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den301095 [7]
3 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]3 years ago
8 0

Try this solution, answers are marked with red colour.

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Which of the following is the best example of a primary circular reaction?
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Primary Circular Reactions (1-4 months): This substage involves coordinating sensation and new schemas. For example, a child may suck his or her thumb by accident and then later intentionally repeat the action. These actions are repeated because the infant finds them pleasurable.

4 0
3 years ago
A trip is taken that passes through the following points in order
IgorLugansk [536]

Answer:

The displacement from point B to point E is 25.0 m left

8 0
3 years ago
Find the sum 5.24 g, 43.261 g, and 7.3458 g. Write your answer with the correct amount of significant figures.
lapo4ka [179]

Answer:

This is how I do it:

  • 5.24 rounded to one significant figure is 5
  • 43.261 rounded to one significant figure is 40.
  • 7.3458 rounded to one significant figure is 7.
  • 5 + 40 + 7 is 52 g

Hope this helps you.

Explanation:

7 0
3 years ago
An oil droplet is sprayed into a uniform electric field of adjustable magnitude. The 0.11 g droplet hovers
ohaa [14]

Answer:

The direction of the field is downward, and negatively charged particles will experience an upwards force due to the field.

F = N e E     where E is the value of the field and N e the charge Q

M g = N e E      and M g is the weight of the drop

N = M g / (e E)

N = 1.1E-4 * 9.8 / (1.6E-19 * 370) = 1.1 * 9.8 / (1.6 * 370) * E15 = 1.82E13

.00011 kg is a very large drop

Q = N e = M g / E = .00011 * 9.8 / 370 = 2.91E-6 Coulombs

Check:     N = Q / e = 2.91E-6 / 1.6E-19 = 1.82E13   electrons

7 0
2 years ago
Suppose that a pane of crown glass, with a refractive index of 1.52, is immersed in water, which has a refractive index of 1.33.
mylen [45]

Answer:

the angle made by this ray with normal will be 45°

Explanation:

given,                                                      

refractive index of crown glass = 1.52

refractive index of the water = 1.33        

ray of light in water strikes glass at 45°                      

angle subtended by the light when it reenter water = ?                  

When light enter in the glass from the water it get bend toward normal because refractive index of glass is more than water.              

And when ray comes out of the glass it is parallel to the initial light ray.

hence, the angle made by this ray with normal will be 45°

3 0
3 years ago
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