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kondor19780726 [428]
2 years ago
10

Write answers with significant figures:a) 17.35 g +8.498 g​

Physics
1 answer:
Genrish500 [490]2 years ago
6 0

Answer:

25.9 g

Explanation:

= 17.35

8.498

________+

= 25.848 g = 25.85 g = 25.9 g

*so sorry if wrong

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Read the paragraph below and answer the question that follows:
elena-s [515]

B. Helium because it is constantly being made in the sun

Explanation:

From the given paragraph, we can conclude that helium is the most common element in a star such as the sun because it is constantly being made.

In the nuclear fusion process that results in the production of helium, hydrogen nuclei are the reactants and helium is the product.

  • since the reactants are constantly being used in the core of the sun, this suggests that the products must be more.
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learn more:

Transmutation brainly.com/question/3433940

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8 0
2 years ago
A stone is dropped from from rest at the top of a mine shaft. It takes 95 seconds for the stone to fall to the bottom of the min
Marianna [84]

Distance of fall from rest,
without air resistance              =  (1/2) (gravity) (time)²

                                             = (1/2) (9.8 m/s²) (95 sec)²

                                             =  (4.9 m/s²) (9,025 sec²)

                                             =        44,222.5 meters  .

The depth of the mine shaft is five times the height of Mt. Everest !


6 0
3 years ago
Which explicit definition matches this sequence? an={10,if n=1an−1−4,if n>1 A. an = 10 + 4(n – 1) B. an = 10 – 4(n – 1) C. an
Step2247 [10]

Answer:

Añ=10+4(n-1)

Explanation:

3 0
3 years ago
Sarah wants to install a wind turbine on her farm, which would convert wind energy to electricity to power her home. She can onl
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Answer:

The best choice would be c

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6 0
2 years ago
Read 2 more answers
A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

Q' = + 42 nC = 42\times 10^{- 9} C

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E} = 708.03 N/C

Now,

Electric field at the mid-point due to charge Q' is given by:

\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E'} = 418.84 N/C

Now,

The net Electric field is given by:

\vec{E_{net}} = \vec{E} - \vec{E'}

\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C

5 0
3 years ago
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