Answer:
d = 52 μm
Explanation:
given,
wavelength of the light source (λ)= 550 nm
distance to form interference pattern(D) = 1.5 m
y = 1.6 cm = 0.016 m
width of the slits = ?
now, using displacement formula
for the first maxima, m = 1
d = 5.2 x 10⁻⁶ m
d = 52 μm
hence, the width of her slits is equal to d = 52 μm
We have that for the Question "A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that the minimum force that must be applied on the <em>book is</em>
From the question we are told
A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal
Generally the equation for the Force is mathematically given as

F=44N
Therefore
the minimum force that must be applied on the <em>book is</em>
F=44N
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Answer:

Explanation:
<u>Given:</u>
- Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
- Distance of separation between the plates, d = 1.0 cm = 0.01 m.
- Minimum value of electric field that produces spark,

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

where,
= surface charge density of the plate of the capacitor =
.
= magnitude of the charge on each of the plate.
= surface area of each of the plate =
= electrical permittivity of free space, having value = 
For the minimum value of electric field that produces spark,

It is the maximum value of the magnitude of charge which can be added up to each of the plates of the capacitor.
Answer:
(a) 17634.24 Ω
(b) 0.0068 A
Explanation:
(a)
The formula for inductive inductance is given as
X' = 2πFL................... Equation 1
Where X' = inductive reactance, F = frequency, L = inductance
Given: F = 60 Hz, L = 46.8 H, π = 3.14
Substitute into equation 1
X' = 2(3.14)(60)(46.8)
X' = 17634.24 Ω
(b)
From Ohm's law,
Vrms = X'Irms
Where Vrms = Rms Voltage, Irms = rms Current.
make Irms the subject of the equation
Irms = Vrms/X'...................... Equation 2
Given: Vrms = 120 V, X' = 17634.24 Ω
Substitute into equation 2
Irms = 120/17634.24
Irms = 0.0068 A
The answer is decompression melting