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Likurg_2 [28]
3 years ago
14

A satellite's speed is 10,000 m/s. After 1 min, it is 5,000 m/s. What is the satellite's acceleration?

Chemistry
1 answer:
Pani-rosa [81]3 years ago
7 0

Answer:

- 83.33 m per second square

Explanation:

Initial velocity u = 10,000 m/s

Final Velocity v = 5,000 m/s

Time t = 1 minutes = 60 seconds

Acceleration a =?

\because \: a =  \frac{v - u}{t}  \\  \\  =  \frac{5000 - 10000}{60}  \\  \\  =  \frac{ - 5000}{60}  \\  \\ a =  - 83.33 \: m {s}^{ - 2}

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Put the following names in the correct alphabetic indexing order:(1) Topper & Casey Plumbing(2) KST Enterprises(3) Leland an
dangina [55]

Answer: the correct option is 2, 3, 4, 1.

Explanation:alphabetic indexing order is the order in which files or names are being arranged according to the alphabet. In the following names:

2) KST Enterprises

3)Leland and Son Graphics

4)Lucinda Topper

1) Topper & Casey Plumbing.

While arranging alphabetically, the first letters are usually considered but in a scenario where alphabet occurs twice( 3 And 4) the second letter is considered. I hope this helps, thanks

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3 years ago
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A) How many grams of water are produced when 2.50 g of aluminum
egoroff_w [7]
The correct answer is 13 I just did it bro
5 0
3 years ago
How does an adaptation begin?
Nataly [62]

In evolutionary theory, adaptation is the biological mechanism by which organisms adjust to new environments or to changes in their current environment.

6 0
3 years ago
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The standard enthalpy of formation for glucose [c6h12o6(s)] is −1273.3 kj/mol. what is the correct formation equation correspond
balu736 [363]
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is 
     C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is 
     6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)

Using the equation for the standard enthalpy change of formation 
     ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
     ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}

C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
     ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
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8 0
3 years ago
You have a solution of 600 mg of caffeine dissolved in 100 mL of water. The partition coefficient for aqueous caffeine extracted
klio [65]

Answer:

159 mg caffeine is being extracted in 60 mL dichloromethane

Explanation:

Given that:

mass of caffeine in 100 mL of water =  600 mg

Volume of the water = 100 mL

Partition co-efficient (K) = 4.6

mass of caffeine extracted = ??? (unknown)

The portion of the DCM = 60 mL

Partial co-efficient (K) = \frac{C_1}{C_2}

where; C_1= solubility of compound in the organic solvent and C_2 = solubility in aqueous water.

So; we can represent our data as:

K=(\frac{A_{(g)}}{60mL} ) ÷ (\frac{B_{(mg)}}{100mL} )

Since one part of the portion is A and the other part is B

A+B = 60 mL

A+B = 0.60

A= 0.60 - B

4.6=(\frac{0.6-B(mg)}{60mL} ) ÷ (\frac{B_{(mg)}}{100mL})

4.6 = \frac{(\frac{0.6-B(mg)}{60mL} )}{(\frac{B_{(mg)}}{100mL})}

4.6 × (\frac{B_{(mg)}}{100mL}) = (\frac{0.6-B(mg)}{60mL} )

4.6 B *\frac{60}{100} = 0.6 - B

2.76 B = 0.6 - B

2.76 + B = 0.6

3.76 B = 0.6

B = \frac{0.6}{3.76}

B = 0.159 g

B = 159 mg

∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.

4 0
3 years ago
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