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jenyasd209 [6]
3 years ago
11

Team infinite dimensions canoed 15 3/4 miles in 3 hours. what was the average rate of speed in miles per hour?

Mathematics
2 answers:
pentagon [3]3 years ago
5 0
Mi/h and mph. same thing I believe. I'm from Canada so I don't know what unit works but the ppl were going an average of 5.25miles per hour

Anestetic [448]3 years ago
4 0

Answer: Average speed would be 5\dfrac{1}{4}\ mph

Step-by-step explanation:

Since we have given that

Distance traveled by team infinite = 15\dfrac{3}{4}=\dfrac{63}{4}\ miles

Time taken = 3 hours

We need to find "Speed".

As we know the relation between distance, time and speed.

Speed=\dfrac{Distance}{Time}\\\\Speed=\dfrac{63}{4\times 3}\\\\Speed=\dfrac{21}{4}\\\\Speed=5\dfrac{1}{4}\ mph

Hence, Average speed would be 5\dfrac{1}{4}\ mph

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Let a_n denote the given sequence with n\ge1:

\{a_n\}_{n\ge1}=\{1,5,11,19,29,41,\ldots\}

Let b_n be the sequence of the forward differences of a_n, so that b_n=a_{n+1}-a_n for n\ge1:

\{b_n\}_{n\ge1}=\{4,6,8,10,12,\ldots\}

b_n follows an arithmetic progression with a difference of 2 between terms, so that

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Then we have

2n+2=a_{n+1}-a_n\implies a_{n+1}=a_n+2n+2

so that a_n is given recursively by

\begin{cases}a_1=1\\a_{n+1}=a_n+2(n+1)&\text{for }n\ge1\end{cases}

By substitution, we can try to find a pattern:

a_2=a_1+2\cdot2

a_3=a_2+2\cdot3=a_1+2(2+3)

a_4=a_3+2\cdot4=a_1+2(2+3+4)

a_5=a_4+2\cdot5=a_1+2(2+3+4+5)

and so on, with the general pattern

a_n=a_1+2(2+3+4+\cdots+n)

and since a_1=1 we can write this as

a_n=1+2(2+3+4+\cdots+n)

a_n=-1+2+2(2+3+4+\cdots+n)

a_n=2(1+2+3+4+\cdots+n)-1

Recall that

\displaystyle\sum_{i=1}^ni=1+2+3+\cdots+n=\dfrac{n(n+1)}2

Then

a_n=2\dfrac{n(n+1)}2-1\implies\boxed{a_n=n^2+n-1}

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Step-by-step explanation:

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3 years ago
Que presion manométrica debe de producir una bomba para subir agua del fondo del gran cañon ( elevación 730 m ) a indian gardens
Alexxandr [17]

Answer:

La presión manométrica debe ser 6,272,000 Pa = 61.9 atm

Step-by-step explanation:

La presión absoluta se refiere al valor de presión referido al cero absoluto o al vacío, sin presión atmosférica. Es decir, la presión absoluta es la presión con respecto al vacío total.

La presión manométrica es la que ejerce un medio distinto al de la presión atmosférica y representa la diferencia entre la presión real o absoluta y la presión atmosférica. Es decir, la presión manométrica se mide en relación a la presión atmosférica y se define como la diferencia entre presión absoluta (Pabs) y presión atmosférica predominante (Patm). La presión manométrica sólo se aplica cuando la presión es superior a la atmosférica.  

Entonces,  Pmanometrica = Pabs – Pamb

Para encontrar la presión a una profundidad h en un líquido sin movimiento, expuesto al aire cerca de la superficie de la Tierra, la presión manométrica puede ser calculada mediante:

Pmanometrica=ρ*g*Δh

donde ρ es la densidad del fluido, g la gravedad y Δh la variación en la altura.

En este caso:

  • ρ=1000 \frac{kg}{m^{3} } (densidad del agua)
  • g= 9.8  \frac{m}{s^{2} }
  • Δh=hfinal - hinitial= 1370 m - 730 m= 640 m

Reemplazando:

Pmanometrica=  1000 \frac{kg}{m^{3} } *9.8 \frac{m}{s^{2} } * 640 m

Resolviendo:

Pmanometrica= <u><em>6,272,000 Pa = 61.9 atm</em></u> (siendo 1 pascal = 9.869*10⁻⁶ atmósferas )

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