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Sloan [31]
3 years ago
13

Who trying to talk?????

Physics
2 answers:
Art [367]3 years ago
7 0

Answer:

Me! :)

Explanation:

Elden [556K]3 years ago
7 0
Well I do have school but sure later tho
You might be interested in
An elevator manufacturing company is stress-testing a new elevator in an airless test shaft. The elevator is traveling at an unk
devlian [24]

Answer:

3.192 m/s

Explanation:

t = Time taken = 0.900 seconds

u = Initial velocity

v = Final velocity

s = Displacement = 1.1 meters

a = Acceleration due to gravity = 9.81 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{1.1-\frac{1}{2}\times 9.81\times 0.9^2}{0.9}\\\Rightarrow u=-3.192\ m/s

Velocity of the elevator when it snapped is 3.192 m/s

4 0
3 years ago
Hello please help i’ll give brainliest
ser-zykov [4K]

The correct answer is B.Antartica and Australia were one landmass millions of years ago.

5 0
3 years ago
Which jovian planet should have the most extreme seasonal changes? a. Saturn b. Neptune c. Jupiter d. Uranus
Maksim231197 [3]

Answer:

D). Uranus.

Explanation:

Jovian planets are described as the planets which are giant balls of gases and located farthest from the sun which primarily include Jupiter, Saturn, Uranus, and Neptune.

As per the question, 'Uranus' is the jovian planet that would have the most extreme seasonal changes as its tilted axis leads each season to last for about 1/4 part of its 84 years orbit. The strong tilted axis encourages extreme changes in the season on Uranus. Thus, <u>option D</u> is the correct answer.

3 0
3 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
2: Assume we have created a packet-switched internet. Using the TCP/IP protocol suite, we
yaroslaw [1]

Answer:

Shown below as 'The advantage is that the router will always choose the faster route to send the packet and each connection is independent.'

Explanation:

The advantage is that the router will always choose the faster route to send the packet and each connection is independent.

Hope this helps!

Kind regards

Farah Syed

3 0
3 years ago
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