Ask question

Ask question

All categories

3 years ago

12

4. How do ordinary objects behave in microgravity?

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.

You might be interested in

An astronaut weighs 863 N on Earth. What is the astronaut’s mass?

Alecsey [184]

86.3 pounds.... I think

6 0

3 years ago

Read 2 more answers

A toy car moves 8 min 4 s at the constant velocity. What is the car's velocity?

Schach [20]

Explanation:

<u>Formula:</u>

$velocity = (d \div t)$

<u>d = distance given</u>

<u>t</u><u> </u><u>=</u><u> </u><u>the amount of time </u><u>given</u>

<u>Substitute the given values into the formula for velocity</u><u>:</u>

$v = 8 \div 4$

velocity is shortened for v.

8 (distance) divided by 4 (time) equals the velocity.

<u>Solve:</u>

$2 = 8 \div 4$

The velocity of the toy car equals: B. 2 m/s.

6 0

2 years ago

Any tips on how to get a good grasp of physics and chemistry?

Helga [31]

Answer: Take notes and research into the subject more get extra help like tutoring if needed.

Explanation: ...

4 0

2 years ago

Which isotope is use to date ancient artifacts such as fossils?

meriva

the isotope they use is carbon-14

5 0

3 years ago

ANSWER One end of a string is attached to an object of mass M, and the other end of the string is secured so that the object is

qaws [65]

Answer:

Explanation:

It is a case of oscillation by simple pendulum . Expression for simple pendulum is given as follows

T = $2\pi\sqrt{\frac{l}{g} }$

where T is time period , l is length of pendulum and g is acceleration due to gravity .

$\frac{1}{f} =2\pi\sqrt{\frac{l}{g} }$ , f is frequency of oscillation

For the given case

$\frac{1}{f_o} =2\pi\sqrt{\frac{l}{g} }$

subsequently length becomes half so

$\frac{1}{f} =2\pi\sqrt{\frac{l}{2g} }$

dividing

$\frac{f}{f_o} = \sqrt{\frac{2}{1} }$

f = $\sqrt{2} f_o$

frequency of oscillation becomes √2 times.

4 0

3 years ago