The force of a particle moving in a circular motion is:
F=ma
where m is the mass and a is the centripetal acceleration, which is expressed as:
a=v²/r
<span>where v is the speed and r the distance between the particles. You use these equations, by substituting the values given in order to solve the problem.</span>
Answer:
Linear charge density (I) = Q/2πR
Explanation:
Linear charge density (I) = charge (Q) per unit length(L)
I = Q/L
For a thin ring of charge with radius R, the length will be equal to the circumference of a circle.
Circumference of a circle = 2πR
Then, the length of the thin ring of charge is 2πR
Linear charge density (I) = Q/2πR
Therefore, for a thin ring of charge of radius R, which lies in the x-y plane and is centered on the z-axis. The linear charge density, l, of the loop is given by Q/2πR
Answer:
a
The total distance traveled is 
b
The average velocity is 
Explanation:
From the question we are told that
The time taken for first part 
The speed for the first part is 
The time taken for second part is 
The speed for the second part is 
The time taken for third part is 
The speed for the third part is 
Generally
Therefore the total distance traveled is
substituting values
Generally the average velocity is mathematically represented as
Where 
is the total time taken which is mathematically represented as
The average velocity is
Answer:
The frequency heard y the ship is 70 MHz.
Explanation:
True frequency, f = 50 MHz
speed of source, vs = 0
speed of observer, vo = 0.4 c
Let the frequency heard by the ship is f'.
Use the formula of doppler's effect
Answer:
(a) The power wasted for 0.289 cm wire diameter is 15.93 W
(b) The power wasted for 0.417 cm wire diameter is 7.61 W
Explanation:
Given;
diameter of the wire, d = 0.289 cm = 0.00289 m
voltage of the wire, V = 120 V
Power drawn, P = 1850 W
The resistivity of the wire, ρ = 1.68 x 10⁻⁸ Ω⋅m
Area of the wire;
A = πd²/4
A = (π x 0.00289²) / 4
A = 6.561 x 10⁻⁶ m²
(a) At 26 m of this wire, the resistance of the is
R = ρL / A
R = (1.68 x 10⁻⁸ x 26) / 6.561 x 10⁻⁶
R = 0.067 Ω
Current in the wire is calculated as;
P = IV
I = P / V
I = 1850 / 120
I = 15.417 A
Power wasted = I²R
Power wasted = (15.417²)(0.067)
Power wasted = 15.93 W
(b) when a diameter of 0.417 cm is used instead;
d = 0.417 cm = 0.00417 m
A = πd²/4
A = (π x 0.00417²) / 4
A = 1.366 x 10⁻⁵ m²
Resistance of the wire at 26 m length of wire and 1.366 x 10⁻⁵ m² area;
R = ρL / A
R = (1.68 x 10⁻⁸ x 26) / 1.366 x 10⁻⁵
R = 0.032 Ω
Power wasted = I²R
Power wasted = (15.417²)(0.032)
Power wasted = 7.61 W
