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2 years ago

Can Hotspots can be found at the edges of tectonic plates?

Physics

1 answer:

PolarNik [594]2 years ago

6 0

Answer:

No.

Explanation:

Hot spots don't appear at the barriers of tectonic plates, instead, they originate at hot centres known as mantle plumes. Mantle plumes exist below the tectonic plates and may develop a string of volcanoes on the Earths surface.

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On a boat ride, the skipper gives you an extra-large life preserver filled with lead pellets. When he sees the skeptical look on

Lelu [443]

Answer:

True.

Explanation:

He is correct but what he does not tell is that you will be drown. Your life preserver will submerge and displace more water than those of your friends who float at the surface. Although buoyant force on you will be greater , the net downward force on you will still be greater. Hence you will be drown inside the water.

8 0

2 years ago

The weighted average of the masses of the isotopes of an element

solmaris [256]

Answer:

Atomic mass

Explanation:

The weighted average of the masses of the various isotopes of an element makes the atomic mass. This is why most atomic mass values usually have decimals.

In order to calculate the atomic mass of an element using the weighted average masses, we use the expression below:

Atomic mass = ∑ $m_{i}$ $α_{i}$

Where $m_{i}$ is the mass of isotope i

$α_{i}$ is the abundance of isotope i

The abundance or geonormal abundance is the proportion by which each of the isotope occurs in nature.

4 0

3 years ago

A soccer ball of diameter 22.6cm and mass 426g rolls up a hill without slipping, reaching a maximum height of 5m above the base

-Dominant- [34]

Answer with Explanation:

We are given that

Diameter=d=22.6 cm

Mass,m=426 g= $426\times 10^{-3} kg$

1 kg=1000 g

Radius,r= $\frac{d}{2}=\frac{22.6}{2}=11.3 cm=11.3\times 10^{-2} m$

1m=100 cm

Height,h=5m

$I=\frac{2}{2}mr^2$

a.By law of conservation of energy

$\frac{1}{2}I\omega^2+\frac{1}{2}mv^2=mgh$

$\frac{1}{2}\times \frac{2}{3}mr^2\omega^2+\frac{1}{2}mr^2\omega^2=mgh$

$v=\omega r$

$gh=\frac{1}{3}r^2+\frac{1}{2}r^2=\frac{5}{6}r^2\omega^2$

$\omega^2=\frac{6}{5r^2}gh$

$\omega=\sqrt{\frac{6gh}{5r^2}}=\sqrt{\frac{6\times 9.8\times 5}{5(11.3\times 10^{-2})^2}}=67.86 rad/s$

Where $g=9.8m/s^2$

b.Rotational kinetic energy= $\frac{1}{2}I\omega^2=\frac{1}{2}\times \frac{2}{3}mr^2\omega^2=\frac{1}{2}\times \frac{2}{3}(426\times 10^{-3})(11.3\times 10^{-2})^2(67.86)^2=8.35 J$

Rotational kinetic energy=8.35 J

6 0

3 years ago

What is the distance between a(-6,3),b(-2,4),c(-8,3)

Schach [20]

Answer:

Explanation:

Appears to be the vertexes of a triangle.

AB = √(-6 - (-2))² + (3 - 4)²) = √17

AC = √(-6 - (-8))² + (3 - 3)²) = 2

BC = √(-2 - (-8))² + (4 - 3)²) = √37

8 0

2 years ago

If the velocity of a body varied uniformly from 10 m s-1 to 25 m s-1

BaLLatris [955]

Answer:

3m/s2

Explanation:

a = (v-u) / t

a = (25 - 10) / 5

a = 15 / 5

a = 3

4 0

2 years ago