Find the derivative of x ​

Answer:

<:

Explanation:

Answer:

d

Explanation:

Answer:

20368.917N

Explanation:

Frictional force (F) is the product of the Coefficient of friction and the normal reaction.

F = μN

Coefficient of friction, μ = 0.2

Normal reaction = MgCosθ

Mass, m = 12 tonnes = 12 * 1000 = 12000 kg

N = 12000 * 9.8 * cos30

N = 101844.58

F = 0.2 * 101844.58

F = 20368.917N

Answer:

a.

y[n] = x[n] x[n-1]  x[n+1]

(i) Memory-less - It is not memory-less because the given system is depend on past or future values.

(ii) Causal - It is non-casual because the present value of output depend on the future value of input.

(iii) Invertible - It is invertible and the inverse of the given system is [tex]\frac{1}{x[n] . x[n-1] x[n+1]}[/tex]

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.

b.

y[n] = cos(x[n])

(i) Memory-less - It is memory-less because the given system is not depend on past or future values.

(ii) Causal - It is casual because the present value of output does not depend on the future value of input.

(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .

For example - for x[n] = 0 , y[n] = cos(0) = 1

                       for x[n] = 2[tex]\pi[/tex] , y[n] = cos(2[tex]\pi[/tex]) = 1

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.

Answer:

answer is option (d) all of the above

Solution :

Finding the cohesion of the soil(c) using the relation:

[tex]$c = \frac{q_u}{2}$[/tex]

Here, [tex]$q_u$[/tex] is the unconfined compression strength of the soil;

[tex]$c = \frac{800}{2}$[/tex]

   = 400 psf

∴ The cohesion value is greater than 0

So the use of the angle of internal friction is 0

Referring to the table relation between bearing capacity factors and angle of internal friction.

For the angle of inter friction [tex]$0^\circ$[/tex]

    [tex]$N_c = 5.14$[/tex]

   [tex]$N_q = 1.0$[/tex]

   [tex]$N_r = 0$[/tex]

Therefore,

[tex]$q_{ult} = (400 \times 5.14 )+(110 \times 3 \times 1.0)+(0.5 \times 100 \times 13 \times 0)$[/tex]

     =  2386 psf

∴ Allowable bearing capacity [tex]$q_{a} = \frac{Q_{allow}}{A}$[/tex]

                                                     [tex]$=\frac{30}{B^2}$[/tex]

∴ [tex]$q_a = \frac{q_{ult}}{F.O.S}$[/tex]

  [tex]$\frac{30}{B^2} = \frac{2386}{3}$[/tex]

∴ B = 0.2 ft

Therefore, the dimension of the square footing is 0.2 ft x 0.2 ft

                                                                                 [tex]$=0.04 \ ft^2$[/tex]

Answer:

Explanation:

A scenario is an actual sequence of interactions (i.e., an instance) describing one specific situation; a use case is a general sequence of interactions (i.e., a class) describing all possible scenarios associated with a situation. Scenarios are used as examples and for clarifying details with the client. Use cases are used as complete descriptions to specify a user task or a set of related system features.

Answer:

Both

Explanation:

Answer

The skateboarder applies pressure to the trucks and gives/releases pressure on the levers. Second, the wheels and the axles are also examples of simple machines. They help the skater ride, spin, grind, and do a bunch of other radical movements on a skateboard.:

Explanation:

Solution :

Assuming that the wire has an uniform temperature, the equivalent convective heat transfer coefficient is given as :

[tex]$h_T= \epsilon \sigma (T_s+T_{surr})(T_s^2 +T^2_{surr})$[/tex]

[tex]$h_T= 0.20 \times 5.67 \times 10^{-8} (1473+323)(1473^2 +323^2)$[/tex]

[tex]$h_T=46.3 \ W/m^2 .K$[/tex]

The total heat transfer coefficient will be :

[tex]$h_T=(250+46.3) \ W/m^2 .K$[/tex]

    [tex]$=296.3 \ W/m^2 .K$[/tex]

Now calculating the maximum volumetric heat generation :

[tex]$\dot {q}_{max}=\frac{2h_t}{r_0}(T_s-T_{\infty})$[/tex]

[tex]$\dot {q}_{max}=\frac{2\times 296.3}{0.0005}(1200-50)$[/tex]

        [tex]$= 1.362 \times 10^{9} \ W/m^3$[/tex]

The heat generation inside the wire is given as :

[tex]$\dot{q} = \frac{I^2R}{V}$[/tex]

Here, R is the resistance of the wire

         V is the volume of the wire

∴ [tex]$\dot{q} = \frac{I^2\left( \rho \times \frac{L}{A} \right)}{A \times L}$[/tex]

     [tex]$=\frac{I^2 \rho}{\left(\frac{\pi}{4}D^2 \right)}$[/tex]

where, ρ is the resistivity.

[tex]$I_{max}= \left(\frac{\dot{q}_{max}}{\rho} \right)^{1/2} \times \frac{\pi}{4}D^2$[/tex]

[tex]$I_{max}= \left(\frac{1.36 \times 10^9}{10^{-6}} \right)^{1/2} \times \frac{3.14}{4}(1 \times 10^{-3})^2$[/tex]

        = 28.96 A

Now considering the relation for the current flow through the finite potential difference.

[tex]$E=I_{max} \times R$[/tex]

[tex]$E=I_{max} \times \rho \times \frac{L}{A}$[/tex]

[tex]$L=\frac{AE}{I_{max} \ \rho}$[/tex]

[tex]$L=\frac{\frac{\pi}{4} \times (1 \times 10^{-3})^2 \times 110}{28.96 \times 10^{-6}}$[/tex]

   = 2.983 m

Now calculating the power rating of the heater:

[tex]$P= E \times I_{max}$[/tex]

[tex]$P= 110 \times 28.96}$[/tex]

   = 3185.6 W

   = 3.1856 kW

Answer:

AC

Explanation:

One situation when alternating current would work better than direct current is if the operator is encountering magnetic arc blow.

Current works best when the operator  encounters magnetic arc blow is AC

Magnetic arc blow is simply defined as the arc deflection due to the warping of the magnetic field that is produced by electric arc current.

This is caused as a result of the following;

- if the material being welded has residual magnetism at an intolerable level

- When the weld root is being made, and the welding current is direct current which indicates constant direction and maintains constant polarity (either positive or negative).

Thus, Current works best when the operator  encounters magnetic arc blow is AC

Read more at; brainly.in/question/38789815?tbs_match=1