Answer:
A. the tendency of moving objects to stay in motion
Explanation:
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Answer:
The electrons in oxygen are paired while in nitrogen, they are not.
Explanation:
To analyse this we start with writing out the ground state electronic configurations for both elements.
Oxygen: 1s²2s²2p4 meaning the p subshell has the following arrangement of electrons ↑↓ ↑ ↑
Nitrogen : 1s²2s²2p³ meaning the p subshell has the following arrangement of electrons ↑ ↑ ↑
Clearly the paired electron in oxygen will be experiencing repulsion from the electron it shares an orbital with causing it to be removed easily. The electrons in nitrogen are unpaired, each orbital is singly occupied
Answer:
F = 39.2 N
Explanation:
Since, the object is in uniform motion. Therefore, the frictional force on object will be:
Frictional Force = μk N = μk mg
where,
μk = coefficient of kinetic friction = 0.2
m = mass of crate = 10 kg
g = 9.8 m/s²
Therefore,
Frictional Force = (0.2)(10 kg)(9.8 m/s²)
Frictional Force = 19.6 N
The horizontal component of force must be equal to this frictional force to continue the uniform motion:
F Sin 30° = 19.6 N
F = 19.6 N/Sin 30°
<u>F = 39.2 N</u>
Answer:
speed of the bullet before it hit the block is 200 m/s
Explanation:
given data
mass of block m1 = 1.2 kg
mass of bullet m2 = 50 gram = 0.05 kg
combine speed V= 8.0 m/s
to find out
speed of the bullet before it hit the block
solution
we will apply here conservation of momentum that is
m1 × v1 + m2 × v2 = M × V .............1
here m1 is mass of block and m2 is mass of bullet and v1 is initial speed of block i.e 0 and v2 is initial speed of bullet and M is combine mass of block and bullet and V is combine speed of block and bullet
put all value in equation 1
m1 × v1 + m2 × v2 = M × V
1.2 × 0 + 0.05 × v2 = ( 1.2 + 0.05 ) × 8
solve it we get
v2 = 200 m/s
so speed of the bullet before it hit the block is 200 m/s
Answer:
D. Increases from pole to equator
Explanation:
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