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never [62]
3 years ago
9

when astronauts work in space, they are sometimes attached to the spacecraft by a line called an umbilical. Why do you think the

line has been given this name?
Physics
2 answers:
wariber [46]3 years ago
3 0
Probably for the umbilical cord that connects babies (from their early stages in the womb to their removal) to their mothers. The cord is cut, forming the belly button. This is analogous to astronauts in space.
Zigmanuir [339]3 years ago
3 0
Because it is like the umbilical cord on a baby to its mother this is why they call it umbilical
You might be interested in
I WILL MARK YOU THE BRAINLIEST
mr Goodwill [35]

Explanation:

700N right

to get the net force

you gotta let one direction be the negative ( the smaller force)

so the total force towards the left is 100N ( 60 + 40= 100)

which is smaller than the right force which is 800 N so you let 100 N be negative

so without even calculating , you can know that it will be moving towards the right because right force > left force

your add both forces ( remember 100 N is negative)

so 800N + ( - 100N)

= 700N

towards the right

hope this helps

this is just one method that helped me understand

please mark it brainliest

3 0
3 years ago
Read 2 more answers
The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1
telo118 [61]
<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life h of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.

In this case, we are given the half life of two elements:

beryllium-13: h_{B-13}=5(10)^{-10}s=0.0000000005s

beryllium-15: h_{B-15}=2(10)^{-7}s=0.0000002s

As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?

We can find it out by the following expression:

h_{B-15}=X.h_{B-13}

Where X is the amount we want to find:

X=\frac{h_{B-15}}{h_{B-13}}

X=\frac{2(10)^{-7}s}{5(10)^{-10}s}

Finally:

X=400

Therefore:

The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.

8 0
2 years ago
Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinalandtransverse moduli
Masja [62]

Answer:

Not possible

Explanation:

E_{cl} = longitudinal modulus of elasticity = 35 Gpa

E_{ct} = transverse modulus of elasticity = 5.17 Gpa

E_m = Epoxy modulus of elasticity = 3.4 Gpa

V_{\rho l} = Volume fraction of fibre (longitudinal)

V_{\rho t} = Volume fraction of fibre (transvers)

E_f = Modulus of elasticity of aramid fibers = 131 Gpa

Longitudinal modulus of elasticity is given by

E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764

Transverse modulus of elasticity is given by

E_{ct}=\frac{E_mE_f}{(1-V_{\rho t})E_f+V_{\rho t}E_m}\\\Rightarrow 5.17=\frac{3.4\times 131}{(1-V_{\rho t})131+V_{\rho t}3.4}\\\Rightarrow \frac{3.4\times 131}{5.17}-131=-127.6V_{\rho t}\\\Rightarrow V_{\rho t}=\frac{\frac{3.4\times 131}{5.17}-131}{-127.6}\\\Rightarrow V_{\rho t}=0.35148

V_{\rho l}\neq V_{\rho t}

Hence, it is not possible to produce a continuous and oriented aramid fiber.

5 0
3 years ago
How does the electric force vary with distance
wlad13 [49]
The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value.
3 0
1 year ago
A spacecraft and a staellite are at diametrically opposite position in the same circular orbit of altitude 500 km above the eart
Tanya [424]

Answer:

Hello the diagram related to your question is attached below

answer: a) 851 m/s

             b)  8506.1 secs

Explanation:

calculate the periodic time of the satellite using the equation below

t = \frac{2\pi }{R} \sqrt{\frac{(R+h)^{3} }{g} }  --  ( 1 )

where ; R = 6370 km

h = 500 km

g = 9.81 m/s^2

input given values into equation 1

t = 5670.75 secs

next calculate the periodic time taken by the space craft  

<u>a) determine the increase in speed </u>

V = v - \sqrt{\frac{gR^2}{R + h} }  

where ; v = 8463 m/s , R = 6370 km, h = 500 km

V = 851 m/s

b) Determine the periodic time for the elliptic orbit

τ = \frac{3t}{2}

 = \frac{3*5670.76}{2}  =  8506.1 secs

attached below is the remaining part of the detailed solution

5 0
3 years ago
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