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den301095 [7]
4 years ago
15

In a bicycle race on a straight road, the leader is 72.0 m from the finish line and continues to travel to the finish line with

a constant speed of 12 m/s. When the leader is 72.0 m from the finish line, the second-place rider is 5.0 m behind him and traveling with a speed of 11.0 m/s when she decides to accelerate, keeping a constant acceleration for the rest of the race. The two riders end up crossing the finish line at exactly the same time.(a) Calculate the value of the second rider?s constant acceleration that allows her to cross the finish line at exactly the same time as the first rider.m/s2(b) What is the speed of the second rider as she crosses the finish line?m/s(c) What is the second rider's average speed over the final 77 m of the race?m/s

Physics
1 answer:
dem82 [27]4 years ago
8 0

Try this solution, answers are marked with red colour.

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An object has a mass of 10.2 kg, what is its weight in Newton's?
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The answer is 98 N

Explanation:

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Help me with this question Please
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Answer:

W has the lowest density and Y has the greatest density

Explanation:

Density of W = mass/volume = 11/24 = 0.45

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From these we can find the answer......

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3 years ago
What is the temperature 32 degrees Fahrenheit in degrees Celsius? A. 0°C B. 20°C C. 10°C D. –10°C
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The answer would 0. The reasoning of this is because freezing point in celsius is always 0 degrees but in fahrenheit the freezing point is 32 degrees.
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3 years ago
Please help with this question.
EastWind [94]

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41.16 Joules

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4 0
3 years ago
A plastic rod 1.3 m long is rubbed all over with wool, and acquires a charge of -3e-08 coulombs. we choose the center of the rod
Anestetic [448]
(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
L/8=1.3m/8=0.1625 m

(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:
c_6 = 0.0812m+0.1625m=0.2437 m

(c) The total charge is Q=-3 \cdot 10^{-8}C. To get the charge on each piece, we should divide this value by 8, the number of pieces:
Q/8=-3\cdot 10^{-8}C/8=-3.75\cdot 10^{-9}C

(d) We have to calculate the electric field at x=0.7 generated by piece 6. The charge on piece 6 is the value calculated at point (c):
q= -3.75\cdot 10^{-9}C
If we approximate piece 6 as a single  charge, the electric field is given by
E=k_e  \frac{q}{d^2}
where k_e=8.99\cdot 10^9Nm^2C^{-2} and d is the distance between the charge (center of piece 6, located at 0.2437m) and point a (located at x=0.7m). Therefore we have
E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m
poiting towards the center of piece 6, since the charge is negative.

(e) missing details on this question.
5 0
3 years ago
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