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3 years ago

Is a bath tub renewable or not?

2 answers:

6 0

The water being poured into the bathtub is renewable and can be recycled, filtered, and renewed. But, metaphorically, each time you use electricity you drain a little water from the "bathtub" or the electric grid.

Mandarinka [93]3 years ago

5 0

The bathtub is not the water inside is questionable some people say is nonrenewable some say it is

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Question 5 (Multiple Choice Worth 3 points) (02.07 MC) Rachel needs to eat fewer carbohydrates to improve her health. Which of t

patriot [66]

Milk, apples, and beans don't have much carbohydrate. So if you
cut down on those, you don't really cut down much on carbohydrates.

If Rachel needs to reduce her intake of carbohydrates, she should
cut down on bread. (Also cake, sugar, corn, pasta, and potatoes.)

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

For the following questions consider a piece of copper wire. a. What type of bond is formed between the copper atoms? b. Describ

yarga [219]

Answer:

a. metallic bond

b. the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “cloud” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions.

c. due to the presence of free electrons in its outer energy levels

6 0

3 years ago

Read 2 more answers

We are running late for school and we want to make our 0.5 kg tea (it’s at 90 C) colder. Let’s assume we can drink tea when it’s

PSYCHO15rus [73]

Answer:

x=0.154kg

Explanation:

(x*L)+(0.5kg*4200*50)+(x*4200*(-50)=0

(x*333 000J/kg*c)+(0.5kg*4200J/kg*C*(-40C))+(x*4200J/kg*C*50C)=0

6 0

3 years ago

Question 7 (2 points)

Arlecino [84]

Answer:

Tire

Explanation:

3 0

3 years ago