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Arisa [49]
3 years ago
8

An aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 30mm in diameter and i

s 100mm long. If the modulus of elasticity for the aluminium is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180kN.

Physics
1 answer:
larisa86 [58]3 years ago
8 0

Answer: 1.228mm

Explanation:

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A mass m = 550 g is hung from a spring with spring constant k = 2.8 N/m and set into oscillation at time t = 0. A second, identi
riadik2000 [5.3K]

Answer:

The second system must be set in motion t=0.70s seconds later

Explanation:

The oscillation time, T, for a mass, m, attached to spring with Hooke's constant, k, is:

T=2\pi\sqrt(\frac{m}{k} )

One oscillation takes T secondes, and that is equivalent to a 2π phase. Then, a difference phase of π/2=2π/4, is equivalent to a time t=T/4.

If the phase difference π/2 of the second system relative to the first oscillator. The second system must be set in motion t=\frac{\pi}{2}\sqrt(\frac{m}{k})=\frac{\pi}{2}\sqrt(\frac{0.55}{2.8}= 0.70s) seconds later

6 0
3 years ago
Electricity and magnetism
stellarik [79]

Answer:

Electricity is which is possessed by the flow of electrons.

Magnitism is the process in which the magnet attract the things which consist magnetic substance.

6 0
2 years ago
If a car is taveling with a speed 6 and comes to a curve in a flat road with radius ???? 13.5 m, what is the minumum value the c
Lemur [1.5K]

Answer:

\mu_s \geq 0.27

Explanation:

The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:

f_s=\frac{mv^{2}}{R}

But we know that:

f_s\leq \mu_s N

And the normal force is given by the sum of the forces in the vertical direction:

N-mg=0 \implies N=mg

Finally, we have:

f_s=\frac{mv^{2}}{R}  \leq \mu_s mg\\\\\implies \mu_s\geq \frac{v^{2}}{gR}  \\\\\mu_s\geq \frac{(6\frac{m}{s}) ^{2}}{(9.8\frac{m}{s^{2}})(13.5m) }\\\\\mu_s\geq0.27

So, the minimum value for the coefficient of friction is 0.27.

4 0
3 years ago
An object has a mass of 300 g. (a) What is its weight on Earth? (b) What is its mass on the Moon?
Neko [114]

Answer:

The mass of object is 300g.

Its weight on earth is W×0.3kg×9.8m/s2=2.94 N

Its weight on moon is A 300 g would be 48 g on the moon

GOOD LUCK!

7 0
3 years ago
I’m kinda stuck right now…
kirza4 [7]
This is easy
Number 2 is 20 N to the right
Number 3 is 30 N up
Both are unbalanced
4 0
2 years ago
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