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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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An object weighs 32 newtons. What is its mass if a gravitometer indicates that g = 8.25 m/s2?
Nata [24]

Answer:

3.88kg

Explanation:

From the question given, we obtained the following data:

W = 32N

g = 8.25 m/s2

M =?

W = M x g

M = W/g

M = 32/8.25

M = 3.88Kg

The mass of the object is 3.88kg

3 0
4 years ago
If I keep F constant in F=ma, what is the relationship between m and a?
Jobisdone [24]

Answer:

If F is a constant, we can take f = 1

f = m×a

ma = 1

therefore we can say that force is hence proportinal to the product of mass and acceleration.

6 0
3 years ago
What is the mass of a liquid having a density of 1.50 g/ml and a volume of 3.5 liters?
PIT_PIT [208]

We know the formula for density = Mass/ volume

So     Mass, M = Volume * Density

         Volume = 3.5 L= 0.0035m^3

         Density = 1.50 g/ml  = 1500 kg/m^3

         Mass, M =  0.0035*1500 = 5.25 kg

So mass of liquid having density 1.50 g/ml and volume 3.5 liters is 5.25 kg.

6 0
4 years ago
Find two words in a vocabulary list that descirbe the bending of a wave
Paraphin [41]
Reflection and refraction 


7 0
4 years ago
Read 2 more answers
A large, 60 turn circular coil of radius 10.0 cm carries a current of 4.2 A. At the center of the large coil is a small 20 turn
Eddi Din [679]

To solve this problem we apply the concepts related to the electric torque generated by the electromagnetic field. Mathematically this Torque can be written under the following relation

\tau = NIAB sin\theta

Here,

N = Number of Turns

I = Current

A = Area

B = Magnetic Field

The maximum torque will be reached when the angle is 90 degrees, then we will have the following relation,

\theta = 90\°C

Magnetic Field is given at function of the number of loops, permeability constant at free space at the perimeter, then

B = \frac{N\mu_0 I}{2\pi r}

B = \frac{(60)(4\pi * 10^{-7})(4.2)}{2\pi (0.1)}

B = 5.04*10^{-4}T

Replacing at the first equation we have,

T = (20)(\pi (0.005)^2)(1)(5.04*10^{-4})

T = 7.91*10^{-7}N\cdot m

4 0
3 years ago
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