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natka813 [3]
3 years ago
5

A football player runs from his own goal line to the opposing team's goal line, returning to his ten-yard line, all in 26.6 s. C

alculate his average speed and the magnitude of his average velocity. (Enter your answers in yards/s.) HINT
Physics
1 answer:
Advocard [28]3 years ago
8 0

Answer:

a) Average Speed = 190yards/26.6s = 7.14 yards/s

b) Average Velocity = 10/26.6 = 0.38 yard/s

Complete Question;

A football player runs from his own goal line to the opposing team's goal line, returning to his ten-yard line, all in 26.6 s. Calculate his average speed and the magnitude of his average velocity. (Assume a 100 yard football field; note that the player's X-yard line is X yards from his own goal line. Enter your answers in yards/s.)

Explanation:

a) Speed = distance/time

time t = 26.6s

distance;

d1 = 100 yards (own goal line to the opposing team's goal line)

d2 = 90 yards (returning to his ten-yard line)

d = d1 + d2

d = 100 + 90 = 190 yards

Speed = 190yards/26.6s = 7.14 yards/s

b) velocity = displacement/time

time = 26.6s

displacement = 100-90 yards = 10yards

Velocity = 10/26.6 = 0.38 yard/s

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Consider two children sitting on a merry-go-round, with one closer to the outer edge and one closer to the center. show answer N
dolphi86 [110]

Answer:

They both have the same angular speed.

Explanation:

The mathematical formula for angular speed is:

w=\frac{2\pi}{T}

where w is angular speed, 2\pi is a constant, and T is the period (the time it takes the marry-go-round to complete a lap).

What we can see from the formula is that, since the 2\pi does not change its value, the angular speed depends only on the period T.

In this case for both the children closer to the outher edge and for the children closer to the center, the time to complete a lap is the same, because the time does not depend on where they are sitting in the marry go round. This means that the period for both is the same.

Thus, since the period for both is the same, the angular speed given by

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4 0
4 years ago
ou have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 660.0 kg and was trav
Montano1993 [528]

Answer:

    vₐ₀ = 29.56 m / s

Explanation:

In this exercise the initial velocity of car A is asked, to solve it we must work in parts

* The first with the conservation of the moment

* the second using energy conservation

let's start with the second part

we must use the relationship between work and kinetic energy

             W = ΔK                             (1)

for this part the mass is

             M = mₐ + m_b

the final velocity is zero, the initial velocity is v

friction force work is

              W = - fr x

the negative sign e because the friction forces always oppose the movement

we write Newton's second law for the y-axis

              N -W = 0

              N = W = Mg

friction forces have the expression

              fr =μ N

              fr = μ M g

we substitute in 1

               -μ M g x = 0 - ½ M v²

             v² = 2 μ g x

let's calculate

              v² = 2  0.750  9.8  6.00

              v = ra 88.5

              v = 9.39 m / s

Now we can work on the conservation of the moment, for this part we define a system formed by the two cars, so that the forces during the collision are internal and therefore the tsunami is preserved.

Initial instant. Before the crash

         p₀ = + mₐ vₐ₀ - m_b v_{bo}

instant fianl. Right after the crash, but the cars are still not moving

         p_f = (mₐ + m_b) v

         p₀ = p_f

         + mₐ vₐ₀ - m_b v_{bo} = (mₐ + m_b) v

           

         mₐ vₐ₀ = (mₐ + m_b) v + m_b v_{bo}

let's reduce to the SI system

          v_{bo} = 64.0 km / h (1000m / 1km) (1h / 3600s) = 17.778 m / s

let's calculate

         660 vₐ₀ = (660 +490) 9.39 + 490 17.778

         vₐ₀ = 19509.72 / 660

         vₐ₀ = 29.56 m / s

we can see that car A goes much faster than vehicle B

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Answer:

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Explanation:

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Hope this answer can help you,

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