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stellarik [79]
3 years ago
13

40ml of Liquid A are poured into a beaker, and 40.0ml of Liquid B are poured into an identical beaker. Stirrers in each beaker a

re connected to motors, and the forces FA and FB needed to stir each liquid at a constant rate are measured.
a. FA will be greater than F B
b. FA will be less than FB
c. FA will be equal to FB
d. It's impossible to predict whether FA or FB will be greater without more information.
Physics
1 answer:
lakkis [162]3 years ago
7 0

Answer:

d

Explanation:

there is not enough information about the liquid to know the force required for each. ie. stirring a cup of water is different than stirring a cup of pudding.

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Answer:

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Explanation:

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Under steady state, a compressor is used to increase the pressure of an ideal gas air from 100 kPa to 1 MPa. Meanwhile the tempe
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3 years ago
The gravitational constant G was first measured accurately by Henry Cavendish in 1798. He used an exquisitely sensitive balance
belka [17]

The gravitational force between the spheres is

F_{\rm g}=\dfrac{G(188\,\mathrm{kg})(0.93\,\mathrm{kg})}{(0.27\,\mathrm m)^2}\approx1.6\times10^{-7}\,\mathrm N

where <em>G</em> = 6.674 x 10⁻¹¹ N m²/kg².

The weight of the lighter sphere is

F_{\rm w}=(0.93\,\mathrm{kg})g\approx9.1\,\mathrm N

where <em>g</em> = 9.80 m/s².

The ratio between the two forces is then

\dfrac{F_{\rm g}}{F_{\rm w}}\approx1.8\times10^{-8}

7 0
3 years ago
4. Suppose the observed motion of a moving airplane, illustrated by a distance vs. time graph, is nearly a straight line with ze
jekas [21]
It shows that the airplane covers equal distance in equal time interval, that's it has a straight line from the origin.

The plane is moving at uniform speed.
7 0
3 years ago
A planet has been observed orbiting a nearby star. This star has a mass of 3.5 solar masses, and the planet is 4.2 AU from the s
kondaur [170]

Answer:

4.6 years

Explanation:

This is solved using Kepler's third law which says:

T^2=\frac{4\pi ^2}{GM} a^2

Where

T = Orbital period of the planet (in seconds)

a = Distance from the star (in meters)

G = Gravitational constant

M = Mass of the parent star (in kg)

From the information given

M = 3.5M_{sun} = 6.96*10^{30} kg

a=4.2AU = 6.28*10^{11} meters

G = 6.67*10^{-11}m^3kg^{-1}s^{-2}

We put this into Kepler's law and get:

T=\sqrt{\frac{4\pi ^2}{6.67*10^{-11}*6.96*10^{30}} (6.28*10^{11})^3}=145,128,196 seconds.

This when converted to years is 4.6 years.

4 1
3 years ago
Read 4 more answers
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