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spin [16.1K]
2 years ago
11

What is true weightlessness ?​

Physics
1 answer:
alukav5142 [94]2 years ago
3 0

Answer:

weightlessness, condition experienced while in free-fall, in which the effect of gravity is canceled by the inertial (e.g., centrifugal) force resulting from orbital flight. ... Excluding spaceflight, true weightlessness can be experienced only briefly, as in an airplane following a ballistic (i.e., parabolic) path.

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How do scientists classify species?
serious [3.7K]

Answer:

D. By comparing traits

Explanation:

Because age isn't genetic, as well as names, as well as who discovered, but traits are genetic.  

6 0
3 years ago
Read 2 more answers
The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the st
Aleonysh [2.5K]

Answer:

a = 5.83 \times 10^{-4} m/s^2

Explanation:

Since the system is in international space station

so here we can say that net force on the system is zero here

so Force by the astronaut on the space station = Force due to space station on boy

so here we know that

mass of boy = 70 kg

acceleration of boy = 1.50 m/s^2

now we know that

F = ma

F = 70(1.50) = 105 N

now for the space station will be same as above force

F = ma

105 = 1.8 \times 10^5 (a)

a = \frac{105}{1.8 \times 10^5}

a = 5.83 \times 10^{-4} m/s^2

3 0
3 years ago
Faiyez wrote the problem below
Ilya [14]

445/100 - 5/4 = 445/100 - 125/100 = 320/100 = 16/5 = 3 1/5.

6 0
3 years ago
3. The car's mass is 400 kg. It moves at a velocity of 20 m/s. Calculate the car's momentum. *
Orlov [11]

Answer:

momentum=mass×velocity

momentum =400kg×20m/s=8000kg.m/s

7 0
3 years ago
On a Saturday afternoon, you decide to pay a neighborhood kid to mow your lawn. The kid usesa manual push lawn mower with a mass
Mars2501 [29]

Answer with Explanation:

We are given that

A.Mass,m=12 kg

\theta=53^{\circ}

\mu_k=0.16

Speed,v=1.5m/s

Net force in x direction must be zero

F_{net}=0

Fsin\theta-f=0

Fsin\theta=f

Net force in y direction

N-mg-Fcos\theta=0

N=mg+Fcos\theta

f=\mu_kN=\mu_k(mg+Fcos\theta)

Fsin\theta=\mu_k(mg+Fcos\theta)

Fsin\theta=\mu_kmg+\mu_kFcos\theta

Fsin\theta-\mu_kFcos\theta=\mu_kmg

F(sin\theta-\mu_kcos\theta)=\mu_kmg

F=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}

Power,P=Fv

P=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}v

Where g=9.8m/s^2

B.Substitute the values

P=\frac{0.16\times 12\times 9.8}{sin53-0.16cos53}\times 1.5

P=40.17W

6 0
3 years ago
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