Answer:
<em><u>M</u></em><em><u>a</u></em><em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u>m</u></em><em><u>a</u></em><em><u>t</u></em><em><u>i</u></em><em><u>c</u></em><em><u>a</u></em><em><u>l</u></em><em><u>l</u></em><em><u>y</u></em><em><u>:</u></em>
That will be
<em>=</em><em> </em><em>1</em><em>5</em><em>0</em><em>0</em><em> </em><em>x</em><em> </em><em>1</em><em>5</em><em> </em><em>x</em><em> </em><em>4</em><em>5</em><em>0</em><em>0</em>
<em>=</em><em> </em><em><u>1</u></em><em><u>0</u></em><em><u>1</u></em><em><u>,</u></em><em><u>2</u></em><em><u>5</u></em><em><u>0</u></em><em><u>,</u></em><em><u>0</u></em><em><u>0</u></em><em><u>0</u></em>
I would say option D, it depends on the size of the star
The centripetal acceleration a is 4.32
10^-4 m/s^2.
<u>Explanation:</u>
The speed is constant and computing the speed from the distance and time for one full lap.
Given, distance = 400 mm = 0.4 m, Time = 100 s.
Computing the v = 0.4 m / 100 s
v = 4
10^-3 m/s.
radius of the circular end r = 37 mm = 0.037 m.
centripetal acceleration a = v^2 / r
= (4
10^-3)^2 / 0.037
a = 4.32
10^-4 m/s^2.
ITS C
This element tends to lose 2 electrons to become a 2+ ion, is the correct statement regarding the element calcium. Calcium has 2 electrons in its outer shell and it is easier to lose them than it is to gain enough to become stable. When stable it has 2 more protons than electrons forming a 2+ ion.
Answer:
The distance travel by block before coming to rest is 0.122 m
Explanation:
Given:
Mass of block
kg
Initial speed of block

Final speed of block

Coefficient of kinetic friction 
Ramp inclined at angle
28.4°
Using conservation of energy,
Work done by frictional force is equal to change in energy,

Where 



m
Therefore, the distance travel by block before coming to rest is 0.122 m