Answer:
Depends
Explanation:
The modem connects you to the Internet via ISP. Without a modem, your router will only allow you to connect to a LAN. A modem will provide connections for just a single wired device. If you want to go wireless you need a router.
Answer:
critical stress required for the propagation is 27.396615 ×
N/m²
Explanation:
given data
specific surface energy = 0.90 J/m²
modulus of elasticity E = 393 GPa = 393 ×
N/m²
internal crack length = 0.6 mm
to find out
critical stress required for the propagation
solution
we will apply here critical stress formula for propagation of internal crack
( σc ) = 
.....................1
here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm = 0.3 ×
m
so now put value in equation 1 we get
( σc ) =
( σc ) = 
( σc ) = 27.396615 × N/m²
so critical stress required for the propagation is 27.396615 × N/m²
Answer:
778.4°C
Explanation:
I = 700
R = 6x10⁻⁴
we first calculate the rate of heat that is being transferred by the current
q = I²R
q = 700²(6x10⁻⁴)
= 490000x0.0006
= 294 W/M
we calculate the surface temperature
Ts = T∞ + 
Ts = 
The surface temperature is therefore 778.4°C if the cable is bare
Answer:
The pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its original value.
Explanation:
For a fully developed laminar flow in a circular pipe, the flowrate (volumetric) is given by the Hagen-Poiseulle's equation.
Q = π(ΔPR⁴/8μL)
where Q = volumetric flowrate
ΔP = Pressure drop across the pipe
μ = fluid viscosity
L = pipe length
If all the other parameters are kept constant, the pressure drop across the circular pipe is directly proportional to the viscosity of the fluid flowing in the pipe
ΔP = μ(8QL/πR⁴)
ΔP = Kμ
K = (8QL/πR⁴) = constant (for this question)
ΔP = Kμ
K = (ΔP/μ)
So, if the viscosity is halved, the new viscosity (μ₁) will be half of the original viscosity (μ).
μ₁ = (μ/2)
The new pressure drop (ΔP₁) is then
ΔP₁ = Kμ₁ = K(μ/2)
Recall,
K = (ΔP/μ)
ΔP₁ = K(μ/2) = (ΔP/μ) × (μ/2) = (ΔP/2)
Hence, the pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its value.
Hope this Helps!!!
