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3 years ago

7. A 1.50 x 103 kg car accelerates uniformly from rest to 10.0 m/s in 3.00 s.

Chemistry

1 answer:

MArishka [77]3 years ago

8 0

75,000‬J

25,000W

Explanation:

Given parameters:

Mass of car = 1500kg

Velocity = 10m/s

time taken = 3s

Unknown:

Work done = ?

Power delivered by engine = ?

Solution:

Work done here is the kinetic energy possessed by the car as it covers the distance.

Work done and energy are related quantities.

Work done = Kinetic energy = $\frac{1}{2}$ mv²

m is the mass of the body

v is the velocity

Therefore, inputting the parameters gives:

K.E = $\frac{1}{2}$ x 1500 x 10² = 75,000‬J

The work done is 75,000‬J

Power is the rate at which work is done:

Power = $\frac{work done }{time}$

Power = $\frac{75000}{3}$ = 25000W

learn more:

Kinetic energy brainly.com/question/9059731

Power supply brainly.com/question/4322575

#learnwithBrainly

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If 1.02 g of nickel reacted with 750. mL of 0.112 M hydrobromic acid, how much of each will be present at the end of the reactio

kati45 [8]

Answer:

35.1% is percent yield

Explanation:

Full question: Assume no volume change. If you formed 0.0910 atm of gas, what is the percent yield?

The reaction that is occurring is:

Ni + 3HBr → NiBr₃ + 3/2H₂(g)

First, we will determine moles of Ni and HBr to determine limiting reactant and theoretical yield

Using ideal gas law, we can determine the moles of hydrogen formed. Thus, we can find percent yield:

Moles Ni (Molar mass: 58.69g/mol):

1.02g * (1mol / 58.69g) = 0.01738moles Ni

Moles HBr:

0.750L * (0.112mol/L) = 0.084 moles of HBr.

For a complete reaction of the 0.084 moles of HBr you need:

0.084mol HBr * (1 mole Ni / 3 moles HBr) = 0.028 moles of Ni.

As there are just 0.01738 moles of Ni, the Ni is limiting reactant. Assuming a theoretical yield, moles of H₂ produced are:

0.01738moles Ni * (3/2 H₂ / 1 mol Ni) = 0.02607 moles H₂

Now, moles of H₂ produced are:

PV = nRT

PV/RT = n

Where P is pressure (0.0910atm)

V is volume (2.50L)

R is gas constant (0.082atmL/molK)

T is absolute temperature in Kelvin (30°C + 273.15 = 303.15K)

And n are moles

PV/RT = n

0.0910atm*2.50L/0.082atmL/molK*303.15K = n

0.00915 moles = n

And percent yield (Produced moles / Theoretical moles * 100) is:

0.00915 moles / 0.02607moles =

<h3>35.1% is percent yield</h3>

8 0

3 years ago

Can anyone answer this?? Thanks in advance AlCl3 + Cu = ?

Kazeer [188]

Reaction won't occur
Because the activity of Cu is less than Al

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3 years ago

Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow

andreyandreev [35.5K]

Explanation:

(a) The given data is as follows.

Pressure on top ( $P_{o}$ ) = 140 bar = $1.4 \times 10^{7} Pa$ (as 1 bar = $10^{5}$ )

Temperature = $15^{o}C$ = (15 + 273) K = 288 K

Density of gas = $\frac{PM}{ZRT}$

$\frac{dP}{dZ} = \rho \times g$

$\frac{dP}{dZ} = \int \frac{PM}{ZRT}$

$\int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ$

$ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}$

= 0.4548

$P_{1} = P_{o} \times e^{0.4548}$

= $1.4 \times 10^{7} Pa \times 1.5797$

= $2.206 \times 10^{7} Pa$

Hence, pressure at the natural gas-oil interface is $2.206 \times 10^{7} Pa$ .

(b) At the bottom of the tank,

$P_{2} = P_{1} + \rho \times g \times h$

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

= $309.8 \times 10^{5} Pa$

= 309.8 bar

Hence, at the bottom of the well at $15^{o}C$ pressure is 309.8 bar.

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3 years ago

Dose your heart muscle get tired? Yes or no.

WARRIOR [948]

Answer:

no 

Explanation:

our heart muscles never get tired, because it has to pump blood in our body 72 times a minute, it is made of special cardiac muscles which helps it to perform it's function without getting tired ....

i hope it helped.....

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3 years ago

Is gravity a matter??

algol13

Answer:

No, gravity isn't matter

Explanation:

Gravity is a force that attracts matter towards the center of a physical body with mass.

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3 years ago