Answer:
D
Explanation:
The longer handles distribute the force across a longer distance.
Answer:
Static Friction.
Explanation:
Friction is the force that resists the relative motion between the surfaces sliding against each other.
Static friction is friction between objects that are not in relative motion with each other.
The coefficient of static friction, typically denoted as μs,
Static friction arises due to surface roughness( relative term)
The static friction force can be overcome by an applied maximum force
F max = μs x N
N= normal force
Any force smaller than F max attempting to slide one surface over the other is opposed by a frictional force of equal magnitude and opposite direction.
Any force larger than F max overcomes the force of static friction and causes sliding to occur.
This maximum force is sometimes called the limiting value also. Here that value is 75 N.
Answer:
N - 1s²2s²2p³
Explanation:
Nitrogen is located in the p-block of the periodic table (groups 13-18) and is on the 2nd period.
The 2nd period tells us the principal energy level (a quantum number) is n = 2. Therefore, it must have already filled up the 1s sublevel.
The groups 13-18 on period 2 tells us that the 2s sublevel is also filled.
Nitrogen is located in Group 15. That means that there are 3 electrons that have filled the 2p sublevel, out of a possible 6.
Therefore, our electron configuration is 1s²2s²2p³
2p³ (Shorthand Config)
[He] 2s²2p³ (Noble Gas Config)
Answer:
In a chemical reaction, reactants that are not used up when the reaction is finished are called excess reagents. The reagent that is completely used up or reacted is called the limiting reagent, because its quantity limits the amount of products formed.
Explanation:
Answer:
The atomic mass of element is 65.5 amu.
Explanation:
Given data:
Abundance of X-63 = 50.000%
Atomic mass of X-63 = 63.00 amu
Atomic mass of X-68 = 68.00 amu
Atomic mass of element = ?
Solution:
Abundance of X-68 = 100-50 = 50%
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (50×63)+(50×68) /100
Average atomic mass = 3150 + 3400 / 100
Average atomic mass = 6550 / 100
Average atomic mass = 65.5 amu.
The atomic mass of element is 65.5 amu.
