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2 years ago

11

A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a velocity of 42 m/s. What is the a

pproximate acceleration of the train during this time?

1 answer:

shepuryov [24]2 years ago

4 0

Answer:

0.16 m/s²

Step-by-step explanation:

Given:

Δx = 5600 m

v₀ = 0 m/s

v = 42 m/s

Find: a

v² = v₀² + 2aΔx

(42 m/s)² = (0 m/s)² + 2a (5600 m)

a = 0.16 m/s²

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Dafna1 [17]

$2.40714$

1) Let's evaluate that expression, given that a=4.9, b=-7, and c=-0.5

$\begin{gathered} a^3\div b^2+c^3 \\ (4.9)^3\div(-7)^2+(-0.5)^3= \\ \frac{4.9^3}{\left(-7\right)^2+\left(-0.5\right)^3} \\ \\ \frac{4.9^3}{48.875} \\ \\ \frac{117.649}{48.875} \\ \\ 2.40714 \end{gathered}$

Note that we have rewritten it as a fraction so that we can easily operate. Also, we have applied here the PEMDAS order of operations, prioritizing the exponents.

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1 year ago

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Step-by-step explanation:

5/18 + (-1/6)

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3 years ago

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3 years ago

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andrezito [222]

Answer:

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Step-by-step explanation:

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