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3 years ago

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The circumference of the Earth is approximately 40,000 km. The Earth has a magnetic field that protects it from solar particle r

adiation. The major element comprising the solid inner core of the Earth is iron. The Earth is approximately 4.6 million years old.a. trueb. false

1 answer:

olasank [31]3 years ago

6 0

Answer:

All are correct except the age of earth. The age of Earth is approximately 4.543 Billion Years.

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If the temperature of water warming on a hot plate rises 10 degrees Celsius [°C], what is the change in temperature in units of

Luba_88 [7]

Answer:

Delta_temp = 18[°F]

Explanation:

°F = 9/5*(10)

°F = 18

Note: It is important to clarify that it is only a temperature increase, that it is only a temperature increase. The question is not related to converting from 10°C to fahrenheit degrees

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3 years ago

PLEASE ANSWER-Why are loose electrons needed for heat conduction?

dolphi86 [110]

Answer:

Why do metals conduct heat so well? The electrons in metal are delocalised electrons and are free moving electrons so when they gain energy (heat) they vibrate more quickly and can move around, this means that they can pass on the energy more quickly.

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2 years ago

Two identical charges q placed 2.0 mapart exert forces of magnitude 4.0 N on each other What is the value of the charge q? a) q

katen-ka-za [31]

Answer:

c) $4.2*10^{-5}C$

Explanation:

Coulomb's law says that the force exerted between two charges is inversely proportional to the square of distance between them, and is given by the expression:

$F=\frac{kq_{1}q_{2}}{r^{2}}$

where k is a proportionality constant with the value $k=9*10^{9}\frac{Nm^{2}}{C^{2}}$

In this case $q_{1}=q_{2}=q$ , so we have:

$F=\frac{kq^{2}}{r^{2}}$

Solving the equation for q, we have:

$kq^{2}=Fr^{2}$

$q^{2}=\frac{Fr^{2}}{k}$

$q=\sqrt{\frac{Fr^{2}}{k}}$

Replacing the given values:

$q=\sqrt{\frac{4.0N*(2.0m)^{2}}{9*10^{-9}\frac{Nm^{2}}{C^{2}}}}$

$q=4.2*10^{-5}C$

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3 years ago

Will mark BRAINLIEST!!!!

REY [17]

Answer:

Atoms found in nature are either stable or unstable. ... An atom is unstable (radioactive) if these forces are unbalanced; if the nucleus has an excess of internal energy. Instability of an atom's nucleus may result from an excess of either neutrons or protons

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2 years ago

Consider the points below. P(1, 0, 1), Q(−2, 1, 4), R(6, 2, 7) (a) Find a nonzero vector orthogonal to the plane through the poi

kozerog [31]

Answer:

a) (0, -33, 12)

b) area of the triangle : 17.55 units of area

Explanation:

<h2>a) </h2>

We know that the cross product of linearly independent vectors $\vec{A}$ and $\vec{B}$ gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.

Luckily for us, we know that vectors $\vec{A} = \vec{P}-\vec{Q}$ and $\vec{B} = \vec{R} - \vec{Q}$ are living in the plane through the points P, Q, and R, and are linearly independent.

We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).

If they weren't linearly independent, we will obtain vector zero as the result of the cross product.

So, for our problem:

$\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)$

$\vec{B} = \vec{R} - \vec{Q} \\\\\vec{B} = (6,2,7) - (-2,1,4)\\\\\vec{B} = (6 +2,2-1,7-4)\\\\\vec{B} = (8,1,3)$

$\vec{A} \times \vec{B} = (A_y B_z - B_y A_z) \ \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( (-1) * 3 - 1 * (-3) ) \ \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( - 3 + 3 ) \ \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}$

$\vec{A} \times \vec{B} = 0 \ \hat{i} - 33 \ \hat{j} + 12 \ \hat{k}$

$\vec{A} \times \vec{B} =(0, -33, 12)$

<h2>B)</h2>

We know that $\vec{A}$ and $\vec{B}$ are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:

$|\vec{A} \times \vec{B} | = 2 * area_{triangle}$

so:

$\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}$

$\sqrt{1233} = 2 * area_{triangle}$

$35.114= 2 * area_{triangle}$

$17.55 \ units \ of \ area = area_{triangle}$

5 0

2 years ago