The oil additive that collects particles that can block the system is called a(n):

find magnitude of the resultant force, if 30N,40N,50N and 60N forces are acting along the line joining the centr of a square to

viktelen [127]

Answer:

56 feet

Explanation:

c 20 n north

7 0

3 years ago

Naomi needs to test an application that will be accessed by multiple users simultaneously. She is planning to use LoadRunner to

EastWind [94]

D bc i did the test and king

8 0

3 years ago

Giải giúp mình câu này được không ạ.em cảm ơn

sp2606 [1]

Answer:

çâdßèöñbvjsjaushdhshs

4 0

3 years ago

Air flows through a device such that the stagnation pressure is 0.4 MPa, the stagnation temperature is 400°C, and the velocity i

RoseWind [281]

To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.

The stagnation temperature can be defined as

$T_0 = T+\frac{V^2}{2c_p}$

Where

T = Static temperature

V = Velocity of Fluid

$c_p =$ Specific Heat

Re-arrange to find the static temperature we have that

$T = T_0 - \frac{V^2}{2c_p}$

$T = 673.15-(\frac{528}{2*1.005})(\frac{1}{1000})$

$T = 672.88K$

Now the pressure of helium by using the Adiabatic pressure temperature is

$P = P_0 (\frac{T}{T_0})^{k/(k-1)}$

Where,

$P_0$ = Stagnation pressure of the fluid

k = Specific heat ratio

Replacing we have that

$P = 0.4 (\frac{672.88}{673.15})^{1.4/(1.4-1)}$

$P = 0.399Mpa$

Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa

<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>

3 0

3 years ago

The clepsydra, or water clock, was a device that the ancient Egyptians, Greeks, Romans, and Chinese used to measure the passage

Nookie1986 [14]

Suppose a tank is made of glass and has the shape of a right-circular cylinder of radius 1 ft. Assume that h(0) = 2 ft corresponds to water filled to the top of the tank, a hole in the bottom is circular with radius in., g = 32 ft/s2, and c = 0.6. Use the differential equation in Problem 12 to find the height h(t) of the water.

Answer:

Height of the water = √(128)/147456 ft

Explanation:

Given

Radius, r = 1 ft

Height, h = 2 ft

Radius of hole = 1/32in

Acceleration of gravity, g = 32ft/s²

c = 0.6

Area of the hold = πr²

A = π(1/32)² ---- Convert to feet

A = π(1/32 * 1/12)²

A = π/147456 ft²

Area of water = πr²

A = π 1²

A = π

The differential equation is;

dh/dt = -A1/A2 √2gh where A1 = Area of the hole and A2 = Area of water

A1 = π/147456, A2 = π

dh/dt = (π/147456)/π √(2*32*2)

dh/dt = 1/147456 * √128

dh/dt = √128/147456 ft

Height of the water = √(128)/147456 ft

3 0

4 years ago