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2 years ago

What can engineers do to prevent structures from collapsing during earthquakes

Physics

1 answer:

vredina [299]2 years ago

3 0

Explanation:

1. The structures should be made with good quality steel and concrete so that they can bear load and deformation.

2. Buildings should be made laterally strong so that they can overcome dominant lateral forces.

3. In High seismic activity areas devices such as energy absorbers should be used. Base isolation must necessarily be done.

4. Use of hybrid Fiber Reinforced Concrete to increase the toughness of a structure in a Moment Resisting Frame. Moment resisting frames are frames that resist moment of any devices.

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A mass is attached to a vertical spring, which then goes into oscillation. At the high point of the oscillation, the spring is i

andrew-mc [135]

Answer:

0.34 sec

Explanation:

Low point of spring ( length of stretched spring ) = 5.8 cm

midpoint of spring = 5.8 / 2 = 2.9 cm

Determine the oscillation period

at equilibrum condition

Kx = Mg

g= 9.8 m/s^2

x = 2.9 * 10^-2 m

k / m = 9.8 / ( 2.9 * 10^-2 ) = 337.93

note : w = $\sqrt{k/m}$ = $\sqrt{337.93}$ = 18.38 rad/sec

Period of oscillation = $2\pi / w$

= 0.34 sec

8 0

2 years ago

1. the most suitable cutting process to use on ferrous metals such as straight carbon steel is

dangina [55]

Answer:

plasma arc

cutting

Explanation:

It is plasma arc cutting because i think it is

6 0

2 years ago

Read 2 more answers

How to convert 250 newton to pounds? It is likely to be difficult to move?

maxonik [38]

I have no idea I need the answer too

3 0

2 years ago

1. I’m in the 2nd column, 4th row, and I’m a metal. Who am I? ________________ 2. I’m a very lonely nonmetal. Who am I? ________

Ghella [55]

Answer:

Explanation:

I'm in 17th column , a nometal, and a solid at room temperature. What am i

6 0

2 years ago

A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to

Karolina [17]

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

3 0

2 years ago