How does cardiovascular fitness allow you to exercise longer

natulia [17]

Answer: In a divergent plate boundary, seafloor spreading taking place. It leads to the formation of oceans as new materials are added here along the mid-oceanic ridge. There occur volcanism and shallow-focus earthquakes.

In a convergent plate boundary, two plates collide to form high mountain belts and also volcanic eruptions take place. There occur long chains of volcanic as well as island arcs, in association with deep-focus earthquakes.

In a transform plate boundary, two plates slide past each other, conserving the plates. Shallow-focus earthquakes are generated here.

The earth has experienced various geological processes, such as weathering and erosion of rocks, earthquakes, volcanic eruptions, mass extinction events, plate tectonic movements and many more. These continuous processes have configured the present shape of the earth's surface.

For example, the breaking up of the supercontinent Pangea divided into Laurasia and Gondwanaland and subsequently formed the present scenario. This separation of continents has taken place due to the convection current that generates in the mantle.

5 0

4 years ago

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Uranium-235 decays to thorium-231 with a half-life of 700 million years. When a rock was formed, it contained 6400 million urani

Dahasolnce [82]

Answer:

proof in explanation

Explanation:

First, we will calculate the number of half-lives:

$n = \frac{t}{t_{1/2}}$

where,

n = no. of half-lives = ?

t = total time passed = 2100 million years

$t_{1/2}$ = half-life = 700 million years

Therefore,

$n = \frac{2100\ million\ years}{700\ million\ years}\\\\n = 3$

Now, we will calculate the number of uranium nuclei left ( $n_u$ ):

$n_u = \frac{1}{2^{n} }(total\ nuclei)\\\\n_u = \frac{1}{2^{3} }(6400\ million)\\\\n_u = \frac{1}{8}(6400\ million)\\\\n_u = 800\ million$

and the rest of the uranium nuclei will become thorium nuclei ( $u_{th}$ )

$n_{th} = total\ nuclei - n_u\\n_{th} = 6400\ million-800\ million\\n_{th} = 5600\ million$

dividing both:

$\frac{n_{th}}{n_u}=\frac{5600\ million}{800\ million} \\\\n_{th} = 7n_u$

Hence, it is proven that after 2100 million years there are seven times more thorium nuclei than uranium nuclei in the rock.

6 0

3 years ago

0.10-kilogram model rocket’s engine is designed to deliver an impulse of 6.0 newton-seconds. If the rocket engine burns for 0.75

UkoKoshka [18]

Answer:

8.0 N

Explanation:

Force: This can be defined as the mass of a body and its acceleration. The S.I unit of Force is Newton (N).

Mathematically, Fore is expressed as

F = ma ........................... equation 1

Where F = force, m = mass, a = acceleration.

and

I = mΔv

Δv = I/m ............................ Equation 2

Where I = impulse, m = mass, Δv = change in velocity

Given: I = 6.0 Newton-seconds, m = 0.1 kilogram.

Substituting into equation 2

Δv = 6.0/0.1

Δv = 60 m/s.

But

a = Δv/t

where t = time = 0.75 seconds.

a = 60/0.75

a = 80 m/s²

Substitute the values of a and m into equation 1.

F = 0.1(80)

F = 8.0 N.

Thus the average force produced = 8.0 N

6 0

3 years ago

A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude

Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$ .

Now, let us call $\lambda$ the charge per unit length, then we know that

$dQ = \lambda Rd\theta$ ;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda }{R}.$

Now, we know that

$\lambda = 3.0*10^{-9}C/m,$

$k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},$

and the radius of the semicircle is

$\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };$

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9}) }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0

3 years ago

Find the voltage across the 15 Ω resistor. . 35 Ω 20 Ω 15 Ω 10 V

almond37 [142]

Answer:

35 Ω

Explanation:

3 0

3 years ago