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3 years ago

How many planets are there in the Milkyway?

Physics

1 answer:

BabaBlast [244]3 years ago

3 0

Answer:

here you go

Explanation:

100 billion planets

brainliest pls

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What can always be said about a negatively-charged ion?

KiRa [710]

Answer: A negatively-charged ion always has more electrons than protons

Explanation:

First, we know that the elementary negative charge is the electron, while the positive one is the proton. Such that both have the same charge in magnitude, but a different sign. Such that if we have the same number of electrons and protons in an atom, the charge of this atom will be neutral.

And an ion is an atom with a different number of electrons and protons, so the charge of the atom is not neutral.

Then if we have a negatively-charged ion, the charge of this atom is negative. Then we must have a larger number of electrons (the negative ones) than protons (the positive ones)

Then the correct option is:

A negatively-charged ion always has more electrons than protons

5 0

3 years ago

What is the ratios of sodium hydroxide

Katena32 [7]

Clarify what you mean by ratios?

8 0

4 years ago

A gas occupies a volume of 1.0 m3 in a cylinder at a pressure of 120kPa. A piston compresses the gas until the volume is 0.25m3,

Hoochie [10]

Answer:

Approximately $480\; \rm kPa$ , assuming that this gas is an ideal gas.

Explanation:

Let $V(\text{Initial})$ and $P(\text{Initial})$ denote the volume and pressure of this gas before the compression.
Let $V(\text{Final})$ and $P(\text{Final})$ denote the volume and pressure of this gas after the compression.

By Boyle's Law, the pressure of a sealed ideal gas at constant temperature will be inversely proportional to its volume. Assume that this gas is ideal. By this ideal gas law:

$\displaystyle \frac{P(\text{Final})}{P(\text{Initial})} = \frac{V(\text{Initial})}{V(\text{Final})}$ .

Note that in Boyle's Law, $P$ is inversely proportional to $V$ . Therefore, on the two sides of this equation, "final" and "initial" are on different sides of the fraction bar.

For this particular question:

$V(\text{initial}) = 1.0\; \rm m^3$ .
$P(\text{Initial}) = 120\; \rm kPa$ .

$V(\text{final}) = 0.25\; \rm m^3$ .
The pressure after compression, $P(\text{Final})$ , needs to be found.

Rearrange the equation to obtain:

$\displaystyle P(\text{Final}) = \frac{V(\text{Initial})}{V(\text{Final})} \cdot P(\text{Initial})$ .

Before doing any calculation, think whether the pressure of this gas will go up or down. Since the gas is compressed, collisions between its particles and the container will become more frequent. Hence, the pressure of this gas should increase.

$\begin{aligned}P(\text{Final}) &= \frac{V(\text{Initial})}{V(\text{Final})} \cdot P(\text{Initial})\\ &= \frac{1.0\; \rm m^{3}}{0.25\; \rm m^{3}} \times 120\; \rm kPa = 480\; \rm kPa\end{aligned}$ .

4 0

4 years ago

A simple hydraulic lift is made by fitting a piston attached to a handle into a 3.0-cm diameter cylinder. The cylinder is connec

stiv31 [10]

Answer:

Approximately $3.1 \times 10^4 \; \rm N$ (assuming that the acceleration due to gravity is $g = 9.81\; \rm kg \cdot N^{-1}$ .)

Explanation:

Let $A_1$ denote the first piston's contact area with the fluid. Let $A_2$ denote the second piston's contact area with the fluid.

Similarly, let $F_1$ and $F_2$ denote the size of the force on the two pistons. Since the person is placing all her weight on the first piston:

$F_1 = W = m \cdot g = 50\; \rm kg \times 9.81 \; \rm kg \cdot N^{-1} =495\; \rm N$ .

Since both pistons fit into cylinders, the two contact surfaces must be circles. Keep in mind that the area of a square is equal to $\pi$ times its radius, squared:

$\displaystyle A_1 = \pi \times \left(\frac{1}{2} \times 3.0\right)^2 = 2.25\, \pi\;\rm cm^{2}$ .
$\displaystyle A_2 = \pi \times \left(\frac{1}{2} \times 24\right)^2 = 144\, \pi\;\rm cm^{2}$ .

By Pascal's Law, the pressure on the two pistons should be the same. Pressure is the size of normal force per unit area:

$\displaystyle P = \frac{F}{A}$ .

For the pressures on the two pistons to match:

$\displaystyle \frac{F_1}{A_1} = \frac{F_2}{A_2}$ .

$F_1$ , $A_1$ , and $A_2$ have all been found. The question is asking for $F_2$ . Rearrange this equation to obtain:

$\displaystyle F_2 = \frac{F_1}{A_1} \cdot A_2 = F_1 \cdot \frac{A_2}{A_1}$ .

Evaluate this expression to obtain the value of $F_2$ , which represents the force on the piston with the larger diameter:

$\begin{aligned}F_2 &= F_1 \cdot \frac{A_2}{A_1} \\ &= 495\; \rm N \times \frac{2.25\, \pi\; \rm cm^2}{144\, \pi \; \rm cm^2} \approx 3.1 \times 10^4\; \rm N\end{aligned}$ .

6 0

3 years ago

Approximately, What is the value of the Hubble Constant, as measured by scientists? Hypothetically, if the value of the Hubble C

Serhud [2]

Answer:

The current value of the Hubble's constant = 73 km/sec/Mpc.

t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc

Explanation:

The current value of the Hubble's constant = 73 km/sec/Mpc. However, recent discoveries in the cosmology contradicts the idea of Hubble constant as being fixed. Some scientists are not agreeing on this value and the debate is going on.

Hubble law states that how fast universe is expanding or in other words, galaxies are expanding separating with a speed directly proportional to the distance of galaxies to the earth.

Hence,

v is directly proportional to d

where, v = apparent velocity

d = distance

if we equate velocity and distance then there comes Hubble constant.

v = $H_{0}$ x d

$H_{0}$ = 73 km/sec/Mpc

where, Mpc = Mega Parsec = 1 Mpc = 3.086 x $10^{19}$ km

We can use Hubble constant to tell the age of universe.

t = d/v

t = d/( $H_{0}$ xd)

t = 1/ $H_{0}$

Scientist calculated the age of universe by using Hubble constant, which is 13.4 billion years.

Now, if we hypothetically change the value of Hubble constant,

from $H_{0}$ = 73 km/sec/Mpc to $H_{0}$ = 700 km/sec/Mpc

then the age of universe will be:

t = 1/ $H_{0}$

first convert the units of new $H_{0}$ into 1/s

$H_{0}$ = (700) x (/3.08 x $10^{19}$ )

$H_{0}$ = 227.27 x $10^{-19}$ = 2.27 x $10^{-21}$ 1/s

So,

Age of universe will be:

t = 1/ $H_{0}$ = 1/2.27x $10^{-21}$ 1/s

t = 2.27 x $10^{21}$ s

t = 71.9 trillion years

t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc

6 0

3 years ago