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aksik [14]
3 years ago
15

Please post detailed answers to the following questions. Please use complete sentences. What are some of the challenges that you

might face if you wanted to observe and study stars from Earth? NEED TWO PARAGHRAPHS
Physics
1 answer:
Degger [83]3 years ago
8 0
As what Douglas Adams says, "<span>Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is. I mean, you may think it's a long way down the road to the chemist, but that's just peanuts to space."

The fact that the stars are very far from Earth, one can't fully understand the stars as it would require hundred of years to closely travel from earth to the stars because it is located million miles away from the Earth. Aside from its location, the stars are also hard to closely study because it emits lights that are harm for people when exposed longer. </span>
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How does the heat enhance <br>the rate of chemical reaction ? Explain in short.​
densk [106]
When the reactants are heated, the average kinetic energy of the molecules increases. This means that more molecules are moving faster and hitting each other with more energy. If more molecules hit each other with enough energy to react, then the rate of the reaction increases.
3 0
3 years ago
You are given five transparent objects: a calcite crystal, a diamond, a piece of window glass, a sample of quartz, and a piece o
blsea [12.9K]

Answer:

Explanation:

The calcite Crystal can be identified by carrying out an acid test on it. This is done by bringing it in contact with a weak acid which cause crack in it structure. Hence a little of carbondioxide gas is released.

Since diamond is the hardest object,it can be identified on a Mohs scale. Likewise it can be tested by bringing it in contact with a newspaper, if the letters on the paper are not seen, that shows it is a pure diamond.

Window glass is identified by the code on it.

While Quartz Crystal is identified by scratch test. When it is used to scratch all other softer stones and metal, it leaves mark on them.

5 0
3 years ago
What is the uses of echoes​
Anestetic [448]

Answer:

Echoes are the reflection of sound from relatively flat object that is far enough away that you can discern the time difference. Echoes are used to measure distance, velocity, and the shape of objects. Echoes off gratings result in an unusual pinging sound

6 0
3 years ago
What is 7 to the power of 5
LekaFEV [45]
7⁵ = 7 × 7× 7× 7× 7

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5 0
3 years ago
Read 2 more answers
Two point charges are placed on the x axis.The firstcharge, q1= 8.00 nC, is placed a distance 16.0 mfromthe origin along the pos
deff fn [24]

Answer:

E = (0, 0.299) N

Explanation:

Given,

  • Charge q_1\ =\ 8.0\ nC
  • Charge q_2\ =\ 6.0\ nC
  • Distance of the first charge from the origin = (16m, 0)
  • Distance of the second charge from the origin = (-9, 0)
  • Point where the electric field required = (0, 12m)

Let \theta_1\ and\ theta_2 be the angle of the electric fields by first and second charge at the point A.

\therefore sin\theta_1\ =\ \dfrac{12}{20}\\\Rightarrow \theta_1\ =\ sin^{-1}\left (\dfrac{12}{20}\ \right )\\\Rightarrow \theta_1\ =\ 36.87^o\\\\\therefore sin\theta_1\ =\ \dfrac{12}{9}\\\Rightarrow \theta_1\ =\ sin^{-1}\left (\dfrac{12}{9}\ \right )\\\Rightarrow \theta_1\ =\ 53.13^o\\

Electric field by charge q_1 at point A,

F_1\ =\ \dfrac{kq_1}{r_1^2}\\\Rightarrow F_1\ =\ \dfrac{9\times 10^9\times 8\times 10^{-9}}{20^2}\\\Rightarrow F_1\ =\ 0.18\ N/C

Electric field by the charge q_2 at point A,

F_1\ =\ \dfrac{kq_1}{r_1^2}\\\Rightarrow F_1\ =\ \dfrac{9\times 10^9\times 6.0\times 10^{-9}}{16^2}\\\Rightarrow F_1\ =\ 0.24\ N/C

Now,

Net electric field in horizontal direction at point AF_x\ =\ F_{1x}\ +\ F_{2x}\\\Rightarrow F_x\ =\ F_1cos\theta_1\ +\ F_2cos\theta_2\\\Rightarrow F_x\ =\ 0.18\times( -cos36.87^o)\ +\ 0.24\times cos53.13^o\\\Rightarrow F_x\ =\ -0.144\ +\ 0.144\ N/C\\\Rightarrow F_x\ =\ 0\ N/C

Net electric field in vertical direction at point A.

F_y\ =\ F_{1y}\ +\ F_{2y}\\\Rightarrow F_y\ =\ F_1sin\theta_1\ +\ F_2sin\theta_2\\\Rightarrow F_y\ =\ 0.18\times sin36.87^o\ +\ 0.24\times sin53.13^o\\\Rightarrow F_y\ =\ 0.180\ +\ 0.192\\\Rightarrow F_y\ =\ 0.299\ N/C

Hence, the net electric field  at point A,

F\ =\ ( 0, 0.299 )\ N/C.

5 0
3 years ago
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