An atomic nucleus is composed of

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

A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of

balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, $N=mg$

∴ $f=\mu mg=0.490\times 90\times 9.8=432.18\ N$

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

$W=-fd=-432.18d$

Now, change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J$

Therefore, from work-energy theorem,

$W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m$

Hence, the skater covers a distance of 15 m before stopping.

7 0

3 years ago

If the atmosphere is like a greenhouse, what parts of it function as the glass

prohojiy [21]

The ozone layer is like the glass. the light from the sun enters the atmosphere and when the heat bounces back it is hard to escape the ozone layer therefore the earth's temperature rises

5 0

3 years ago

Read 2 more answers

You wad up a piece of paper and throw it into the wastebasket. How far will

vitfil [10]

The range of the piece of paper is C) 1.4 m

Explanation:

The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:

- A uniform motion along the horizontal direction, with constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, $g=9.8 m/s^2$ )

From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

$d=\frac{u^2 sin 2\theta}{g}$