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2 years ago

Why aren’t the Appalachian Mountains still as tall as the Himalayas?

Physics

2 answers:

IRISSAK [1]2 years ago

4 0

Answer:

It probably won't get any taller, though. From a geological standpoint, the Appalachians haven't seen much growth in quite a while. Since the dawn of the dinosaurs about 225 million years ago, this range has been getting whittled down by weathering forces.

Explanation:

stealth61 [152]2 years ago

3 0

Answer:

mountains are limited in their theoretical height by several processes. First is isostasy: the bigger a mountain gets, the more it weighs down its tectonic plate, so it sinks lower. ... Bottom line: mountains can get taller than Mount Everest in earth gravity, like the Appalachians probably did—but not much taller.

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When a person jumps up he or she is able to come down because of what?

Ainat [17]

Answer:

<h2>due to gravity........hope its helpful </h2>

7 0

3 years ago

Read 2 more answers

A physics student of mass 51.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriend

balandron [24]

Answer:

Explanation:

The wheel and falling student will have common acceleration .

For rotational motion of wheel

Tx r = I α , T is tension in the crank , α is angular acceleration of wheel , I is moment of inertia , r is radius of the wheel.

= I a / r

T = I a / r²

For motion of student

Mg - T = Ma , M is mass of the wheel.

Mg - I a / r² = Ma

Mg = Ma +I a / r²

Mg = (M +I / r²)a

a = Mg / (M +I / r²)

= 51 x 9.8 / ( 51 + 9.6 / .3² )

499.8 / (51+ 106.67 )

= 499.8 / 157.67

= 3.17 m / s².

If time t is taken to fall by 12 m

12 = 1/2 a t²

24 / a = t²

24 / 3.17 =t²

t²= 7.57

t = 2.75 s

velocity to reach sidewalk

v = u + at

= 3.17 x 2.75

= 8.72 m / s

5 0

3 years ago

Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24

Oksi-84 [34.3K]

Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
$E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}$
Now At xp we have:
$\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0$
$\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0$
Which is a second order equation, using the quadratic formula to solve for xp would give us:
$xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
or
$xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.