Answer:
0.65 m/min
Explanation:
The volume of material to be removed is
80*8*10 = 650 cm^3
The tool has a diameter of 4 mm and a maximum axial cutting capacity of 50 mm, so its cross section normal to advance is
0.4*5 = 2 cm^2
If the groove have to be made in T = 5 minutes the advance speed would be
V/(S * T)
650/(2 * 5) = 65 cm/min = 0.65 m/min
Answer:
I'm pretty sure it's false
Explanation:
Brainstorm is part of a problem-solving method. you can't solve a problem with nothing but brainstorming
Answer:
The three major network categories are LAN (Local Area Network), MAN (Metropolitan Area Network) and WAN (Wide Area Network)
Explanation:
A Local Area Network (LAN) is the simplest form of a network. It is usually limited to a geographical area and consists of a group of computers in the same organisation linked together. In most cases, the same technology is used for all computers in this network (eg. ethernet).
A Metropolitan Area Network (MAN) can connect multiple LANs as long as they are close to each other (geographically). MANs allow high speed remote connection as if the devices were on the same LAN.
A Wide Area Network (WAN) can connect multiple LANs across large geographical distances. Unlike MANs, the speed might differ based on factors such as distance. The internet is the most popular example of a WAN.
Answer:
The amperage draw of the condensing unit will be low.
Explanation:
A condensing unit is made up of a compressor and condenser, while an evaporating unit is made up of an evaporator coil.
A split AC system is a type of air conditioner system that has a condensing unit which is placed separately from the evaporative coil unit. Then the two units are connected to each other via a copper tube containing refrigerants.
The liquid line connects the condenser to the evaporator, and if this liquid line is restricted, the amp consumed by the condensing unit will be low.
Answer:
A(t) = 160 - 130 e^(-t/40)
Explanation:
At the start, the tank contains A(0) = 30 g of salt.
Salt flows in at a rate of
(1 g/L) * (4 L/min) = 5 g/min
and flows out at a rate of
(A(t)/160 g/L) * (4 L/min) = A(t)/40 g/min
so that the amount of salt in the tank at time t changes according to
A'(t) = 4 - A(t)/40
Solve the ODE for A(t):
A'(t) + A(t)/40 = 4
e^(t/40) A'(t) + e^(t/40)/40 A(t) = 4e^(t/40)
(e^(t/40) A(t))' = 4e^(t/40)
e^(t/40) A(t) = 160e^(t/40) + C
A(t) = 160 + Ce^(-t/40)
Given that A(0) = 30, we find
30 = 160 + C
C = -130
so that the amount of salt in the tank at time t is
A(t) = 160 - 130 e^(-t/40)