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2 years ago

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

1 answer:

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:

You might be interested in

Which object will have greater acceleration? Why?

dexar [7]

Answer:

Object D

Explanation:

Use Newton's Second Law to determine the acceleration that each object has.

F = ma

The force applied in both cases is 50 N, but the mass for object C and object D is different.

Let's start with object C first:

F = ma
50 N = 10 kg · a
50 = 10a
5 = a

The acceleration object C undergoes is 5 m/s².

Now let's calculate object D next:

F = ma
50 N = 2 kg * a
50 = 2a
25 = a

The acceleration object D undergoes is 25 m/s².

Object D has greater acceleration because it has a smaller mass. The object with a smaller mass will accelerate more in order to satisfy Newton's 2nd Law.

7 0

3 years ago

[01.02]

julia-pushkina [17]

That information describes the plane's speed.
If he had also told them what direction they were flying, then
they would have been able to put the two pieces of information
together, and they would know the plane's velocity.

6 0

4 years ago

Read 2 more answers

A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric

jeka94

Answer:

a) F₃₁ = 63.0 μN

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

$F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}} = 63.0 \mu N (1)$

Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

$F_{32} = F_{3} - F_{31} = 49.0\mu N - 63.0\mu N = -14.0 \mu N (2)$

Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

$q_{2} = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC (3)$

6 0

3 years ago

Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$