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3 years ago

11

The car has a mass of 1200kg. The car has a forward force of 3400N but friction has a force of 400N going against it. What is th

e acceleration of the car?

1 answer:

jonny [76]3 years ago

6 0

The acceleration of the car : 2.5 m/s²

<h3>Further explanation</h3>

Newton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object

∑F = m. a

<h3 />

Mass of the car = 1200 kg

Forward force = 3400 N

Friction force = 400 N

The direction of motion of the friction force will be opposite to the forward force, so that both of them can be subtracted to get the value of the net force (net force: combination / sum of all forces acting on an object)

$\tt \sum F=3400-400=3000~N$

So the acceleration :

$\tt a=\dfrac{\sum F}{m}\\\\a=\dfrac{3000}{1200}\\\\a=2.5~m/s^2$

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What is the net force acting on the object below

sp2606 [1]

Answer:

10 N

Explanation:

The pair of 5 N is pressing on 1 side of the square

8 0

4 years ago

What mass of water is needed to raise the temperature by 15 degrees C using 165 Joules of energy

Lady bird [3.3K]

Mass of the water is 2.63 g.

<u>Explanation:</u>

Mass of the water, m = ? g

Temperature, ΔT = 15 °C

Heat absorbed, q = 165 J

Specific heat capacity, c = 4.18 J / g °C

q = m × c × ΔT

Now, we have to find the mass of the water by rewriting the above equation as,

m = $$\frac{q}{c\times dT }$

Now Plugin the above values in the equation as,

m = $$\frac{165}{4.18\times15}$

= 2.63 g

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8 0

3 years ago

Read 2 more answers

Use the following equation to answer the questions and please show all work.

DIA [1.3K]

Answer:

a.36 g of water is produced.

b.64 g of $O_{2}$ is consumed.

Explanation:

The reaction is $2H_{2} + O_{2}$ ⇒ $2H_{2}O$

a.

Given,

Weight of $H_{2}$ reacted = 4g

Weight of 1 mole of $H_{2}$ = 2 $\times$ 1 = 2g

Therefore no. of moles of $H_{2}$ reacted = $\frac{4}{2}$ = 2 moles;

Also given,

Weight of $O_{2}$ reacted = 32 g

Weight of 1 mole of $O_{2}$ = 2 $\times$ 16 = 32 g

Therefore no. of moles of $O_{2}$ reacted = $\frac{32}{32}$ = 1

We know that 2 moles of Hydrogen reacts with 1 mole of Oxygen to give 2 moles of water,

As we took 2 moles of Hydrogen and 1 mole of Oxygen,

Directly,from the equation we can tell 2moles of water will be produced.

Therefore no. of moles of $H_{2} O$ produced = 2

Weight of 1 mole of water = $2\times 1+16$ = 18

Therefore weight of $H_{2}O$ produced = $2\times 18$ = 36gm

b.

Given ,

72 g of $H_{2}O$ is produced.

So,

no. of moles of $H_{2}O$ produced = $\frac{72}{18}$ = 4 moles

From equation For every 2 moles of water formed , 1 mole of oxygen must be required.

So for producing 4 moles of water,

No. of moles of Oxygen required = 2 moles.

Therefore weight of $O_{2}$ reacted = $2\times32$ = 64 g

Method 2:

Given,

8 g of $H_{2}$ has reacted.

So,

no. of moles of $H_{2}$ reacted = $\frac{8}{2}$ = 4 moles.

From equation , we know that For every 2 moles of $H_{2}$ reacted,1 mole of $O_{2}$ will react.

Therefore,

No. of moles of $O_{2}$ that reacts with 4 moles of $H_{2}$ = $2\times1$ = 2 moles

Therefore the weight of $O_{2}$ reacted = $2\times 32$ = 64 g

6 0

3 years ago

Is one gram equal to one mL?

VladimirAG [237]

Only for water it is

7 0

4 years ago

Read the given equation. 2Na + 2H2O ? 2NaOH + H2 During a laboratory experiment, a certain quantity of sodium metal reacted with

emmasim [6.3K]

Answer:

The number of moles of Na metal that used initially = 0.70 mol.

The quantity of Na metal used initially to produce 7.80 of H₂ gas = 16.02 g.

Explanation:

It is a stichiometry problem.

<em>2Na + 2H₂O → 2NaOH + H₂,</em>

The balanced equation shows that <em>2.0 moles of Na metal </em>react with 2.0 moles of water to produce 2.0 moles of NaOH and <em>1.0 mole of H₂</em>,

Firstly, we need to convert the volume of H₂ (7.80 L) produced to no. of moles (n) using the ideal gas law: <em>PV = nRT</em>,

where, P is the pressure of the gas in atm<em> (P at STP = 1.0 atm)</em>,

V is the volume of the gas in L <em>(V = 7.80 L)</em>,

n is the number of moles in mole,

R is the general gas constant<em> (R = 0.082 L.atm/mol)</em>,

T is the temperature of the gas in K <em>(T at STP = 0.0 °C + 273 = 273.0 K)</em>.

∴ The number of moles of H₂ gas (n) = PV / RT = [(1.0 atm)(7.80 L)] / [(0.082 L.atm/mol.K)(273.0 K)] = 0.35 mol.

<em>Using cross multiplication:</em>

2.0 moles of Na will produce → 1.0 mole of H₂, from the stichiometrey.

??? moles of Na will produce → 0.35 mole of H₂.

∴ The number of moles of Na metal that used initially = (2.0 mol)(0.35 mol) / (1.0 mol) = 0.70 mol.

Now, we can get the quantity of Na metal using the relation:

∴ mass = n x molar mass = (0.70 mol)(22.989 g/mol) = 16.02 g.

6 0

3 years ago