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4 years ago

6

At a boundary, plates move away from each other as magma pushes them from below. When this change happens beneath the ocean floo

r, it results in .

2 answers:

Rasek [7]4 years ago

7 0

The plates under the ocean squeeze really hard and they can start an earthquake.

slava [35]4 years ago

6 0

The first one would Be Divergent, and the Second One would be, Ridges

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(b) Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a c

andreyandreev [35.5K]

Answer:

$\frac{dA}{dt} \approx 245.044\,\frac{m^{2}}{s}$

Explanation:

The formula for the surface of the circle is:

$A(r) = \pi\cdot r^{2}$

The rate of change of the spill area is obtained by deriving the previous formula in terms of time:

$\frac{dA}{dt} = 2\pi\cdot r\cdot \frac{dr}{dt}$

Finally, variables are replaced by known data:

$\frac{dA}{dt} = 2\pi\cdot (39\,m)\cdot (1\,\frac{m}{s} )$

$\frac{dA}{dt} \approx 245.044\,\frac{m^{2}}{s}$

4 0

4 years ago

Read 2 more answers

Which statement supports the idea that the Earth rotates on its axis?

Ivenika [448]

Answer:

c

Explanation:

because the other would not make sense

3 0

3 years ago

When the sun heats up the suface of the water, the water

nadya68 [22]

Answer:

In the water cycle, evaporation occurs when sunlight warms the surface of the water. The heat from the sun makes the water molecules move faster and faster, until they move so fast they escape as a gas. Once evaporated, a molecule of water vapor spends about ten days in the air.

Explanation:

5 0

3 years ago

A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) Ho

Crank

Answer:

a)N = 3.125 * 10¹¹

b) I(avg) = 2.5 × 10⁻⁵A

c)P(avg) = 1250W

d)P = 2.5 × 10⁷W

Explanation:

Given that,

pulse current is 0.50 A

duration of pulse Δt = 0.1 × 10⁻⁶s

a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles

N = Δq/e

charge is given by Δq = IΔt

so,

N = IΔt / e

$N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1$

N = 3.125 * 10¹¹

b) Q = nqt

where q is the charge of 1puse

n = number of pulse

the average current is given as I(avg) = Q/t

I(avg) = nq

I(avg) = nIΔt

= (500)(0.5)(0.1 × 10⁻⁶)

= 2.5 × 10⁻⁵A

C) If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,

eV = K

V = K/e

the power is given by

P = IV

P(avg) = I(avg)K / e

$P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}$

= 1250W

d) Final peak=

P= Ik/e

= $= P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W$

P = 2.5 × 10⁷W

5 0

3 years ago

Read 2 more answers

deff fn [24]

Answer:

31.28m/s

Explanation:

We can use the third key equation of accelerated motion

Δd = viΔt+1/2aΔt^2

subtitute in the values we know

41.6 = vi(1.89)+1/2(-9.8)(1.89)^2

41.6 = vi(1.89) - 17.52

59.12 = vi(1.89)

vi = 31.28

4 0

3 years ago