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2 years ago

4 kg of water was

Physics

1 answer:

morpeh [17]2 years ago

6 0

Given :

Mass =5kg

=20

∘

C,T

=100

∘

ΔT=100−20=80

∘

Q=m×C×ΔT

where C= specific heat capacity of water

=4200J/(kgK)

Q=5×4200×80

=1680000 Joule.

=1680KJ

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Answer:

Explanation:

Given the following

Distance S = 70.6m

Time t = 0.32secs

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Required

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2 years ago

What is the total wavelength if one-half of the wave is 3?

seraphim [82]

Answer:

Explanation:

Half the wave = 3

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2 years ago

Sammy feels an ocean breeze as he plays volleyball at the beach. Why do ocean winds or sea breezes blow toward shore during the

DedPeter [7]

The heat coming from the sun warms the land more quickly than the sea. As a result of these, the air near the land warm up and rises and the cooler air from the sea moves in to replace the risen air. The correct answer is option A

There will be heat transfer from a region of higher temperature to the region of lower temperature. But in the case of land and sea breeze, the transfer of heat are the result of convectional current in nature. Because the land is a better absorber of heat and also has a lower specific heat capacity compare to sea, during the day, the heat coming from the sun warms the land more quickly than the sea. As a result of these, the air near the land warm up and rises.

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Learn more here : brainly.com/question/1114842

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2 years ago

If it takes Ashley 10 seconds to run from the batter's box to first base at an average speed

abruzzese [7]

Answer:

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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:

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:

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

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

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

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

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

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

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

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