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2 years ago

By increasing the temperature, the solubility of _____.

Physics

2 answers:

kati45 [8]2 years ago

5 0

The answer is A my man

svetoff [14.1K]2 years ago

3 0

The answer is A - when liquid is heated it evaporates (Decreases) and the vapor (Gas) increases.

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The lift does 3000 J of work in 5 seconds. What is the power of the lift?

diamong [38]

Answer:

600

Explanation:

p=Work/time

3000/5=600 is power

8 0

2 years ago

A 71 kW radio station broadcasts its signal uniformly in all directions. - What is the average intensity of its signal at a dist

marshall27 [118]

Answer:

Explanation:

Energy of signal being radiated per second on all sides = 71 x 10³ J .

At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.

So energy crossing per unit area

= $\frac{71\times10^3}{4 \times \pi\times(220)^2}$

= 11.67 x 10⁻² Wm⁻²s⁻¹.

This is the intensity of the signal.

At 2200 m this intensity will further reduce by 100 times

So there it becomes equal to

11.67 x 10⁻⁴ Wm⁻² s⁻¹.

3 0

2 years ago

A molecular cloud fragments as it collapses because Choose one: A. the interstellar wind is stronger in some places than others.

kondaur [170]

The cause for a molecular cloud forming fragments when it collapses is indicated correctly by option D. density variations from place to place grow larger as the cloud collapses.

Molecular cloud:

A molecular cloud, also known as a stellar nursery, is a specific kind of interstellar cloud, whose density and size allow the development of molecules, absorption nebulae, and H II regions. In contrast, some regions of the interstellar medium mostly consist of ionized gas.

Molecular clouds are cold, dense areas of space where stars form. The cloud collapses into a proto-star when the gravitational force pulling it in outweighs the internal pressure pushing it in.

When a molecular cloud collapses, it is observed that the density varies from place to place with the variation increasing with collapse. As a result, the collapse is characterized by fragmentation of the cloud.

Thus the correct option is: D. density variations from place to place grow larger as the cloud collapses.

Learn more about molecular clouds,

brainly.com/question/3459894

#SPJ4

6 0

1 year ago

An electric grinder uses a grinding wheel

luda_lava [24]

(1500 rev/min)(min / 60 s) / (3.0 s) = 8.33 rev/s²

(B) 
(1/2)(8.33 rev/s²)(3.0 s)² = 37.5 rev 

(C) 
(1500 rev/min)(min / 60 s)[2π(0.12 m) / rev] = 18.8 m/s

3 0

2 years ago

Read 2 more answers

I NEED HELP PLEASE, THANKS! :)

mrs_skeptik [129]

Answer:

1. Largest force: C; smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

$F=K\dfrac{ q_{1}q_{2}}{r^{2}}$

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write

$F=\dfrac{k}{r^{2}}$

1. Net force on each particle

Let's

Call the distance between adjacent charges d.
Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

$\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(b) Force on B

$\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(C) Force on C

$\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(d) Force on D

$\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

$F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\$

2. Ratio of largest force to smallest

$\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}$

7 0

2 years ago