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katrin2010 [14]
3 years ago
15

An airplane dropped a flare from a height of 2860 feet above a lake. How many seconds did it take for the flare to reach the wat

er?
Physics
2 answers:
KATRIN_1 [288]3 years ago
8 0

Answer: 13.2 seconds.

Explanation: using equation of motion; S= ut +1/2at² where u = initial velocity=0

S= distance travelled

a = acceleration due gravity

t= time.

1 foot = 0.305m so,

S= 2860 feet =872.3m

S= ut+1/2 at²

872.3 = 0×t + 1/2×10 × t²

872.3 =0 + 5t²

T²= 872.3/5

T²= 174.46

Take the square root of T we then have;

t = 13.2 seconds to one decimal place.

QveST [7]3 years ago
7 0

Answer:

Aircraft crashed and sank into the water ~ 50 yards off shore, in 45 feet water, ... On his second flight, Charles reached an altitude of 2,700 metres (8,900 ft) ... carrying 16 personnel made an emergency landing on Lake Qadisiyah in Al ... The 777 will reach 500 airplanes delivered faster than any other twin-aisle airplane in ...

Explanation:

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Earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.
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The Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x 10^{24} Kg

<h3>Relationship between Linear and angular speed</h3>

Linear speed is the product of angular speed and the maximum displacement of the particle. That is,

V = Wr

Where

  • V = Linear speed
  • W = Angular speed
  • r = Radius

Given that the earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.

a) To determine the Earth’s average orbital speed, we will make use of the below formula to calculate angular speed

W = 2\pi/T

W = (2 x 3.143) / (365.26 x 24)

W = 6.283 / 876624

W = 7.2 x 10^{-4} Rad/hr

The Earth’s average orbital speed V = Wr

V = 7.2 x  10^{-4} x 149.6 x 10^{6}

V = 107225.5 kilometers per hours.

b) Based on the information given in this question, to calculate the approximate mass of the Sun, we will use Kepler's 3rd law

M = (4\pi ^{2}r^{3}) / GT^{2}

M = (4 x 9.8696 x 3.35 x 10^{24}) / (6.67 x 10^{-11} x 7.68 x 10^{11}<em>)</em>

<em>M = 1.32 x </em>10^{26} / 51.226

M = 2.58 x 10^{24} Kg

Therefore, the Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x 10^{24} Kg

Learn more about Orbital Speed here: brainly.com/question/22247460

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