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AnnZ [28]
3 years ago
14

Describe any major landmarks (buildings, bridges, historical sites, etc.) that were destroyed during the valdivia earthquake.

Physics
1 answer:
yawa3891 [41]3 years ago
6 0

Answer:

Homes, Workplaces, Statues, Meuseums

Explanation:

if you did look into it you would know after the earthquake there was a tsunami which flooded alot of places, this made many homeless, jobless, and many lossing vaklueable things

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4 for sure because you dont want anything spilling on you or others that is harmful.
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3 years ago
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2)It is known that the connecting rodS exerts on the crankBCa 2.5-kN force directed down andto the left along the centerline ofA
11111nata11111 [884]

Answer:

M_c = 100.8 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two cases shown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

                        M_c = F_a*( 42 / 150 ) *144

                        M_c = 2.5*( 42 / 150 ) *144

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- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

5 0
3 years ago
Astronaut X of mass 50kg floats next to Astronaut Y of mass 100kg while in space, as shown in the figure. The positive direction
jonny [76]

Answer:

C

Explanation:

The change in momentum of x has to be the opposite of the change in momentum of Y because the momentum is just transferred from one to another. But I'm still trying to figure it out how to calculate.

5 0
3 years ago
A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar
m_a_m_a [10]

Answer:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

#Where u is in meters and t in seconds.

Explanation:

Given that :m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s

From \omega=kL we have:

k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2

From F_d(t)=-\gamma u\prime(t) we have that:

\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m

Now,given that the initial value problem is given by:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

Hence,the position of u at time t is given by

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\, u in meters,t in seconds.

4 0
3 years ago
In the lesson a thermos is presented as an example of an isolated energy system. How could you change the thermos into an open e
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If you want to change the thermos into an open energy system, you have to remove the lid. Once the lid is removed, the energy is no longer contained inside the thermos bottle. From the bottle, the energy dissipates to the environment.
7 0
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