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3 years ago

11

A farmer in western Minnesota is retiring. He wants to stop farming his 3,000 acres and turn the land into a prairie. He begins

to plant grasses that are native to a short-grass prairie. Is this a good or bad idea?
Question 9 options:

A)

It is a good idea because we need more short-grass prairie in Minnesota

B)

It is a good idea because prairie land is good for animal habitats and migrations

C)

It is a bad idea because he won't be able to afford the restoration

D)

It is a bad idea because Minnesota has never had a naturally occurring short-grass prairie; only tall-grass prairies exist here.

1 answer:

Snowcat [4.5K]3 years ago

8 0

B b is the right answer

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3 years ago

2.86 A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight

Rashid [163]

Answer:

Distance Below the top = 6,02 m

Explanation:

To get the Final velocity of the first ball (that will be the intial velocity of the second) you need to solve the kinematic equation for Velocity:

(1) $V_{1f} =V_{1O} - gt$

As the ball is dropped from rest $V_{1O} = 0$ , so:

(2) $V_{1f} = - gt$

Note that the velocity is going to be negative as the ball is going down. To get the time it would take the ball to reach de base you can use the kinematic equation for position:

(3) $h_{1} = h_{1O} + V_{1O} *t - \frac{1}{2} gt^{2}$

We need the answer when h=0, and from the initial conditions $V_{1O} = 0, h_{1O} = 24m$ , you get:

(4) $24m = \frac{1}{2} gt_{1f}^{2}$

Solving for t, with $9,8\frac{m}{s^{2}}$ and :

(5) $t_{1f} = \sqrt{\frac{24m*2}{g} } = 2,21 s$

Replacing this time in (2), the final velocity is:

(6) $V_{1f} = -21,66 \frac{m}{s}$

So the initial velocity of ball 2 is equial to this but oppossite in direction so: $V_{2O}= -V_{1f} = 21,66 \frac{m}{s}$

The general position equation for ball 2 is (considering $h_{2O} = 0$ :

(7) $h_{2} = V_{2O} *t - \frac{1}{2} gt^{2}$

They cross paths when $h_{1}=h_{2}$ so:

(8) $h_{1O} - \frac{1}{2} gt^{2} = V_{2O} *t - \frac{1}{2} gt^{2}$

Rearranging:

(9) $t_{cross} =\frac{h_{1O}}{V_{2O} }$

Replacing values:

$t_{cross} = 1,108 s$

To get the absolute position replace $t_{cross}$ on equation (3) or (7):

$h_{cross} = 17,98 m$

To get it below the top of te cliff:

$h_{cross, from above} = 24 m - 17,98 m$

$h_{cross, from above} = 6,02 m$

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3 years ago