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nalin [4]
2 years ago
6

How much horse power does a Lamborghini have

Engineering
2 answers:
4vir4ik [10]2 years ago
5 0

Answer:

  • Countach - 375
  • Huracan - 610 to 630
  • Aventador -  729 to 759
  • Urus - 641
  • Gallardo - 543 to 562
  • Centenario - 770
  • SCV12 - 830

Explanation:

It really all depends, it varies from 375 to 830, you can't mark one as " Lamborghinis have this much hp always " seeing it fluctuates so much car to car

statuscvo [17]2 years ago
4 0
The Lamborghini SCV12 has 830 horse power.
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2 years ago
What flow type occurs when the heat exchanger has multiple passes and baffles?
Vanyuwa [196]
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4 0
3 years ago
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
A 1-w, 350-ω resistor is connected to 24 v. Is this resistor operating within its power rating?
seraphim [82]

Answer:

No.

Explanation:

P_r = Power rating = 1 W

R = Resistance = 350\ \Omega

V = Voltage = 24\ \text{V}

Power is given by

P=\dfrac{V^2}{R}\\\Rightarrow P=\dfrac{24^2}{350}\\\Rightarrow P=1.65\ \text{W}

1.65\ \text{W}>1\ \text{W}

So

P>P_r

Hence, the resistor is not operating within its power rating.

8 0
2 years ago
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