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yan [13]
3 years ago
8

Wha happens to the weight of astronauts when they are in orbit

Physics
2 answers:
Lynna [10]3 years ago
5 0

That depends on the size of the orbit.  

In "low Earth orbit", like the Mercury, Gemini, early Apollo flights, and aboard the ISS, the astronauts' weight is about 85% of their weight when they're standing on the Earth's surface.  

BUT ... since everything in orbit is in "free fall", they don't <em>feel </em>the weight of their own body or anything else.  The feeling is like being completely without any weight.    

Bingel [31]3 years ago
4 0

Microgravity. It is sometimes called zero- gravity, but this is wrong. What happens is they go into free fall. On earth, gravity causes all things to fall, it can essentially be called a 'vacuum' if you will. So, I'm not to sure about it, but that's kind of what happens.

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How much work do you do on a 15 N book in lifting it straight up for a distance of<br> 0.40 meters?
astra-53 [7]

Answer:

Work done, W = 6 J

Explanation:

It is given that,

Force of gravity acting on the book, weight of the book is 15 N

We need to find the work done in lifting the book straight up for a distance of  0.4 meters.

The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :

W=Fd\cos180\\\\W=15\times 0.4\times \cos180\\\\W=-6\ J

So, the magnitude of work done in lifting the book is 6 joules.

7 0
3 years ago
Which of the following biomes might be found at a latitude of 33° South?
cupoosta [38]
The answer is going be desert. 
5 0
3 years ago
Describe one practical use of a concave mirror.
Tatiana [17]

Answer:

Here are some uses...

Explanation:

Uses of convex mirror:

It is used in supermarkets and stores for surveillance.

It it is used as rear view mirror in automobiles.this is due to the reason that a convex mirror provides a wider field of view than a plane or concave mirror.

Uses of concave mirror:

It is used in torches.

They are used in headlights of vehicles to send parallel rays to infinity , in shaving mirror to get an enlarged image of the face and also by dentists to see a bigger image of the tooth / teeth .

Mark as brainliest if helped thanks!

7 0
3 years ago
Read 2 more answers
A golf ball is struck at ground level. The speed of the golf ball as a function of the time is shown in Figure 4-36, where t = 0
lana66690 [7]
In your question where as a golf ball is struck at a ground level and the speed of the ball as a function of time is in the figure where time t=0 and va = 16m/s and vb=32m/s. The following is the answer: 
a) How far does the golf ball travel horizontally before returning to ground level? 
-<span>80m</span>
<span>(b) What is the maximum height above ground level attained by the ball?
</span>-39.87m
3 0
3 years ago
Read 2 more answers
a solenoid that is 98.6 cm long has a cross-sectional area of 24.3 cm2. There are 1310 turns of a wire carrying a current of
Natalija [7]

Complete question:

A solenoid that is 98.6 cm long has a cross-sectional area of 24.3 cm2. There are 1310 turns of a wire carrying a current of 6.75 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy stored in the magnetic field there (neglect end effects).

Answer:

(a) the energy density of the magnetic field inside the solenoid is 50.53 J/m³

(b) the total energy stored in the magnetic field is 0.121 J

Explanation:

Given;

length of the solenoid, L = 98.6 cm = 0.986 m

cross-sectional area of the solenoid, A = 24.3 cm² = 24.3 x 10⁻⁴ m²

number of turns of the solenoid, N = 1310 turns

The magnitude of the magnetic field inside the solenoid is given by;

B = μ₀nI

B = μ₀(N/L)I

Where;

μ₀ is permeability of free space, = 4π x 10⁻⁷ m/A

B = \frac{4\pi*10^{-7}*1310*6.75}{0.986} \\\\B = 0.01127 \ T

(a) Calculate the energy density of the magnetic field inside the solenoid

u = \frac{B^2}{2 \mu_o}\\\\u = \frac{(0.01127)^2}{2*4\pi *10^{-7}} \\\\u = 50.53 \ J/m^3

(b) Find the total energy stored in the magnetic field

U = uV

U = u (AL)

U = 50.53 (24.3 x 10⁻⁴  x 0.986)

U = 0.121 J

8 0
3 years ago
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