The average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.
<h3>How to calculate the Average speed?</h3>
The average speed can be calculated by adding the speed of each trial divided by the number of trials,
For 0.25 m the average speed will be:

For the 0.50 m, the average speed will:

Therefore, the average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.
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Answer:
6 m/s
Explanation:
12m / 2s = 6 m/s
Hope that's the answer you seek.
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The final velocity is 2.7 m/s
Explanation:
We can solve this problem by using the principle of conservation of momentum: in fact, in absence of external forces, the total momentum of the system must be conserved before and after the collision.
Therefore we can write:
where:
is the mass of the putty
is the initial velocity of the putty (we take its direction as positive direction)
is the mass of the ball
is the initial velocity of the ball (at rest)
is the final combined velocity of the two putty+ball
Re-arranging the equation and substituting the values, we find the final combined velocity:
And the positive sign indicates their final direction is the same as the initial direction of the putty.
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Recall the definition of the cross product with respect to the unit vectors:
i × i = j × j = k × k = 0
i × j = k
j × k = i
k × i = j
and that the product is anticommutative, so that for any two vectors u and v, we have u × v = - (v × u). (This essentially takes care of part (b).)
Now, given a = 8i + j - 2k and b = 5i - 3j + k, we have
a × b = (8i + j - 2k) × (5i - 3j + k)
a × b = 40 (i × i) + 5 (j × i) - 10 (k × i)
… … … … - 24 (i × j) - 3 (j × j) + 6 (k × j)
… … … … + 8 (i × k) + (j × k) - 2 (k × k)
a × b = - 5 (i × j) - 10 (k × i) - 24 (i × j) - 6 (j × k) - 8 (k × i) + (j × k)
a × b = - 5k - 10j - 24k - 6i - 8j + i
a × b = -5i - 18j - 29k