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3 years ago

6

You throw a balloon that floats in the air with a velocity of 2 m / s south . If the wind speed is 5 m / s west , how far south

will the balloon travel after 2 seconds ?

1 answer:

zvonat [6]3 years ago

4 0

Answer:

The distance traveled by the balloon is 10.77 m

Explanation:

velocity of the ball, $v_b$ = 2 m/s south

velocity of the air, $v_a$ = 5 m/s west

To determine the distance the balloon will travel after 2 seconds, first determine the resultant velocity of the balloon.

| 2m/s

|

|

↓

5m/s ←------------------

the two velocities forms a right angled triangle and the resultant will be the hypotenuses side of the triangle.

R² = 5² + 2²

R² = 29

R = √29

R = 5.385 m/s

The distance traveled by the balloon is calculated as;

d = R x t

where;

t is time of the motion = 2 seconds

d = 5.385 x 2

d = 10.77 m

Therefore, the distance traveled by the balloon is 10.77 m.

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Ostrovityanka [42]

Answer:

a. 0.5307 sec

b. 0.4458 m

c. = $v_x=0.84\ m/s\ ,\ v_y=5.2\ m/s$

Explanation:

Horizontal Motion

It describes the dynamics of an object thrown horizontally in free air. The initial horizontal velocity is maintained all the time since no horizontal forces are acting. The initial vertical velocity is zero at launch time, but it grows downwards powered by the acceleration of gravity.

The object hits the ground at a distance x from the point of launching, after having traveled a vertical distance $y_o$ , taking a time t to complete the travel. The formulas who relate the different magnitudes are

$v_x=v_o$

The horizontal velocity $v_x$ is the same regardless of the elapsed time

$v_y=gt$

$x=v_ot$

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The plate of apple pie left the counter at a speed

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The counter is $y_o=1.38\ m$ high.

a.

Knowing that

$y_o=1.38\ m$

We use this formula to compute t

$\displaystyle y=y_o-\frac{gt^2}{2}$

At the moment when the plate hits the floor y=0

$\displaystyle 0=y_o-\frac{gt^2}{2}$

$\displaystyle y_o=\frac{gt^2}{2}$

Solving for t

$\displaystyle t=\sqrt{\frac{2y_o}{g}}$

$\displaystyle t=\sqrt{\frac{2(1.38)}{9.8}}$

$t=0.5307\ sec$

b.

$x=(0.84)(0.5307)$

$x=0.4458\ m$

c.

$v_x=0.84\ m/s$

$v_y=(9.8)(0.5307)$

$v_y=5.2\ m/s$

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How many seconds will elapse between seeing lightning and hearing the thunder if the lightning strikes 1mi (5280 ft) away and th

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Answer:

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Explanation:

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