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Arisa [49]
2 years ago
9

An Airbus A380 airliner can takeoff when its speed reaches 80 m/s. Suppose its engines together can produce an acceleration of 3

m/s. (a) How long does it take the plane to accelerate from 0 m/s to 80 m/s? (b) What is the distance traveled during this period?
Physics
1 answer:
Tomtit [17]2 years ago
7 0

Answer:

(a). The time is 26.67 sec.

(b). The distance traveled during this period is 1066.9 m.

Explanation:

Given that,

Speed = 80 m/s

Acceleration = 3 m/s

Initial velocity = 0

(a). We need to calculate the time

Using equation of motion

v = u+at

t = \dfrac{v-u}{a}

Put the value into the formula

t = \dfrac{80-0}{3}

t =26.67\ sec

The time is 26.67 sec.

(b). We need to calculate the distance traveled during this period

Using equation of motion

s = ut+\dfrac{1}{2}at^2

s = 0+\dfrac{1}{2}\times3\times(26.67)^2

s =1066.9\ m

The distance traveled during this period is 1066.9 m.

Hence, This is the required solution.

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T =\dfrac{mv^2}{r}                                

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Read 2 more answers
Glycerin is poured into an open U-shaped tube until the height in both sides is 20cm. Ethyl alcohol is then poured into one arm
Liula [17]

Answer:

7.5 cm

Explanation:

In the figure we can see a sketch of the problem. We know that at the bottom of the U-shaped tube the pressure is equal in both branches. Defining \rho_A: Ethyl alcohol density and \rho_G: Glycerin density , we can write:

\rho_A\times g \times h_1 + \rho_G \times g \times h_2 = \rho_G \times g \times h_3

Simplifying:

\rho_A\times h_1 = \rho_G \times (h_3 - h_2) (1)

On the other hand:

h_1 + h_2 = \Delta h + h_3

Rearranging:

h_1 - \Delta h = h_3 - h_2 (2)

Replacing  (2) in (1):

\rho_A\times h_1 = \rho_G \times (h_1 - \Delta h)

Rearranging:

\frac{h_1 \times (\rho_A - \rho_G)}{- \rho_G} = \Delta h

Data:

h_1 = 20 cm; \rho_A = 0.789 \frac{g}{cm^3}; \rho_G = 1.26 \frac{g}{cm^3}

\frac{20 cm \times (0.789 - 1.26) \frac{g}{cm^3}}{- 1.26\frac{g}{cm^3}}  = \Delta h

7.5 cm =  \Delta h

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3 years ago
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