Answer:
can you send me a picture of what layer a and look b look like and then I will be able to evaluate
Yo sup??
we can solve this problem by applying Newton's 2nd law
F*t=Δp
p=momentum
pi=mu=1500*30
pf=mv=m*0=0
Therefore
F*3=1500*30
F=15000 N
Hope this helps.
Answer:
Aluminum atoms are in 3.78 g of aluminum
02214076x10^23)8.4367659370640769254888x10^22 atoms, therefore there are 8.4367659370640769254888x10^22 atoms of Aluminum(Al) present in 3.78 grams of Aluminum(Al).
Explanation:
Answer:
The number of atoms of aluminum contained in one square inch of the foil=6
Explanation:
Interpretation:
The number of atoms of aluminum contained in one square inch of the foil is to be calculated.
Concept introduction: relation to calculate a number of atoms contained in one square inch of the foil is given by:
NA/c=mNA/AA1
Where,
M=mass of aluminum
NA= Avogadro's number
AA= Atomic mass of aluminum
Given :
Mass of aluminum foil=0.3 g
M= 0.39g
Relation to calculate a mass of atoms contained in one square inch of the foil is given by:
NA/c= mNA/AA1
M= Mass of aluminum
NA= Avogadro's number
AA= Atomic mass of aluminum
Since,
Avogadro's number
=6
