The band of stability curves upward at high atomic numbers due to the fact that excess of neutrons are required due to the repulsion between protons.
Atomic number is the number of protons. As the number of protons (atomic number) increase, the electrical repulsion force, due to the same sign of the protons inside the nucleus, becomes more dominant compared to the nuclear force attractions, then the nucleus needs more neutrons to gain stability.The presence of more neutrons decrease the density of protons which decreases the repulsion among them.
Answer:
Option (A)
Explanation:
A flowchart can be described as a representation of a process or an algorithm in a sequential manner (stepwise) in the form of diagrams. These steps are made of symbols that are connected to one another with the help of arrows. These arrows are used to show the direction of the process.
There are mainly four symbols that are used to describe a flow chart, which includes-
- Oval or Pill Shape- This type of symbol is used to depict the start or the end of a process.
- Rectangle Shape- This type of symbols is used to describe the process
.
- Diamond Shape- This type of symbol is used to represent decision
s.
- Parallelogram- This type of symbol is used for the representation of an input or an output.
Thus, the correct answer is option (A).
Answer:
(b) To get m3 to slide, m1 must be increased, never decreased.
Explanation:
Lab experiments require attentiveness. If there is one thing missed or not taken seriously whole experiment could go wrong. In this case to slide m3 there should be more weight at m1. If the weight of m1 is lesser than m3 then the object will not slide. It will remain at the point where there is more weight. To slide an object there must be less frictional surface and more weight placed at the desired end point.
Answer:
Explanation:
90 rpm = 90 / 60 rps
= 1.5 rps
= 1.5 x 2π rad /s
angular velocity of flywheel
ω = 3π rad /s
Let I be the moment of inertia of flywheel
kinetic energy = (1/2) I ω²
(1/2) I ω² = 10⁷ J
I = 2 x 10⁷ / ω²
=2 x 10⁷ / (3π)²
= 2.2538 x 10⁵ kg m²
Let radius of wheel be R
I = 1/2 M R² , M is mass of flywheel
= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .
1/2 πR⁴ x t x d = 2.2538 x 10⁵
R⁴ = 2 x 2.2538 x 10⁵ / πt d
= 4.5076 x 10⁵ / 3.14 x .1 x 7800
= 184
R= 3.683 m .
diameter = 7.366 m .
b ) centripetal accn required
= ω² R
= 9π² x 3.683
= 326.816 m /s²
0.29 m/s (wave velocity = wavelength (lamda)/period (T) in metres)
35 / 1.2 = 29.16
29.16 ÷ 100 = 0.29
Wave velocity in string:
The properties of the medium affect the wave's velocity in a string. For instance, if a thin guitar string is vibrated while a thick rope is not, the guitar string's waves will move more quickly. As a result, the linear densities of the two strings affect the string's velocity. Linear density is defined as the mass per unit length.
Instead of the sinusoidal wave, a single symmetrical pulse is taken into consideration in order to comprehend how the linear mass density and tension will affect the wave's speed on the string.
Learn more about density here:
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