Question 8

A circular hole in an aluminum plate is 2.739 cm in diameter at 0.000°C. What is the change in its diameter when the temperature

prisoha [69]

Answer:

$L = 2.746 cm$

Explanation:

As we know that thermal expansion coefficient of aluminium is given as

$\alpha = 24 \times 10^{-6} per ^oC$

now we also know that after thermal expansion the final length is given as

$L = L_o(1 + \alpha \Delta T)$

here we know that

$L_o = 2.739 cm$

$\alpha = 24 \times 10^{-6}$

$\Delta T = 108 - 0= 108^oC$

now we will have

$L = 2.739(1 + 24 \times 10^{-6} (108))$

$L = 2.746 cm$

3 0

3 years ago

The electrostatic attractions in crystalline salts, such as sodium chloride, are formed by __________ between positive and negat

nikklg [1K]

D transferring electrons because that causes electricity

8 0

3 years ago

Read 2 more answers

I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block

borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

$T = 2\pi \sqrt{\frac{m}{k} }$

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

$T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\$

Now, determine the amplitude of oscillation, A;

$E = \frac{1}{2} kA^2$

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

$E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm$

Therefore, the amplitude of the oscillation is 2.82 cm

8 0

3 years ago

Does research lessen its value for its being cyclical? Enumerate Two Possible advantages of a cyclical research

Tanya [424]

Answer:

Research usually is a cyclical process because it starts with a problem and ends with a problem.

This is bad or lessens the value of the research?

No, because there are problems that in nature are cyclical, (for example the related ones to action research).

And it also may be a good thing, solving one "problem" leads to another problem, but in the process of solving the first one we may win a lot of knowledge, and the same happens with the next one, and so on.

Two possible advantages of cyclical research are:

Knowing beforehand that the research will be cyclical, will allow us to estimate better the amount of time and money needed because we already know (more or less) where to aim.

It also may lead to a better end product, as we already know that we must focus in solving one thing and then we can focus in the next one.

Another possible advantage may be that we know that after the work, there will be a new thing to research, and it is fun, so if you are curious enough this may be a good thing (especially in scientific areas, physics, chemistry, etc).

7 0

3 years ago

A rifle fires a bullet at a velocity of 78 m/s, 40 degrees above the horizontal. Determine the height (h) above the starting pos

qwelly [4]

Answer:

H = Vy t - 1/2 g t^2 height of an object with an initial "vertical" velocity

at t sec after firing

Vy = 78 m/s * sin 40 = .643 * 78 m/s = 50.1 m/s

H = 50.1 * 6 - 1/2 * 9.8 * 6^2 = 300 m - 176 m = 124 m

7 0

3 years ago