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Gelneren [198K]

3 years ago

12

What discovery lead J.J. Thomson to decide that atoms were not indivisible, but are actually composed of smaller parts? Thomson

used a scanning tunneling micrograph to capture images of atoms. Thomson used a beam of negatively charged particles. Thomson used only the experimental results of other scientists.

2 answers:

Margaret [11]3 years ago

8 0

Thomson used a beam of negatively charged particles.

Oksi-84 [34.3K]3 years ago

7 0

Thomson used a beam of negatively charged particles.

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Write an essay about the Electromagnetic spectrum and its many uses in real life practices. Answer these questions in the essay:

vampirchik [111]

Answer:

90% of people marry there 7th grade love. since u have read this, u will be told good news tonight. if u don't pass this on nine comments your worst week starts now this isn't fake. apparently if u copy and paste this on ten comments in the next ten minutes you will have the best day of your life tomorrow. you will either get kissed or asked out in the next 53 minutes someone will say i love you

Explanation:

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6 0

3 years ago

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Which of the following is not a natural resource? a. time b. water c. land d. air Please select the best answer from the choices

AleksAgata [21]

Answer: Option (a) is the correct answer.

Explanation:

Resources which are created by human beings are known as man-made resources.

For example, glass, rayon, nylon etc are all man-made resources.

Whereas resources which are naturally created are known as natural resources.

For example, wind, air, water etc are all natural resources.

Thus, we can conclude that out of the given options time is not a natural resource.

3 0

3 years ago

When the light on the moon is increasing, what is it called

nataly862011 [7]

Answer:

Waxing Gibbous phase occurs when the Moon is mostly lit and the illuminated portion is egg-shaped (gibbous) with the eastern edge shaded. The amount of illuminated area visible is increasing from one day to the next which is what is meant by "waxing"

hope this helps

have a good day :)

Explanation:

6 0

3 years ago

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Two asteroids collide and stick together. The first asteroid has a mass of 15\times 10^3\,\mathrm{kg}15×10 3 kg and is initially

statuscvo [17]

Answer:

Final speed is 900.06 m/s at $0.2215^{\circ}$

Solution:

As per the question:

Mass of the first asteroid, m = $15\times 10^{3}\kg$

Mass of the second asteroid, m' = $20\times 10^{3}\kg$

Initial velocity of the first asteroid, v = 770 m/s

Initial velocity of the second asteroid, v' = 1020 m/s

Angle between the two initial velocities, $\theta = 20^{\circ}$

Now,

Since, the velocities and hence momentum are vector quantities, then by the triangle law of vector addition of 2 vectors A and B, the resultant is given by:

$\vec{R} = \sqrt{A^{2} + 2ABcos\theta + B^{2}}$

Thus applying vector addition and momentum conservation, the final velocity is given by:

$(m + m')v_{final} = \sqrt{(mv)^{2} + 2(mv)(m'v')cos20^{\circ} + (m'v')^{2}}$ (1)

Now,

$(m +m')v_{final} = (35\times 10^{3})v_{final}$

$(mv)^{2} = (15\times 10^{3}\times 770)^{2} = 1.334\times 10^{14}$

$(m'v')^{2} = (20\times 10^{3}\times 1020)^{2} = 4.16\times 10^{14}$

$2(mv)(m'v')cos20^{\circ} = 2(15\times 10^{3}\times 770)(20\times 10^{3}\times 1020)cos20^{\circ} = 4.43\times 10^{14}$

Now, substituting the suitable values in eqn (1), we get:

$v_{final} = 900.06\ m/s$

Now, the direction for the two vectors is given by:

$\theta = sin^{- 1} \frac{m'v'sin20^{\circ}}{(m + m')v_{final}}$

$\theta = sin^{- 1} \frac{20\times 10^{3}\times 1020sin20^{\circ}}{(35\times 10^{3})\times 900.06} = 0.2215^{\circ}$

5 0

3 years ago

A small asteroid with a mass of 1500 kg moves near the earth. At a particular instant the asteroid’s velocity is ⟨3.5 × 104, −1.

zalisa [80]

Answer:

$P_{f}$ =(5.7 x $10^{7$ i - 2.24 x $10^{7$ j) kgm/s

Explanation:

Due to earths gravity, force on asteroid is given by:

$F= \frac{Gm_{1}m_{2} }{r^{2} }$ r^

Plugging in the values, we have

F= [(6.67x $10^{-11}$ )(1500)(5.97 x $10^{24}$ )(8x $10^{6}$ i + 9x $10^{6$ j)] / ((8x $10^{6}$ )² + (9x $10^{6$ )² $)^{1.5}$

F= 2736 i^ + 3078 j^

In order find the final momentum of the Asteroid, apply impulse momentum theorem

$P_{f}$ = $P_{i$ + FΔt

$P_{f}$ = 1500(3.5 x $10^{4$ i - 1.8x $10^{4$ j) + (2736i + 3078j)(1.5x $10^{3$ )

$P_{f}$ =(5.7 x $10^{7$ i- 2.24 x $10^{7$ j)kgm/s

4 0

4 years ago