A plant takes in nutrients from the soil, gasses from the air, and water from the rain. What spheres are interacting?

Have some fr/ee points because i finished highschool today :)

Aleks04 [339]

Answer:

Thank you :)

Explanation:

And congrats on finishing !!

3 0

3 years ago

a projectile whose masss is 10g is fired directly upward from ground level with an initial velocity of 1000m/s neglecting the ef

sammy [17]

Answer:

The speed of the projectile when it impacts the ground is 1000 m/s

Explanation:

Vertical Launch

When an object is launched vertically and upwards it starts to move at an initial speed vo, then the acceleration of gravity makes that speed to reduce until it reaches 0. The object has reached its maximum height. Then, it starts to move downwards in free fall, with initial speed zero and gradually increasing it until it reaches the ground level. We will demonstrate that the speed it has when impacts the ground is the same (and opposite) as the initial speed vo.

The speed when the object is moving upwards is given by

$v_f=v_o-g.t$

The time it takes to reach the maximum height is when vf=0, i.e.

$0=v_o-g.t$

solving for t

$\displaystyle t=\frac{v_o}{g}$

The maximum height reached is

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

Then, the object starts to fall. The object's height is given by

$\displaystyle y=\frac{v_o^2}{2g}-\frac{gt'^2}{2}$

where t' is the time the object has traveled downwards. The height will be 0 again when

$\displaystyle \frac{v_o^2}{2g}=\frac{gt'^2}{2}$

Solving for t'

$\displaystyle t'=\frac{v_o}{g}$

We can see the time it takes to reach the maximum height is the same it takes to return to ground level. Of course, the speed when it happens is

$v_f=g.t'=v_o$

Thus, the speed of the projectile when it impacts the ground is 1000 m/s

4 0

4 years ago

PLZ ANSWER THESE ASAPPP INEED YOUR HELP !!

Burka [1]

Where the air has low humidity
idk ;-; you can do it!

4 0

4 years ago

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The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC

enyata [817]

Answer:

The y-component of the electric force on this charge is $F_y = -1.144\times 10^{-6}\ N.$

Explanation:

Given:

Electric field in the region, $\vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.$
Charge placed into the region, $q = 10.4\ nC = 10.4\times 10^{-9}\ C.$

where, $\hat i,\ \hat j$ are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

$\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.$

Thus, the y-component of the electric force on this charge is $F_y = -1.144\times 10^{-6}\ N.$

3 0

3 years ago

Describe your motion in terms of velocity and acceleration as you ride in a car down the street, come to a red light, wait for t

aleksandr82 [10.1K]

Ride in a car down the street: constant velocity,
come to a red light: negative acceleration,
wait for the green light: zero velocity,
start off again: positive acceleration

7 0

3 years ago

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