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3 years ago

10 properties of metals?

Engineering

2 answers:

Mrrafil [7]3 years ago

6 0

Answer:

1. they are malleable

2. they are ductile

3. they are good conductor of electricity and heat

4. they are solid at room temperature except mercury

5. they are sonorous

6. they are hard

7. they can be hammered into small sheets

8. they are lustrous

9. they are strong

10. they have high density

Misha Larkins [42]3 years ago

4 0

Answer:

Physical Properties of Metals:

Metals can be hammered into thin sheets. ...

Metals are ductile. ...

Metals are a good conductor of heat and electricity.

Metals are lustrous which means they have a shiny appearance.

Metals have high tensile strength. ...

Metals are sonorous. ...

Metals are hard.

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If particleboard is used as wall sheathing, the grade mark with type _____ or _____ should be stamped on it.

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2 years ago

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3 years ago

Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you

viktelen [127]

Answer:

$\theta_1=15^o\\\theta_2=75^o$

Explanation:

<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

$V_x=V_{ox}=V_ocos\theta$

Where $\theta$ is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

$V_y=V_{oy}-gt=V_osin\theta-gt$

The horizontal and vertical distances are, respectively:

$x=V_{o}cos\theta t$

$\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}$

The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

$\displaystyle t_f=\frac{2V_osin\theta}{g}$

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

$\displaystyle R=\frac{V_o^2sin2\theta}{g}$

Let's solve for $\theta$

$\displaystyle sin2\theta=\frac{R.g}{V_o^2}$

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

$\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}$