Answer: the answer is storing properly in refrigeration.
Explanation:
Answer:
Explanation:
Given a ball of mass m= 0.5kg
The ball moves in as circle of radius r= 0.4m
Speed of the ball is v = 4m/s
Centripetal force is exerted on ball and it is given as
Fc = m•ac
ac is centripetal acceleration and it is given as
ac = v²/r
Then,
Fc = mv²/r
Fc = 0.5 × 4²/0.4
Fc = 20N.
The force exerted on the ball is 20N
Answer:
100
Explanation:
by dividing 2000N and 1000kg.
Answer:
Option b. Effective nuclear charge increases as we move to the right across a row in the periodic table
Explanation:
The <em>effective nuclear charge </em>is a measure of how strong the protons in the nucleus of an atom attract the outermost electrons of such atom.
The <em>effective nuclear charge</em> is the net positive charge experienced by valence electrons and is calculated (as an approximation) by the equation: Zeff = Z – S, where Z is the atomic number and S is the number of shielding electrons.
The shielding electrons are those electrons in between the interesting electrons and the nucleus of the atom.
Since the shielding electrons are closer to the nucleus, they repel the outermost electrons and so cancel some of the attraction exerted by the positive charge of the nucleus, meaning that the outermost electrons feel less the efect of attraction of the protons. That is why in the equation of Zeff, the shielding electrons (S) subtract the total from the atomic number Z.
The <em>effective nuclear charge</em>, then, is responsible for some properties and trends in the periodic table. Here, you can see how this explains the trend of the atomic radius (size of the atom) accross a row in the periodic table.
- As the<em> effective nuclear charge</em> is larger, in a same row of the periodic table, the shielding effect is lower, the outermost electrons are more strongly attracted by the nucleus, and the size of the atoms decrease. That is why as we move to the right in the periodic table, the size of the atoms decrease.
Answer:
Solution given:
No of waves[N] =20crests & 20 troughs
=20waves
Time[T]=4seconds
distance[d]=3cm=0.03m
Now
<u>Wave</u><u> </u><u>length</u><u>=</u>3cm=3 × 
<u>Frequency</u>=
=
=5Hertz
and
Wave speed:wave length×frequency=3 × ×5=1.5 × 
.
