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tatyana61 [14]
3 years ago
15

Temperature is the measure of the average kinetic energy of the particles in an object. True False

Physics
1 answer:
NemiM [27]3 years ago
3 0

Answer:

true

Explanation:

with increased temperature particles move faster as they gain kinetic energy

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You blow across the open mouth of a test tube and produce the fundamental standing wave of the air column inside the test tube i
WITCHER [35]

Answer:

B: 980 Hz

Explanation:

8 0
3 years ago
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How many atoms of oxygen are there in one molecule of trinitrotoluene, if the chemical formula is C6H3(NO2)3?
kvasek [131]

。☆✼★ ━━━━━━━━━━━━━━  ☾  

2 x 3 = 6

Thus there are 6 atoms of oxygen in this molecule.

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Jack and Jill are on two different floors of their high rise office building and looking out of their respective windows. Jack s
Maru [420]

Answer:

a) speed when Jack sees the pot : 12.92 meters per second

b) height difference 163.115 meters

Explanation:

First to calculate te initial speed we use the acceleration formula:

a= v1-v0/t

Acceleration being gravity's acceleration (9.8 m/s^2)

v1 being the speed when Jill sees the pot

v0 when Jack sees it

and t the time between

Solving for v0 it would be

v1 - a*t = v0

replacing

58 m/s -  9.8 m/s^2 *4.6 s = v0 = 12.92 m/s

For the second question we use the position formula setting y0 and t0 as the position and time when jack sees the pot. (and setting the positive axis downward I.E. one meter below jack would be 1m not -1m)

The formula is

y0 + v0*t + 1/2 g *t^2 = yt

replacing

0m + 12.92m/s* 4.6 s + 1/2 * 9.8 m/s^2 * (4.6 s)^2 = 163.115 m

5 0
3 years ago
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
3 years ago
Of all the radio waves, these have the shortest wavelengths and the the highest frequency?
Julli [10]
If by chance you mean any wave, the wave with the shortest wavelength/highest frequency is gamma rays.
4 0
3 years ago
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