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3 years ago

Plss help

Physics

1 answer:

Genrish500 [490]3 years ago

3 0

Answer:

Explanation:

imagine urself on an elevator dont you feel lighter

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The fast train known as the TGV (Train à Grande Vitesse) that runs south from Paris, France, has a scheduled average speed of 21

andrezito [222]

Answer:

Explanation:

Given

average speed of train $(v_{avg})=216 kmph\approx 60 m/s$

Maximum acceleration=0.05g

Now centripetal acceleration is

$a_c=\frac{v^2}{r}$

$0.05\times 9.8=\frac{60^2}{r}$

r=7346.93 m

(b)Radius of curvature=900 m

therefore $a_c=\frac{v^2}{r}$

$v=\sqrt{a_cr}$

$v=\sqrt{0.05\times 9.8\times 900}$

$v=\sqrt{441}=21 m/s$

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4 years ago

Durning which type of process does pressure remain consistent

Simora [160]

Answer:

$\fbox {D. Isobaric}$

Explanation:

The process during which pressure remains constant is called an isobaric process.

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2 years ago

Read 2 more answers

Freezing Point Depression: Can someone explain this formula to me? ΔTf = Kfcm

Leya [2.2K]

If the solution is treated as an ideal solution, the extent of freezing point depression depends only on the solute concentration that can be estimated by a simple linear relationship with the cryoscopic constant: ΔTF = KF · m · i ΔTF, the freezing point depression, is defined as TF (pure solvent) - TF (solution). KF, the cryoscopic constant, which is dependent on the properties of the solvent, not the solute. Note: When conducting experiments, a higher KF value makes it easier to observe larger drops in the freezing point. For water, KF = 1.853 K·kg/mol.[1] m is the molality (mol solute per kg of solvent) i is the van 't Hoff factor (number of solute particles per mol, e.g. i = 2 for NaCl).

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3 years ago

At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building

Inessa [10]

Answer:

The thrown rock strike 2.42 seconds earlier.

Explanation:

This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

$x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration$

So now we have an equation and unkown value.

for the thrown rock

$\frac{1}{2}(9.8)*t^2+29*t-300=0$

for the dropped rock

$\frac{1}{2}(9.8)*t^2+0*t-300=0$

solving both equation with the quadratic formula:

$\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}$

we have:

the thrown rock arrives on t=5.4 sec

the dropped rock arrives on t=7.82 sec

so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)

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3 years ago

The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a

likoan [24]

Answer:

14869817.395 m

Explanation:

$\theta$ =22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

$22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians$

From Rayleigh Criterion

$\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m$

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

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3 years ago