Find the steady-state temperature distribution for the semi-infinite plate problem

Which statement describes why scientific notation is useful? It makes very small numbers into whole numbers. It converts fractio

disa [49]

It makes calculations with very large and small numbers easier.

Scientific notation is a system used in order to It makes calculations with very large and small numbers easier. It is useful as it allows very large number that would take a lot of space to write otherwise, and it allows them to be calculated easier. $10^{23}$ for example is a incredible large number, but written in this form is immediately understandable and useful for calculation.

3 0

3 years ago

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What building materials do you believe would work well to build a home in that area? Explain why you chose these materials.

Luden [163]

Depends on what the area is. If it’s a rural place, Wood is cheep & easy to build. If there’s a lot of corrosion, strong weather/hurricane, bricks.

8 0

3 years ago

A centrifuge in a medical research laboratory rotates at an angular speed of 3,650 rev/min. When switched off, it rotates 46.0 t

erma4kov [3.2K]

Answer:

The constant angular acceleration of the centrifuge = -252.84 rad/s²

Explanation:

We will be using the equations of motion for this calculation.

Although, the parameters of this equation of motion will be composed of the angular form of the normal parameters.

First of, we write the given parameters.

w₀ = initial angular velocity = 2πf₀

f₀ = 3650 rev/min = (3650/60) rev/s = 60.83 rev/s

w₀ = 2πf₀ = 2π × 60.83 = 382.38 rad/s

θ = 46 revs = 46 × 2π = 289.14 rad

w = final angular velocity = 0 rad/s (since the centrifuge come rest at the end)

α = ?

Just like v² = u² + 2ay

w² = w₀² + 2αθ

0 = 382.38² + [2α × (289.14)]

578.29α = -146,214.4644

α = (-146,214.4644/578.29)

α = - 252.84 rad/s²

Hope this Helps!!!

6 0

3 years ago

Read 2 more answers

As part

umka21 [38]

Answer:

Part a)

$a = 3.68 m/s^2$

Part b)

$a = 11.8 m/s^2$

Explanation:

Part a)

For force conditions of two blocks we will have

$m_1g - T = m_1 a$

$T - m_2g = m_2 a$

now from above equations we have

$(m_1 - m_2) g = (m_1 + m_2) a$

$a = \frac{m_1 - m_2}{m_1 + m_2} g$

now we know that

$m_1 = \frac{908}{9.8} = 92.65 kg$

$m_2 = \frac{412}{9.8} = 42 kg$

now from above equation we have

$a = \frac{92.65 - 42}{92.65 + 42}(9.8)$

$a = 3.68 m/s^2$

Part b)

When heavier block is removed and F = 908 N is applied at the end of the string then we have

$F - mg = ma$

$908 - 412 = 42 a$

$a = 11.8 m/s^2$

8 0

3 years ago

Find the angle formed by two forces of 7N and 15N respectively if its result is worth 20N

nadezda [96]

First, you need to make certain assumptions before solving this question. Why? Because there are no information given about the direction of these forces. In such questions as above, ALWAYS make the following assumptions:

1) Take first force, say $F_{1}$ , and assume that it is pointing towards the x-direction.

Let us take the 7N force! By keeping the above assumption in our minds, the force vector would be like:
$F_{1}$ = $7i$ , where $i$ = Unit vector in the x-direction.

2) Take the second force, say $F_{2}$ , and assume that it is making an angle $\alpha$ with the first force $F_{1}$ .

Let us take the 15N force! By keeping the above assumption in our minds, the forces vector would be like:

$F_{2} = (15*cos \alpha)i + (15*sin \alpha )j$

Now from simple vector addition, we know that,
$F_{R} = F_{1} + F_{2}$ --- (A)

Where $F_{R}$ = Resultant vector.
NOTE: In equation (A), all forces are in vector notation. Assume that there is an arrow head on top of them.

Let us find $F_{1}+F_{2}$ first!
$F_{1}+F_{2}$ =   $7i+(15*cos \alpha)i + (15*sin \alpha )j$

=> $F_{1}+F_{2}$ =   $(7+15*cos \alpha)i + (15*sin \alpha )j$

Now the magnitude of $F_{1}+F_{2}$ is,
| $F_{1}+F_{2}$ | = $\sqrt{ (7+ 15*cos \alpha)^{2} + (15*sin \alpha )^{2}}$

=> | $F_{1}+F_{2}$ | = $\sqrt{ 49 + 225*(cos \alpha)^{2} + 210*(cos \alpha)+ 255*(sin \alpha )^{2}}$

Since $(sin \alpha)^{2} + (cos \alpha)^{2} = 1$ , therefore,

=> | $F_{1}+F_{2}$ | = $\sqrt{ 49 + 225 + 210*(cos \alpha)}$

Since  | $F_{1}+F_{2}$ | = | $F_{R}$ |, and the magnitude of the resultant force is 20N, therefore,

| $F_{R}$ | = | $F_{1}+F_{2}$ |
20 = $\sqrt{ 49 + 225 + 210*(cos \alpha)}$

Take square on both sides,
400 = $49 + 225 + 210*(cos \alpha)$
$(cos \alpha) = \frac{3}{5}$

$\alpha = 53.13^{o}$

Ans: Angle formed by the two forces, 7N and 15N, is: 53.13°

-israr

4 0

3 years ago