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dezoksy [38]
2 years ago
12

What can happen if water is removed from an aquifer faster than it can be replaced?

Physics
1 answer:
Mandarinka [93]2 years ago
7 0

Answer:If there may be greater water extracted than replaced, you're depleting the ones resources. It's like taking extra money from your bank account than you earn.

Explanation:

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The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c
eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

X_f = X_i +\frac{1}{2}(V_i+V_f)t

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

X_f = \frac{1}{2} (V_i+V_f)t

Clearing for time,

t = \frac{2X_f}{(V_i+V_f)}

t = \frac{2*1.2}{24+0}

t = 0.1s

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

a= \frac{\Delta V}{t}

Then

F = m\frac{\Delta V}{t}

F = m\frac{V_f-V_i}{t}

F = m\frac{-V_i}{t}

F = \frac{(1600kg)(-24m/s)}{(0.1s)}

F = -384000N

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

V_f^2=V_i^2-2ax

0 = (24m/s)^2-2*a(1.2m)

a = \frac{(24m/s)^2}{1.2m}

a=480m/s^2

The gravity is equal to 0.8, then the acceleration is

a = 480*\frac{g}{9.8}

a = 53.3g

3 0
3 years ago
The height of a circular tower is 179 meters and its diameter is 48 meters. (Assume that the building is a perfect cylinder). Wh
EastWind [94]

The definition of a scale of 1: 166 will mean that the scale of 1 in the model will be equivalent to 166 times the measurement in the real model, therefore we will have that the height would be 166 times smaller than the 179m given:

h = \frac{179m}{166} = 1.078m = 107.8cm

The same for the diameter,

\phi = \frac{48m}{166}= 0.2891m = 28.91cm

The volume of a cylinder is given as

V = (\pi r^2)(h)

V = (\pi (\frac{d}{2})^2)(h)

V = (\pi (\frac{ 28.91}{2})^2)(107.8)

V = 70762.8cm^3

Therefore the volume would be V = 70762.8cm^3

6 0
4 years ago
Suppose you are working on green house, which light you would<br> use for their growth?
nikitadnepr [17]

Answer:

Fluorescent lighting is usually used.

Explanation:

6 0
3 years ago
Read 2 more answers
The chart show the masses and velocities of two colliding objects that stick together after a collision.
finlep [7]

Answer:

<u><em></em></u>

  • <u><em>1,500 kg.m/s</em></u>

Explanation:

First, arrange the information in a table:

Object        Mass (kg)           Velocity (m/s)

  A                    200                      15

  B                     150                    - 10

After the collision, the two objects are stick together, thus you talk aobut one object and one momentum.

According to the law of convervation of momentum, the momentum after the collision is equal to the momentum before the collision.

<u>Momentum before the collision, P₁</u>:

          P_1=m_{A,1}\times v_{A,1}+m_{B,1}\times v_{B,1}

         P_1=200kg\times 15m/s+150kg\times (-10m/s)\\\\  P_1=3000kg.m/s-1500kg.m/s=1500kg.m/s

<u>Momentum after the collision</u>:

  • As stated, it es equal to the momentum before the collision: 1,500 kg . m/s
6 0
3 years ago
Gravitational attraction depends on the mass of the objects as well as their distance. The gravitational force between objects i
shepuryov [24]
<h2>Answer: Gravitational attraction will be the same</h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

F=G\frac{m_{1}m_{2}}{r^2}    (1)

Where:

F is the module of the force exerted between both bodies

G is the universal gravitation constant.

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

Now, if we double both masses and the distance also doubles, this means:

m_{1} and m_{2} will be now 2m_{1} and 2m_{2}

r will be now 2r

Let's rewrite the equation (1) with this new values:

F=G\frac{(2m_{1})(2m_{2})}{(2r)^2}    (2)

Solving and simplifying:

F=4G\frac{m_{1}2m_{2}}{4r^2}    

F=G\frac{m_{1}m_{2}}{r^2}     (3)

As we can see, equation (3) is the same as equation (1).

So, if the masses both double and the distance also doubles the <u>Gravitational attraction between both masses will remain the same.</u>

7 0
3 years ago
Read 2 more answers
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