if an average cloud has a density of 0.5 g/m3 and has a volume of 1,000,000,000 m3,what is the weight of an average cloud

1 answer:

3 0

If you have (1 x 10⁹) cubic meters of volume, and 1/2 gram of mass
in each cubic meter, then you have

(1/2 x 10⁹) grams = (5 x 10⁸) grams = (5 x 10⁵) kilograms of mass .

On Earth, that mass weighs 4,900,000 newtons .

(about 1,102,300 pounds)

(about 551 tons)

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find a magnitude of the force such that if the act at right angle there resultant is √10N but if the act of 50° the resultant is

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Explanation:

Let magnitude of the two forces be x and y.

Resultant at right angle R1= √15N) and at

60 degrees be R2= √18N.

Now, R1 = √(x² + y²) = √15,

R2= √(x² + y² +2xycos50) = √18.

So x² + y² = 15,

and x² + y² + 1.29xy = 18,

therefore 1.29xy = 3,

y = 3/1.29x.

y = 2.33/x

Now, x2 + (2.33/x)2 = 15,

x² + 5.45/x² = 15

multiply through by x²

x⁴ + 5.45 = 15x²

x⁴ - 15x2 + 5.45 = 0

Now find the roots of the equation, and later y. The two values of x will correspond to the

magnitudes of the two vectors.

Good luck

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4 years ago

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3 years ago

A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min, clockwise (when viewed from above). When swit

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Answer:

The magnitude of angular acceleration is $232.38\ rad/s^2$ .

Explanation:

Given that,

Initial angular velocity, $\omega_i=3500\ rev/min=366.5\ rad/s$

When it switched off, it comes o rest, $\omega_f=0$

Number of revolution, $\theta=46=289.02\ rad$

We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{-(366.51)^2}{2\times 289.02}\\\\\alpha =-232.38\ rad/s^2$