Physics, 11.06.2021 04:30 GaryCarmine

To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeoff velocity, depends on several factors, including the weight of the aircraft and the wind velocity. Required:
a. A plane accelerates from rest at a constant rate of 5.00m/s^2 along a runway that is 1800m long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the time tTO needed to take off?
b. What is the distance dfirst traveled by the plane in the first second of its run?
c. What is the distance dfirst traveled by the plane in the first second of its run?

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Answer from: davidgarcia522083

(a)

67.1 s

(b) 2.5 m

Explanation:

initial speed, u = 0 m/s

final speed, v = 70 m/s

acceleration, a = 5 m/s2

distance, s = 1800 m

(a) Use third equation of motion

$v^2= u^2 + 2 a s \\\\v^2 = 0 + 2 \times 5\times 1800\\\\v =134.2 m/s$

Let the time is t.

Use first equation of motion

v = u + at

134.2 = 0 + 5 t

t = 67.1 s

(b) Use second equation of motion

$s = u t +0.5 at^2\\\\s = 0 +0.5\times 5\times 1 \\\\s = 2.5 m$

Answer from: Quest

answer: the correct answer id option(d).

explanation:

snell's law refraction:

$\frac{n_2}{n_1}=\frac{sin \theta_1}{sin\theta_2}$

the angle of refraction = $\theta_2=38 ^o$

the angle of incidence = $\theta_1=?$

refractive index of air = $n_1$ = 1

refractive index of glycerol $n_2$ = 1.47

$\frac{1.47}{1}=\frac{sin\theta_1}{sin 38^o}, (sin38^o=0.61)$

$sin\theta_2=0.8967$

$\theta_2=24.2^o, (sin^{-1}0.8967=63.72^o)$

from the given options, the most closest answer to 63.72° is 65° of option (d).

Answer from: Quest

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