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Pavel [41]
3 years ago
6

2

Engineering
1 answer:
aksik [14]3 years ago
5 0

Answer:tech A

Explanation:

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Explain how smart materials can be used by manufacturers to improve health and safety for children's products and goods.​
Ierofanga [76]

...simplify devices, reducing weight and the chance of failure.

6 0
1 year ago
What do you think of this schematic diagram?​
Sholpan [36]

Explanation:

A schematic diagram is a picture that represents the components of a process, device, or other object using abstract, often standardized symbols and lines. ... Schematic diagrams do not include details that are not necessary for comprehending the information that the diagram was intended to convey.

5 0
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}

\displaystyle \theta_2=\frac{180^o-asin\left(\frac{R.g}{V_o^2}\right)}{2}

Or equivalently:

\theta_2=90^o-\theta_1

Given Vo=37 m/s and R=70 m

\displaystyle \theta_1=\frac{asin\left(\frac{70\times 9.8}{37^2}\right)}{2}

\theta_1=15^o

And

\theta_2=90^o-15^o=75^o

5 0
3 years ago
Kerosene flows through 3/4 standard type K drawn copper tube. The pressure drop measured at two points 50 m apart is 130 kPa. De
Anettt [7]

Answer:

Q=4.98\times 10^{-3}\ m^3/s

Explanation:

Given that

L= 50 m

Pressure drop = 130 KPa

For Copper tube is 3/4 standard type K drawn tube

Outside diameter=22.22 mm

Inside diameter=18.92 mm

Dynamic viscosity for kerosene

\mu =0.00164\ Pa.s

Pressure difference given as

\Delta P=\dfrac{128\mu QL}{\pi d_i^4}

Where

L is length of tube

μ is dynamic viscosity

Q is volume flow rate

d is inner diameter of tube

ΔP is pressure drop

Now by putting the values

\Delta P=\dfrac{128\mu QL}{\pi d_i^4}

130\times 1000=\dfrac{128\times 0.00164\times 50\times Q}{\pi\times 0.0189^4}

Q=4.98\times 10^{-3}\ m^3/s

So flow rate is Q=4.98\times 10^{-3}\ m^3/s

7 0
3 years ago
When the Challenger spacecraft exploded, which design step had not been thoroughly explored?
Llana [10]

Answer:

The escape systems not working

8 0
2 years ago
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