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3 years ago

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

1 answer:

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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-g With what tension must a rope with length 2.50 m and mass 0.120 kg be stretched for transverse waves of frequency 40.0 Hz to

Alex17521 [72]

Answer:

43.2 N

Explanation:

$\lambda$ = Wavelength = 0.75 m

f = Frequency = 40 Hz

m = Mass of string = 0.12 kg

L = Length of string = 2.5 m

$\mu$ = Linear density = $\dfrac{m}{L}$

Velocity of wave is given by

$v=\lambda f\\\Rightarrow v=0.75\times 40\\\Rightarrow v=30\ m/s$

The tension in string is given by

$T=v^2\mu\\\Rightarrow T=30^2\times \dfrac{0.12}{2.5}\\\Rightarrow T=43.2\ N$

The tension in the string is 43.2 N

8 0

3 years ago

During the contraction of the heart, 65 cm3 blood is ejected from the left ventricle into the aorta with a velocity of approxima

vova2212 [387]

Answer:

The value $F = 0.1396 \ N$

Explanation:

From the question we are told that

The volume blood ejected is $V = 65 \ cm^3 = 65*10^{-6} \ m^3$

The velocity of the blood ejected is $v = 103 \ cm/s = \frac{103}{100} = 1.03 \ m/s$

The density of blood is $\rho = 1060 \ kg/m^3$

The heart beat is $R = 59 \ bpm(beats \ per \ minute) = \frac{59}{60}= 0.9833\ bps$

The average force exerted by the blood on the wall of the aorta is mathematically represented as

$F = 2 * \rho * V * R * v$

=> $F = 2 * 1060 * 65*10^{-6} * 0.9833 * 1.03$

=> $F = 0.1396 \ N$

8 0

3 years ago

The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which

emmainna [20.7K]

Answer:

Explanation:

A ) angular velocity ω = 2π / T

= 2 x 3.14 / 60

= .10467 rad / s

linear velocity v = ω R

= .10467 x 50

= 5.23 m / s

centripetal force = m v² / R

= mg v² / gR

= 834 x 5.23² / 9.8 x 50

= 46.55 N

B )

apparent weight

= mg - centripetal force

= 834 - 46.55

= 787.45 N

C ) apparent weight

= mg + centripetal force

= 834 + 46.55

= 880.55 N.

D )

For apparent weight to be zero

centripetal force = mg

mg = mv² / R

v² = gR

= 9.8 x 50

= 490

v = 22.13 m /s

time period of revolution

= 2π R /v

2 x 3.14 x 50 / 22.13

= 14.19 s

8 0

3 years ago

What happens to ar object in free fall?

Inessa05 [86]

Energy from the gravitational potential store in converted to kinetic energy. Air friction acts against the object, dissipating some energy as heat or sound. The object will continuously accelerate until the acceleration is equal to the air friction acting against it. This is when it reaches terminal velocity

5 0

2 years ago

When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for

sveta [45]

Answer:

F = -4567.40 N

Explanation:

Given that,

The power developed by the engine, P = 196 hp

1 hp = 746 W

196 hp = 146157 W

Speed of the car, v = 32 m/s

Let F is the total friction force acting on the car. The product of force and velocity is called the power developed by the engine. It is given by :

$P=-F\times v$

$F=\dfrac{-P}{v}$

$F=\dfrac{-146157\ W}{32\ m/s}$

F = -4567.40 N

So, the total frictional force acting on the car is 4567.40 N. Hence, this is the required solution.

7 0

3 years ago