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3 years ago

BRAINLIEST/BRAINLIEST/ Please help

Physics

1 answer:

Alexxx [7]3 years ago

8 0

Answer:

1. the pressure inside any enclosures on the moon as the sun hits the exterior and the temperature begins to rise is going to increase, you could use an air pump to lower the pressure in the room.

2. you could put them in jars, I once saw people selling Jarred air supposedly from Honolulu, Hawaii, but they didn't seem to do that well for some reason XD. or if you are staying long term in space, you could bring a couple of plants and plant them if you want, that what NASA did.

3. you could try removing the carbon dioxide with canisters that contain powdered lithium hydroxide. When air containing carbon dioxide gets passed through the canister, it combines with the lithium hydroxide to form lithium carbonate and water. you could dump the lithium carbonate and water into space and make space trash if you don't want them taking up too much space, water will probably freeze or something if you launch it into space due to the temperature out there.

Explanation:

can I have brainiest pls? this took me like half an hour

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Arrange the following substances in order of decreasing magnitude of lattice energy.

kolbaska11 [484]

The formula is F = ( q1 * q2 ) / r ^ 2
where: q is the individual charges of each ion 
r is the distance between the nuclei 
The formula is not important but to explain the relationship between the atoms in the compounds and their lattice energy. 

From the formula we can first conclude that compounds of ions with greater charges will have a greater lattice energy. This is a direct relationship. 
For example, the compounds BaO and SrO, whose ions' charges are ( + 2 ) and ( - 2 ) respectively for each, will have greater lattice energies that the compounds NaF and KCl, whose ions' charges are ( + 1 ) and ( - 1 ) respectively for each. 

So Far: ( BaO and SrO ) > ( NaF and KCl ) 

The second part required you find the relative distance between the atoms of the compounds. Really, the lattice energy is stronger with smaller atoms, an indirect relationship. 
For example, in NaF the ions are smaller than the ions in KCl so it has a greater lattice energy. Because Sr is smaller than Ba, SrO has a greater lattice energy than BaO. 

Therefore: 
Answer: SrO > BaO > NaF > KCl

5 0

4 years ago

Read 2 more answers

Here is something to blow your mind. In one slice of "everything" pizza there are 650 Calories (1000 chemistry calories). That e

amid [387]

Answer:

Number of slices of pizza is 993.

Explanation:

It is given that, in one slice of "everything" pizza there are 650 Calories. The conversion factor from calories to joules is :

1 calorie = 4.184 joules

650 calories = 2719.6 joules

Total energy in the pizza, E = 2700000 J

Let there are n number of slices of pizza. It is given by :

$n=\dfrac{2700000}{2719.6}$

n = 992.79

n = 993

So, there are 993 slices of pizza. Hence, this is the required solution.

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3 years ago

A mass m = 0.6 kg is released from rest at the top edge of a hemispherical bowl with radius = 1.1 meters. The mass then slides w

Aneli [31]

Answer:

The angle is $\theta = 36.24 ^o$

Explanation:

From the question we are told that

The mass is $m = 0.6 \ kg$

The radius is $r = 1.1 \ m$

The speed is $v = 3.57 \ m /s$

According to the law of energy conservation

The potential energy of the mass at the top is equal to the kinetic energy at the bottom i.e

$m * g * h = \frac{1}{2} * m * v^2$

=> $h = \frac{1}{2 g } * v^2$

Here h is the vertical distance traveled by the mass which is also mathematically represented as

$h = r * sin (\theta )$

$\theta = sin ^{-1} [ \frac{1}{2* g* r } * v^2]$

substituting values

$\theta = sin ^{-1} [ \frac{1}{2* 9.8* 1.1 } * (3.57)^2]$

$\theta = 36.24 ^o$

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4 years ago

Suppose a drinking straw is in a glass of water. Someone blows air across the top of the straw. The water level in the straw ___

enyata [817]

The answer is A. drops! Suppose a drinking straw is in a glass of water. Someone blows air across the top of the straw. The water level in the straw will be drops.

Hope it helped!

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4 years ago

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an astronaut weighs 8.00x10^2 on the surface of earth. what is the weight of the astronaut 6.37x10^6 meters above the surface of

Marta_Voda [28]

To solve this problem, we use the Law of Universal Gravitation:

F = Gm1m2/d^2

where m1 and m2 are two objects. In this case, earth and man. d is the distance between the objects. Lastly, G is the gravitational constant. Since the mass of the earth and man are constant, this is lumped up with G into k. The equation would be:

F = k/d^2
k = Fd^2
$F_{1} d_{1} ^{2} =F_{2} d_{2} ^{2}$

The radius of earth, d1, is equal to 6.371E+6 m. Thus, d2 = 2d1

(8E+2)(d1)^2 = F2(2d1)^2

(8E+2)(d1)^2 = 4F2(d1)^2

(8E+2)=4F2
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3 years ago