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3 years ago

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Suppose the earth suddenly came to halt and ceased revolving around the sun. The gravitational force would then pull it directly

to the sun. What would be the earth's speed as it crashed?

1 answer:

AleksAgata [21]3 years ago

3 0

Answer:

613373.65233 m/s

Explanation:

M = Mass of Sun = $1.989\times 10^{30}\ kg$

m = Mass of Earth

v = Velocity of Earth

r = Distance between Earth and Sun = $147.12\times 10^{9}\ m$

$r_e$ = Radius of Earth = $6.371\times 10^6\ m$

$r_s$ = Radius of Sun = $695.51\times 10^6\ m$

In this system it is assumed that the potential and kinetic energies are conserved

$\dfrac{1}{2}Mv_2-\dfrac{GMm}{r_e+r_s}=0-\dfrac{GMm}{r}\\\Rightarrow v=\sqrt{2GM(\dfrac{1}{r_e+r_s}-\dfrac{1}{r})}\\\Rightarrow v=\sqrt{2\times 6.67\times 10^{-11}\times 1.989\times 10^{30}(\dfrac{1}{6.371\times 10^6+695.51\times 10^6}-\dfrac{1}{147.12\times 10^{9}})}\\\Rightarrow v=613373.65233\ m/s$

The velocity of Earth would be 613373.65233 m/s

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Explanation :

The points regarding the gravitational waves are as follows :

(1) Gravitational waves are predicted to travel through space at the speed of light. So, the speed of gravitational waves is also equals $c=3\times 10^8\ m/s$ .

(2) The first direct detection of gravitational waves came in 2015. It is detected using a advanced LIGO detectors.

(3) The existence of gravitational waves is predicted by Einstein's general theory of relativity.

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According to astronomers, how many galaxies are distributed across the observable universe?

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B: 1 trillion distributed across the universe. If I remember correctly.

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Which disadvantage of nuclear power use poses the greatest threat to the environment?

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The correct answer is The storage and management of radioactive wastes

Explanation:

In general, nuclear reactions (changes in the nucleus of an atom such as fission) release a lot of energy including a lot of heat. Moreover, this heat is used by humans to obtain electricity and other types of energy, which is known as a nuclear power. This type of power is considered positive because it does not emit carbon and it is quite efficient.

However, in most cases, it is a threat to the environment and living beings because storing and managing the wastes of this type of power is difficult and expensive. Indeed, dealing with the wastes of nuclear power requires complex infrastructure, and any accident or leaking leads to serious consequences from the death of those exposed to the wastes to permanent loss of diversity or changes in nearby areas.

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An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude

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1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

$\omega_f = \omega_i + \alpha t$

where

$\omega_i = 0.300 rev/s$ is the initial angular velocity

$\alpha = 0.895 rev/s^2$ is the angular acceleration

Substituting t = 0.200 s, we find

$\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s$

Let's now convert it into rad/s:

$\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s$

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

$r=\frac{0.800}{2}=0.400 m$

And so now we can find the tangential speed at t = 0.200 s:

$v=\omega_f r =(3.01)(0.400)=1.2 m/s$

2) $2.25 m/s^2$

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

$a_t = \alpha r$

where $\alpha$ is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

$\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2$

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

$a_t = (5.62)(0.400)=2.25 m/s^2$

3) $3.6 m/s^2$

The centripetal acceleration is given by

$a=\frac{v^2}{r}$

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

$a=\frac{1.2^2}{0.400}=3.6 m/s^2$

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3 years ago

How do you convert seconds to minutes, seconds to hours , minutes to seconds and hours to seconds

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Answer:

You could memorize conversions, or use conversion charts, or do a quick internet search for help.

Explanation:

I'll do a few of the conversions for you

<u>Seconds to minutes: </u>

there are 60 seconds in one minute. So if you are wondering how many seconds are in 3 and 1/2 minutes, you would do this conversion:

60 sec/min <em>times </em>x sec/3.5 min

which can be written as 60 x 3.5 = 210. so "x" seconds would be 210 seconds in 3 1/2 minutes.

<u>you could do the same thing in the opposite direction for minutes to seconds:</u>

1 min has 60 seconds. So in 7.25 minutes, how many seconds are there?

1 min/60 sec <em>times </em>7.25 min/x sec

which can be written as 7.25 x 60 = 435. so "x" seconds would be 435 seconds in 7.25 minutes.

<u>hours to seconds:</u>

this one is slightly more complicated

In one hour there are 60 minutes, and in one minute there are 60 seconds.

so to convert from hours to seconds you would do this conversion:

1 hr/60 min times 1 min/60 sec. then the "min" would cancel out, and you would be left with the label "hr/sec". to do the math, it would be 1 hr / 60 x 60.

60x60 = 3600. so you would have 1 hr/3600 sec. So in one hour there are 3600 seconds.

so if you want to know how many seconds are in 6.75 hours:

6.75 hr/x sec <em>times </em>3600 sec/1 hr

6.75 x 3600 = 24,300 so there are 24,300 seconds in 6.75 hours.

I hope this helps :)

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