The blade tension is correct when you can hear a<br>O Thunk<br>O Twang<br>O Neither

Disadvantages of bezier curve and b-spline

Gnesinka [82]

Answer:

The main disadvantage of the Bezier-curves is the global influence of the control points on the whole curve.

This has two major drawbacks: (1) every change of a control point (insert, move, delete) changes the look of the curve in all points of the curve, and (2) the computation time needed for a big set of control points is comparatively high. The reason is the weighting-function form.

hope it helps,pls mark me as brainliest

5 0

3 years ago

A thick aluminum block initially at 26.5°C is subjected to constant heat flux of 4000 W/m2 by an electric resistance heater whos

Yanka [14]

Given Information:

Initial temperature of aluminum block = 26.5°C

Heat flux = 4000 w/m²

Time = 2112 seconds

Time = 30 minutes = 30*60 = 1800 seconds

Required Information:

Rise in surface temperature = ?

Answer:

Rise in surface temperature = 8.6 °C after 2112 seconds

Rise in surface temperature = 8 °C after 30 minutes

Explanation:

The surface temperature of the aluminum block is given by

$T_{surface} = T_{initial} + \frac{q}{k} \sqrt{\frac{4\alpha t}{\pi} }$

Where q is the heat flux supplied to aluminum block, k is the conductivity of pure aluminum and α is the diffusivity of pure aluminum.

After t = 2112 sec:

$T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (2112)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.51098)\\\\T_{surface} = 26.5 + 8.6\\\\T_{surface} = 35.1\\\\$

The rise in the surface temperature is

Rise = 35.1 - 26.5 = 8.6 °C

Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.

After t = 30 mins:

$T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (1800)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.4717)\\\\T_{surface} = 26.5 + 7.96\\\\T_{surface} = 34.5\\\\$

The rise in the surface temperature is

Rise = 34.5 - 26.5 = 8 °C

Therefore, the surface temperature of the block will rise by 8 °C after 30 minutes.

5 0

3 years ago

People tend to self-disclose to others that are in age, social status, religion, and personality.

pickupchik [31]

Answer:

??????yeh

Explanation:

8 0

3 years ago

Read 2 more answers

Consider a junction that connects three pipes A, B and C. What can we say about the mass flow rates in each pipe for steady flow

Elis [28]

Answer:

The statement regarding the mass rate of flow is mathematically represented as follows $\Rightarrow \rho \times Q_{3}=\rho \times Q_{1}+\rho \times Q_{2}$

Explanation:

A junction of 3 pipes with indicated mass rates of flow is indicated in the attached figure

As a basic sense of intuition we know that the mass of the water that is in the pipe junction at any instant of time is conserved as the junction does not accumulate any mass.

The above statement can be mathematically written as

$Mass_{Junction}=Constant\\\\\Rightarrow Mass_{in}=Mass_{out}$

this is known as equation of conservation of mass / Equation of continuity.

Now we know that in a time 't' the volume that enter's the Junction 'O' is

1) From pipe 1 = $V_{1}=Q_{1}\times t$

1) From pipe 2 = $V_{2}=Q_{2}\times t$

Mass leaving the junction 'O' in the same time equals

From pipe 3 = $V_{3}=Q_{3}\times t$

From the basic relation of density, volume and mass we have

$\rho =\frac{mass}{Volume}$

Using the above relations in our basic equation of continuity we obtain

$\rho \times V_{3}=\rho \times V_{1}+\rho \times V_{2}\\\\Q_{3}\times t=Q_{1}\times t+Q_{2}\times t\\\\\Rightarrow Q_{3}=Q_{1}+Q_{2}$

Thus the mass flow rate equation becomes $\Rightarrow \rho \times Q_{3}=\rho \times Q_{1}+\rho \times Q_{2}$

6 0

4 years ago

A pool of contaminated water is lined with a 40 cm thick containment barrier. The contaminant in the pit has a concentration of

konstantin123 [22]

This question is incomplete, the complete question is;

A pool of contaminated water is lined with a 40 cm thick containment barrier. The contaminant in the pit has a concentration of 1.5 mol/L, while the groundwater circulating around the pit flows fast enough that the contaminate concentration remains 0. There is initially no contaminant in the barrier material at the time of installation. The governing second order, partial differential equation for diffusion of the contaminant through the barrier is:

dC/dt = D( d²C / dz²)

where c(z,t) represent the concentration of containment of any depth into the barrier at anytime and D is the diffusion coefficient (a constant) for the containment in the barrier material.

a) write all boundary and initial conditions needed to solve this equation for C(z, t)

b) Find the steady state solution (infinite time) for C(z)

Answer:

a) At t = 0, z= 0, c = 1.5 mol/L

at t =0, z = 0.4m, c = 0 mol/L

b) C(z) = z² - 4.15z + 1.5

Explanation:

a)

The boundary and initial conditions are as follows

At t = 0, z= 0, c = 1.5 mol/L

at t =0, z = 0.4m, c = 0 mol/L

b)

The governing second order, partial differential equation for diffusion of the contaminant through the barrier is :

(dC/dt) = D*(d²C/dz²) ..............equ(1)

For steady state, above equation becomes,

(d²C/dz²) =0

Integrating above equation,

(dC/dz) = Z + C1 { where C1 is integration constant) }

again integrating above equation,

C = z² + C1*z + C2 ...................equ(2)

applying boundary condition : at t =0, z= 0, c = 1.5 mol/L, to above equation

C = z² + C1*z + C2

1.5 = 0 + 0*0 + c2

C2 = 1.5

applying boundary condition : at t =0, z= 0.4m, c = 0 mol/L, to equation (2) ,

0 = 0.4² + C1*0.4 + 1.5

0 = 0.16 + 0.4C1 + 1.5

0.4C1 = - 1.66

C1 = -1.66/0.4

C1 = -4.15

So, the steady state solution for C(z) is:

C(z) = z² - 4.15z + 1.5

6 0

3 years ago

The blade tension is correct when you can hear aO ThunkO TwangO Neither​

The blade tension is correct when you can hear aO ThunkO TwangO Neither