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1 year ago

Which of the following can be used to measure temperature accurately

Physics

1 answer:

Illusion [34]1 year ago

4 0

Answer:

Any of those terms can be converted to either of the other terms, so either term is correct. People are accustomed to everyday temperatures in Fahrenheit. The ideal gas law specifies that

P V = N R T where T is in Kelvin which is Celsius + 273 deg.

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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

6 0

2 years ago

Sound waves, water waves, and light waves are all alike in that they all

Paha777 [63]

Sound and water waves are longitudinal waves, they require a medium to travel through and occilate particles 90 degrees to the wave motion
Light is a transverse wave. It doesnt require a medium to travel through.
All three reflect, refract and diffract

Light is difficult to think of because it acts in ways which waves cannot explain in some cirumstances. It acts like a particle (called photons) in some conditions, but acts like a normal sound or water wave does in others. Try not to get too caught up in light being a wave or a particle because even physists dont know how to explain it yet.

7 0

2 years ago

In a science museum, a 140 kg brass pendulum bob swings at the end of a 16.8 m -long wire.

Mice21 [21]

The period of the pendulum is 8.2 s

Explanation:

The period of a simple pendulum is given by the equation:

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration of gravity

T is the period

We notice that the period of a pendulum does not depend at all on its mass, but only on its length.

For the pendulum in this problem, we have

L = 16.8 m

and

$g=9.8 m/s^2$ (acceleration of gravity)

Therefore the period of this pendulum is

$T=2\pi \sqrt{\frac{16.8}{9.8}}=8.2 s$

#LearnWithBrainly

3 0

2 years ago

A wire loop of radius 0.37 m lies so that an external magnetic field of magnitude 0.35 T is perpendicular to the loop. The field

Galina-37 [17]

Answer:

168.57 mV

Explanation:

Initial magnetic flux = BA , B magnetic field and A is area of loop

= .35 x 3.14 x .37²

= .15 Weber

Final magnetic flux

= - .2 x 3.14 x .37²

= - .086 Weber

change in flux

.15 + .086

= .236 Weber

rate of change of flux

= .236 / 1.4

= .16857 V

= 168.57 mV

5 0

2 years ago

Which of the following examples illustrates static friction in action?

lilavasa [31]

I think its d. but im not sure

3 0

2 years ago

Read 2 more answers