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Irina-Kira [14]
2 years ago
15

What is the acceleration of a Porsche that can go from 15 mi/hr to 75 mi/hr in 4 seconds?

Physics
2 answers:
Elodia [21]2 years ago
8 0

Hi there!

Acceleration = change in velocity / change in time = Δv/Δt

Thus:

a = (75 - 15)/4 = 60/4 = 15 mi/hr²

AveGali [126]2 years ago
6 0
Equation you need to use is a=Vf-Vi/t
Vi=initial velocity
Vf= Final Velocity
Write out the variables and their numbers
Vf=75 mi/hr Vi=15 mi/hr t=4 seconds
Plug it all in
75-15/4
60/4
a=15 mi/hr^2
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A Christmas light is made to flash via the discharge of a capacitor. The effective duration of the flash is 0.25 s (which you ca
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Answer:

The correct solution is:

(a) 1.375\times 10^{-2} \ J

(b) 4.43\times 10^{-3} \ C

(c) 1.42\times 10^{-3} \ F

(d) 178.57 \ \Omega

Explanation:

The given values are:

Effective duration of the flash,

ζ = 0.25 s

Average power,

P_{avg}=55 \ mW

       =55\times 10^{-3} \ W

Average voltage,

V_{avg}=3.1 \ V

Now,

(a)

⇒ E=P_{avg}\times \zeta

On substituting the values, we get

⇒     =55\times 10^{-3}\times 0.25

⇒     =1.375\times 10^{-2} \ J

(b)

⇒ E=Q\times V_{avg}

then,

⇒ Q=\frac{E}{V_{avg}}

On substituting the values, we get

⇒     =\frac{1.375\times 10^{-2}}{3.1}

⇒     =4.43\times 10^{-3} \ C

(c)

⇒ C=\frac{Q}{V}

⇒     =\frac{4.43\times 10^{-3}}{3.1}

⇒     =1.42\times 10^{-3} \ F

(d)

As we know,

⇒ R=\frac{1}{4C}

⇒     =\frac{1}{4\times 1.42\times 10^{-3}}

⇒     =\frac{1000}{5.6}

⇒     =178.57 \ \Omega

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2 years ago
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Andre45 [30]
D.<span>Wave 3 resulted from constructive interference, and Wave 4 resulted from destructive interference.

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Answer:

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A particular circuit toolbox contains only 80 Ω resistors and switches, which can be open or closed. Construct a circuit, fillin
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Answer:

Here we need to make parallel connection of two 80 ohm resistors to achieve 40 ohm net resistance.

Explanation:

As we know that the resistances in series add up directly and here we are given with only the resistors of 80 Ω.

So when we connect two resistors of 80 ohm in parallel we get the resultant of 40 ohm.

Mathematically:

\frac{1}{R_p} =\frac{1}{R} +\frac{1}{R}

\frac{1}{40} =\frac{1}{R} +\frac{1}{R}

\frac{1}{40} =\frac{2}{R}

R=80\Omega gives us the only combination of two resistors in parallel.

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