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vlabodo [156]
3 years ago
8

99 points on the line please I really need help with these four questions. Please don't just put random stuff I will report you.

thank you for the help :)

Physics
2 answers:
sukhopar [10]3 years ago
8 0

Answer:

1. Each station can detect how far away the epicenter was. So each station basically has a circle made of possible epicenters. When you have three, you narrow it down to one, final point.

2. Below the tectonic plate is a magma chamber that is shooting up mafic magma. As it emerges, it creates an island. The plate that the magma is puncturing is also happening to be moving, so it makes one island, and it floats on. Think of it as a chimney. It blows a puff of smoke, the wind blows it away. Then another puff emerges, and so on. Those other islands no longer sit on the source of the volcano.

3. Metamorphism occurs when there is enough pressure and heat in order to rearrange the structures of each mineral into a new mineral. Contact metamorphism occurs when lava shoots through and touches other rocks, which heats them up and driving metamorphism. In contact, there is no pressure. In regional, there is a large area of both heat and pressure being applied on the rocks.

bearhunter [10]3 years ago
6 0
<span>1. Each station can detect how far away the epicenter was. So each station basically has a circle made of possible epicenters. When you have three, you narrow it down to one, final point.

2. Below the tectonic plate is a magma chamber that is shooting up mafic magma. As it emerges, it creates an island. The plate that the magma is puncturing is also happening to be moving, so it makes one island, and it floats on. Think of it as a chimney. It blows a puff of smoke, the wind blows it away. Then another puff emerges, and so on. Those other islands no longer sit on the source of the volcano.

3. Metamorphism occurs when there is enough pressure and heat in order to rearrange the structures of each mineral into a new mineral. Contact metamorphism occurs when lava shoots through and touches other rocks, which heats them up and driving metamorphism. In contact, there is no pressure. In regional, there is a large area of both heat and pressure being applied on the rocks. </span>
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Jet001 [13]

Answer: a all the outer planets are large

Explanation: because i have the smarts

3 0
3 years ago
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A parallel-plate capacitor with circular plates of radius R is being discharged. The displacement current through a central circ
Dvinal [7]

Answer:

The discharging current is I_d  =  36.8 \ A

Explanation:

From the question we are told that  

     The radius of each circular plates is  R

     The displacement current is  I  = 9.2 \ A

      The radius of the central circular area is  \frac{R}{2}

The discharging current is mathematically represented as

       I_d  =  \frac{A}{k} *  I

where A is the area of each plate which is mathematically represented as

       A  =  \pi R ^2

and   k is central circular area which is mathematically represented as

     k  =  \pi [\frac{R}{2} ]^2

So  

     I_d  =  \frac{\pi R^2 }{\pi * [ \frac{R}{2}]^2 } *  I

     I_d  =  \frac{\pi R^2 }{\pi *  \frac{R^2}{4} } *  I

     I_d  =  4 *  I

substituting values

     I_d  =  4 *  9.2

     I_d  =  36.8 \ A

     

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B%5Csqrt%7B4%7D%20%3D87" id="TexFormula1" title="x^{2} +\sqrt{4} =87" alt="x^{
Alja [10]

Answer:

<h2><em><u>ᎪꪀsωꫀᏒ</u></em></h2>

➪x= √ 85

Explanation:

x²+√4 = 87

=> x²+2 = 87

=> x² = 87-2

=> x²= 85

=> x= √85

7 0
3 years ago
A 2kg object is dropped from height of 10m. ignoring air resistance calculate:
lbvjy [14]

Answer:

ME= 196.2 J

KE= 136.2

Explanation:

potential energy=mgh 2*9.81*10

Our ME is quivalent to PE as that is the toal amount of energy in the system

Kinetic energy= 1/2 mv^{2}

to solve for kinetic enrgy we need to use a kinaetmtic equation that help us find velocity

vf= vi+at

but we need to find time first

d=vi+1/2(accelretaion)t^{2}

7=0+1/2(9.81)t^{2}

t= 1.19 s

vf= 0+ 9.81*1.19

vf= 11.67 m/s

Now

1/2 mv^{2}

1/2*2*11.67^{2}

= 136. 2

or we could just (PE/10)*7

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5 0
3 years ago
A 0.75 kg book is pushed across the table with an acceleration of 0.3 m/s2. What force is being applied to the
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Answer:

\boxed {\boxed {\sf 0.225 \ Newtons}}

Explanation:

We are asked to find the force being applied to a book. According to Newton's Second Law of Motion, force is the product of mass and acceleration.

F= ma

The mass of the book is 0.75 kilograms and the acceleration is 0.3 meters per square second. Substitute these values into the formula.

  • m= 0.75 kg
  • a= 0.3 m/s²

F= 0.75 \ kg * 0.3 \ m/s^2

Multiply.

F =0.225 \ kg * m/s^2

1 kilogram meter per second squared is equal to 1 Newton. Therefore, our answer of 0.225 kilogram meters per second squared is equal to 0.225 Newtons.

F= 0.225  \ N

<u>0.225 Newtons of force</u> are applied to the book.

5 0
3 years ago
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