answers
The _of air resistance
Anettt [7]

Answer:

A)friction

Explanation:

The friction of air resistance slows down the acceleration due to gravity for real-world objects.

Air resistance is a form of friction found in atmosphere.

Frictional force is a force that opposes the motion of a body. It acts in the opposite direction of motion. Friction slows down real-world objects moving through the air.

Without friction, motion will not be possible.

8 0
3 years ago
A wire as a length of 1.50m, diameter 0.60mm and resistance of 2ohms. calculate the resistance R of a wire of the same materials
Otrada [13]
  1. R=ρ×L/A

ρ=R×A/L

A=pie(r²)=pie(0.6²)=pie(0.36)=1.13

ρ=2ohm×1.13mm²/1500mm

ρ=0.0015ohm.mm

2. we were told that the wires were made of the same substance so the resistivity(ρ)is the same for both wires so:

R=ρ×L/A

R=0.0015ohm.mm×500mm/0.09mm²

R=8.3'ohm

so our resistance for the second wire is :

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>R</em><em>=</em><em>8</em><em>.</em><em>3</em><em>'</em><em>ohm</em>

3 0
3 years ago
Introduction: The specific heat capacity of a substance is the amount of energy needed to change the temperature of that substan
ki77a [65]

Answer:

A) 8,368 J

B) ) 0.893 J/gºC

Explanation:

A)

  • The heat gained by the water can be obtained solving the following equation:

       q_{g} = c_{w} * m *  \Delta T (1)

  • where cw = specific heat of water = 4.184 J/gºC
  • m= mass of water = 1,000 g
  • ΔT = 2ºC
  • Replacing these values in (1) we get:

       q_{g} = c_{w} * m *  \Delta T = 4.184 J/gºC*1,000 g* 2ºC = 8,368 J (2)

B)

  • Assuming that the heat energy gained by the water is equal to the one lost by the aluminum, we can use the same equation, taking into account that the energy is lost by the aluminum, so the sign is negative:  -8,368 J.
  • Replacing by the mass of aluminum (125 g), and the change in temperature (-74.95ºC), in (1), we can solve for the specific heat of aluminum, as follows:

       q_{l} = c_{Al} * m_{Al} *  \Delta T  (3)

⇒    -8,368 J = c_{Al}* 125 g * (-74.95ºC) (4)

       c_{Al} = \frac{-8,368J}{125g*(-74.95ºC} = 0.893 J/gºC (5)

  • which is pretty close to the Aluminum's accepted specific heat value of 0.900 J/gºC.

8 0
3 years ago
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