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sergey [27]
3 years ago
8

8. Which element will form a 2-ion? sulfur Calcium potassium helium

Physics
2 answers:
Nookie1986 [14]3 years ago
8 0
The Group 16 elements
Hoochie [10]3 years ago
4 0

Answer:

I believe it's helium.

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How fast is the relay stick moving relative to Ingrid
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3 years ago
How does proximity to oceans affect Temperatures and<br> rainfall amounts?
vampirchik [111]

Answer:

Because they absorb and emit heat more steadily than land, oceans help moderate temperatures in coastal areas, making winters warmer and summers cooler. Winds that blow in from across the oceans bring more rainfall and higher humidity.

Explanation:

4 0
3 years ago
1.
Mumz [18]

Answer

Procedural bias occurs when the unfair amount of pressure is applied to complete their response quickly.

Measurement bias occurs when there is systematic or random error occur while taking the reading.

Design bias occur when the research fails to encounter all the aspects of the projects before presenting conclusion.

1) The correct answer is option B.

  Balance is miscalibrated hence there will be systematic error which is a type of measurement bias.

2)  The correct answer is option A.

    Limiting response time for participant is an example of Procedural bias.

3) The correct answer is option A

    Here, Jackie has not surveyed the diverse group so it is the case of design bias

8 0
4 years ago
I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block
borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

T = 2\pi \sqrt{\frac{m}{k} }

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\

Now, determine the amplitude of oscillation, A;

E = \frac{1}{2} kA^2

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A =  \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm

Therefore, the amplitude of the oscillation is 2.82 cm

8 0
3 years ago
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