Explanation:
Given that,
Mass of Nichrome, m = 0.5 g
The resistance of the wire, R = 0.673 ohms
Resistivity of the nichrome wire, 
Density, 
(A) The length of the wire is given by using the definition of resistance as :
Volume,

Area,
....(1)

(b) Equation (1) becomes :

Hence, this is the required solution.
Answer:
The frequency of a sound wave is what your ear understands as pitch. A higher frequency sound has a higher pitch and the lower the period
The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.
<h3>How to determine the friction factor</h3>
Using the formula
μ = viscosity = 0. 06 Pas
d = diameter = 120mm = 0. 12m
V = velocity = 1m/s and 3m/s
ρ = density = 0.9
a. Velocity = 1m/s
friction factor = 0. 52 × 
friction factor = 0. 52 × 
friction factor = 0. 52 × 0. 55
friction factor 
b. When V = 3mls
Friction factor = 0. 52 × 
Friction factor = 0. 52 × 
Friction factor = 0. 52 × 0. 185
Friction factor 
Loss When V = 1m/s
Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter
Head loss = 0. 289 ×
×
× 
Head loss = 1. 80 × 10^8
Head loss When V = 3m/s
Head loss =
×
×
× 
Head loss = 5. 3× 10^8
Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.
Learn more about friction here:
brainly.com/question/24338873
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Answer:
B) waves speed up
C) waves bend away from the normal
Explanation:
The index of refraction of a material is the ratio between the speed of light in a vacuum and the speed of light in that medium:

where
c is the speed of light in a vacuum
v is the speed of light in the medium
We can re-arrange this equation as:

So from this we already see that if the index of refraction is lower, the speed of light in the medium will be higher, so one correct option is
B) waves speed up
Moreover, when light enters a medium bends according to Snell's Law:

where
are the index of refraction of the 1st and 2nd medium
are the angles made by the incident ray and refracted ray with the normal to the interface
We can rewrite the equation as

So we see that if the index of refraction of the second medium is lower (
), then the ratio
is larger than 1, so the angle of refraction is larger than the angle of incidence:

This means that the wave will bend away from the normal. So the other correct option is
C) waves bend away from the normal