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Zina [86]
2 years ago
6

A forklift pushes a box with a force of 500 N across the floor for a distance of 5.0 m, then turns around and pushes with the sa

me force 5.0 m back to its original position. How much work did the forklift do during this time?
Physics
1 answer:
yanalaym [24]2 years ago
4 0

Answer:

5,000J

Explanation:

Work = Force x Distance

Distance back and forth is canceled out, so either the answer is + or -

5.0m + 5.0m = 10.0m

500N x 10.0m = 5,000J

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In a $100$ meter track event, Alice runs at a constant speed and crosses the finish line $5$ seconds before Beatrice does. If it
Liono4ka [1.6K]

Answer:

10s

Explanation:

If it took Beatrice 25 seconds to complete the race

Distance = 100 meter

Beatrice speed = 100/25

                          = 4m/s

If Alice runs at a constant speed and crosses the finish line $5$ seconds, she must have completed the race in 20s (25 -5).

Her speed where constant

= 100/20

= 5 m/s

It would take Alice

= 50/5

= 10s

It would take Alice 10s to run $50$ meters.

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3 years ago
What the planet after Venus
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Mercury and Venus are therefore closer to each other most of the time. But Earth is the planet closest to Venus. And that's why from here on Earth, Venus looks so big and luminous. Venus is the brightest thing in the night sky after the sun and the moon.

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3 years ago
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What happens when you "crack/pop your knuckles?
Art [367]

It could result in it not being good for your joints, as well as in the long run but shouldn't cause problems when your a child. I hope this helps your question!

6 0
2 years ago
A first order reaction, A -> products, has a rate reaction of .00250 Ms-1 when [A] = . 484 M. (a) What is the rate constant,
tamaranim1 [39]

Answer: a)  The rate constant, k, for this reaction is 0.00516s^{-1}

b) No t_{\frac{1}{2}} does not depend on concentration.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

A\rightarrow products

Given: Order with respect to A = 1

Thus rate law is:

a) Rate=k[A]^1

k= rate constant

0.00250=k[0.484]^1

k=0.00516s^{-1}

The rate constant, k, for this reaction is 0.00516s^{-1}

b) Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant  

a - x = amount left after decay process  

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{2.303}{k}\log\frac{100}{50}

t_{\frac{1}{2}}=\frac{0.69}{k}

Thus t_{\frac{1}{2}} does not depend on concentration.

8 0
2 years ago
Enrico Fermi (1901–1954) was a famous physicist who liked to pose what are now known as Fermi problems, in which several assumpt
Katarina [22]

Answer:

Explanation:

(a)

Since the earth is assumed to be a sphere.

Volume of atmosphere = volume of (earth +atm osphere) — volume of earth

= \frac{4}{3}\pi(6400+ 50)^3 -  \frac{4}{3}\pi (6400)
^3\\\\=  \frac{4}{3}\pi(6192125000) km’^3\\= 2.6\times 10^{19} m^3

Hence the volume of atmosphere is 2.6\times 10^{19} m^3

(b)

Write the ideal gas equation as foll ows:

PV = nRT\\\\n\frac{0.20atm\times 2.6\times10^{19} m^3}{0.08206L\, atm/mok\, K \times (15+273+15)K}\times \frac{1L}{10^{-3}m^3}\\\\= 2.20\times 10^{20} moles

no.\, of\, molecules = 2.20\times 10^{20} moles \times \frac{6.022\times10^{23}\,molecules}{1mole}= 13.3\times10^{43} molecules


Hence the required molecules is 13.3\times10^{43} molecules


(c)

Write the ideal gas equation as follows:

PV =nRT
\\\\n=\frac{1.0 atm \times 0.5L
}{0.08206 L\, atm/mol\,K \times (37 +273.1 5)K} = 0.0196 moles

no.\, of\, molecules = 0.0196 moles \times\frac{6.022\times10^{23} molecules}
{Imole}= 1.2\times 10^{23} molecules

Hence the required molecules in Caesar breath is 1.2\times 10^{23} molecules

(d)

Volume fraction in Caesar last breath is as follows:  

Fraction,\, X =\frac{12\times 10 molecules}{13.3\times 10^{43} \,molecules}= 9.0\times 10\, molecule/air\, molecule}

(e)

Since the volume capacity of the human body is 500 mL.

Volume\, of\, Caesar\, nreath\, inhale\, is =\frac{ 12\times 10^{22}\, molecules}{breath}\times \frac{9.0\times10^{-23} molecule}{air\, molecule}\\\\= 1.08 molecule/breath

5 0
2 years ago
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