Explanation:
Given that,
Length of gold wire, l = 4 m
Voltage of battery, V = 1.5 V
Current, I = 4 mA
The resistivity of gold, 
Resistance in terms of resistivity is given by :

Also, V = IR
So,

A is area of wire,
, r is radius, r = d/2 (diameter=d)

Out of four option, near option is (C) 17 μm.
The correct answer is D. Newton
Answer:
From the data we know that runner A and runner B are 11 km apart from the start because (6+5) km
So the runner from the east direction has distance as unknown km, rate= 9 k/h ; time= d/r=x/9 hr
So runner towards the west will be
distance = 11-x, rate= 8 k/h, time = d/r = (11-x)/8
So equating east and west time we have
x/9= (11-x)/8
8x=99-9x
17x=99
x=5.92 km
That is the distance covered by runner towards the east and he will meet the runner toward the west at
6-5.92=0.08 km west of the flagpole.
Answer: Acceleration due to gravity
Explanation: Force, F = mg, is a vector quantity because the acceleration due to gravity, g, is a vector quantity. Explanation: F = mg Where m is the mass (in kilograms) of the object in question and g is the acceleration due to gravity. Mass is a scalar quantity; mass has no dependence on direction whatsoever.
Answer:
a) that laser 1 has the first interference closer to the central maximum
c) Δy = 0.64 m
Explanation:
The interference phenomenon is described by the expression
d sin θ = m λ
Where d is the separation of the slits, λ the wavelength and m an integer that indicates the order of interference
For the separation of the lines we use trigonometry
tan θ = sin θ / cos θ = y / x
In interference experiments the angle is very small
tan θ = sin θ = y / x
d y / x = m λ
a) and b) We apply the equation to the first laser
λ = d / 20
d y / x = m d / 20
y = m x / 20
y = 1 4.80 / 20
y = 0.24 m
The second laser
λ = d / 15
d y / x = m d / 15
y = m x / 15
y = 0.32 m
We can see that laser 1 has the first interference closer to the central maximum
c) laser 1
They ask us for the second maximum m = 2
y₂ = 2 4.8 / 20
y₂ = 0.48 m
For laser 2 they ask us for the third minimum m = 3
In this case to have a minimum we must add half wavelength
y₃ = (m + ½) x / 15
m = 3
y₃ = (3 + ½) 4.8 / 15
y₃ = 1.12 m
Δy = 1.12 - 0.48
Δy = 0.64 m