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andre [41]
3 years ago
6

The images formed by diverging (concave) lenses are always _____. real virtual upright inverted magnified diminished

Physics
2 answers:
andre [41]3 years ago
8 0
The answer is virtual xox
amid [387]3 years ago
5 0
The answer is virtual.
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Explain why sound is more clear at night than the day​
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because of god we can hear better at night.

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What measurements will be made to determine the magnitude of the test-mass centripetal acceleration?
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Define the term dimension<br>​
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Answer:

Q1. A measurable extent of a particular kind, such as length, breadth, depth, or height.

Q2. A dimensional constant is a physical quantity that has dimensions and has a fixed value. Some of the examples of the dimensional constant are Planck's constant, gravitational constant, and so on.

Q3. Physical quantities which posses dimensions and have variable are called dimensional variables. Examples are length, velocity, and acceleration etc.

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Q7. Mass, length, time, temperature, electric current, amount of light, and amount of matter.

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Hope that helps. x

4 0
2 years ago
calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame
kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2

0 - p × Area for trachea =

\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

⇒-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}

           \frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

3 0
3 years ago
Read 2 more answers
An empty semi-truck sits on the scale of a weighing station. It is then loaded with 100 irritated pigeons. These two events then
mina [271]

Answer:

The reading in the scale is going to be the same, or if it experiences some change it would be minimum. The reason is because the truck will act as a big closed cage, therefore, when the pigeons fly, the air they move with their wings in order to keep flying,  exerts the same force on the closed cage, that if they were standing on the ground.

If the truck however, allows the air flow, the weight might change, because under this scenario, the air flowing could represent less force exerted on the balance.

Explanation:

4 0
3 years ago
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