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Afina-wow [57]
3 years ago
12

Which of the following is most likely to be the best conductor of electricity?

Physics
1 answer:
anzhelika [568]3 years ago
4 0

Answer:

I think its oxygen, a conductor is something that allows heat very easily.

Explanation:

Please mark brainliest and have a great day!

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Which characteristic surface of the moon contains the same rock as the ocean floor
Ratling [72]
The answer is Maria or Lunar mare because they are both made of basalt. Hope this helps.
3 0
3 years ago
Two conducting spheres are each given a charge Q. The radius of the larger sphere is three times greater than that of the smalle
padilas [110]

Answer:

Explanation:

If E₀ is the electric field outside the smaller sphere and r is the radius of larger sphere.

E₀ = kQ/r²

The radius of the larger sphere is 3r and the charge on both sphere is same then the electric field outside the larger sphere is given as

E = kQ/(3r)² = kQ/9r² = 1/9 (kQ/r²)= 1/9 x E₀

hence the correct option is e.

5 0
3 years ago
A 5.50-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical
iris [78.8K]

Answer:

17.71N/m

Explanation:

The period of the spring is expressed according to the expression;

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

m is the mass of the object

k is the force constant

Given

m = 5.50kg

T = 3.50s

Substitute into the formula;

T = 2 \pi \sqrt{\frac{m}{k} } \\3.5 = 2 (3.14) \sqrt{\frac{5.5}{k} } \\3.5 = 6.28 \sqrt{\frac{5.5}{k} } \\\frac{3.5}{6.28} =  \sqrt{\frac{5.5}{k} } \\0.557 = \sqrt{\frac{5.5}{k} } \\square \ both \ sides\\0.557^2 = (\sqrt{\frac{5.5}{k} })^2 \\0.3106 = \frac{5,5}{k}\\k = \frac{5.5}{0.3106}\\k =  17.71N/m

Hence the force constant of the spring is 17.71N/m

4 0
3 years ago
Question 2 of 10
tangare [24]

Explanation:

everything can be found in the picture

8 0
2 years ago
ASTM A229 oil-tempered carbon steel is used for a helical coil spring. The spring is wound with D = 50 mm d = 10.0 mm, and a pit
horrorfan [7]

Answer

given,

D = 50 mm = 0.05 m

d = 10 mm = 0.01 m

Force to compress the spring

F = \dfrac{d^4G\delta}{8D^3N}

\dfrac{\delta}{N} = p - d = 14 - 10 = 4 mm

F = \dfrac{d^4G}{8D^3}\times 0.004

F = \dfrac{0.1^4\times 79\times 10^9}{8\times 0.05^3}\times 0.004

     F = 3160 N

stress correction factor from stress correction curve is equal to 1.1

now, calculation of corrected stress

\tau = \dfrac{8FDk_s}{\pi d^3}

\tau = \dfrac{8\times 3160 \times 0.05 \times 1.1}{\pi \times 0.01^3}

              = 442.6 Mpa

The tensile strength of the steel material of  ASTM A229 is equal to 1300 Mpa

now,

\tau_s \leq 0.45 S_u

\tau_s \leq 0.45 \times 1300

\tau_s \leq 585\ Mpa

since corrected stress is less than the \tau_s

hence, spring will return to its original shape.

6 0
3 years ago
Read 2 more answers
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