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ra1l [238]
2 years ago
9

How much water makes up earth in percentages? thx?

Physics
1 answer:
Bond [772]2 years ago
7 0

Answer:

71%

Explanation:

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A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th
melomori [17]

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

5 0
3 years ago
A basketball leaves a player's hands at a height of 2.00 m above the floor. The basket is 3.05 m above the floor. The player lik
shutvik [7]

Answer:(10.69, 11.436)

Explanation:

Given

initial height of ball is 2 m

height of basket is 3.05 m

Launching angle=40^{\circ}

x =12\pm 0.27

y=1.05

equation of trajectory of ball is given by

y=xtan\theta -\frac{gx^2}{2u^2cos^2\theta }

for x=12.27

1.05=12.27\times tan40-\frac{g12.27^2}{2u^2cos^{2}40 }

u=10.69

for x=11.73

1.05=11.73\times tan40-\frac{g11.73^2}{2u^2cos^{2}40 }

u=11.436 m/s

Thus range of speed is (10.69, 11.436)

3 0
2 years ago
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