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mezya [45]
2 years ago
7

A dragster race car can accelerate from rest to incredible speeds. In one case a dragster is able to finish the 305 m run in 3.6

4 s. What was the average acceleration during this run?
Physics
1 answer:
Dafna1 [17]2 years ago
4 0

Answer:

45.89m/s²

Explanation:

Given

Distance S = 305m

Time t = 3.64s

To get the acceleration during this run, we will apply the equation of motion:

S = ut+1/2at²

Substitute the given parameters into the formula and calculate the value of a

305 = 0+1/2 a(3.64)²

304 = 1/2(13.2496)a

304 = 6.6248a

a = 304/6.6248

a = 45.89m/s²

Hence the average acceleration during this run is 45.89m/s²

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ehidna [41]
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Describe the full water cycle
sergey [27]

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5 0
3 years ago
Compute the expected shell-model quadrupole moment of 209Bi () and compare with the experimental value, - 0.37 b
Over [174]

Answer:

0.22 b

Explanation:

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Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2}

And also,

R^{2}=R^{2} _{0}A^{\frac{2}{3} }

And, R _{0}=1.2\times 10^{-15}m

Now,

Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2} _{0}A^{\frac{2}{3} }

For Bismuth j=\frac{9}{2} and A is 209.

Q=-\frac{2\frac{9}{2} -1}{2(\frac{9}{2} +1)}\frac{3}{5}(1.2\times 10^{-15}) ^{2}(209)^{\frac{2}{3} }\\Q=0.628\times 35.28\times 10^{-30} \\Q=22.15\times 10^{-30} m^{2} \\Q=0.2215\times 10^{-28} m^{2} \\Q=0.22 barn

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3 0
3 years ago
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vodomira [7]

Answer:

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Explanation:

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         a = dv / dt

they give us the function of speed

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         a = - sin (t²/2) -  (t-1) cos (t²/2)  2t / 2

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the acceleration for t = 4 s

          a = - sin (4²/2) - 4 (4-1) cos (4²/2)

          a = -sin 8 - 12 cos 8

remember that the angles are in radians

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the problem does not indicate the units, but to be correct they must be m/s²

We see that the acceleration is positive therefore the speed is increasing

6 0
2 years ago
Imagine that you drop an object of 10 kg, how much will be the acceleration and
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4 0
2 years ago
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