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cestrela7 [59]
2 years ago
5

Look at the graph above. It shows how three runners ran a 100 meter race

Physics
2 answers:
Rzqust [24]2 years ago
8 0

Answer:

1. Albert

2. Charlie

3. 05 Seconds

4. 14 seconds

5. 8.33 m/s

Explanation:

The problems in the given scenarios can be solved just by analyzing the given the graph. Here the x-axis of the graph shows the time in seconds and y-axis shows the distance covered.

1. We can see that Albert finished the race in 12 seconds. This is lesser in comparison with Bob (14 seconds) and Charlie (17 seconds). Thus Albert is the winner of the race.

2. We can see that when Charlie reached at 50 meters, the time is increasing while the distance is same. It implies that he stopped there for some rest.

3. This rest was of 05 seconds.

4. Bob completed the race in 14 seconds.

5. Albert covered 100 m in 12 seconds, so his speed (v) will be,

v = \frac{100}{12}  = 8.33 m/s

anastassius [24]2 years ago
7 0

Answer:

1. Albert

2. Charlie

3. 5 seconds

4. 14 seconds

5. 8.33... meters per second

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Answer:

B

Explanation:

The impulse experienced by an object is the force•time.

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3 years ago
A textbook of mass 2.05 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose d
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Answer:

a. 7.38 N b. 40.87 N c. 0.113 kg-m²

Explanation:

a. Let T be the tension in the cord. For the textbook, T = ma since no other force acts on it and it is an horizontal force, and m = mass = 2.05 kg and a = acceleration. We find the acceleration from s = ut + 1/2at² where u = initial speed = 0 (since it starts from rest),  s = distance moved = 1.30 m and t = time = 0.850 s.

Substituting these values into s,

1.30 m = 0 × 0.850 + 1/2a × 0.850² = 0 + 0.36125a

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Substituting this into T, we have

T = ma = 2.05 kg × 3.6 m/s² = 7.38 N

b.  Let T be the tension in the cord attached to the book. The book has the only vertical forces acting on it as the tension, T(acting upwards) and its weight mg (acting downwards). So the net force acting on it is

T - mg = ma

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α = a/r = 3.6 m/s² ÷ 0.100 m = 36 rad/s²

Now, the torque on the pulley τ = Tr = Iα where I = moment of inertia of pulley about its rotational axis and T = tension in cord attached to book and r = radius of pulley = 0.200 m/2 = 0.100 m

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Answer:

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τ = L/R ,  represent the time requiered for the crrent to be 63 % of its final value

4 0
3 years ago
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