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kvasek [131]
3 years ago
12

What is a caldera? plz help me

Physics
2 answers:
Alexeev081 [22]3 years ago
7 0

Answer:

A caldera is a large depression formed when a volcano erupts and collapses.

<em>Hope this helps!!!</em>

Sofia

Fudgin [204]3 years ago
5 0
A caldera is a large cauldron-like depression that forms following the evacuation of a magma chamber or reservoir. When large volumes of magma are erupted over a short time, structural support for the crust above the magma chamber is lost, and the entire structure caves in and forms a large bowl-like shape.
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On a roller coaster, describe the kinetic and potential energy of the car at the top of the first hill, the bottom of the first
vaieri [72.5K]
The total amount of energy stays the same, but throughout the ride, the kinetic energy and the potential energy change, still adding up to the same number. At the top of the ride it has potential energy, and as it goes down the potential energy decreases and the kinetic energy increases. When it’s at the bottom of the first drop it has maxed out its kinetic energy, and minimized its potential energy. Friction slows down the car, and pushes on the cart with a force that is equal and opposite to the force being exerted in the track. The reason the track keeps going is because though it exerts and equal and opposite force the momentum of the objects is different, allowing the car to continue moving, however friction will slow it down until eventually it comes to a stop.
8 0
3 years ago
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g calculate the effectiveness radiation dosage in sieverts for a 79 kg person who is exposed 6.8x10^9
Elza [17]

Answer:

The answer is "\bold{dosage = 0.031 rem}"

Explanation:

please find the complete question in the attached file.

Given value:

m = 79\  kg  \\\\n = 3.4 \times  10^9 \\\\E = 5.5  \times  10^{-13} \\\\ RBE = 15

\to E = n E\\

        = 3.4  \times  10^9  \times  5.5  \times  10^{-13} \\\\      = 1.87 \times  10^{-3}

\to E(absorbed) = 1.87  \times 10^{-3}  \times  0.87 = 1.63  \times  10^{-3}

calculating the radiation absorbed per kg:

= \frac{1.63  \times  10^{-3}}{79}  \\\\ = 2.06  \times  10^{-5} \\\\ = 0.00206 \  rad

\to Dosage = 0.00206  \times  15 \\

                 = 0.031 \ \ rem

4 0
3 years ago
Four point charges are placed at the corners of a square. Each charge has the identical value +Q. The length of the diagonal of
kvv77 [185]

Answer:V_{net}=4\frac{kQ}{a}    

Explanation:

Given

charge on each Particle is Q

Length of diagonal of the square is 2a

therefore distance between center and each charge is \frac{2a}{2}=a

Electric Potential of charged Particle is given by

For First Charge

V_1=\frac{kQ}{a}

V_2=\frac{kQ}{a}

V_3=\frac{kQ}{a}

V_4=\frac{kQ}{a}

total Electric Potential At center is given by

V_{net}=V_1+V_2+V_3+V_4

V_{net}=4\times \frac{kQ}{a}

V_{net}=4\frac{kQ}{a}            

7 0
3 years ago
The Periodic Table of Elements is organized by the number of:
QveST [7]
The answer to your question is Protons!
5 0
3 years ago
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Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0° with the horizo
Aleks [24]

Answer: Given mass of man = 85kg

Distance crate was pushed = 4m

Force exerted on crate = 500N

Angle inclined = 20°

Let Total force exerted Ft = (mg * sin(20)) + 500

Ft =( 85*9.81* sin20°) + 500

Ft = 758N

Work done Wd = Ft * distance

Wd = 785 * 4 = 3.14 * 10^3J

3 0
3 years ago
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