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antoniya [11.8K]

2 years ago

5

Differences between weightlessness in space and weightlessness in earth

2 answers:

zhuklara [117]2 years ago

6 0

Answer:

it depends on a person's own weight

ExtremeBDS [4]2 years ago

3 0

Answer:

The gravity is roughly the same. The inverse square law applies to gravity and that means that being 60 miles above Earth is about the same as being on Earth. The thing is, they are going really fast in an orbit and thus falling around the Earth. The weightlessness comes from a sort of cancellation of gravity pulling them down and their going perpendicular to that force. The forces are cancelling and making them relatively “weightless”. It’s not really weightlessness but just an equilibrium reached between the two accelerations of gravity and their motion against the pull of the Earth.

Explanation:

You might be interested in

What is the moment of inertia of a cube with mass M=0.500kg and side lengths s=0.030m about an axis which is both normal (perpen

prisoha [69]

Answer:

The moment of inertia I is

I = 2.205x10^-4 kg/m^2

Explanation:

Given mass m = 0.5 kg

And side lenght = 0.03 m

Moment of inertia I = mass x radius of rotation squared

I = mr^2

In this case, the radius of rotation is about an axis which is both normal (perpendicular) to and through the center of a face of the cube.

Calculating from the dimensions of the the box as shown in the image below, the radius of rotation r = 0.021 m

Therefore,

I = 0.5 x 0.021^2 = 2.205x10^-4 kg/m^2

8 0

3 years ago

An object moves on a trajectory given by Bold r left parenthesis t right parenthesis equals left angle 10 cosine 6 t comma 10 si

spayn [35]

Answer:

10

Explanation:

(r) = <10 cos 6t, 10 Sin 6t>

The distance traveled by the object is the magnitude of vector r.

The magnitude of vector r is given by

$r = \sqrt{(10 Cos 6t)^{2}+(10 Sin 6t)^{2}}$

$r =10 \sqrt{(Cos^{2} 6t)+(Sin^{2} 6t)$

r = 10

5 0

2 years ago

3. The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second i

Darya [45]

Answer:

(a) Angular velocity will be 125.6 rad/sec

(b) Linear velocity will be 144.44 m /sec

(c) Centripetal acceleration = 1849.3031 g

Explanation:

We have given diameter d = 2.30 m

So radius r = $\frac{d}{2}=\frac{2.30}{2}=1.15m$

(a) Speed is given as 1200 rev/min

We know that angular velocity is given by $\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 1200}{60}=125.6rad/sec$

(b) Linear speed is given by $v=\omega r=125.6\times 1.15=144.44m/sec$

(c) Centripetal acceleration is given by $a_c=\frac{v^2}{r}=\frac{144.44^2}{1.15}=18141.664m/sec^2$

We know that $g=9.81m/sec^2$

So $18141.66m/sec^2=\frac{18141.664}{9.81}=1849.3031g$

6 0

2 years ago

A hop is a springing from one foot but landing on the other foot? Question 4 options: True False

Rama09 [41]

Answer:

i thinc it's False

Explanation:

sorry im 6

6 0

2 years ago

Read 2 more answers

A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×10^{4} 4 m/s^{2} 2 , and 1.85 ms (1

Paha777 [63]

Answer:

u = - 38.85 m/s^-1

Explanation:

given data:

acceleration = 2.10*10^4 m/s^2

time = 1.85*10^{-3} s

final velocity = 0 m/s

from equation of motion we have following relation

v = u +at

0 = u + 2.10*10^4 *1.85*10^{-3}

0 = u + (21 *1.85)

0 = u + 38.85

u = - 38.85 m/s^-1

negative sign indicate that the ball bounce in opposite directon

6 0

2 years ago