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katrin [286]
2 years ago
8

Question 1 of 15

Physics
1 answer:
Ierofanga [76]2 years ago
8 0

Answer:

d 75 miles per hour

Explanation:

60 ÷ 20

3 × 25 = 75

hoped that helped!!

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The ray diagram shows a vase that is placed beyond the center of curvature of a concave mirror.
SIZIF [17.4K]
The image will form in the vicinity of F. Its nature will be small and inverted 
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3 years ago
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How can you drop two eggs the fewest amount of times, without them breaking?
MrMuchimi

Answer:

Both eggs are identical. The aim is to find out the highest floor from which an egg will not break when dropped out of a window from that floor. If an egg is dropped and does not break, it is undamaged and can be dropped again. However, once an egg is broken, that's it for that egg.

4 0
2 years ago
Arrange the following substances in order of decreasing magnitude of lattice energy.
kolbaska11 [484]
The formula is F = ( q1 * q2 ) / r ^ 2 
<span>where: q is the individual charges of each ion </span>
<span>r is the distance between the nuclei </span>
<span>The formula is not important but to explain the relationship between the atoms in the compounds and their lattice energy. </span>

<span>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. </span>
<span>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. </span>

<span>So Far: ( BaO and SrO ) > ( NaF and KCl ) </span>

<span>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. </span>
<span>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. </span>

<span>Therefore: </span>
<span>Answer: SrO > BaO > NaF > KCl </span>
5 0
3 years ago
Read 2 more answers
FREE BRAINIEST IF YOU ANSWER THIS
Inessa [10]

The watt is a rate, similar to something like speed (miles per hour) and other time-interval related measurements.

Specifically, watt means Joules per Second. We are given that the electrical engine has 400 watts, meaning it can make 400 joules per second. If we need 300 kJ, or 3000 Joules, then we can write an equation to solve the time it would take to reach this amount of joules:

w * t = E

w: Watts

t: Time

E: Energy required

(Watts times time is equal to the energy required)

<u>Input our values:</u>

400 * t = 3000

(We need to write 3000 joules instead of 300 kilojoules, since Watts is in joules per second. It's important to make sure your units are consistent in your equations)

<u>Divide both sides by 400 to isolate t:</u>

<u />\frac{400t}{400} = \frac{3000}{400}

t = 7.5 (s)

<u>It will take 7.5 seconds for the 400 W engine to produce 300 kJ of work.</u>

<u></u>

If you have any questions on how I got to the answer, just ask!

- breezyツ

6 0
3 years ago
Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ
bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; H_0 = 51km/s/Mly

Age of the universe; t = \ ?

We know that, the reciprocal of the Hubble's constant ( H_0 ) gives an estimate of the age of the universe ( t ). It is expressed as:

Age\ of\ Universe; t = \frac{1}{H_0}

Now,

Hubble's constant; H_0 = 51km/s/Mly

We know that;

1\ light\ years = 9.46*10^{15}m

so

1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m

Therefore;

H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

Now, we input this Hubble's constant value into our equation;

Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0
2 years ago
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