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Diano4ka-milaya [45]
3 years ago
9

A group of runners complete a 26.2 mile marathon in 3.4 hours. The distance between the start and finish lines is 12.2 miles. Wh

at was the average velocity for the run?
Physics
2 answers:
NNADVOKAT [17]3 years ago
8 0

26.2/3.4 would be the average velocity for the run.

7.7 miles/hr

wlad13 [49]3 years ago
5 0

Answer:

\vec v = 3.6 mph

Explanation:

As we know that the total path length of the Marathon is given as

L = 26.2 miles

while the straight distance between start and finish lines is given as

d = 12.2 miles

so here we can say that the displacement is the straight line distance between initial and final position which is 12.2 miles

so it is given as

\vec d = 12.2 miles

now the average velocity is defined as the ratio of displacement and time

\vec v = \frac{\vec d}{time}

\vec v = \frac{12.2 miles}{3.4 hr}

\vec v = 3.6 mph

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Explain what parallel medium boundaries do to the path of the wave after it travels through both boundaries. (air and water)
Brilliant_brown [7]

Answer:

The boundaries cause the waves to change direction an effect called <u>refraction.</u>

Explanation:

When a wave crosses a boundary between different materials, the speed of the wave and its wavelength changes.When passing from air to water the two properties (speed and wavelength) decreases, and the wave is observed to change direction as it crosses the boundary between the two material.The bending of the wave is called refraction.

7 0
3 years ago
In pairs figure-skating competition, a 65-kg man and his 45-kg female partner stand facing each other both at rest on the ice. I
Lina20 [59]

Answer:

The final velocity of her partner is approximately -1.04 m/s or 1.04 m/s in the opposite direction to her direction of motion

Explanation:

The given parameters are;

The mass of the man, m₁ = 65 kg

The mass of the woman, m₂ = 45 kg

Taking the relative initial velocity of the man and the woman as 0 m/s, we have;

The initial velocity of the man, v₁₁ = 0 m/s

The initial velocity of the man, v₁₂ = 0 m/s

The final velocity of the woman, v₂₂ = 1.5 m/s

The final velocity of the man = v₂₁

Therefore, we have, by the conservation of momentum principle;

The total initial momentum = The total final momentum

Which gives;

m₁ × v₁₁ + m₂ × v₁₂ = m₁ × v₂₁ + m₂ × v₂₂

Substituting the known values;

65 × 0 + 45 × 0 = 65 × v₂₁ + 45 × 1.5

∴ 65 × v₂₁ + 45 × 1.5 = 0

45 × 1.5 = - 65 × v₂₁

v₂₁ = 45 × 1.5/(-65) ≈ -1.04 m/s

The final velocity of the man, her partner = v₂₁ ≈ -1.04 m/s.

6 0
3 years ago
two masses are separated by 1 m. suppose the masses are moved so they are 2 m apart how will the gravitational force change
Verdich [7]
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5 0
3 years ago
Scenario
Anvisha [2.4K]

Answer:

1) t = 23.26 s,  x = 8527 m, 2)   t = 97.145 s,  v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

         v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

         y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

         y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t2

          t = \sqrt{ \frac{2y_o}{g} }

          t = √(2 2650/9.8)

          t = 23.26 s

this is the horizontal scrolling time

          x = v₀ t

          x = 366.67  23.26

          x = 8527 m

the speed at the point of arrival is

         v_y = v_{oy} - g t = 0 - gt

         v_y = - 9.8 23.26

         v_y = -227.95 m / s

Module and angle form

        v = \sqrt{v_x^2 + v_y^2}

         v = √(366.67² + 227.95²)

        v = 431.75 m / s

         θ = tan⁻¹ (v_y / vₓ)

         θ = tan⁻¹ (227.95 / 366.67)

         θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

           cos θ = v₀ₓ / v₀

           sin θ = v_{oy} / v₀

            v₀ₓ = v₀ cos θ

            v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

            y = y₀ + v_{oy} t - ½ g t²

            0 = y₀ + vo sin θ t - ½ g t²

            0 = -50 + vo sin 50 t - ½ 9.8 t²

            x = x₀ + v₀ₓ t

            0 = x₀ + vo cos θ t

            0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

          50 = 0,766 vo t – 4,9 t²

          400 =   0,643 vo t

resolved

          50 = 0,766 ( \frac{400}{0.643 \ t}) t – 4,9 t²

          50 = 476,52 t – 4,9 t²

          t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

         t = [97.25 ± \sqrt{97.25^2 - 4 \ 10.2}] / 2

         t = 97.25 ±97.04] 2

         t₁ = 97.145 s

         t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

         t = 97.145 s

let's find the initial velocity

         x = x₀ + v₀ cos 50 t

         0 = -400 + v₀ cos 50 97.145

         v₀ = 400 / 62.44

         v₀ = 6.4 m / s

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