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kumpel [21]
4 years ago
12

What is the drag force exerted on a runner running at 7 min/mi pace (3.83 m/s) on a day with no wind? Assume a coefficient of dr

ag of 0.8, an air density of 1.204 kg/m3, and a frontal surface area of 1.0 m2.
Physics
1 answer:
labwork [276]4 years ago
4 0

Answer:

7.06454224 Newtons

Explanation:

C = Drag coefficient = 0.8

ρ = Density of air = 1.204 kg/m³

A = Surface area = 1 m²

v = Velocity of wind = 7 min/mi = 3.83 m/s

The Drag force is given by

F_d=\frac{1}{2}\rho CAv^2 \\\Rightarrow F_d=\frac{1}{2}\times 1.204 \times 0.8\times 1\times 3.83^2\\\Rightarrow F_d=7.06454224\ N

The drag force exerted on the runner is 7.06454224 Newtons

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Determine the kinetic energy of the ball immediately after it is hit. (You must provide an answer before moving to the next part
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The question is incomplete. Here is the complete question.

A baseball palyer hits a 5.1 oz baseball with an initial velocity of 140ft/sat an angle of 40° with the horizontal as shown. Determine

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b) The kinetic energy of the ball when it reaches its maximum height

c) The maximum height above the ground reached by the ball.

Answer: a) KE = 131.64 J

              b) KE = 0

              c) h = 126 ft

Explanation: <u>Kinetic</u> <u>energy</u> is the energy an object posses due to its motion. It can be calculated as KE=\frac{1}{2}mv^{2}

a) Kinetic energy's unit is Joule. So, we have to transform ounce in kg and ft/s in m/s for the units to correspond:

m = 5.1(0.02835)

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v = 140 ft = 42.67 m/s

Then, kinetic energy is

KE=\frac{1}{2}(0.1445)(42.67)^{2}

KE = 131.64 J

Kinetic energy immediately after the ball is hit is 131.64 J.

b) At its maximum height, the ball has its highest potential energy. Because of the law of conservation of energy, when potential energy is maximum, kinetic energy is minimum and vice-versa. So, at the maximum height, kinetic energy is 0.

c) This type of motion is <u>projectile</u> <u>motion</u>. The maximum height on a projectile motion can be determined by

v_{y}^{2}=v_{0y}^{2}-2g\Delta y

When h is maximum, v_{y}=0

Velocity of the ball has an angle with the horizontal, so initial velocity at the y-axis is

v_{0y}=v_{0}sin(\theta)

Substituting and solving

v_{y}^{2}=v_{0}^{2}sin^{2}(\theta)-2gh

0=(42.67)^{2}sin^{2}(40)-2(9.8)h

19.6h=(42.67)^{2}(0.643)^{2}

h=\frac{(1820.73)(0.4132)}{19.6}

h = 38.4 m

Transforming into ft: h = 126 ft

The maximum height above the ground reached by hte ball is 126 feet.

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