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3 years ago

11

Why are objects that fall near earth’s surface rarely in free fall?

1 answer:

Sloan [31]3 years ago

3 0

Answer:

Because of the presence of air resistance

Explanation:

When an object is in free fall, ideally there is only one force acting on it:

- The force of gravity, W = mg, that pushes the object downward (m= mass of the object, g = acceleration of gravity)

However, this is true only in absence of air (so, in a vacuum). When air is present, it exerts a frictional force on the object (called air resistance) with upward direction (opposite to the motion of free fall) and whose magnitude is proportional to the speed of the object.

Therefore, it turns out that as the object falls, its speed increases, and therefore the air resistance acting against it increases too; as a result, the at some point the air resistance becomes equal (in magnitude) to the force of gravity: when this happens, the net acceleration of the object becomes zero, and so the speed of the object does not increase anymore. This speed reached by the object is called terminal velocity.

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Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water

Ksenya-84 [330]

To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

$v_2^2 = v_1^2 + 2gh$

Where,

$V_i =$ Velocity in each state

g= Gravity

h = Height

Our values are given as,

$A_1 = 2.4*10^{-4} m^2$

$v_1 = 0.8 m/s$

$h = 0.11m$

Replacing at the kinetic equation to find $V_2$ we have,

$v_2 = \sqrt{v_1^2 + 2gh}$

$v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}$

$v_2= 1.67 m/s$

Applying the concepts of continuity,

$A_1v_1 = A_2v_2$

We need to find A_2 then,

$A_2= \frac{A_1v_1 }{v_2}$

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

$A_2= \frac{A_1v_1 }{v_2}$

$A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}$

$A_2= 1.14*10^{-4} m2$

Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is $1.14*10^{-4} m2$

8 0

3 years ago

1. (30 pts) Let x(t) = cos(πt/2) be a continuous-time signal,

posledela

Given that the function of the wave is f(x) = cos(π•t/2), we have;

a. The graph of the function is attached

b. 4 units of time

c. Even

d. 4.935 J/kg

e. 1.234 W/kg

<h3>How can the factors of the wave be found?</h3>

a. Please find attached the graph of the signal created with GeoGebra

b. The period of the signal, T = 2•π/(π/2) = <u>4</u>

c. The signal is <u>even</u>, given that it is symmetrical about the y-axis

d. The energy of the signal is given by the formula;

$\frac{1}{2} \cdot \mu^{2} \cdot \omega ^{2} \cdot \: {a}^{2} \times \lambda$

Which gives;

E = 0.5 × 1.571² × 1² × 4 = <u>4.935 J/kg</u>

e. The power of the wave is given by the formula;

E = 0.5 × 1.571² × 1² × 4 × 0.25 = <u>1.234 W/</u><u>kg</u>

Learn more about waves here:

brainly.com/question/14015797

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Mashutka [201]

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never [62]

Answer:

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timama [110]

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