As there is no postive or negative assigned so
Initial velocity= -2.8759
Displacement= 0.5at^2+ut
= 0.5(-1.77)(3.33)^2+(-2.8759)(3.33)=-19.4m
Answer:
A negatively charged particle -q is placed at the center of a uniformly charged ring, where the ring has a total positive charge Q as shown in the following figure. The particle, confined to move along the x axis, is moved a small distance x along the axis ( where x << a) and released. Show that the particle oscillates in simple harmonic motion with a frequency given by,
Answer:
normal force = 10 N
Explanation:
Given data
frictional force = 0.400 N
coefficient of kinetic friction = 0.04
Solution
we get here normal force that is express as
normal force = 
............1
put here value and we will get value
normal force = 
solve it we get
normal force = 10 N
The plastic jars of the air capacitor represent the parallel conducting plates.
<h3>What is Air capacitor?</h3>
Air capacitor is a type of capacitor that uses air as its dielectric. The simplest air capacitors will contain two conductive plates separated by an air gap.
This capacitor stores and releases electricity in the circuit using;
- air as the electrical source,
- balloon as the insulator and
- the plastic jar as the parallel conducting plates.
Thus, the plastic jars of the air capacitor represent the parallel conducting plates.
Learn more about air capacitor here: brainly.com/question/15755974
Answer:
The total surface are of the bowl is given by: 0.0532*pi m² (approximately 0.166533 m²)
Explanation:
The total surface area of the semi-spherical bowl can be decomposed in three different sections: 1) an outer semi-sphere of radius 12 cm, 2) an inner semi-sphere of radius 10 cm, and 3) the edge, which is a 2-dimensional ring with internal radius of 10 cm and external radius of 12 cm. We will compute the areas independently and then sum them all.
a) Outer semi-sphere:
A1 = 2*pi*r² = 2*pi*(12 cm)² = 288*pi cm² = 904.78 cm²
b) Inner semi-sphere:
A2 = 2*pi*(10 cm)² = 200*pi cm² = 628.32 cm²
c) Edge (Ring):
A3 = pi*(r1² - r2²) = pi*((12 cm)²-(10 cm)²) = pi*(144-100) cm² = 44*pi cm² = 138.23 cm²
Therefore, the total surface area of the bowl is given by:
A = A1 + A2 + A3 = 288*pi cm² + 200*pi cm² + 44*pi cm² = 532*pi cm² (approximately 1665.33 cm²)
Changing units to m², as required in the problem, we get:
A = 532*pi cm² * (1 m² / 10, 000 cm²) = 0.0532*pi m² (approximately 0.166533 m²)
