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Basile [38]
2 years ago
7

A car starts from rest and moves along the positive x-axis with a constant acceleration of 5 m/s/s for 8 seconds. It then contin

ues with constant velocity, i.e. the velocity it attains at 8 seconds for the next 5 seconds. What will be the distance covered by the car in the first 12 seconds of its motion?
Physics
2 answers:
boyakko [2]2 years ago
8 0
Answer is 340 plz believe
tino4ka555 [31]2 years ago
5 0

Answer:

340

Explanation:

Because you are moving at 5 m/s^2 for 8 seconds, you need to do some arithmetic to get the distance covered in the beginning. TO do this, you do 5 + 5*2 + 5*3 + 5*4 + 5*5 + 5*6 + 5*7 + 5*8, and once you have passed eight seconds, you need 4 more seconds with constant velocity, or 4(5*8). Your final equation would be 5+10+15+20+25+30+35+40+40+40+40+40. Hope this helps!

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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
Sonbull [250]

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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<span><span>D.</span><span>Measurements are taken in a way that is the same every time.- apex

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The line graph below shows the number of downloads of two songs after
daser333 [38]

Answer: D

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Explanation:

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