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antoniya [11.8K]
3 years ago
12

The tendency for an object at rest to remain at rest

Physics
1 answer:
Zarrin [17]3 years ago
7 0
Inertia is the force in play here
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An astronaut decides to perform an experiment to monitor how much weight he loses during his stay on the International Space Sta
kumpel [21]

Answer:

The first one is: His weight on the Earth before take-off and the weight after take-off back on Earth once he gets back should be recorded as his Independent variable and his dependent variable.

The second one is: If he gained the weight back that he had lost while on the trip then you should disregard them unless that was the weight he was when he weighed himself after he got back.

The Third one is: The mass of an object is the amount of matter it contains, regardless of its volume or any forces acting on it. … Gravity is a force that attracts objects toward the Earth. The weight of the object is defined as the force caused by gravity on a mass.

Explanation:

I took  the quiz earlier. Hope this Helps you.

4 0
3 years ago
Outside controlled airspace, the minimum flight visibility requirement for vfr flight above 1,200 feet agl and below 10,000 feet
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The least possible flight visibility prerequisite for VFR flight above 1200 feet AFL and below 10000 feet MSL for the duration of daylight hours is 1 mile in the external controlled airspace. In aeronautics, visual meteorological conditions or as called as VMC is an aviation flight group in which visual flight rules or VFR is allowed that is in circumstance in which pilot have adequate visibility to fly the aircraft sustaining visual leave-taking from terrain and other aircraft. They are the contradictory of instrument meteorological conditions. The border standards among instrument meteorological conditions and visual meteorological circumstances are recognized as the visual meteorological conditions minima and are well-defined by visibility, cloud ceilings for take offs and landing and cloud allowances.  
3 0
3 years ago
#26 question
4vir4ik [10]

So, the time that taken for the astronaut to fall to the surface of the moon is <u>2.5 s.</u>

<h3>Introduction</h3>

Hi ! In this question, I will help you. In this question, you will learn about the fall time of the free fall motion. Free fall is a downward vertical motion without being preceded by an initial velocity. When moving in free fall, the time required can be calculated by the following equation:

\sf{h = \frac{1}{2} \cdot g \cdot t^2}

\sf{\frac{2 \cdot h}{g} = t^2}

\boxed{\sf{\bold{t = \sqrt{\frac{2 \cdot h}{g}}}}}

With the following condition :

  • t = interval of the time (s)
  • h = height or any other displacement at vertical line (m)
  • g = acceleration of the gravity (m/s²)

<h3>Problem Solving</h3>

We know that :

  • h = height = 5.00 m
  • g = acceleration of the gravity = 1.6 m/s²

What was asked :

  • t = interval of the time = ... s

Step by step :

\sf{t = \sqrt{\frac{2 \cdot h}{g}}}

\sf{t = \sqrt{\frac{2 \cdot 5}{1.6}}}

\sf{t = \sqrt{6.25}}

\boxed{\sf{t = 2.5 \: s}}

<h3>Conclusion</h3>

So, the time that taken for the astronaut to fall to the surface of the moon is 2.5 s.

<h3>See More</h3>
  • Time that needed for hearing the splash of fallen rock in the well brainly.com/question/26485521
  • The speed of the object at a certain height (free fall motion) brainly.com/question/26377041
  • The relationship between acceleration and the change in velocity and time in free fall brainly.com/question/26486625
3 0
3 years ago
Divers found two substances on the bottom of the ocean. At room temperature, both substances are liquid. Scientists then transfe
shtirl [24]

They have different melting points

3 0
3 years ago
Read 2 more answers
A cyclist intends to cycle up a 7.70o hill whose vertical height is 126m. Assuming the mass of bicycle plus person is 75.0kg, ca
Ksivusya [100]

Answer:

92704.5 J

596.44737 N

Explanation:

m = Mass of person + bicycle = 75 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Vertical height = 126 m

\theta = Angle = 7.7°

d = Diameter = 0.388 m

Work done against gravity is given by

P=mgh\\\Rightarrow P=75\times 9.81\times 126\\\Rightarrow P=92704.5\ J

Work done is 92704.5 J

Force required is given by

F=\dfrac{mgrsin\theta}{\pi d}\\\Rightarrow F=\dfrac{75\times 9.81\times sin7.7\times 5.12}{\pi\times 0.388}\\\Rightarrow F=596.44737\ N

The force is 596.44737 N

7 0
3 years ago
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