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seropon [69]
3 years ago
7

A scientist is trying to determine the identity of an element. It is highly reactive to water and forms an ionic bond with chlor

ine. It is also very shiny. The element is likely A. from Group 17. B. from Group 1. C. from Group 15. D. a noble gas.
Physics
1 answer:
wel3 years ago
8 0

The answer is noble gas. Since noble gas are constant and unreactive. They can still shape compounds with other elements. 
Group 15 is also group 5A and Group 17 is also group 7A. Elements in these sets do not typically form ionic bonds; they are more on creating covalent bonds since they're non-metals. 
Therefore, that leaves us with B. from Group 1. They are metals (but Hydrogen) which respond violently with water, and they form ionic bonds, for they drop outer electrons easily.

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Is 2/2 1 or 0? please help lol
Komok [63]

Answer:

1.

Explanation:

Hello!

In this case, for such mathematical operations, we can wee that the slash represents a fraction or a division, say 8 ÷ 4 = 2, 6 ÷ 3 = 2, 20 ÷ 4 = 5, etc. In such a way, since the operation 2/2, represents 2 ÷ 2, it is clear that two is once in 2, therefore, the result is:

2 ÷ 2 = 1.

Best regards!

4 0
3 years ago
A 100-kg astronaut is floating in outer space. If the astronaut throws a 2-kilogram wrench at a speed of 10 meters per second, w
sergey [27]

Since Astronaut and wrench system is isolated in the space and there is no external force on it

So here momentum of the system will remain conserved

so here we can say

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

initially both are at rest

so here plug in all values

0 = 100 v_{1f} + 2\times 10

v_{1f} = -0.20 m/s

so here the astronaut will move in opposite direction and its speed will be equal to 0.20 m/s

3 0
3 years ago
Light of wavelength 580 nm is incident on a slit of width 0.30 mm. An observing screen is placed 2.0 m past the slit. Find the d
uysha [10]

Answer:

Y = 3.87 x 10⁻³ m = 3.87 mm

Explanation:

This problem can be solved by using Young's double-slit experiment formula:

Y = \frac{\lambda L}{d}

where,

Y = fringe spacing = ?

L = slit to screen distance = 2 m

λ = wavelength of light = 580 nm = 5.8 x 10⁻⁷ m

d = slit width = 0.3 mm = 3 x 10⁻⁴ m

Therefore,

Y = \frac{(5.8\ x\ 10^{-7}\ m)(2\ m)}{3\ x\ 10^{-4}\ m}

<u>Y = 3.87 x 10⁻³ m = 3.87 mm</u>

3 0
3 years ago
Match the letters to the numbers for 30 points!
liq [111]

-- 'Ca' (Calcium) is an element.
-- The proton has a positive charge.
-- Nuclear fusion results in the synthesis of atoms of new elements.
-- H₂O (water) is a chemical compound.
-- Nuclear fission is a decay of the nucleus.
-- The atomic number of an element is the number of protons
      in each atom of it.
-- I suppose you're using the Greek letter  <span>η  ('eta', not 'nu')
      to represent the neutron.
-- I suppose you're using ' e ' to represent the electron.
</span>
7 0
3 years ago
Read 2 more answers
A phonograph record accelerates from rest to 28.0 rpm in 5.73 s.
Arturiano [62]

Answer:

a) \alpha=0.5117\ rad.s^{-2}

b) \theta=8.4\ rad

Explanation:

Given:

  • initial rotational speed of phonograph, \omega_i=0\ rad.s^{-1}
  • final rotational speed of phonograph, N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}
  • time taken for the acceleration, t=5.73\ s

a)

Now angular acceleration:

\alpha=\frac{\omega_f-\omega_i}{t}

\alpha=\frac{2.932-0}{5.73}

\alpha=0.5117\ rad.s^{-2}

b)

Using eq. of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2

\theta=8.4\ rad

5 0
3 years ago
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