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Svetradugi [14.3K]
3 years ago
6

A paratrooper is initially falling downward at a speed of 27.6 m/s before her parachute opens. When it opens, she experiences an

upward instantaneous acceleration of 74 m/s^2. (a) If this acceleration remained constant, how much time would be required to reduce the paratrooper's speed to a safe 4.95 m/s? (Actually the acceleration is not constant in this case, but the equations of constant acceleration provide an easy estimate.) (b) How far does the paratrooper fall during this time interval?
Physics
1 answer:
nika2105 [10]3 years ago
8 0

Answer:

a) 0.31 s

b) 19.77 m

Explanation:

We will need the following two formulas:

V_{f} = V_{0}+at\\\\X=V_{0}t + \frac{at^{2}}{2}

We first use the final velocity formula to find the time that it takes to decelerate the paratrooper:

4.95\frac{m}{s}=27.6\frac{m}{s}-74\frac{m}{s^{2}}t\\\\-22.65\frac{m}{s}=-74\frac{m}{s^{2}}t\\\\t= \frac{22.65\frac{m}{s}}{74\frac{m}{s^{2}}}=0.31s

Now that we have the time, we can use the distance formula to calculate the distance travelled by the paratrooper:

X=27.6\frac{m}{s}*0.31s - \frac{74\frac{m}{s^{2}}*(0.31 s)^{2}}{2}=19.77 m

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To solve this problem we can use the concepts related to the change of flow of a fluid within a tube, which is without a rubuleous movement and therefore has a laminar fluid.

It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law.

The mathematical equation that expresses this concept is

\dot{Q} = \frac{\pi r^4 (P_2-P_1)}{8\eta L}

Where

P = Pressure at each point

r = Radius

\eta = Viscosity

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Of all these variables we have so much that the change in pressure and viscosity remains constant so the ratio between the two flows would be

\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}

From the problem two terms are given

R_A = \frac{R_B}{2}

L_A = 2L_B

Replacing we have to

\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}

\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_B^4}{16*r_B^4}\frac{L_B}{2*L_B}

\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{1}{32}

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6 0
3 years ago
A well lagged copper calorimeter of mas 120g contains 70g of water and 10g ice both at 0°C . Dry steam at 100°C is passed in unt
Lina20 [59]

Answer:

7.6 g

Explanation:

"Well lagged" means insulated, so there's no heat transfer between the calorimeter and the surroundings.

The heat gained by the copper, water, and ice = the heat lost by the steam

Heat gained by the copper:

q = mCΔT

q = (120 g) (0.40 J/g/K) (40°C − 0°C)

q = 1920 J

Heat gained by the water:

q = mCΔT

q = (70 g) (4.2 J/g/K) (40°C − 0°C)

q = 11760 J

Heat gained by the ice:

q = mL + mCΔT

q = (10 g) (320 J/g) + (10 g) (4.2 J/g/K) (40°C − 0°C)

q = 4880 J

Heat lost by the steam:

q = mL + mCΔT

q = m (2200 J/g) + m (4.2 J/g/K) (100°C − 40°C)

q = 2452 J/g m

Plugging the values into the equation:

1920 J + 11760 J + 4880 J = 2452 J/g m

18560 J = 2452 J/g m

m = 7.6 g

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Explanation:

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