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3 years ago

13

How do volcanoes above hot spots differ from volcanoes above subduction zones?

2 answers:

miss Akunina [59]3 years ago

4 0

Option (C)

Explanation:

The magma in the hot-spot region are basaltic, they can move fast because of their low viscosity. Whereas, the magma present in the subduction zone are rhyolitic, and moves slowly due to its high viscosity. Viscosity of a magma is directly proportional to its silica content.

This is the difference between the volcanoes over a hot-spot region and over subduction zone.

Answer: Thus, the correct answer is option (C).

BARSIC [14]3 years ago

3 0

C. Volcanoes above subduction zones have lava that is not basaltic.

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Which type of energy is water held back by a dam?

stepladder [879]

Answer: hydroelectric dams

Explanation: Enormous energy is stored in water held back by hydroelectric dams. The energy is transformed into (a form of kinetic energy) as it falls across the dam. The falling water strikes the blades of a turbine and makes them spin, and the then turns the shaft of a generator.

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What is the acceleration of a proton moving with a speed of 6.5 m/s at right angles to a magnetic field of 1.5 T?

Brilliant_brown [7]

Answer:

The acceleration of the proton is 9.353 x 10⁸ m/s²

Explanation:

Given;

speed of the proton, u = 6.5 m/s

magnetic field strength, B = 1.5 T

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F = ma = qvB(sin90°)

ma = qvB

where;

m is mass of the proton, = 1.67 x 10⁻²⁷ kg

charge of the proton, q = 1.602 x 10⁻¹⁹ C

The acceleration of the proton is given by;

$a = \frac{qvB}{m}\\\\a = \frac{(1.602*10^{-19})(6.5)(1.5)}{1.67*10^{-27}}\\\\a = 9.353*10^8 \ m/s^2$

Therefore, the acceleration of the proton is 9.353 x 10⁸ m/s²

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3 years ago

what is the kinetic energy of an object that has mass of 30 kilograms and move with a velocity of 20 m/s

kotegsom [21]

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3 years ago

SIZIF [17.4K]

C is the first & the second question is A

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3 years ago

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A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}$

so that the <em>x</em>-component of $\vec v_{\rm ave}$ is

$\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}$

and its <em>y</em>-component is

$\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}$

Solve for $x_2$ and $y_2$ , which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

$x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}$

$y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}$

Note that I'm reading the given details as

$x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}$

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

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3 years ago