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alukav5142 [94]
3 years ago
5

A long distance runner running a 5.0km track is pacing himself by running 4.5km at 9.0km/h and the rest at 12.5km/h. What is his

average speed?
Physics
1 answer:
sweet [91]3 years ago
6 0

Answer:

9.26 km/h

Explanation:

Applying,

V' = D'/t'............... Equation 1

Where V' = Average speed, D' = Total distance, t' = total time.

Given: D' = 5 km

But,

v = d/t............ Equation 2

Where v = speed , d = distance, t = time

t = d/v............ Equation 3

Given: d = 4.5 km, v = 9 km/h, and d = 0.5 km, v = 12.5 km/h

Therefore,

t₁ = 4.5/9 = 0.5 hours

t₂ = 0.5/12.5

t₂ = 0.04 hours

Therefore,

V' = 5/(0.5+0.04)

V' = 5/0.54

V' = 9.26 km/h

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Vladimir79 [104]
Upside down if its further than 1 focal point you have seen this with a spoon and enlarged right side up if closer than 1 focal point
5 0
3 years ago
La estrella mas proxima a la tierra esta a 2 años de luz. Calcula esta distancia en unidades del SI
tino4ka555 [31]

Answer:

2\ ly=1.89\times 10^{16}\ m

Explanation:

The question says that, "The closest star to the earth is 2 light years away. Calculate this distance in SI units".

Given that,

The distance between the closest star and the Earth is 2 light years.

The SI unit of distance is m. It means we need to convert light years to meters. We know that,

1\ ly=9.461\times 10^{15}\ m

2 light-years means,

2\ ly=2\times 9.461\times 10^{15}\ m\\\\=1.89\times 10^{16}\ m

So, the required distance is equal to 1.89\times 10^{16}\ m.

5 0
3 years ago
Which of the followings are true about Vmax? A. The higher the [enzyme], the higher the Vmax B. Vmax is proportional to k2 C. Vm
Vera_Pavlovna [14]

Answer:

Correct answer is A.

The higher the enzyme, the higher the Vmax

Explanation:

Although, in the absence of enzyme, the rate of a reaction(Vmax) increase linearly with substrate concentration. The reaction rate is given as dp/dt.

The rate of a reaction involving enzyme also increases.

At low enzyme concentrations or high substrate concentrations, all of the available enzyme active sites could be occupied with substrates. Therefore, increasing the substrate concentration further will not change the rate of diffusion. In other words, there is some maximum reaction rate (Vmax) when all enzyme active sites are occupied. The reaction rate will increase with increasing substrate concentration, but must asymptotically approach the saturation rate, Vmax. Vmax is directly proportional to the total enzyme concentration, E

5 0
3 years ago
GIVING BRAINLIEST TO FIRST RIGHT ANSWER!! Help as soon as possible please!
crimeas [40]

Answer:

thx for the points lolol

5 0
4 years ago
Read 2 more answers
3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag
aleksandr82 [10.1K]

Answer:

(a) W=-19.25J

(b) W=-52.8J

Explanation:

Hello.

(a) In this case, since the initial volume is 18.5 dm³ and the final volume is 21 dm³ (18.5 +2.5), we can compute the work at constant pressure as shown below:

W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J

Which is negative as it expands against the given pressure.

(b) Moreover, of the process is carried out reversibly, the pressure can change, therefore, we need to compute the work via:

W=nRTln(\frac{V_1}{V_2} )

Whereas the moles are computed from the given mass of argon:

n=6.56g*\frac{1mol}{39.95g}=0.164mol

Thus, the work is:

W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J

Regards.

4 0
3 years ago
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