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4 years ago

What is the mass of an object with an acceleration of 10 meters per second and a force of 90 newtons?

Chemistry

1 answer:

fenix001 [56]4 years ago

7 0

Answer:

9kg

Explanation:

According to Newton's second law, Force=Mass x Acceleration.

Since we have the Force and the Acceleration, we can plug the values in and solve for mass using this equation:

90N = Mass x 10m/s

Divide both sides by 10:

90/10=Mass x 10/10

Mass = 9

The Mass of the object is 9kg.

Hope this helped!

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What is chlorine's atomic number and mass

makkiz [27]

$\red{⭐hello my \: friend!}$

chlorine's atomic number and mass :

35,453 u

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☝️follow the rules of brainly!

8 0

3 years ago

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m

koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

$R_{in}$ $= \frac{0.04}{100}*2000$

$R_{in}$ = 0.8

The rate-out

$R_{out}$ = $\frac{A}{6000}*2000$

$R_{out}$ = $\frac{A}{3}$

We can say that:

$\frac{dA}{dt}=$ $0.8-\frac{A}{3}$

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

$\frac{dA}{dt} +\frac{A}{3} =0.8$

Integration of the above linear equation =

$e^{\int\limits \frac {1}{3}dt } =$ $e^{\frac{1}{3}t$

so we have:

$e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A$ $= 0.8e^{\frac{1}{3}t$

$\frac{d}{dt}[e^{\frac{1}{3}t}A]$ $= 0.8e^{\frac{1}{3}t$

$Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C$

∴ $A(t) = 2.4 +Ce^{-\frac{1}{3}t$

Since A(0) = 12

Then;

$12 =2.4 + Ce^{-\frac{1}{3}}(0)$

$C= 12-2.4$

$C =9.6$

Hence;

$A(t) = 2.4 +9.6e^{-\frac{t}{3}}$

$A(0) = 2.4 +9.6e^{-\frac{10}{3}}$

$A(t) = 2.74$

∴ the concentration at 10 minutes is ;

= $\frac{2.74}{6000}*100$ %

= 0.0456667 %

= 0.046% to three decimal places

7 0

4 years ago

Enthalpies of reaction calculated from bond energies and from enthalpies of formation are often, but not always, close to each o

jolli1 [7]

The enthalpy change in a reaction is given by-

ΔH°rxn = ∑nΔH°f,products - ∑nΔH°f,reactants

This can be expressed in terms of bond energy as-

ΔH°rxn = BEreactants - BEproducts

Therefore, the calculated bond energy according to the above equation will be-

ΔH°rxn = [ (C-C) + 2(C-O) + 4(C-H) + 2(O-H) ] - [ (C-C) + 2(C-O) + 4(C-H) + 2(O-H) = 0 kJ/mol

<h3>What is enthalpy change?</h3>

Enthalpy change is a measure of the energy emitted or consumed in a reaction. This can be determined using the following equation which involves standard enthalpy of reactant and product formation:

ΔH°rxn = ∑nΔH°f,products - ∑nΔH°f,reactants

<h3>What is bond energy?</h3>

Bond energy is defined as the amount of energy needed to dissociate a mole of molecules into their individual atoms.

Learn more about the Enthalpy Change here:

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2 years ago

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pantera1 [17]

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3 years ago

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MAVERICK [17]

Answer:

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3 years ago