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zlopas
2 days ago
10

A wave is a disturbance that carries energy through space. Which of these characterizes the fundamental difference between mecha

nical waves and electromagnetic waves
Physics
You might be interested in
A sphere of radius 0.03m has a point charge of q= 7.6 micro C located at it’s centre. Find the electric flux through it?
Sav [3153]

Answer:

The electric flux going through the sphere is 8.58 *10^{5} \frac{Nm^2}{C}

Explanation:

Given

Radius,\ r = 0.03m\\Charge,\ q =7.6\µC

Required

Calculate the electric flux

Electric flux can be computed using the formula;

Ф = q/ε

Where ε stands for the electric constant permittivity

ε = 8.8542 * 10^{-12}

Substituting ε = 8.8542 * 10^{-12} and q =7.6\µC; the formula simplifies to

Ф = \frac{7.6\µC}{8.8542 * 10^{-12}}

Ф = \frac{7.6 * 10^{-6}}{8.8542 * 10^{-12}}

Ф = \frac{7.6}{8.8542} *\frac{10^{-6}}{10^{-12}}

Ф = \frac{7.6}{8.8542} *10^{12-6}}

Ф = 0.85834970974 *10^{12-6}}

Ф = 0.85834970974 *10^{6}}

Ф = 8.5834970974 *10^{5}}

Ф = 8.58 *10^{5} \frac{Nm^2}{C}

Thus, the electric flux through the sphere is 8.58 *10^{5} \frac{Nm^2}{C}

7 0
3 months ago
12.Two ice skaters are initially at rest. The 78.2 kg male ice skater pushes his 48.5 kg female partner forward and away from hi
Yuliya22 [3333]
The male skater reaches a velocity of 13.71 m/s. According to the principle of conservation of momentum, m1u1 = (m1 + m2)u2, where m1 signifies the mass of the male skater at 78.2 kg, m2 is the mass of the female partner at 48.5 kg, u1 is the male skater's resulting velocity from the push, and u2 is the velocity imparted to the female skater, which was 8.46 m/s. Through the formula, we find u1 = [(78.2 + 48.5) × 8.46] ÷ 78.2, which calculates to 1071.882 ÷ 78.2 resulting in u1 = 13.71 m/s.
3 0
1 month ago
Read 2 more answers
15. Three semicircles of radius 1 are constructed on diameter AB of a semicircle of radius 2. The centers of the small semicircl
inna [3103]

Answer:

The area that remains is 4.201 m²

Explanation:

Provided that

AB=D=  4 m  (R=2 m)

The area of the half-circle AB is

A=\pi R^2/2

A=2π

The area of the smaller half-circle is

a=5π/6 + 2√3/4  m²

a=5π/6 + √3/2  m²

<pThus, the remaining area = A - a

                                     = 2π - (5π/6 + √3/2) m²

The remaining area is expressed as  2π - (5π/6 + √3/2) m²

5 0
3 months ago
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