It looks like it's wanting you to find speed and velocity
these formula's should help
speed=distance/time
velocity=overall distance/time
Answer:
![1.99\cdot 10^{-7} rad/s](https://tex.z-dn.net/?f=1.99%5Ccdot%2010%5E%7B-7%7D%20rad%2Fs)
Explanation:
The average angular speed of the Earth about the Sun is given by:
![\omega = \frac{2 \pi}{T}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7B2%20%5Cpi%7D%7BT%7D)
where
rad is the total angle corresponding to one revolution of the Earth around the Sun
T is the orbital period of the Earth
The orbital period of the Earth is 365.25 d. We must convert it into seconds first:
![T=365.25 d \cdot (24 h/d) \cdot (60 min/h) \cdot (60 s/min)=3.16\cdot 10^7 s](https://tex.z-dn.net/?f=T%3D365.25%20d%20%5Ccdot%20%2824%20h%2Fd%29%20%5Ccdot%20%2860%20min%2Fh%29%20%5Ccdot%20%2860%20s%2Fmin%29%3D3.16%5Ccdot%2010%5E7%20s)
And by substituting into the equation above, we find the average angular speed:
![\omega=\frac{2 \pi rad}{3.16\cdot 10^7 m/s}=1.99\cdot 10^{-7} rad/s](https://tex.z-dn.net/?f=%5Comega%3D%5Cfrac%7B2%20%5Cpi%20rad%7D%7B3.16%5Ccdot%2010%5E7%20m%2Fs%7D%3D1.99%5Ccdot%2010%5E%7B-7%7D%20rad%2Fs)
Moon and Earth have different gravitational constants- g. g on the Earth is <span>equal to 9.8 m/s^2 while g on the moon is equal to 1.6m/s^2.
</span><span>Let assume that two balls are thrown: one on Earth and one on Moon and that their kinetic energy is the same.
The kinetic energy is E=(m*v^2)/2.
So, this means that t</span><span>he ball would fall much more slowly on Moon than on Earth. Every aspect of the ball's fall would be slower.
</span>
![L = 2.4 \; \text{kg} \cdot \text{m}^{2} \cdot \text{s}^{-1}](https://tex.z-dn.net/?f=L%20%3D%202.4%20%5C%3B%20%5Ctext%7Bkg%7D%20%5Ccdot%20%5Ctext%7Bm%7D%5E%7B2%7D%20%5Ccdot%20%5Ctext%7Bs%7D%5E%7B-1%7D)
<h3>Explanation</h3>
The angular momentum of a rolling body is the product of the body's moment of inertia and its angular velocity.
![L = I \cdot \omega](https://tex.z-dn.net/?f=L%20%3D%20I%20%5Ccdot%20%5Comega)
where
is the angular momentum of a rolling body;
is the body's moment of inertia; And
is the body's angular moment.
What's the moment of inertia of this bowling ball?
Assuming that the ball is a solid sphere. For a solid sphere,
.
where
is the moment of inertia of the sphere;
is the mass of the sphere; and
is the radius of the sphere.
for this sphere.
.
![I = \dfrac{2}{5} \; m \cdot r^{2}\\\phantom{I} = \dfrac{2}{5} \times 5.5 \times 0.120^{2}\\\phantom{I} = 0.0317 \; \text{kg}\cdot m^{2}](https://tex.z-dn.net/?f=I%20%3D%20%5Cdfrac%7B2%7D%7B5%7D%20%5C%3B%20m%20%5Ccdot%20r%5E%7B2%7D%5C%5C%5Cphantom%7BI%7D%20%3D%20%5Cdfrac%7B2%7D%7B5%7D%20%5Ctimes%205.5%20%5Ctimes%200.120%5E%7B2%7D%5C%5C%5Cphantom%7BI%7D%20%3D%200.0317%20%5C%3B%20%5Ctext%7Bkg%7D%5Ccdot%20m%5E%7B2%7D)
What's the angular momentum of this bowling ball?
.
.