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lozanna [386]
3 years ago
11

How does a catapult incorporate technology?

Engineering
1 answer:
storchak [24]3 years ago
6 0
I’m sure the answer is correct and it is below
“ A catapult works because energy can be converted from one type to another and transferred from one object to another. ... This energy is stored in the launching device as potential, or stored, energy. The catapult you are about to make uses elastic potential energy stored in a wooden stick as you bend it.”
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How is the difference between science and engineering Best stated?
stiv31 [10]

Answer:Science is the body of knowledge that explores the physical and natural world. Engineering is the application of knowledge in order to design, build and maintain a product or a process

Explanation:

8 0
2 years ago
HELP PLEASE!!!!!!!!!!!
MAXImum [283]
C is your answers!!!!!$3&2)//
7 0
3 years ago
Read 2 more answers
For a cylindrical annulus whose inner and outer surfaces are maintained at 30 ºC and 40 ºC, respectively, a heat flux sensor mea
miskamm [114]

Answer:

k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

where r_1 and r_2 be the inner radius, outer radius of the annalus.

Explanation:

Let r_1, r_2 and L be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, T_1=30^{\circ}C\\ and at the outer surface, T_2=40^{\circ}C.

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

40 W/m^2.

\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40

\Rightarrow q=2.4\pi L \;W

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

q=-kA\frac{dT}{dr}\;\cdots(i)

where,

K=Thermal conductivity of the material.

T= temperature at any radial distance r.

A=Area through which heat transfer is taking place.

Here, A=2\pi rL\;\cdots(ii)

Variation of temperature w.r.t the radius of the annalus is

\frac {T-T_1}{T_2-T_1}=\frac{\ln(r/r_1)}{\ln(r_2/r_1)}

\Rightarrow \frac{dT}{dr}=\frac{T_2-T_1}{\ln(r_2/r_1)}\times \frac{1}{r}\;\cdots(iii)

Putting the values from the equations (ii) and (iii) in the equation (i), we have

q=\frac{2\pi kL(T_1-T_2)}{\LN(R_2/2_1)}

\Rightarrow k= \frac{q\ln(r_2/r_1)}{2\pi L(T_2-T_1)}

\Rightarrow k=\frac{(2.4\pi L)\ln(r_2/r_1)}{2\pi L(10)} [as q=2.4\pi L, and T_2-T_1=10 ^{\circ}C]

\Rightarrow k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.

7 0
3 years ago
If anyone knows manufacturing plz help
sveta [45]

Answer:

I don't know ask my dad he would

Explanation:

but I can't ask him because he went to get milk and forgot to come back

8 0
3 years ago
The modulus of elasticity for a ceramic material having 6.0 vol% porosity is 303 GPa. (a) Calculate the modulus of elasticity (i
Phantasy [73]

Answer:

modulus of elasticity for the nonporous material is 340.74 GPa

Explanation:

given data

porosity = 303 GPa

modulus of elasticity = 6.0

solution

we get here  modulus of elasticity for the nonporous material Eo that is

E = Eo (1 - 1.9P + 0.9P²)    ...............1

put here value and we get Eo

303 = Eo ( 1 - 1.9(0.06) + 0.9(0.06)² )  

solve it we get

Eo = 340.74 GPa

8 0
3 years ago
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