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ELEN [110]
3 years ago
15

Studen

Physics
1 answer:
Vesna [10]3 years ago
7 0

Complete Question

A football coach walks 18 meters westward, then 12 meters eastward, then 28 meters westward, and finally 14 meters eastward.

a

From this motion what is the distance covered

b

What is the magnitude and direction of the displacement

Answer:

a

  D =   72 \  m

b

Magnitude

             d =  - 20 \ m

 Direction  

West

Explanation:

From the question we are told that

The first distance covered westward is d_w_1  =  18 \  m /tex]     The  first distance covered eastward is [tex]d_e1 =  12 \  m /tex]      The second distance covered westward is [tex]d_w_2  =  28 \  m /tex]      The  second distance covered eastward is [tex]d_e2 =  14 \  m /tex]   Generally the distance covered is mathematically represented as       [tex]D =  d_w1 + d_w2 + d_e1 + d_e2

=> D =   18 + 28 + 12 +  14

=> D =   72 \  m

For the second question eastward is in the direction of the positive x-axis so it would be positive and westward is in the direction of the negative x-axis so it would be negative

The magnitude of the displacement is

d =  -d_w1 -d_w2 + d_e1 + d_e2

=>  d =  -18-28 + 12 + 14

=> d =  - 20 \ m

The direction is west

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Answer:

The answer to the question is;

The total potential energy of the mass on the spring when the mass is at either endpoint of its motion is 5.0255 Joules.

Explanation:

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That is the total potential energy of the mass on the spring when the mass is at either endpoint of its motion is equal to the maximum kinetic energy.

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3 years ago
1. Describe what is happening at point A when these 3 vectors act on point A. 2. Describe what happen to the resultant vector at
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Your friend's 10.8 g graduation tassel hangs on a string from his rearview mirror. When he accelerates from a stoplight, the tas
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The tension in the string holding the tassel and the vertical will the tension in the string

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  • Ф = 34.4 °

<h3>What is the tension in the string holding the tassel. ?</h3>

Generally, the equation for Tension is  mathematically given as

TCos\theta = mg

Therefore

TCos6.58^{o} = 19.8*10^{-3}*9.8

T = 0.1953 N

b).

Where

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T* sinФ = ma

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T = 0.1953 N

Ф = 34.4 °

Read more about tension

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Answer:

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