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saul85 [17]
2 years ago
8

Some runners train with parachutes that trail behind them to provide a large drag force. These parachutes are designed to have a

large drag coefficient. One model expands to a square 1.8 mm on a side, with a drag coefficient of 1.4. A runner completes a 240 mm run at 6.0 m/s with this chute trailing behind.
Required:
How much thermal energy is added to the air by the drag force?
Physics
1 answer:
Svetradugi [14.3K]2 years ago
6 0

Answer:

by the drag force, 2.4004512 × 10⁻⁵ J is added to the air.

Explanation:

Given the data in the question;

drag coefficient of Cd = 1.4

speed v = 6.0 m/s

One model expands to a square 1.8 mm on a side

Area A = 1.8 × 1.8 = 3.24 mm² = 3.24 × 10⁻⁶ m²

distance travelled s = 240 mm = 0.24 m

we know that; density of air e = 1.225 kg/m³

Now,

Dragging force F_D = ( Cd × e × v² × A  ) / 2

thermal energy = F_D × s

so

thermal energy = ( 1.4 × 1.225  × (6)² × (3.24 × 10⁻⁶) × 0.24  ) / 2

thermal energy = ( 4.8009024 × 10⁻⁵ ) / 2

thermal energy = 2.4004512 × 10⁻⁵ J

Therefore,  by the drag force, 2.4004512 × 10⁻⁵ J is added to the air.

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marusya05 [52]

The system's tension is 616 N and acceleration is 5.6 m / s^{2}

<u>Explanation:</u>

From newton’s second law of motion which state that net force acting on a body is product of mass of a body and acceleration of a body which is given as,

             F_{n e t}=m_{t o t} \times a

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                a=\frac{F_{n e t}}{m_{\mathrm{tot}}}=\frac{m \times g}{m_{\mathrm{tot}}}

Put the value for m = hanging mass = 40 kg and g=9.8 \mathrm{m} / \mathrm{s}^{2}, we get

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The tension in the ropes,  T=(m \times g)+(m \times a)

Here, m as hanging mass

T = tension, N or  k g m / s^{2}

m = mass, kg  

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