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Anna71 [15]
3 years ago
12

4. How do ordinary objects behave in microgravity?

Physics
1 answer:
Nikolay [14]3 years ago
8 0

Answer:

Bubbles paused

Explanation:

the air bubble doesn't rise because it is no lighter than the water around it—there's no buoyancy. The droplet doesn't fall from the leaf because there's no force to pull it off. It's stuck there by molecular adhesion.

for instance, onto the International Space Station, gravity becomes negligible, and the laws of physics act differently than here on Earth

On Earth, the buoyancy of the air bubbles causes them to rise to the top together, creating a segregation between air and water. However, in microgravity, nothing forces the air bubbles to interact and thus rise together, Green said.

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A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.
babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8  + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

    T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

 T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

4 0
3 years ago
How can you tell one element from another?
3241004551 [841]
Their atomic number?? (easy guess)
6 0
3 years ago
The activity of a radioisotope is found to decrease 40% of its original value in 2.59 x 10 s.
Rainbow [258]

Answer: 0.0353\ s^{-1}

Explanation:

Given

Radioactive material is found to decrease 40% of its original value in 2.59\times 10\ s

Sample at any time is given by

N=N_oe^{-\lambda t}

where, \lambda=\text{decay constant}

Put values

\Rightarrow 0.4N_o=N_oe^{-\lambda\cdot 2.59\times 10}\\\Rightarrow 0.4=e^{-\lambda\cdot 2.59\times 10

Taking natural logarithm both side

\Rightarrow \lambda=\dfrac{\ln 2.5}{25.9}\\\\\Rightarrow \lambda =0.0353\ s^{-1}

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Blizzard [7]

Answer:

it would help if we knew the question and other answers

Explanation:

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ElenaW [278]

Answer:

1.knife

2.flashlight

3.batteries for flashlight

4.flares

5.matches

###### sorry UwU

7.first aid

8.12 foot plastic tarp

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