Answer:
4.384 * 10^13
Explanation:
Given the expression :
[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]
Applying the laws of indices
[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]
13.2733 * 10^19 ÷ 3.0276 * 10^6
(13.2733 / 3.0276) * 10^(19 - 6)
4.3840996 * 10^13
= 4.384 * 10^13
You can find muscle fibers, nerves, connective tissue, and blood vessels in every skeletal system. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.
This question involves the concepts of density, volume, and mass.
The approximate diameter of a magnesium atom is "3.55 x 10⁻¹⁰ m".
<h3>STEP 1 (FINDING MASS OF INDIVIDUAL ATOM)</h3>
It is given that:
Mass of one mole = 24 grams
Mass of 6 x 10²³ atoms = 24 grams
Mass of 1 atom = 
= 4 x 10⁻²³ grams
<h3>STEP 2 (FINDING VOLUME OF A SINGLE ATOM)</h3>
where,
= density = 1.7 grams/cm³- m = mass of single atom = 4 x 10⁻²³ grams
- V = volume of single atom = ?
Therefore,
V = 2.35 x 10⁻²³ cm³
<h3>STEP 3 (FINDING DIAMETER OF ATOM)</h3>
The atom is in a spherical shape. Hence, its Volume can be given as follows:
![V =\frac{\pi d^3}{6}\\\\d=\sqrt[3]{ \frac{6V}{\pi}}\\\\d=\sqrt[3]{ \frac{6(2.35\ x\ 10^{-23}\ cm^3)}{\pi}}](https://tex.z-dn.net/?f=V%20%3D%5Cfrac%7B%5Cpi%20d%5E3%7D%7B6%7D%5C%5C%5C%5Cd%3D%5Csqrt%5B3%5D%7B%20%5Cfrac%7B6V%7D%7B%5Cpi%7D%7D%5C%5C%5C%5Cd%3D%5Csqrt%5B3%5D%7B%20%5Cfrac%7B6%282.35%5C%20x%5C%2010%5E%7B-23%7D%5C%20cm%5E3%29%7D%7B%5Cpi%7D%7D)
d = 0.355 x 10⁻⁷ cm = 3.55 x 10⁻¹⁰ m
Learn more about density here:
brainly.com/question/952755
In a projectile, the horizontal acceleration is zero. The velocity remains constant at all times. However, the <u>vertical acceleration</u> is -9.81m/s^2.
Hope this helps!
Answer:
1.67 m/s
Explanation:
Momentum is conserved.
Initial momentum = final momentum
(30 kg) (10 m/s) + (35 kg) (-10 m/s) = (30 kg) v + (35 kg) (0 m/s)
300 - 350 = 30v
v = -5/3 m/s
Linus will move at 1.67 m/s in the direction opposite that he started.
