The best answer is that it reduces the level of ground water
To answer the two questions, we need to know two important equations involving centripetal movement:
v = ωr (ω represents angular velocity <u>in radians</u>)
a = 
Let's apply the first equation to question a:
v = ωr
v = ((1800*2π) / 60) * 0.26
Wait. 2π? 0.26? 60? Let's break down why these numbers are written differently. In order to use the equation v = ωr, it is important that the units of ω is in radians. Since one revolution is equivalent to 2π radians, we can easily do the conversion from revolutions to radians by multiplying it by 2π. As for 0.26, note that the question asks for the units to be m/s. Since we need meters, we simply convert 26 cm, our radius, into meters. The revolutions is also given in revs/min, and we need to convert it into revs/sec so that we can get our final units correct. As a result, we divide the rate by 60 to convert minutes into seconds.
Back to the equation:
v = ((1800*2π)/60) * 0.26
v = (1800*2(3.14)/60) * 0.26
v = (11304/60) * 0.26
v = 188.4 * 0.26
v = 48.984
v = 49 (m/s)
Now that we know the linear velocity, we can find the centripetal acceleration:
a = 
a = 
a = 9234.6 (m/
)
Wow! That's fast!
<u>We now have our answers for a and b:</u>
a. 49 (m/s)
b. 9.2 *
(m/
)
If you have any questions on how I got to these answers, just ask!
- breezyツ
Answer:
The length of the solar day will get shorter.
Explanation:
- The blue planet Earth not only rotates around it's own axis but also rotates around the Sun and everyday it moves a little bit around the axis.
- Since the speed of the Earth's rotation on it's own axis and around the Sun is constant we don't feel the effects of the rotation.We can only feel the motion if the earth changes it's rotation speed.
- If by any means or chance the Earth stopped spinning (stopped rotation) then the atmosphere surrounding the Earth would be in motion and all the Earth's land would be scoured clean.
The weight of the meterstick is:

and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance

from the pivot.
The torque generated by the weight of the meterstick around the pivot is:

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:

from which we find the value of d2:

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
Gravity obeys the inverse square law. At 6400 km above the center of the Earth (Earth's surface) you weigh x. Twice that reduces your weight to 1/4th. Four times that height reduces your weight to 1/16th. 4 times 6400 km is 25,600 km. But that is above the center of the earth, and the question requests the height above the surface, so we deduct 6400 km to arrive at our final answer: 19,200 km.
Incidentally, it doesn't exactly work the opposite way. At the center of the Earth the mass would be equally distributed around you, and you would therefore be weightless.