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A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand

enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

$v=gt$

And the distance traveled downwards is:

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

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

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

Replacing into the first equation:

$\displaystyle v=g\sqrt{\frac{2y}{g}}$

Rationalizing:

$\displaystyle v=\sqrt{2gy}$

Let's call v1 the final speed of the package dropped from a height H. Thus:

$\displaystyle v_1=\sqrt{2gH}$

Let v2 be the final speed of the package dropped from a height 4H. Thus:

$\displaystyle v_2=\sqrt{2g(4H)}$

Taking out the square root of 4:

$\displaystyle v_2=2\sqrt{2gH}$

Dividing v2/v1 we can compare the final speeds:

$\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}$

Simplifying:

$\displaystyle v_2/v_1=2$

The final speed of the second package is twice as much as the final speed of the first package.

5 0

3 years ago

A soccer ball kicked with a force of 12.5 N accelerates at 6.2 m/s’ to the right. What is the mass of the ball? Answer in units

stepladder [879]

Answer:

m = 2.01[kg]

Explanation:

This problem can be solved using Newton's second law which tells us that the force applied on a body is equal to the product of mass by acceleration.

$F =m*a$

where:

F = force = 12.5 [N]

m = mass [kg]

a = acceleration = 6.2 [m/s²]

$12.5=m*6.2\\m = 2.01[kg]$

3 0

3 years ago

In a race, you run 3000 meters east in 21 minutes. What is your velocity in km/min?

STALIN [3.7K]

velocity = distance / time v= d/t firstly you change 3000 m into km distance D = 3000/1000 = 3 km secondly time t = 21 min using velocity formula v = d/t = 3 km/21min =0.142km/min

6 0

3 years ago

You throw a baseball directly upward at time ????=0 at an initial speed of 14.9 m/s. What is the maximum height the ball reaches

xeze [42]

Answer:

Maximum height, h = 11.32 meters

Explanation:

It is given that,

The baseball is thrown directly upward at time, t = 0

Initial speed of the baseball, u = 14.9 m/s

Ignoring the resistance in this case and using a = g = 9.8 m/s²

We have to find the maximum height the ball reaches above where it leaves your hand. Let the maximum height is h. Using third equation of motion as :

$v^2-u^2=2ah$

At maximum height, v = 0

and a = -g = -9.8 m/s²

$h=\dfrac{v^2-u^2}{2a}$

$h=\dfrac{0-(14.9\ m/s)^2}{2\times -9.8\ m/s^2}$

h = 11.32 meters

Hence, the maximum height of the baseball is 11.32 meters.

8 0

3 years ago

How can a large construction project impact water quality, supply or flow?

Korolek [52]

Answer:

Construction also impacts the community’s water quality. During the construction phase, pollutants can end up in the water supply. Once construction is over, the water quality can still be a problem due to an increased demand for water, as well as increased pollution due to additional residents.

Explanation:

Hope This Could be of some use

3 0

3 years ago