How many GT2RS cars were made in 2019

Determine the period of each of the following discrete-time signals (if a signal is not periodic, denote its period by infinity)

sergiy2304 [10]

Answer:

a) it is periodic

N = (20/3)k = 20 { for K =3}

b) it is Non-Periodic.

N = ∞

c) x(n) is periodic

N = LCM ( 5, 20 )

Explanation:

We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.

then the period of the signal is given as

N = ( 2π/w₀)K

k is least integer for which N is also integer

Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2

now

a) cos(2π(0.15)n)

w₀ = 2π(0.15)

Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3

so, it is periodic

N = (20/3)k = 20 { for K =3}

b) cos(2n);

w₀ = 2

Now, 2π/w₀ = 2π/2) = π

so, it is Non-Periodic.

N = ∞

c) cos(π0.3n) + cos(π0.4n)

x(n) = x1(n) + x2(n)

x1(n) = cos(π0.3n)

x2(n) = cos(π0.4n)

so

w₀ = π0.3

2π/w₀ = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3

∴ N1 = 20

AND

w₀ = π0.4

2π/w₀ = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5

∴ N² = 5

so, x(n) is periodic

N = LCM ( 5, 20 )

6 0

3 years ago

Two resistors, A and B, individually connect to a 9V battery. A student notices that resistor A is warmer than resistor B. Which

dybincka [34]

Answer:

Resistor B

Explanation:

Since resistance is the opposition to the flow of current in a circuit,

first let assume the two resistors are connected in parallel to the voltage, recall that when connection is in parallel, the different amount of current pass through the resistors depending on the value with the small resistor having a lower resistance effect hence higher current will pass through

The energy dissipated in each resistor can be calculated as

$E=\frac{1}{2}IR^{2}t$ .

from the formula we can conclude that the energy value will be higher for the resistor with small resistance value. hence more heating effect which will cause it to be warm.

Also when connected individually the current flow from the voltage source will pass through the resistor which when we calculate the energy dissipated, the resistor with smaller value will be higher because it will draw more current which will in turn lead to a heating effect and cause the resistor to be warm. Hence we can conclude that the resistance B has greatest resistance value.

4 0

2 years ago

Refrigerant-134a at 700 kPa, 70°C, and 7.2 kg/min is cooled by water in a condenser until it exists as a saturated liquid at the

alex41 [277]

Answer:

The mass flow rate of cooling water required to cool the refrigerant is $123.788\,\frac{kg}{min}$ .

Explanation:

A condenser is a heat exchanger used to cool working fluid (Refrigerant 134a) at the expense of cooling fluid (water), which works usually at steady state. Let suppose that there is no heat interactions between condenser and surroundings.The condenser is modelled after the First Law of Thermodynamics, which states:

$\dot Q_{ref} - \dot Q_{w} = 0$

$\dot Q_{ref} = \dot Q_{w}$

$\dot m_{ref}\cdot (h_{ref, in} - h_{ref,out}) = \dot m_{w}\cdot (h_{w, out} - h_{w,in})$

The mass flow rate of the cooling water is now cleared:

$\dot m_{w} = \dot m_{ref }\cdot \frac{h_{ref,in}-h_{ref,out}}{h_{w,out}-h_{w,in}}$

Given that $h_{ref,in} = 808.34\,\frac{kJ}{kg}$ , $h_{ref, out} = 88.82\,\frac{kJ}{kg}$ , $h_{w,out} = 104.83\,\frac{kJ}{kg}$ and $h_{w,in} = 62.98\,\frac{kJ}{kg}$ , the mass flow of the cooling water is:

$\dot m_{w} = \left(7.2\,\frac{kg}{min} \right)\cdot \left(\frac{808.34\,\frac{kJ}{kg}-88.82\,\frac{kJ}{kg} }{104.83\,\frac{kJ}{kg}-62.98\,\frac{kJ}{kg} } \right)$

$\dot m_{w} = 123.788\,\frac{kg}{min}$

The mass flow rate of cooling water required to cool the refrigerant is $123.788\,\frac{kg}{min}$ .

4 0

3 years ago

What are the parameters that affect life and drag forces on an aerofoil?

Vinil7 [7]

Answer:

1.The velocity of fluid

2.Fluid properties.

3.Projected area of object(geometry of the object).

Explanation:

Drag force:

Drag force is a frictional force which offered by fluid when a object is moving in it.Drag force try to oppose the motion of object when object is moving in a medium.

Drag force given as

$F_D=\dfrac{1}{2}\rho\ A\ V^2$

So we can say that drag force depends on following properties

1.The velocity of fluid

2.Fluid properties.

3.Projected area of object(geometry of the object).

6 0

3 years ago

Determine the speed of sound in air at 400 K. Also determine the Mach number of an aircraft moving in the air at a velocity of 3

Reika [66]

Answer:

$\alpha = \sqrt{1.4 *0.287 \frac{KJ}{Kg K}*\frac{1000J}{1KJ} *400 K}= 400.899 m/s$

$Ma= \frac{310 m/s}{400.899 m/s}= 0.773$

Explanation:

For this case we have given the following data:

$T= 400 K$ represent the temperature for the air

$v = 310 m/s$ represent the velocity of the air

$k = 1.4$ represent the specific heat ratio at the room

$R = 0.287 KJ/ Kg K$ represent the gas constant for the air

And we want to find the velocity of the air under these conditions.

We can calculate the spped of the sound with the Newton-Laplace Equation given by this equation:

$\alpha = \sqrt{\frac{K}{\rho}}=\sqrt{k RT}$

Where K = is the Bulk Modulus of air, k is the adiabatic index of air= 1.4, R = the gas constant for the air, $\rho$ the density of the air and T the temperature in K

So on this case we can replace and we got:

$\alpha = \sqrt{1.4 *0.287 \frac{KJ}{Kg K}*\frac{1000J}{1KJ} *400 K}= 400.899 m/s$

The Mach number by definition is "a dimensionless quantity representing the ratio of flow velocity past a boundary to the local speed of sound" and is defined as:

$Ma=\frac{v}{\alpha}$

Where v is the flow velocity and $\alpha$ the volocity of the sound in the medium and if we replace we got:

$Ma= \frac{310 m/s}{400.899 m/s}= 0.773$

And since the Ma<0.8 we can classify the regime as subsonic.

7 0

3 years ago