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lisov135 [29]
3 years ago
13

What is a cone of depression

Physics
1 answer:
Thepotemich [5.8K]3 years ago
7 0

A Cone of Depression occurs when a well pumps water from a certain area and causes a depression in a cone shape in the water table.


Hope this helps! :)


if you get it correct Brainliest is greatly appreciated

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Identify the arrows that show the movement of carbon from the biosphere to the atmosphere in the carbon cycle.
Illusion [34]

Answer:

All of the arrows pointing up that have a red box NOT the arrow pointing down with a red box. (if the blue squigglies on the water are arrows then they count too, the picture is not too clear)

Explanation:

8 0
3 years ago
What is the potential energy of a 25 kg bicycle resting at the top of a hill 3 m high? (Formula: PE = mgh)
Lisa [10]

Answer:

The answer would be 735J

Explanation:

PE=mgh

=(mass)(force of gravity)(height)

=(25kg)(9.8m/s^2)(3m)

=735J

8 0
3 years ago
Read 2 more answers
HELP PLEASE <br> Find the net force necessary for a 15 kg object to accelerate at 20 m/s/s.
tensa zangetsu [6.8K]

Answer:

<h3>The answer is 300 N</h3>

Explanation:

The force acting on an object given the mass and acceleration we use the formula

<h3>force = mass × acceleration</h3>

We have

force = 15 × 20

We have the final answer as

<h3>300 N</h3>

Hope this helps you

3 0
2 years ago
A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, wh
ladessa [460]

Answer:

a) p₂ = 1.88 kg*m/s

   θ = 273.4 º

b)  Kf = 37% of Ko

Explanation:

a)

  • Assuming no external forces acting during the collision, total momentum must be conserved.
  • Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.
  • The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.
  • So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:

       p_{o1x} = 2.00 kg*m/s (1)

       p_{o1y} = 0 (2)

  • We can do the same for the particle moving along the positive y-axis:

        p_{o2x} = 0 (3)

        p_{o2y} = 4.00 kg*m/s (4)

  • Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.
  • Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:

       p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)

      p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s  (6)

  • Now, the total initial momentum, along these directions, must be equal to the total final momentum.
  • We can write the equation for the x- axis as follows:

       p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x}  (7)

  • We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:

       p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)

  • Now, we can repeat exactly the same process for the y- axis, as follows:

       p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y}  (9)

  • We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:

       p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)

  • Since we have the x- and y- components of the final momentum of  the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:

       p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} }  = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)

  • We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:

       tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)

  • The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.

b)

  • Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:

       \frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)

  • So, the final kinetic energy has lost a 37% of the initial one.

6 0
2 years ago
What mass of steam at 100°C must be mixed with 119 g of ice at its melting point, in a thermally insulated container, to produce
N76 [4]

Answer:M=27.92\ gm

Explanation:

Given

mass of ice m=119\ gm

Final temperature of liquid T_f=57^{\circ}C

Specific heat of water c=4186\ J/kg-K

Latent heat of fusion L=333\ kJ/kg

Latent heat of vaporization L_v=2256\ kJ/kg

Suppose M is the mass of steam at 100^{\circ} C

Heat required to melt ice and convert it to water at 57^{\circ}C

Q_1=mL+mc(T_f-0)

Heat released by steam

Q_2=ML_v+Mc(100-T_f)

Q_1 and Q_2 must be equal as the heat gained by ice is equal to Heat released by steam

Q_1=Q_2

\Rightarrow mL+mc(T_f-0)= ML_v+Mc(100-T_f)

\Rightarrow M=\dfrac{m[L+c\times T_f]}{L_v+c(100-T_f)}

\Rightarrow M=\dfrac{119[333\times 10^3+4186\times 57]}{2256\times 10^3+4186\times (100-57)}

\Rightarrow M=119\times 0.2346

M=27.92\ gm

7 0
3 years ago
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