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3 years ago

What kinds of features would you expect to see at the edges of two plates, like these that are moving apart at their boundary?

1 answer:

8 0

Along these boundaries you will, lava spewing from long fissures and geysers spurting superheated water, have frequent earthquakes strike along the rift.

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Mount Tambora, in Indonesia, erupted in April 1815, spewing huge amounts of fine ash into the air. The ash quickly spread throug

fredd [130]

Crops need sunlight to grow as well as water.

Reduced sunlight means reduced growth rate of crops means reduced crop yield means reduced number of mouths each yield can feed means famine means death.

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4 years ago

An open 1-m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the ce

Mamont248 [21]

Answer:

Explanation:

To find the angular velocity of the tank at which the bottom of the tank is exposed

From the information given:

At rest, the initial volume of the tank is:

$V_i = \pi R^2 h_i --- (1)$

where;

height h which is the height for the free surface in a rotating tank is expressed as:

$h = \dfrac{\omega^2 r^2}{2g} + C$

at the bottom surface of the tank;

r = 0, h = 0

∴

$h = \dfrac{\omega^2 r^2}{2g} + C$

0 = 0 + C

C = 0

Thus; the free surface height in a rotating tank is:

$h=\dfrac{\omega^2 r^2}{2g} --- (2)$

Now; the volume of the water when the tank is rotating is:

dV = 2π × r × h × dr

Taking the integral on both sides;

$\int \limits ^{V_f}_{0} \ dV = \int \limits ^R_0 \times 2 \pi \times r \times h \ dr$

replacing the value of h in equation (2); we have:

$V_f} = \int \limits ^R_0 \times 2 \pi \times r \times ( \dfrac{\omega ^2 r^2}{2g} ) \ dr$

$V_f = \dfrac{ \pi \omega ^2}{g} \int \limits ^R_0 \ r^3 \ dr$

$V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{r^4}{4} \Big]^R_0$

$V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{R^4}{4} \Big] --- (3)$

Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.

Then $V_f = V_i$

Replacing equation (1) and (3)

$\dfrac{\pi \omega^2}{g}( \dfrac{R^4}{4}) = \pi R^2 h_i$

$\omega^2 = \dfrac{4g \times h_i }{R^2}$

$\omega =\sqrt{ \dfrac{4g \times h_i }{R^2}}$

$\omega = \sqrt{\dfrac{4 \times 9.81 \ m/s^2 \times 0.7 \ m}{(0.5)^2} }$

$\omega = \sqrt{109.87 }$

$\mathbf{\omega = 10.48 \ rad/s}$

Finally, the angular velocity of the tank at which the bottom of the tank is exposed = 10.48 rad/s

6 0

3 years ago

If the spring constant is doubled, what value does the period have for a mass on a spring?

soldi70 [24.7K]

if spring constant is doubled, the mass on spring will be doubled as well. according to this formula, F=ke

k stands for spring constant and e stands for the length extended

4 0

3 years ago

Read 2 more answers

A paper clip is pushed horizontally off a table with speed 1.5m/s. If the table has a height of 1.1m how far from the table does

Julli [10]

the answer is D. 0.71 m

3 0

3 years ago

Read 2 more answers

Proposed Kinematic Exercise I

valentinak56 [21]

Answer:

(a) 20 m

(b) 6 m/s²

Between t=2 and t=4, the body moves to the right.

Explanation:

v = 3t² − 6t

x(0) = 4

(a) Position is the integral of velocity.

x = ∫ v dt

x = ∫ (3t² − 6t) dt

x = t³ − 3t² + C

Use initial condition to find value of C.

4 = 0³ − 3(0)² + C

4 = C

x = t³ − 3t² + 4

Find position at t = 4.

x = 4³ − 3(4)² + 4

x = 20

(b) Acceleration is the derivative of velocity.

a = dv/dt

a = 6t − 6

Find acceleration at t = 2.

a = 6(2) − 6

a = 6

v = 3t (t − 2)

The velocity is 0 at t = 0 and t = 2. Evaluate the intervals.

When 0 < t < 2, v < 0.

When t > 2, v > 0.

3 0

3 years ago