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sergij07 [2.7K]
3 years ago
7

Which of the following is true about evaporation from land surface as compared to precipitation over them?(

Physics
1 answer:
ankoles [38]3 years ago
5 0
I’m not 100% sure but I think it’s 3 sorry if it’s wrong
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lightning is actually an enormous display of the concept of A.chemical heating B.solar energy C.magnetism D.grounding
allsm [11]

Answer:

D) Grounding

Explanation:

The potential difference between cloud and ground leads to ionization of the atmosphere and resulting conduction through the air often to ground (although it can be between clouds at different potentials. I would say grounding, like the spark when you touch a hot battery terminal to ground on a car.

5 0
3 years ago
a boy takes 15min to reach his school on bicycle.if the bicycle have speed of 2m/s the what is the distance between the house an
11111nata11111 [884]
When looking for distance you multiply speed by time

So 15 x 2 = 30

30 is the distance between his house and school
6 0
3 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
3 years ago
A delivery drone uniformly slowed down from 45 m/s to 23 m/s in a straight line over a 30-minute duration within a 16-kilometer
Neporo4naja [7]
I think it is letter A
7 0
3 years ago
An athlete is working out in the weight room. he steadily holds 50 kg above his head for ten seconds which statement is true abo
Andrews [41]
Hi. The answer to your question is the first option.

The athlete isn’t doing any work because he doesn’t move the weight.

Hope this helps :))
5 0
3 years ago
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