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2 years ago

What type of plate boundary interactions takes place when tectonic plates move apart?

1 answer:

8 0

Tectonic plates interactions are of three different basic types: Divergent boundaries are areas where plates move away from each other, forming either mid-oceanic ridges or rift valleys. <span />

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Please help me and Can you show your work

RSB [31]

Answer:

7. 1ml = 20 drops

325ml = 6500

( 325 x 20 = 6500)

there are 6500 drops in a pepsi can

8. not sure sorry ; (

9. 1 mile per hour = 88 feet per minute

45 miles per hour = 3960 feet per minute

( 88 x 45 = 3960)

there are 3960 feet per minute in 45 miles per hour

4 0

2 years ago

In a RLC circuit, a second capacitor is connected in parallel with the capacitor previously in the circuit. What is the effect o

Marrrta [24]

Answer:

<h2>Case i) if $\omega L > \frac{1}{\omega c}$ </h2><h2>So initially if the circuit is inductive in nature then its net impedance will decrease after this</h2><h2>Case ii) if $\omega L < \frac{1}{\omega c}$ </h2><h2>So initially if the circuit is capacitive in nature then its net impedance will increase after this</h2>

Explanation:

As we know that the impedance of the circuit is given as

$z = \sqrt{(\omega L - \frac{1}{\omega c})^2 + R^2}$

when we join another identical capacitor in parallel with previous capacitor in the circuit then we will have for parallel combination

$c_{eq} = c_1 + c_2$

so it is

$c_{eq} = 2c$

now we have

$z = \sqrt{(\omega L - \frac{1}{2\omega c})^2 + R^2}$

Case i) if $\omega L > \frac{1}{\omega c}$

So initially if the circuit is inductive in nature then its net impedance will decrease after this

Case ii) if $\omega L < \frac{1}{\omega c}$

So initially if the circuit is capacitive in nature then its net impedance will increase after this

7 0

2 years ago

Suppose earth's mass increased but earth's diameter didn't change. Describe how the gravitational force between Earth and the ob

zheka24 [161]

Answer:

It increases proportionally

Explanation:

The gravitational force between the Earth and an object on its surface is given by

$F=G\frac{Mm}{R^2}$

where

G is the gravitational constant

M is the Earth's mass

m is the mass of the object

R is the Earth's radius

In this problem, the Earth's mass is increased, while the diameter (and therefore, the radius) doesn't change. From the equation, we see that the gravitational force is directly proportional to the Earth's mass: therefore, if the mass is increased, the force will increase as well by the same proportion (for example, if the mass is doubled, the force will double as well)

7 0

2 years ago

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.220 rev/s . The magnitude

matrenka [14]

Answer:

1) The fan's angular velocity after 0.208 seconds is approximately 2.585 rad/s

2) The number of revolutions the blade has travelled in 0.208 s is approximately 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is approximately 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds is approximately 2.312 m/s²

Explanation:

The given parameters are;

The initial velocity of the fan, n = 0.220 rev/s

The magnitude of the angular acceleration = 0.920 rev/s²

The direction of the angular acceleration and the angular velocity = Clockwise

The diameter of the circle formed by the electric ceiling fan blades, D = 0.800 m

1) The initial angular velocity of the fan, ω₀ = 2·π × n = 2·π × 0.220 rev/s = 1.38230076758 rad/s

The angular acceleration of the fan, α = 2·π×0.920 rad/s² = 5.78053048261 rad/s²

The fan's angular velocity, 'ω', after a time t = 0.208 seconds has passed is given as follows;

ω = ω₀ + α·t

From which we have;

ω = 1.38230076758 rad/s + 5.78053048261 rad/s × 0.208 s = 2.58465110796 rad/s

The fan's angular velocity after 0.208 seconds is ω ≈ 2.585 rad/s

2) The number of revolutions the blade has travelled in the given time interval is given from the angle turned, 'θ', in the given time as follows;

θ = ω₀·t + 1/2·α·t²

θ = 1.38230076758 × 0.208 + 1/2 × 5.78053048261 × 0.208² = 0.41256299505 radians

2·π radians = 1 revolution

∴ 0.41256299505 radians = 0.41256299505 radian× 1 revolution/(2·π radian) = 0.06566144 revolution

The number of revolutions the blade has travelled in 0.208 s ≈ 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is given as follows;

The tangential speed, $v_t$ = ω × r = ω × D/2

At t = 0.208 s, ω = 2.58465110796 rad/s, therefore, we have;

$v_t$ = ω × D/2 = 2.58465110796 × 0.800/2 = 1.0338604413

The tangential speed, $v_t$ = 1.0338604413 m/s

The tangential speed ≈ 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds, 'a' is given as follows;

a = α × r = α × D/2

a = 5.78053048261 × 0.800/2 = 2.31221219304

The tangential acceleration, a ≈ 2.312 m/s²

4 0

2 years ago

Which volcanic hazard can block the sunlight and temporarily cool the Earth’s surface?

marissa [1.9K]

Pretty sure its volcanic ash or magma, hope this helps

4 0

2 years ago