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4 years ago

What are the basic building blocks of matter

Physics

2 answers:

AlladinOne [14]4 years ago

4 0

Atoms. Also, don't trust them. They make up everything.

Rashid [163]4 years ago

3 0

The basic building blocks are called atoms

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A 120.0 kg crate is placed on a 15.00°

Citrus2011 [14]

F = 2820.1 N

Explanation:

Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as

Fnet = ma = 0 (a = 0 no sliding)

= F - mgsin15°

= 0

F = mgsin15°

= (120 kg)(9.8 m/s^2)sin15°

= 2820.1 N

7 0

3 years ago

Highlight each sentence that is true about atoms, matter, and electric charge.

lisov135 [29]

Answers B and D are true.

7 0

3 years ago

Tidal Forces near a Black Hole. An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black

arlik [135]

Answer:

$833.4801043*10^6N$ on ear that is closer to Black hole.

$13.83803929*10^6N$ On ear that is farther from Black hole.

Explanation:

This problem can be solved as two masses that are at two different location from a bigger mass whose gravity affects both.

tension is an equal and opposite force that is exerted in response to applied force.

so on ear that is closer to black hole would have tension that is equal in magnitude and opposite in direction to gravitational force that ear experience due to the black hole at that location.

this true for ear that is further away from black hole as well.

(1) Force on ear that is closer to black hole.

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

$m_{1}= 5*1.989*10^30kg$ is mass of Black hole.

$m_{2}= 0.030kg$ is mass of ear that is close to black hole.

$G = 6.679*10^-11m^3*kg^-1*s^-2$

r = $120km-\frac{6}{100000} km=119.99994km=119999.94m$

Note, we have subtracted because ear is closer to black hole.

plugging all this in formula gives.

$F = 833.4801043*10^6N$

That is tension of ear.

(2) Force on ear that is further from black hole.

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

$m_{1}= 5*1.989*10^30kg$ is mass of Black hole.

$m_{2}= 0.030kg$ is mass of ear that is close to black hole.

$G = 6.679*10^-11m^3*kg^-1*s^-2$

this time r is further away from black hole so it would be.

$r = 120km+\frac{6}{100000} km = 120000.06m$

Plugging this all in we get

$F = 13.83803929N$

and that is tension on ear that is further from black hole.

Notice the tension difference, and order of magnitude of tension,it is enormous .

this astronaut is lethally close to black hole.

5 0

3 years ago

A coach tells his Little League players that hitting a 0.275 batting average, within 7% error, means that they had a really grea

Mashutka [201]

Yes. According to the coach's mathematical criteria, Tommy had a great season.

Sadly, Tommy doesn't even know it. His tone-deaf coach decided to describe success in terms that are absurd for 7-yr-olds, as well as for most of their parents.

7 0

3 years ago

calculate the period of a wave whose frequency is 5 Hz and whose wavelength is 1 cm give your answer in decimal form

Pavlova-9 [17]

Period T = 1/f, where f = frequency = 5 Hz

T = 1/5 = 0.20 s

Period = 0.20 seconds.

8 0

3 years ago

Read 2 more answers