A. number of protons increases, and the number of shells of electrons
remains the same.
Explanation:
The number of protons increases as you go from left to right in a period. That increases the number of positive charges in the nucleus. The number of negatively charged electrons in the last shell also increases but the number of shells remains the same. The increasingly positive nuclear charge attracts the increasingly negative shells. The result is the atomic radius decreases as the atomic number increases in a period.
A. number of protons increases, and the number of shells of electrons
remains the same.
Explanation:
The number of protons increases as you go from left to right in a period. That increases the number of positive charges in the nucleus. The number of negatively charged electrons in the last shell also increases but the number of shells remains the same. The increasingly positive nuclear charge attracts the increasingly negative shells. The result is the atomic radius decreases as the atomic number increases in a period.
Moving from left to right in the periodic table, we notice that the magnitude of nuclear charge increases with increasing atomic number.
Recall that across the period, nuclear charge increases without a corresponding increase in the number of shells hence the size of the atom (atomic radius) contracts due to a greater pull of the nuclear charge on the electrons.
Moving from left to right in the periodic table, we notice that the magnitude of nuclear charge increases with increasing atomic number.
Recall that across the period, nuclear charge increases without a corresponding increase in the number of shells hence the size of the atom (atomic radius) contracts due to a greater pull of the nuclear charge on the electrons.
Explanation:
The weight of the two blocks acts downwards.
Let the weight of the two blocks be W. Solving for T₁ and T₂:
w = T₁/cos 60° -----(1);
w = T₂/cos 30° ----(2);
equating (1) and (2)
T₁/cos 60° = T₂/cos 30°;
T₁ cos 30° = T₂ cos 60°;
T₂/T₁ = cos 30°/cos 60°;
T₂/T₁ =1.73.
Therefore, option a is false since T₂ > T₁.
Option B is true since T₁ cos 30° = T₂ cos 60°.
Option C is true because the T₃ is due to the weight of the two blocks while T₄ is only due to one block.
Option D is wrong because T₁ + T₂ > T₃ by simple summation of the two forces, except by vector addition.
D. Heat each cube to the same temperature, place each cube into different containers with 500 grams of water at the same temperature, and measure the temperature of the water.
Explanation:
First find the distance it takes to stop
s = [v^2-u^2]/2a = 0^2 - 8.7^2/2[-2.4] = 8.7^2/4.8
Next find the time it takes to go that distance , s = ut +[1/2] at^2
8.7^2/4.8 = 8.7t +[1/2] [ -2.4]t^2 , rearrange and
t^2 -[8.7/1.2]+ 8.7^2/[(1.2)(4.8)]=0 complete the square
[t - (8.7/2.4)]^2=0
t = 8.7/2.4 = 3.625 secs
At this stage the deceleration will push the object back in the direction it came from for another 3.625 secs when it will be 8.7 m/s again
Total time , T =2t = 7.25 secs.
Note:
The term differential is used in calculus to refer to an infinitesimal (infinitely small) change in some varying quantity. For example, if x is a variable, then a change in the value of x is often denoted Δx (pronounced delta x). The differential dx represents an infinitely small change in the variable x.
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