Answer:
1.74x10⁻⁵ V
Explanation:
n = 85.7 turns/cm => 8570 turns/metre
The field inside the long solenoid is given by B = μ₀ni
B = 4πx10⁻⁷ x 8570 x 0.175t² = 1.884x10⁻³ t²
dB/dt = 3.78x10⁻³ t
Cross-sectional Area'A'= 2.16 cm²=> 2.16 x 
m²
Now, rate of change of flux linkage '|Emf|' is given by:
|Emf| = d(NAB)/dt = NA dB/dt
|Emf| = 5 x 2.16 x x 3.78x10⁻³ t
|Emf| = 4.0824x10⁻⁶ t
Considering time 't' at which the current = 3.2A
, we have
3.2 = 0.175T²
T²
= 3.2/0.175
T = 4.28 s
|emf| = 4.0824x10⁻⁶ t => 4.0824x10⁻⁶ x4.28
|emf|= 1.74x10⁻⁵ V
Therefore,the magnitude of the emf induced in the secondary winding is 1.74x10⁻⁵ V
Answer:FALSE
Explanation: The Central message of a poem is the Theme of the poem, the poem is a poetry done to pass a particular message which may be regarding the present predicament of the people or to entertain. The poem the Desert Run was written in the year 1988 it was meant to express DISCRIMINATION AND HOPE. The poem was written by Mitsuye Yamada an activist of women rights and a Japanese National who lives in Seattle, Washington.
Answer:
0.75Hz
Explanation:
Given parameters:
Speed of the wave = 3m/s
Wavelength = 4m
Unknown:
Frequency of the wave = ?
Solution:
The speed of a wave is given by the expression below:
Speed = frequency x wavelength
Frequency = 
= 
= 0.75Hz
Answer:
a) The strength of gravity decreases if one moved away from Jupiter
b) The strength of gravity increases if one fell into Jupiter
Explanation:
The gravitational attraction is given by Newton law of gravitation as follows;
Where;
G = The universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)
M = The mass of Jupiter
m = The mass of the nearby body
R = The distance between the centers of Jupiter and the body
From the equation, we have that the gravitational strength varies inversely with the square of the separation distance between two bodies
Therefore, as one moves away, R increases, and the strength of gravity reduces
Similarly as the body falls into Jupiter, R, reduces the gravitational strength increases.
In other words a infinitesimal segment dV caries the charge
<span>dQ = ρ dV </span>
<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>
<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>
<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
