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umka21 [38]
2 years ago
13

The proper length of the first lane of a competitive running track is 400 m (1,312.3 ft). A runner runs track from the starting

line four times, stopping at the starting line.
Which statement is TRUE?

Group of answer choices

The runner's distance is 1600 m, his displacement is 0 m, and his speed is 0 m/s.

The runner's distance is 1600 m, his displacement is 0 m, and his velocity is 0 m/s.

The runner's displacement is 1600 m, his distance is 0 m, and his velocity is 0 m/s.

The runner's distance and displacement are 1600 m and his velocity cannot be calculated given this information.
Physics
2 answers:
charle [14.2K]2 years ago
6 0

Answer:

The runner's distance is 1600 m, his displacement is 0 m, and his velocity is 0 m/s.

Explanation:

It is given that,

Length of the truck, l = 400 m

A runner runs track from the starting line four times, stopping at the starting line. In this case, the distance covered by the runner is, 4 l i.e. 4 × 400 m = 1600 meters.

Distance covered by an object is the overall path travelled during its entire journey. So, runner's distance is 1600 meters.

We know that the displacement of an object is equal to the shortest path covered or it is equal to the difference between final position and initial position. As a result, the displacement of runner is 0 as he is coming back to starting line.

We know that the net displacement divided by total time taken is called velocity of an object. So, velocity will be 0.

Hence, the correct option is (b) "The runner's distance is 1600 m, his displacement is 0 m, and his velocity is 0 m/s".

irakobra [83]2 years ago
4 0

Answer:

The runner's distance is 1600 m, his displacement is 0 m, and his speed is 0 m/s.

Explanation:

Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. In this case distance is 400m x 4= 1600m. Displacement is a vector quantity that refers to the distance betwen the first and the last position of an object. In this case is 0 because the runer always track in the same place.

Finally, the velocity is 0 m/s because the runner always tracks a point, and in order to measure the velocity we need a distance between to points and a time.

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Yes, the above-given statement is true

<u>Explanation:</u>

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<u>Answer;</u>

<em>D.  The object’s weight changes, but its mass stays the same.</em>

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Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

F_s = am = 1.31*16 = 20.92 N

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

So the maximum acceleration on the block is

a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2

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F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N

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