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oksano4ka [1.4K]

3 years ago

6

Your friend claims that weather predictions are just guesses. What would you say to change your friend's mind?

1 answer:

VashaNatasha [74]3 years ago

4 0

To change my friends mind, after they've already said weather predictions are just guesses, I would say they are correct, but not only to they use guesses to predict what will happen, but they also, before hand, use atmospheric tools. They then predict from the data collected. They do both.

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Using the law of conservation of angular momentum, estimate how fast a collapsed stellar core would spin if its initial spin rat

Nataly_w [17]

Answer:

$\omega_{f} = 1000000\,\frac{rev}{day}$

Explanation:

The law of conservation of angular momentum states that angular momentum remains constant when there is no external moment or forces applied to the system. Let assume that star can be modelled as an sphere, then:

$\frac{2}{5}\cdot M\cdot R_{o}^{2} \cdot \omega_{o} = \frac{2}{5}\cdot M\cdot R_{f}^{2} \cdot \omega_{f}$

The final angular speed is:

$\omega_{f} = \omega_{o}\cdot (\frac{R_{o}}{R_{f}})^{2}$

$\omega_{f} = (1\,\frac{rev}{day} )\cdot (\frac{10000\,km}{10\,km} )^{2}$

$\omega_{f} = 1000000\,\frac{rev}{day}$

3 0

3 years ago

A ball is attached to a string of length 3 m to make a pendulum. The pendulum is placed at a location that is away from the Eart

Musya8 [376]

1) $0.61 m/s^2$

2) 13.9 s

Explanation:

1)

The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.

It can be calculated using the equation:

$g=\frac{GM}{r^2}$ (1)

where

G is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

r is the distance of the object from the Earth's center

The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is

$r=4R$

where

$R=6.37\cdot 10^6 m$ is the Earth's radius

Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):

$g=\frac{GM}{(4R)^2}=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})0.}{(4\cdot 6.37\cdot 10^6)^2}=0.61 m/s^2$

2)

The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration due to gravity at the location of the pendulum

Note that the period of a pendulum does not depend on its mass.

For the pendulum in this problem, we have:

L = 3 m is its length

$g=0.61 m/s^2$ is the acceleration due to gravity (calculated in part 1)

Therefore, the period of the pendulum is:

$T=2\pi \sqrt{\frac{3}{0.61}}=13.9 s$

4 0

3 years ago

Find the Kinetic Energy of a<br>trailer of mass 30,000 kg moving at a<br>Speed of 60 km/h

olasank [31]

Explanation: Formulas you need to use:

K=0.5mass(Velocity^2)

km/h / 3.6 ===>M/S

Step 1:

convert km/h to m/s

Put the values in the formula :

3 0

3 years ago

What is the physical state of water at 250°C?

mars1129 [50]

It is probably gas as pure water starts boiling at 100d degrees C

7 0

2 years ago

Need help! 10 points

Aleksandr-060686 [28]

Umm do u need both??

3 0

2 years ago