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2 years ago

5

A thermometer is placed into a flask of water with pieces of ice and reads 273.15. Which temperature scale is being used? A. Kel

vin B. Celsius C.Centigrade
D. Fahrenheit

2 answers:

Aleks [24]2 years ago

8 0

A.) Kelvin bc it it wouldnt make sense for the other three options plus options B.) and C.) are the same :)

zzz [600]2 years ago

6 0

A. Kelvin. It is 32 degrees fahrenheit

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Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

Mass inside the orbit in terms of Volume and Density is

$M =V \rho$

Where,

V = Volume

$\rho =$ Density

Now considering the volume of the star as a Sphere we have

$V = \frac{4}{3} \pi R^3$

Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

$a_g = \frac{4}{3} G\pi R\rho$

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

At the same time the general expression for the centripetal acceleration is

$a_c = \frac{\Theta^2}{R}$

Where $\Theta$ is the orbital velocity

Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

$\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}$

$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

$\Theta \propto R$

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

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2 years ago

Witch of the following is the correct definition of convention

iogann1982 [59]

Answer:

A is the answer .(A) is correct

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2 years ago

A 5 kg block is being pushed horizontally across a level surface at a constant velocity. What is the magnitude of the Normal for

luda_lava [24]

Answer:

50 N.

Explanation:

On top of a horizontal surface, the normal force acting on an object is equivalent to the force of gravity acting on the object. That is:

$\displaystyle \begin{aligned} F_N = F_g & = ma \\ & = mg\end{aligned}$

The mass of the block is 5 kg and the given force due to gravity is 10 N/kg. Substitute and evaluate:

$\displaystyle F_N = F_g = (5\text{ kg})(10 \text{ N/kg}) = 50 \text{ N}$

In conclusion, the normal force acting on the block is 50 N.

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1 year ago

Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a po

gregori [183]

Answer:

E = 0 r <R₁

Explanation:

If we use Gauss's law

Ф = ∫ E. dA = $q_{int}$ / ε₀

in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

E = 0 r <R₁

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2 years ago

Why is earth the only planet on which water exists in a liquid state

Gennadij [26K]

This is due to earths location in the solar system. Earth is in the habitat zone or the Goldie locks zone, in this zone it's not too hot or not too cold for water to exist. Other planets in different star systems have liquid oceans due to them being in the habitat zone.

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2 years ago