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NeX [460]
3 years ago
15

Does riding a bike with larger tires help you go faster even if you pedal at the same rate A:if the person moves faster, then th

e person is riding a bike with larger tires B: if a person rides a bike with larger tires ,then the person with move faster than on a bike with smaller tires
Chemistry
1 answer:
Alekssandra [29.7K]3 years ago
8 0

Answer:

Yes, bikes with larger tires help you go faster.

B: if a person rides a bike with larger tires ,then the person with move faster than on a bike with smaller tires

Explanation:

Bikes with larger wheels cover more distances in one revolution, when compared to bikes with a much smaller wheel, if the wheels are turning at the same revolution per minutes. Although more effort is put into turning a big wheel, when compared to that for a smaller wheel, but it will go faster if you can keep it spinning at the same rate as a small one.

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Por que el petroleo es un no recurso renovable
mote1985 [20]

Answer:

Por ejemplo, el petróleo o el carbón son ejemplos de recursos no renovables porque, aunque se forman mediante un proceso natural, este necesita demasiado tiempo. ... Esos combustibles fósiles provienen de materia orgánica, pero tardan cientos de miles de años en producirse.

3 0
3 years ago
Scientists are testing a new drug, which they think might allow people to memorize large amounts of information. Group A is taki
musickatia [10]
The answer to this question is C: the ability to memorize

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6 0
3 years ago
Read 2 more answers
A bike rider approaches a hill with a speed of 8.5 m/s. The total mass of the bike and the rider is 85 kg. Find the kinetic ener
krek1111 [17]

KE=3070.625 J

Height = 3.686 m

<h3>Further explanation</h3>

mass of bike+rider=85 kg

velocity = 8.5 m/s

\tt KE=\dfrac{1}{2}mv^2\\\\KE=\dfrac{1}{2}\times 85\times 8.5^2\\\\KE=3070.625~J

Conservation of energy :

(KE+PE)₁ (downhill) = (KE+PE)₂ (up the hill)

PE₁=0⇒h=0

KE₂=0⇒v=0(stop), so equation becomes :

KE₁=PE₂

\tt KE_1=3070.625\\\\KE_1=mgh_2\\\\3070.625=85\times 9.8\times h\Rightarrow h=3.686~m

5 0
3 years ago
A 50.0 mL sample of gas at 20.0 atm of pressure is compressed to 40.0 atm of pressure at constant temperature. What is the new v
guapka [62]

Ans: Final volume = 25.0 ml

<u>Given:</u>

Initial volume V1 = 50.0 ml

Initial pressure P1 = 20.0 atm

Final pressure P2 = 40.0 atm

<u>To determine:</u>

The final volume V2

<u>Explanation:</u>

Ideal gas equation: PV = nRT

under constant temperature, T and number of moles n we have:

PV = constant

or, P1V1 = P2V2

V2 = P1V1/P2 = 20*50/40 = 25 ml.

6 0
3 years ago
In a sample of UO 2, the uranium has been enriched so that 3.0% of it is 235U. How long could the fissioning of the uranium in t
8090 [49]

Explanation:

Molecular mass of UO_{2} is the sum of molecular mass of U and O_{2} molecule.

             Molar mass of UO_{2} = (238 + 32) g/mol

                                                                  = 270 g/mol

Hence, in 1 kg there are 1000 grams. So, total number of atoms given molecules present in 1 kg will be calculated as follows.

          \frac{1000 g}{270 g/mol} \times 6.023 \times 10^{23}atoms

                     = 2.229 \times 10^{24}

Hence, number of ^{235}U = \frac{3}{100} \times 2.229 \times 10^{24}

                                            = 6.69 \times 10^{22}

Now, let x seconds is required for burning 100 W lamp bulb. As 200 MeV is released per reaction.

             100 x =  6.69 \times 10^{22} \times 200 \times 10^{6} \times 1.6 \times 10^{-19}

                    x = 21.4 \times 10^{9} sec

Converting this value of x into years as follows.

                       x = \frac{21.4 \times 10^{9}}{3600 \times 24 \times 365}

                    x = 6.8 \times 10^{2} years

Thus, we can conclude that fissioning of the uranium in given situation requires 6.8 \times 10^{2} years to keep a 100 W lamp burning.

6 0
3 years ago
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