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3 years ago

6

Why do the passengers on a high-flying airplane not appear weightless, similar to the astronauts on the space station?

1 answer:

sergey [27]3 years ago

4 0

<span>Even in space, there is still presence of gravity. The cause of weightlessness is not how far above the earth the space shuttle is but rather how fast it is travelling. The shuttle is in free fall causing weightlessness, but it is travelling fast enough to miss the earth as it falls. Similarly, the airplane could also provide weightlessness if it went free fall as well. However, that ends as the plane hits the ground. </span>

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A circuit is constructed with six resistors and two batteries as shown. the battery voltages are v1 = 18 v and v2 = 12 v. the po

VladimirAG [237]

Answer:

V4=9.197v

Explanation:

Given:

V1= 18v ,V2= 12v ,r1=r5=58ohms ,r2=r6=124ohms , r3=47ohms ,r4= 125ohms

V4= I4R4 = V2/(R4 + R5)×R4

V4= 12×125 /(125 + 58)

V4=1500/183 =9.197v

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3 years ago

what is the largest and smallest possible resultant force of two force with magnitude of 41N and 14N

MArishka [77]

Explanation:

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3 years ago

Pani-rosa [81]

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4 years ago

Read 2 more answers

El tubo de entrada que suministra presión de aire para operar un gato hidráulico tiene 2 cm de diámetro. El pistón de salida es

topjm [15]

Answer:

P₁ = 219.3 Pa

Explanation:

This fluid mechanics problem, we can use that the pressure is distributed with the same value throughout the system, which is Pascal's principle.

Let's use the subinidce1 for the small diameter and the subscript 2 for the larger diameter.

P₁ = P₂

pressure is defined by

P = F / A

we subtitute

F₁ / A₁ = F₂ / A₂

F₁ = F₂ A₁ / A₂

the area in a circle is

A = π r² = π d² / 4

we substitute

F₁ = F₂ (d₁ / d₂)²

we calculate

F₁ = 17640 (2/32)²

F₁ = 68.9 N

Having the force to be applied we can find the air pressure on the small plunger

P₁ = F₁ / A₁

P₁ = F₁ 4 / π d₁²

let's calculate

P₁ = 68.9 4 / (π 0.02²)

P₁ = 219.3 Pa

3 0

3 years ago

A satellite of mass 5600 kg orbits the Earth and has a period of 6200 s.

Troyanec [42]

Answer:

(a) Radius of orbit will be $=7.32\times10^6m$

(b) Earth gravitational force will be $=4.18\times 10^4N$

(C) Height will be $0.92\times 10^6m$

Explanation:

We have given

Mass of the earth, $M=6\times 10^{24}kg$

Mass of the satellite, m = 5600 kg

Radius of earth, $R=6.4\times 10^6m$

Time period T = 6200 sec

We know that $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{6200}=0.00101rad/sec$

Now

(a) We know that $\omega ^2=\frac{GM}{R^3}$

$R^3=\frac{GM}{\omega ^2}$

$R^3=\frac{6.67\times 10^{-11}\times 6\times 10^{24}}{0.00101 ^2}$

$R^3=3.92\times 10^{20}$

Radius of the orbit $R=7.32\times 10^6m$

(b)

Force $F=\frac{GMm}{R^2}=\frac{6.67\times 10^{-11}\times 6\times 10^{24}\times 5600}{(7.32\times 10^6)^2}=4.18\times 10^4N$

(c)

Altitude $h=radius\ of\ orbit-radius\ of\ earth=7.32\times 10^6-6.4\times 10^6=0.92\times 10^6m$

8 0

3 years ago