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Shkiper50 [21]
3 years ago
15

(02.05 LC)

Physics
2 answers:
In-s [12.5K]3 years ago
8 0
Convection is the right answer!!!
Ps.: hope this helps!!!!
mezya [45]3 years ago
7 0
Convection is the transfer of heat by the circulation or movement of the heated parts of a liquid or gas. So convection is your answer.
You might be interested in
In a meeting of mimes, mime 1 goes through a displacement d1 = (6.00 m)i + (5.74 m)j and and mime 2 goes through a displacement
Blizzard [7]

a) Vector product: |d1 × d2| = (34.2 m) k

b) Scalar product: d1 · d2 = -0.874 m

c) (d1 + d2) · d2 = 16.1 m

d) Component of d1 along direction of d2: -0.21 m

Explanation:

a)

In this part, we want to calculate

|d1 × d2|

Which is the vector product between the two displacements d1 and d2.

The two vectors are:

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

The vector product of two vectors (a_x,a_y,a_z) and (b_x,b_y,b_z) is also a vector which has components:

r=(r_x,r_y,r_z)\\r_x=a_yb_z-a_zb_y\\r_y=a_zb_x-a_xb_z\\r_z=a_xb_y-a_yb_x

We notice immediatly that in this problem ,the two vectors d1 and d2 lie in the x-y plane, so they do not have components in zero. Therefore, the vector product has only one component, which is the one in z, and it is:

r_z=(6.00)(2.9)-(5.74)(-2.92))=34.2 m

Therefore, the vector product of d1 and d2 is:

|d1 × d2| = (34.2 m) k

b)

In this case, we want to calculate

d1 · d2

Which is the scalar product between the two displacements.

The scalar product of two vectors (a_x,a_y,a_z) and (b_x,b_y,b_z) is a scalar given by:

a \cdot b = a_x b_x + a_y b_y + a_z b_z

In this problem,

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

Therefore, the scalar product between the two vectors is:

d_1 \cdot d_2 = (6.00)(-2.92)+(5.74)(2.9)=-0.87m

c)

In this  case, we want to calculate

(d1 + d2) · d2

Which means that first we have to calculate the resultant displacement d1 + d2, and then calculate the scalar product of the resultant vector with d2.

Given two vectors  (a_x,a_y,a_z) and (b_x,b_y,b_z), the resultant vector is also a vector given by

r=(r_x,r_y,r_z)\\r_x=a_x+b_x\\r_y=a_y+b_y\\r_z=a_z+b_z

In this case,

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

So the resultant vector is

r_x=6.00+(-2.92)=3.08 m\\r_y = 5.74+2.9=8.64 m

So

(d_1+d_2)=(3.08 m,8.64 m)

And calculating the scalar product with d2, we find:

(d_1 + d_2)\cdot d_2 = (3.08)(-2.92)+(8.64)(2.9)=16.1 m

d)

The component of a vector a along another vector b is given by

a_b = \frac{a\cdot b}{|b|}

wherea\cdot b is the scalar product between and b

|b| is the magnitude of vector b

In this problem, we have the two vectors

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

We want to find the component of d1 along the direction of d2.

We already calculated the scalar product of the two vectors in part b):

d1 · d2 = -0.874 m

The magnitude of a vector b is given by

|b|=\sqrt{b_x^2+b_y^2+b_z^2}

So, for vector d2,

|d_2|=\sqrt{(-2.92)^2+(2.9)^2}=4.1 m

Now we can calculate the component of d1 along d2:

d_1_{d_2}=\frac{d_1 \cdot d_2}{|d_2|}=\frac{-0.874}{4.1}=-0.21 m

Learn more about operations with vectors:

brainly.com/question/2927458

brainly.com/question/2088577

brainly.com/question/1592430

#LearnwithBrainly

4 0
4 years ago
What is a pure substance
irina [24]
A pure substance is a substance that is made with only one type of molecule
3 0
3 years ago
A 1.0 C charged object and a 2.0 C charged object are separated by 100 m. Where should a -1.0x10-3 C charged object be placed on
horsena [70]

Answer:

x = 41.2 m

Explanation:

The electric force is a vector magnitude, so it must be added as vectors, remember that the force for charges of the same sign is repulsive and for charges of different sign it is negative.

In this case the fixed charges (q₁ and q₂) are positive and separated by a distance (d = 100m), the charge (q₃ = -1.0 10⁻³ C)) is negative so the forces are attractive, such as loads q₃ must be placed between the other two forces subtract

             F = F₁₃ - F₂₃

let's write the expression for each force, let's set a reference frame on the charge q1

           F₁₃ = k \frac{q_1 q_3}{x^2}

           F₂₃ = k \frac{q_2 q_3}{(d-x)^2}

they ask us that the net force be zero

           F = 0

           0 = F₁₃ - F₂₃

           F₁₃ = F₂₃

          k \frac{q_1 q_3}{x^2} =k \frac{q_2 q_3}{(d-x)^2}

          \frac{q_1}{x^2} = \frac{q_2}{(d-x)^2 }q1 / x2 = q2 / (d-x) 2

       

           (d-x)² = \frac{q_2}{q_1} x²

we substitute

           (100 - x)² = 2/1  x²

           100- x = √2  x

           100 = 2.41 x

           x = 41.2 m

6 0
3 years ago
SCIENCE HELP!!!!!!!!!!!!!!!!!!!!!!!
Alex Ar [27]
Electrical > light,sound and thermal
8 0
3 years ago
(NEED HELP BADLY)
spin [16.1K]

Answer:

The answer to your question is below

Explanation:

Data 1

mass 1 = 250

mass 2 = 250 kg

gravity constant = 6.67 x 10⁻¹¹ Nm²/kg²

distance = 8 m

Formula

F = G\frac{m1m2}{r^{2} }

Substitution

F = 6.67 x 10^{-11} \frac{250 x 250}{8^{2} }

Result

F = 0.000000065 N

Data 2

mass 1 = 1000 kg

mass 2 = 1000 kg

distance = 5 m

Substitution

F = 6.67 x 10^{-11} \frac{1000 x 1000 }{5^{2} }

Result

F = 0.000002667 N      

8 0
4 years ago
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