Goodmorning I can’t seem to remember this can someone help me

sasho [114]

The universe is made up of baryonic matter.(neutrons,electrons,protons)

5 0

3 years ago

An astronaut has a mass of 75 kg and is floating in space 500 m from his 125,000 kg spacecraft. What will be the force of gravit

nikdorinn [45]

Answer:

1. 2.5×10¯⁹ N

2. 3.33×10¯¹¹ m/s²

Explanation:

1. Determination of the force of attraction.

Mass of astronaut (M₁) = 75 Kg

Mass of spacecraft (M₂) = 125000 Kg

Distance apart (r) = 500 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

The force of attraction between the astronaut and his spacecraft can be obtained as follow:

F = GM₁M₂ /r²

F = 6.67×10¯¹¹ × 75 × 125000 / 500²

F = 2.5×10¯⁹ N

Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N

2. Determination of the acceleration of the astronaut.

Mass of astronaut (m) = 75 Kg

Force (F) = 2.5×10¯⁹ N

Acceleration (a) of astronaut =?

The acceleration of the astronaut can be obtained as follow:

F = ma

2.5×10¯⁹ N = 75 × a

Divide both side by 75

a = 2.5×10¯⁹ / 75

a = 3.33×10¯¹¹ m/s²

Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²

5 0

2 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

3 years ago

In the lab downstairs physics majors use a rotating mirror to measure the speed of light within a few percent of the actual valu

iris [78.8K]

The number of complete cycles the rotating mirror goes through before the angular velocity gets to zero is approximately 1166.8 revs

<h3>What is angular velocity?</h3>

Angular velocity is the ratio of the angle turned to the time taken.

The kinematic equation for angular velocity are presented as follows;

ω = ω₀ + α·t

θ = θ₀ + ω₀·t + 0.5·α·t²

Where;

θ₀ = The initial angle turned = 0

ω₀ = The initial angular velocity of the mirrors = 115 rad/s clockwise

α = The angular acceleration = (115 - (-115))rad/s/(85 s) = -46/17 m/s²

t = The duration of the motion;

When the angular velocity, ω is zero, we get;

0 = 115 - 46/17·t

t = 85/2

Which indicates;

θ = 0 + 115× (85/2) + 0.5×(46/17) ×(85/2)² = 7331.25

θ = 7331.25 radians

θ = 7331.25/(2×π) ≈ 1166.8 rev

The mirrors would have turned through approximately 1166.8 revolutions when the angular gets to zero

Learn more about angular velocity and acceleration here:

brainly.com/question/13014974

#SPJ1

7 0

9 months ago

Les propriétés de l’air?

crimeas [40]

Answer:

Lorsque l'on détend l'air son volume augmente et sa pression diminue. L'air qui est un mélange de gaz est compressible et expansible. – Lorsque l'on comprime l'air, son volume diminue et sa pression augmente. – Lorsque l'on détend l'air, son volume augmente et sa pression diminue.

4 0

2 years ago