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3 years ago

What type of particle found inside the atom is uncharged

Physics

2 answers:

soldi70 [24.7K]3 years ago

3 0

The neutron is uncharged. That's why it was
given a name built on "neu" from "neutral".

tangare [24]3 years ago

3 0

It is an neutron. That is what the answer is.

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While driving on a rural road, your right wheels run off the pavement. You should hold the steering wheel firmly and

Vsevolod [243]

Answer:

The answer is C. Steer in a straight line while gently slowing down

Explanation:

The following are advised when your cars go off the pavement while driving;

firstly, Do not panic.

ensure you hold on to your steering wheel tightly.

keep Steering straight ahead.

ensure you Stay on the shoulder.

Ease up on the accelerator and brake gently.

When you know you can safely do so, turn back on the road at a much lower speed.

3 0

3 years ago

Because the pressure falls, water boils at a lower temperature with increasing altitude. Consequently, cake mixes and boiled egg

AlexFokin [52]

Answer:

1) The boiling point of water reduces by 3.28°C at 1,000 m above sea-level

2) The boiling point of water reduces by 6.56°C at 2,000 m above sea-level

Explanation:

The variation of the boiling point of water with elevation is given as follows

The boiling point reduces by 0.5°C for every 152.4 meter increase in elevation

At sea-level, the boiling point temperature of water = 100°C

1) At 1,000 m elevation, the boiling point temperature, T = 100 - (1,000/152.4) × 0.5 ≈ 96.72 °C

Therefore, the boiling point of water reduces by 100° - 96.72° = 3.28°C at 1,000 m above sea-level

2) At 2,000 m elevation, the boiling point temperature, T = 100 - (2,000/152.4) × 0.5 ≈ 93.44°C

The boiling point of water reduces by 100° - 93.44° = 6.56°C at 2,000 m above sea-level

7 0

3 years ago

Chord progressions that move to resting points that release tension are called

Sav [38]

Cadences.

These cadences are the resulting tensions that chords release from their resting points. This movement is classified from a unstable chord progression to a stable one. Thank you for your question. Please don't hesitate to ask in Brainly your queries.

6 0

3 years ago

a concave mirror has a focal length of 18 cm. where will an image form if an object is placed 58 cm from the mirror

MAXImum [283]

Answer:

here

Explanation:

8 0

3 years ago

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr

mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = $5 \ * (M \ _{sun})$

where : $M_{sun} = 1.99*10^{33} \ kg$

Then; we have:

$m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg$

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear $m_{ear} = 0.030 \ kg$

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

$\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R - d) \omega^2$

$\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)$

The net force on the ear farther to the black hole is :

$\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R + d) \omega^2$

$\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)$

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

$T = \frac{3GMm_{ear}d}{R^3}$

$T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}$

$T = 2073.9 N\\T = 2.07 KN$

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0

4 years ago