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3 years ago

How can I convert this?Please answer with solution. Thank you.

Physics

1 answer:

fomenos3 years ago

8 0

Answer:

1 hr 45 min

Explanation:

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What is heart sound?

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its the sound that a heart produces when beating, this can help doctors detect abnormalities

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3 years ago

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How do I find the cosine of theta.

NNADVOKAT [17]

The value of cos θ in the given figure is 0.98.

<h3>What is cosine of an angle?</h3>

The cosine of an angle is defined as the sine of the complementary angle.

The complementary angle equals the given angle subtracted from a right angle, 90.

cos θ = sin(90 - θ)

For example, if the angle is 30°, then its complement is 60°

cos 30 = sin(90 - 30)

cos 30 = sin 60

0.866 = 0.866

<h3>Cosine of an angle with respect to sides of a right triangle</h3>

cos θ = adjacent side / hypotenuse side

adjacent side of the given right triangle is calculated as follows;

adj² = 10² - 2²

adj² = 100 - 4

adj² = 96

adj = √96

adj = 9.8

cos θ = 9.8/10

cos θ = 0.98

Thus, the value of cos θ in the given figure is 0.98.

Learn more about cosine of angles here: brainly.com/question/23720007

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5 0

1 year 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

A uniform thin wire is bent into a quarter-circle of radius a = 20.0 cm, and placed in the first quadrant. Determine the coordin

Mashcka [7]

Answer:

$r_{cm}=[12.73,12.73]cm$