Answer:
Q = 5.06 x 10⁻⁸ m³/s
Explanation:
Given:
v=0.00062 m² /s and ρ= 850 kg/m³
diameter = 8 mm
length of horizontal pipe = 40 m
Dynamic viscosity =
μ = ρv
=850 x 0.00062
= 0.527 kg/m·s
The pressure at the bottom of the tank is:
P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²
The laminar flow rate through a horizontal pipe is:
Q = 5.06 x 10⁻⁸ m³/s
Answer: hello some parts of your question is missing attached below is the missing information
The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube
answer : Total surface area = 3/2 * area of old radiator
Explanation:
we will use this relation
K =
change in T = ΔT
therefore New Area ( A ) = 3/2 * area of old radiator
Given that the thermal conductivity is the same in the new and old radiators
Answer:
The operating system
Explanation:
The job of the operating system is to manage system resources allowing the abstraction of the hardware, providing a simple user interface for the user. The operating system is also responsible for handling application's access to system resources.
For this purpose, the operating system allows a user to run applications on their computing device.
Cheers.
Answer:
battery life in year = 9 years and 48 days
Explanation:
given data
Battery Ampere-hours = 1.5
Pulse voltage = 2 V
Pulse width = 1.5 m sec
Pulse time period = 1 sec
Electrode heart resistance = 150 Ω
Current drain on the battery = 1.25 µA
to find out
battery life in years
solution
we get first here duty cycle that is express as
duty cycle = ...............1
duty cycle = 1.5 ×
and applied voltage will be
applied voltage = duty energy × voltage ...........2
applied voltage = 1.5 × × 2
applied voltage = 3 mV
so current will be
current = ................3
current =
current = 20 µA
so net current will be
net current = 20 - 1.25
net current = 18.75 µA
so battery life will be
battery life =
battery life = 80000 hours
battery life in year =
battery life in year = 9.13 years
battery life in year = 9 years and 48 days