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KonstantinChe [14]
3 years ago
6

Which location, 23 degrees or 48 degrees would experience the same earthquake at stronger intensity?Explain why.​

Physics
1 answer:
BaLLatris [955]3 years ago
6 0

Answer:

48 degress

Explanation:

An earthquake causes many different intensities of shaking in the area of the epicenter where it occurs. So the intensity of an earthquake will vary depending on where you are. Sometimes earthquakes are referred to by the maximum intensity they produce. In the United States, we use the Modified Mercalli Scale. Earthquake intensity is a ranking based on the observed effects of an earthquake in each particular place. Therefore, each earthquake produces a range of intensity values, ranging from highest in the epicenter area to zero at a distance from the epicenter.

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Vikentia [17]
If an object that is 86kg on the moon then that is the answer, 86kg.
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3 years ago
Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t
marshall27 [118]

Answer:

        x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

         

the initial position of car a is zero

           x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

     

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

         0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

        v_{ob} = 23.4 km / h = 6.5 m / s

        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

        0.2 t² - 2.5 t - 30 = 0

        t² - 12.5 t - 150 = 0

we solve the quadratic equation

       t = \frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150}  }{2}

       t = \frac{12.5 \  \pm 27.5}{2}

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

        x = 4 20 + ½ 0.8 20²

        x = 240 m

8 0
3 years ago
Two girls stand 500m away from a tall wall
Mama L [17]

From the calculations, the speed of sound in this case is 16.9 m/s.

<h3>What is an echo?</h3>

The term echo has to do with the reflection of sound waves. Sound is a mechanical wave.

we know that the speed of sound is obtained from;

V = 2x/t

x = distance covered

t = time taken

V = 2(500)/59

v = 16.9 m/s

The error in the experiment could come from;

  • Lack of precise time measurement
  • Error can also arise from the environment of the experiment

Learn more about echo:brainly.com/question/9527413

#SPJ1

4 0
2 years ago
Can someone please help me on this
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Answer:

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Explanation:

4 0
3 years ago
If the pipe of the previous exercise is connected to a hose and the speed there is 3.8 m/s, what is the diameter of the hose?
Amiraneli [1.4K]

The diameter of the hose is 6.34 cm.

<em>"Your question is not complete, it seems to be missing the following information";</em>

the flow rate of water in the pipe is 0.012 m³/s

The given parameters;

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  • flow rate of water in the hose, Q = 0.012 m³/s

Volumetric flow rate is directly proportional to the product of the area of the hose through which the water flows and the velocity of the water flowing through the hose.

 Q = Av

where;

<em>Q is the volumetric flow rate</em>

<em>A is the area of the hose</em>

<em>v is the velocity of flow</em>

The area of the hose is calculated as follow;

A = \frac{Q}{v} \\\\A = \frac{0.012}{3.8} \\\\A = 0.00316 \ m^2

The diameter of the hose is calculated as follows;

A = \frac{\pi D^2}{4} \\\\\pi D^2 = 4A\\\\D^2 = \frac{4 \times A}{\pi} \\\\D =  \sqrt{\frac{4 \times A}{\pi} } \\\\D = \sqrt{\frac{4 \times 0.00316}{\pi} } \\\\D = 0.0634 \ m\\\\D = 6.34 \ cm

Thus, the diameter of the hose is 6.34 cm.

Learn more here: brainly.com/question/15061170

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