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Leto [7]
3 years ago
13

Please help me! Some people have proposed a new way to build houses in areas that are likely to experience tsunamis. In this des

ign, a house wouldn’t have solid walls on all four sides. Instead, some of the wall areas would be replaced by substances that water can travel through quickly, as shown in the diagram. How would this design help a house survive a tsunami? What drawbacks might there be to this design?

Physics
2 answers:
Sergeeva-Olga [200]3 years ago
6 0

Answer:

I think some drawbacks are that since there are no solid walls meaning it is weak and if murphy's law is in place, the water will destroy the substance. Tsunami waves also happen very quickly so even if the water can travel thru the substance quickly, it probably won't be quick enough. This design could help if the wave is smaller because less destruction would occur.

Explanation:

yeah

Alla [95]3 years ago
3 0

Answer:

house wouldn't have solid walls on all four sides. Instead, some of the wall areas would be replaced by substances that

water can travel through quickly, as shown in the diagram. How would this design help a house survive a tsunami? What

drawbacks might there be to this design?

Explanation:

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Is this statement true or false? The next generation of nuclear power plants being built in California and South Africa are even
alexira [117]

Answer:

True

Explanation:

Modern safer and cheaper nuclear reactors can not only meet the range of our long term energy demands, they can also fight global warming.

Modern techniques provide ways to reduce radioactive waste amount. "A closed fuel cycle may be switched on for new kinds of nuclear plants. Alternatively, the waste is chemically dissuaded to transform the reusable element into fuel. This implies that nuclear waste would not be buried.

8 0
3 years ago
A 0.683 kg mass moves in SHM at the end of a spring. It takes 1.41 s to move from the position with the spring fully extended to
dsp73

Answer:

Spring constant of the spring will be equal to 9.255 N /m          

Explanation:

We have given mass m = 0.683 kg

Time taken to complete one oscillation is given T = 1.41 sec

We have to find the spring constant of the spring

From spring mass system time period is equal to T=2\pi \sqrt{\frac{m}{K}}, here m is mass and K is spring constant

So 1.41=2\times 3.14\sqrt{\frac{0.683}{K}}

0.2245=\sqrt{\frac{0.683}{K}}

Squaring both side

0.0504=\frac{0.4664}{K}

K=9.255N/m

So spring constant of the spring will be equal to 9.255 N /m

7 0
3 years ago
A 438kg car is accelerating east at 2.55m/s^2. What is the total force acting east on the car
lisabon 2012 [21]

Answer:

<h2>1116.9 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 438 × 2.55

We have the final answer as

<h3>1116.9 N</h3>

Hope this helps you

5 0
3 years ago
How do we determine the conditions that existed in the very early universe? A We can only guess at the conditions, since we have
lakkis [162]

Answer:

D By looking all the way to the cosmological horizon, we can see the actual conditions that prevailed all the way back to the first instant of the Big Bang.

Explanation:

Astrophysicists are able to determine the conditions that existed in the early universe, by using instruments such as telescopes to observe and study cosmic horizons. More ideas about the early universe can be found from the thermal light present in cosmic backgrounds.

Scientists study these details that provide an insight into the conditions that existed so many years ago. They have been able to determine that the Big Bang involved so many collisions from these observations.

5 0
3 years ago
A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-
djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}

\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}

\frac{3-y}{y}=\sqrt{\frac{7}{2}}

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

6 0
3 years ago
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