What is code in Arduino to turn led on and off

The size of an engine is called the engine

Stolb23 [73]

Answer:

a

Explanation:

I guess so

8 0

3 years ago

Read 2 more answers

A hole of diameter D = 0.25 m is drilled through the center of a solid block of square cross section with w = 1 m on a side. The

attashe74 [19]

Answer:

$q=4.013\:\:kW\\\\T_1=278.91\:\:^{\circ}C\\\\T_2=275.82\:\:^{\circ}C$

Explanation:

$R_{conv,1}=(h_1\pi D_1L)^{-1}=(50*0.25*2)^{-1}=0.01273\:\:K/W\\\\R_{conv,2}=(h^2*4wL)^{-1}=(4*4*1)^{-1}=0.0625\:\:K/W\\\\R_{cond(2D)}=(Sk)^{-1}=(8.59*150)^{-1}=0.00078\:\:K/W$

So, heat rate can be calculated as follows:

$q=\frac{T_{\infty,1}-T_{\infty,2}}{R_{conv,1}+R_{conv,2}+R_{cond(2D)}} =\frac{330-25}{0.076} =4.013\:\:kW$

Surface temperatures can be calculated as follows:

$T_1=T_{\infty,1}-qR_{conv,1}=330-51.09=278.91\:\:^{\circ}C\\\\T_2=T_{\infty,2}+qR_{conv,2}=25+250.82=275.82\:\:^{\circ}C$

6 0

4 years ago

To cushion the extend stroke of a cylinder, a cushion

Murljashka [212]

I think it is cap cause

3 0

3 years ago

What is the net torque about on the bar shown in (figure 1) about the axis indicated by the dot? suppose that ϕ = 30 ∘ and θ = 3

AysviL [449]

EDFFTRGJIKHMHNGBTR56YHUNJ BHYUJHMK.

3 0

2 years ago

1. You have a Co-Cr alloy with Young's Modulus: 645 MPa, Poisson ratio 0.28, and yield strength 501 MPa for that alloy when used

bazaltina [42]

Answer:

F_x = 100,200 N

x' = 21.321 cm ... Length

y' =0.7825 cm

z' = 1.565 cm

A' = ( 0.783 x 1.565 ) cm

Explanation:

Given:

- The Modulus of Elasticity E = 645 MPa

- The poisson ratio v = 0.28

- The Yield Strength Y = 501 MPa

- The Length along x-direction x = 12 cm

- The length along y-direction y = 1 cm

- The length along z--direction z = 2 cm

Find:

The maximum tensile load that can be applied in the longitudinal direction of the bar without inducing plastic deformation. b. The length and cross-sectional area of the bar at its tensile elastic limit.

Solution:

- The Tensile forces within the limit of proportionality is given as:

F_i = б_i*A_jk

- A maximum tensile Force F_x along x direction can be given as:

F_x = Y*A_yz

F_x = 501*( 0.01*0.02)*10^6

F_x = 100,200 N

- The corresponding strains in x, y and z direction due to F_x are:

ξ_x = Y / E

ξ_x = 501 / 645 = 0.7767

ξ_y = ξ_z = -v*Y / E

ξ_y = ξ_z = -0.28*501 / 645 = - 0.2175

- The corresponding change in lengths at tensile elastic stress are:

Δx = x*ξ_x = 12*0.7767 = 9.321 cm

Δy = y*ξ_y = - 1*0.2175 = -0.2175 cm

Δz = z*ξ_z = - 2*0.2175 = -0.435 cm

- The final lengths are:

x' = x + Δx = 12 + 9.321 = 21.321 cm

y' = y + Δy = 1 - 0.2175 = 0.7825 cm

z' = z + Δz = 2 - 0.435 = 1.565 cm

3 0

4 years ago