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Viktor [21]
3 years ago
10

A plane has a takeoff speed of 89.3m/s and requires 1465m to reach that speed determine the acceleration of the plane and the ti

me required to reach this speed
Physics
1 answer:
Tems11 [23]3 years ago
4 0

Answer:

a = 2.72 ms⁻²

32.83 s

Explanation:

By using the kinematic equations you get,

v² = u² +2as and v = u + at   where all terms in usual meaning

Using 1st equation,

89.3² = 0² + 2a×1465 ⇒ a = 2.72 ms⁻²

By 2nd equation,

89.3 = 0 + 2.72×t ⇒ t = 32.83 s

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Flammability, I think!
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6. The hole on a level, elevated golf green is a horizontal distance of 150 m from the tee and at an elevation of 12.4 m above t
Georgia [21]

Answer:

u = 104.68 m/s

Explanation:

given,

horizontal distance = 150 m

elevation of  12.4 m

angle = 8.6°

horizontal motion = x = u cos θ. t .............(1)

vertical motion =

y = u sin \theta - \dfrac{1}{2}gt^2................(2)

from equation(1) and (2)

y = x tan \theta - \dfrac{gx^2}{2u^2cos^2\theta}..........{3}

12.4 = 150\times tan (8.6) - \dfrac{9.8\times 150^2}{2u^2cos^2(8.6)}

\dfrac{9.8\times 150^2}{2u^2cos^2(8.6)} = 10.29

\dfrac{9.8\times 150^2}{2\times 10.29\times cos^2(8.6)} = u^2

u = \sqrt{10959.34}

u = 104.68 m/s

The initial speed of the ball is u = 104.68 m/s

8 0
3 years ago
When you sit in a chair, your body exerts a downward force on the chair. What
Ahat [919]

I feel like the answer would be B. The chair pushes down on the floor becuase the question says when you sit in a chair, your body exerts a downward force on the chair so it would be pushing downward meaning that the chair would also go down making it push onto the floor.

7 0
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What force, in newtons, must you exert on the balloon with your hands to create a gauge pressure of 62.5 cm H2O, if you squeeze
yulyashka [42]

Answer:54.70 N

Explanation:

Given

Gauge Pressure of 62.5 cm of H_2O

i.e. h=62.5 cm =0.625 m

Effective area A=51 cm^2

initial Pressure= 1 atm=101.325 kPa

Gauge Pressure P=\rho gh

\rho =density\ of\ water =1000 kg/m^3

P_{gauge}=1000\times 9.8\times 0.625=5.937 kPa

Force creates a pressure of P_1 which will be equal to Gauge Pressure

P_1=\frac{F}{A}

P_1=P_{gauge}

\frac{F}{A}=5.937 kPa

F=5.937\times 51\times 10^{-4}\times 10^3

F=30.27 N

6 0
3 years ago
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

f = f_{0}\frac{v + v_{o}}{v - v_{s}}        

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer  

v: is the speed of the sound = 343 m/s

v_{o}: is the speed of the observer = 0 (it is heard in the town)

v_{s}: is the speed of the source =?

The frequency of the train before slowing down is given by:

f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}  (1)                  

Now, the frequency of the train after slowing down is:

f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}   (2)  

Dividing equation (1) by (2) we have:

\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}   (3)  

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

v_{s_{b}} = 2v_{s_{a}}     (4)

Now, by entering equation (4) into (3) we have:

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}  

\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}

By solving the above equation for v_{s_{a}} we can find the speed of the train after slowing down:

v_{s_{a}} = 11.06 m/s

Finally, the speed of the train before slowing down is:

v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.                        

I hope it helps you!                                                        

7 0
2 years ago
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