Helpppp!!!

a vertical solid steel post 29cm in diameter and 2.0m long is required to support a load of 8200kg, ignore the weight of the pos

erik [133]

Answer:

The stress is $\sigma = 1.218*10^{6} \ N/m^2$

Explanation:

From the question we are told that

The diameter of the post is $d = 29 \ cm = 0.29 \ m$

The length is $L = 2.0 \ m$

The weight of the loading mass

Generally the radius of the post is mathematically represented as

$r = \frac{0.29}{2}$

=> $r = 0.145 \ m$

Generally the area of the post is

$A = \pi r^2$

=> $A = 3.14 * 0.145 ^2$

=> $A = 0.066 \ m^2$

Generally the weight exerted by the load is mathematically represented as

$F = m * g$

=> $F = 8200 * 9.8$

=> $F = 80360 \ N$

Generally the stress is mathematically represented as

$\sigma = \frac{F}{A}$

=> $\sigma = \frac{80360 }{0.066}$

=> $\sigma = 1.218*10^{6} \ N/m^2$

7 0

2 years ago

Which technology is intended to work at a distance of about 5 to 10 centimeters with transmission speeds of 250 Kbps?

Mars2501 [29]

Answer:

NFC Near Field Communication

Explanation:

The Near Field Communication is a communication protocol, for extra short distance with a maximum of 10 centimeters, but usually used in 4 to 5 cm. Its intended to be used in contactless pay systems and in transportation card. Actually has been used to transfer multimedia from cell phones and other devices. The maximum data rate is around 424 kbit/s, with mean in 250 Kbp

5 0

3 years ago

inessss [21]

The popular GPS devices that people use to find directions while driving use "Global Navigation Satellite System (GNSS)".

Explanation:

The umbrella term for all global satellite tracking systems is GNSS i.e Global Satellite Navigation System. This involves satellite constellations circulating over the surface of the earth and continuous signal transmission that allow users to evaluate their location.

A satellite array of 18–30 medium Earth Orbit (MEO) satellites distributed across several orbital planes typically achieves greater coverage for each network. The specific systems differ, but use > 50 ° orbital inclinations and approximately twelve hours orbital cycles.

4 0

3 years ago

Read 2 more answers

A skier leaves the horizontal end of a ramp with a velocity of 31.0 m/s and lands 156.3 m from the base of a ramp how high is th

BartSMP [9]

Answer:

The height of ramp = 124.694 m

Explanation:

Using second equation of motion,

$s = ut + \frac{1}{2}at^2$

From the question,

u = 31 m/s; s = 156.3 m, a=0

substituting values

$156.3 = 31\times t + 0$

t = $\frac{156.3}{31 }$

= 5.042 s

Similary, for the case of landing

t = 5.042 s; initial velocity, u =0

acceleration = acceleration due to gravity, g = 9.81 $m/s^2$

Substituting in $h = ut + \frac{1}{2}gt^2$

$h = 0 + \frac{1}{2} \times 9.81 \times (5.042)^2$

h = 124.694 m

So height of ramp = 124.694 m

3 0

3 years ago

Write the equations of motions

Sav [38]

Answer:

In physics, equations of motion are equations that describe the behavior of a physical system in terms of its motion as a function of time.[1] More specifically, the equations of motion describe the behaviour of a physical system as a set of mathematical functions in terms of dynamic variables. These variables are usually spatial coordinates and time, but may include momentum components. The most general choice are generalized coordinates which can be any convenient variables characteristic of the physical system.[2] The functions are defined in a Euclidean space in classical mechanics, but are replaced by curved spaces in relativity. If the dynamics of a system is known, the equations are the solutions for the differential equations describing the motion of the dynamics.

6 0

3 years ago

Read 2 more answers