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kirza4 [7]
2 years ago
6

Heather and Matt take 34 minutes to walk eastward along a straight road to a store 2.0km away. What is their average velocity in

m/s
Physics
1 answer:
frez [133]2 years ago
7 0

Their average velocity is 0.98 m/s east.

Explanation:

The average velocity of an object is defined as the ratio between the displacement and the time taken:

v=\frac{displacement}{time}

The displacement is a vector connecting the initial point to the final point of motion: so, being displacement a vector, velocity is a vector as well, having the same direction of displacement.

Here, Heather and Matt walk 2.0 km eastward, so their displacement is

d = 2.0 km = 2000 m (east)

While the time they took is

t=34 min \cdot 60 = 2040 s

Therefore, their average velocity is

v=\frac{2000}{2040}=0.98 m/s

And the direction is the same as the displacement (east).

Learn more about average velocity:

brainly.com/question/5248528

#LearnwithBrainly

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Ways in which a teacher plays a role in the literacy development of the learners​
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A mechanic pushes a 3540 kg car from rest to a speed of v, doing 4864 J of work in the process. Find the speed v. Neglect fricti
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Answer:

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Explanation:

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Why are the temperatures on each of the other terrestrial planets more extreme than earth?
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Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible co
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Answer:

 q = 8.61 10⁻¹¹ m

charge does not depend on the distance between the two ships.

it is a very small charge value so it should be easy to create in each one

Explanation:

In this exercise we have two forces in balance: the electric force and the gravitational force

          F_e -F_g = 0

          F_e = F_g

Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.

Let's write Coulomb's law and gravitational attraction

         k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{r^2}

In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.

          k q² = G m²

          q = \sqrt{ \frac{G}{k} }    m

we substitute

           q = \sqrt{ \frac{ 6.67 \ 10^{-11}}{8.99 \ 10^{9}} }   m

            q = \sqrt{0.7419 \ 10^{-20}}   m

           q = 0.861 10⁻¹⁰ m

           q = 8.61 10⁻¹¹ m

This amount of charge does not depend on the distance between the two ships.

It is also proportional to the mass of the ships with the proportionality factor found.

Suppose the ships have a mass of m = 1000 kg, let's find the cargo

            q = 8.61 10⁻¹¹ 10³

            q = 8.61 10⁻⁸ C

             

this is a very small charge value so it should be easy to create in each one

6 0
2 years ago
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