All categories

3 years ago

Name two examples of deposition.

2 answers:

8 0

The reverse of deposition is sublimation and hence sometimes deposition is called desublimation. One example of deposition is the process by which, in sub-freezing air, water vapor changes directly to ice without first becoming a liquid.

TEA [102]3 years ago

4 0

some examples of deposition are when streams deposit sediments into deltas and alluvial fans.

Also, wind deposits sand when wind velocity decreases because it cannot carry enough particles

Next, glaciers deposit all sizes of sediment. They leave a big pile of rocks called moraine. Moraine usually is deposited in lakes called terminal moraines.

Lastly, water deposits sediments in an ocean bedrock into layers. The layers are based on the size of the particle. The bottom layer will be filled with large sediments and the top layer will be filled with loose sediments such as sand and pebbles.

You might be interested in

Putting effort at point IV of this lever offers no mechanical advantage because itA)requires a smaller distance

julsineya [31]

Putting effort at point IV of this lever gives no mechanical advantage because it requires a greater force. So the correct answer is C).

3 0

3 years ago

Read 2 more answers

Scientists treat the number of stars in a given volume of space as a Poisson random variable. The density of our galaxy in the v

vladimir1956 [14]

Answer:

$P(X\ge 1) = 0.9502$

Explanation:

Given

Density = 3 starts in 10 cubic light years.

Required

Determine the probability of 1 or more in 10 cubic light years

Since the number of stars follow a Poisson distribution, we make use of:

$P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}$

$\lambda = density$

$\lambda = \frac{3}{10}$

$\lambda = 0.3$

T = the light years

$T = 10$

Calculating $P(X \ge 1)$

In probability:

$P(X \ge 1) = 1 - P(X = 0)$

Calculating P(X=0)

Substitute 0 for k and the values for $\lambda$ and T in

$P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}$

$P(X=0) = (0.3* 10)^0 * \frac{ e^{-0.3 * 10}}{0!}$

$P(X=0) = (3)^0 * \frac{ e^{-0.3 * 10}}{1}$

$P(X=0) = (3)^0 * e^{-0.3 * 10}$

$P(X=0) = 1 * e^{-0.3 * 10}$

$P(X=0) = 1 * e^{-3}$

$P(X=0) = e^{-3}$

$P(X=0) = 0.04979$

Substitute 0.04979 for P(X=0) in $P(X \ge 1) = 1 - P(X = 0)$

$P(X\ge 1) = 1 - 0.04979$

$P(X\ge 1) = 0.95021$

$P(X\ge 1) = 0.9502$ --- approximated

Hence, the required probability is 0.9502

3 0

3 years ago

An elevator is moving upward at 1.00 m/s when it experiences an acceleration 0.37 m/s2 downward, over a distance of 0.79 m. What

Nastasia [14]

Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

We're given $v_i=1.00\,\frac{\mathrm m}{\mathrm s}$ , $a=-0.37\,\frac{\mathrm m}{\mathrm s^2}$ (so we take the upward direction to be positive), and $\Delta y=0.79\,\mathrm m$ . Then the final velocity $v_f$ satisfies