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Svetach [21]
4 years ago
7

Process of changing a liquid into a gas

Physics
2 answers:
Komok [63]4 years ago
7 0
The process of changing a liquid to a gas is called evaporation. 
lions [1.4K]4 years ago
3 0
The process is called evaporation. Liquid turns into gas. 
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Calculate the force between two objects that have masses of 70 kilograms and 2,000 kilograms separated by a distance of 1 meter.
Makovka662 [10]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

The Gravitational Force between given objects will be ~

  • 9.34 \times  {10}^{ - 6}  \:  \: N

\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}

We know that ~

\huge\boxed{\mathrm{F = \dfrac{ Gm_1m_2}{ r²}}}

where ~

  • F = gravitational force

  • m_1 = mass of 1st object = 70 kg

  • m_2 = mass of 2nd object = 2000 kg

  • G = gravitational constant = 6.674 × {10}^ {-11}

  • r = distance between the objects = 1 m

Let's calculate the force ~

  • F =  \dfrac{6.674 \times 10 {}^{ - 11} \times 70 \times 2000 }{1 {}^{2} }

  • F =6.674 \times 7 \times 2 \times 10 { }^{ - 11} \times 10 {}^{4}

  • 93.436 \times 10 {}^{ - 7}

  • 9.3436  \times 10 {}^{ - 6} \:  \:  newtons
6 0
3 years ago
Please help fast!! I need this in less than 17 hours!
djyliett [7]

SOLUTION is given in attachment below.

7 0
3 years ago
What is photosynthesis, food chain, evaporation and passive smokers?​
navik [9.2K]

Answer:

Photosynthesis: the process where plants convert light energy into sugar, for food

Food chain: a series of organisms showing which organisms eat each other

Explanation:

7 0
3 years ago
The rock is tied to a string and swung in a circular path by a
victus00 [196]

Answer:

6.3

Explanation:

6 0
3 years ago
How much work would it take to push two protons very slowly from a separation of 2.00×10−10m (a typical atomic distance) to 3.00
laiz [17]

Answer:

Work= -7.68×10⁻¹⁴J

Explanation:

Given data

q₁=q₂=1.6×10⁻¹⁹C

r₁=2.00×10⁻¹⁰m

r₂=3.00×10⁻¹⁵m

To find

Work

Solution

The work done on the charge is equal to difference in potential energy

W=ΔU

Work=U_{1}-U_{2}\\ Work=-kq_{1}q_{2}[\frac{1}{r_{2}}-\frac{1}{r_{1}} ]\\Work=(-9*10^{9})*(1.6*10^{-19} )^{2}[\frac{1}{3.0*10^{-15} }-\frac{1}{2*10^{-10} } ]\\  Work=-7.68*10^{-14}J

4 0
4 years ago
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