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3 years ago

10

As an airplane takes off, the air flows across the wings of the airplane. Which of the following would be the best description?

I. The air flow across the top of the wing is greater than the air flows across the bottom of the wing. II. The air flow across the top and bottom of the wing is the same. III. The lift force is present because the pressure across the top is less than the pressure across the bottom of the wing.

1 answer:

antoniya [11.8K]3 years ago

8 0

Answer:

III. The lift force is present because the pressure across the top is less than the pressure across the bottom of the wing.

Explanation:

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The space shuttle usually orbited Earth at altitudes of around 300.0 km. 1) Determine the time for one orbit of the shuttle abou

777dan777 [17]

Answer:

Explanation:

distance of shuttle from centre of the earth = radius of the orbit

= 6300 + 300 = 6600 km

= 6600 x 10³

Formula of time period of the satellite

T = 2π R /v₀ , v₀ is orbital velocity

v₀ = √gR , ( if height is small with respect to radius )

T = 2π R /√gR

= 2π√ R /√g

= 2 x 3.14 x √ 6600 x 10³ / √9.8

= 2 x 3.14 x 256.9 x 10 / 3.13

= 5154.41 s

= 5154.41 / 60 minutes

= 85.91 m

85.9 minutes.

2 ) No of sunrise per day = no of rotation per day

= 24 x 60 / 85.9

= 16.76

or 17 sunrises.

3 0

3 years ago

Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

ddd [48]

Answer:

Al's mass is 102.92 kg

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

$F_{net} = 0$

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

$F_{net} = \frac{dp_{horizontal}}{dt} = 0$

$p_{horizontal_i }= p_{horizontal_f}$

where the suffix i and f means initial and final respectively.

The initial momentum will be:

$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

But, as they are at rest, initially

$p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0$

$p_{horizontal_i} = 0$

So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

5 0

3 years ago

In a milkans apparatus,an oil drop of weight 2.0×10^-15kg accquires two surplus electrons. when a Potential difference of 620 vo

andriy [413]

Answer:

r = 9.92 mm

Explanation:

Given that,

Mass of oil drop, $m=2\times 10^{-15}\ kg$

It acquires 2 surplus electrons, q = +2e $=3.2\times 10^{-19}\ C$

Potential difference, V = 620 V

Thie potential difference is applied between the pair of horizontal metal plates the drop is in equilibrium.

We need to find the distance between the plates.

At equilibrium,

mg = qE

Since, E = V/r (r is distance between plates)

$mg=\dfrac{qV}{r}\\\\r=\dfrac{qV}{mg}\\\\r=\dfrac{3.2\times 10^{-19}\times 620}{2\times 10^{-15}\times 10}\\\\=0.00992\ m\\\\=9.92\ mm$

So, the distance between the plates is 9.92 mm.

6 0

3 years ago

Two vectors of magnitudes 30 units and 70 units are added to each other. What are possible results of this addition? (section 3.

Yanka [14]

Answer:

the correct answer is option C which is 50 units.

Explanation:

given,

two vector of magnitude = 30 units and of 70 units

to calculate resultants vector = \sqrt{a^2+b^2+2 a b cos\theta}

cos θ value varies from -1 to 1

so, resultant vector

= $\sqrt{a^2+b^2-2 a b cos\theta}\ to\ \sqrt{a^2+b^2+ 2 a b cos\theta}$

a = 30 units and b = 70 units

= $\sqrt{30^2+70^2-2\times 30\times 70}\ to\ \sqrt{30^2+70^2+2\times 30\times 70}$

= 40 units to 100 units

hence, the correct answer is option C which is 50 units.

4 0

3 years ago

Neglecting friction, of a pendulum bob has 100 joules of kinetic energy at the bottom of its swing, how much potential energy do

galben [10]

100 joule of kinetic energy

7 0

3 years ago