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2 years ago

A train is accelerating at a rate of 11.8 m/s2. What is this rate in units of nm/ms2?

Chemistry

1 answer:

jasenka [17]2 years ago

7 0

Answer:

E. 1.18x10^4nm/ms^2

Explanation:

The aceleration of the train is given in m/s². We need to conver, first, the meters to nm and then, the seconds to ms as follows:

11.8m/s² * (1x10⁹nm / 1m) = 1.18x10¹⁰nm / s²

Now, as 1s is equal to 1000ms:

1.18x10¹⁰nm / s² * ((1s)² / (1000ms)²)) =

1.18x10⁴nm / ms²

Thus, right option is:

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An isotope undergoes radioactive decay by emitting radiation that has no mass. What other characteristic does the radiation have

Advocard [28]

Answer : Option D) No charge

Explanation : An isotope undergoes radioactive decay by emitting radiation that has no mass. The radiation will not have any charge as it does not has any mass it will not emit a radiative charge.

It is known that there are some unstable radioactive isotopes which has no mass and the radiation thus has no charge in it.

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2 years ago

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Consider the following incomplete reaction. Mg + 2Y ---> MgCl₂ + H₂ Choose the formula for the missing substance Y.

vesna_86 [32]

It would be B as the answer

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2 years ago

Modern commercial airliners are largely made of aluminum, a light and strong metal. But the fact that aluminum is cheap enough t

Dafna11 [192]

Answer:

The plane with aluminium can lift more mass of passangers than the plane of steel.

Explanation:

The total mass the airplane canc lift is:

$m_{tot}=m_{fuselage}+m_{passangers}$

For aluminium:

$m_{tot}=m_{fus-Al}+m_{pas-Al}$

$m_{fus-Al}=\delta _{Al}*V_{fuselage}$

and

$V_{fuselage}=\frac{\pi *L}{4}*[D^2-(D-e)^2]$

where:

L is lenght
D is diameter
e is thickness

$m_{tot}=\delta _{Al}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Al}$

For steel (same procedure):

$m_{tot}=\delta _{Steel}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Steel$

Knowing that the total mass the airplane can lift is constant and that aluminum has a lower density than the steel, we can afirm that the plane with aluminium can lift more mass of passangers.

Also you can estimate an average weight of passanger to estimate a number of passangers it can lift.

5 0

2 years ago

What is the freezing point of a 1.40 m aqueous solution of a nonvolatile un-ionized solute? (the freezing point depression const

Alina [70]

Depression is freezing point is a colligative property. It is mathematically expressed as ΔTf = Kf X m

where Kf = <span>freezing point depression constant = 1.86°c kg /mol (for water)
m = molality of solution = 1.40 m

</span>∴ ΔTf = Kf X m = 1.86 X 1.40 = 2.604 oC

Now, for water freezing point = 0 oC

∴Freezing point of solution = -2.604 oC

6 0

2 years ago

If 10.0 grams of NaHCO3 is added to 10.0 g of HCl, determine the efficiency of baking soda as an antacid if 6.73 g of NaCl was p

Lapatulllka [165]

Answer:

percentage yield of NaCl = 96.64%

Explanation:

The reaction was between NaHCO3 and HCl .The chemical equation can be represented below:

NaHCO3 + HCl → NaCl + H2O + CO2 . The balance equation is

NaHCO3 + HCl → NaCl + H2O + CO2

The question ask us to calculate the percentage yield of NaCl.

The efficiency of NaHCO3 as an antacid , the limiting reactant is NaHCO3

1 mole of NaHCO3 produces 1 mole of NaCl

Therefore,

molar mass of NaHCO3 = 23 +1 + 12 + 48 = 84 g

molar mass of NaCl = 23 + 35.5 = 58.5 g

1 mole of NaHCO3 = 84 g

1 mole of NaCl = 58.5 g

since 84 g of NaHCO3 produces 58.5 g of NaCl

10 g of NaHCO3 will produce ? grams of NaCl

cross multiply

Theoretical yield of NaCl = (10 × 58.5)/84

Theoretical yield of NaCl = 585/84

Theoretical yield of NaCl = 6.9642857143 g

percentage yield of NaCl = actual yield/theoretical yield × 100

percentage yield of NaCl = 6.73/6.9642857143 × 100

percentage yield of NaCl = 673/6.9642857143

percentage yield of NaCl = 96.635897436%

percentage yield of NaCl = 96.64%

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2 years ago

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