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4 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]4 years ago

4 0

<span>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.</span><span>
</span>

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A weightlifter lifts a 250-kg mass 0.5 meters above his head, how much PEg does the mass have (Note: g=9.8 m/s2)? Round your ans

Anna71 [15]

The Gravitationa potential energy of the mass (PEG) is given by:
$U=mgh$
where
m is the mass
g is the gravitational acceleration
h is the heigth of the mass above the reference level (the ground)

In this problem, $m=250 kg$ and $h=0.5 m$ , therefore the gravitational potential energy of the mass is:
$U=mgh=(250 kg)(9.8 m/s^2)(0.5 m)=1225 J$

6 0

4 years ago

Read 2 more answers

If the velocity of blood flow in the aorta is normally about 0.32 m/s, what beat frequency would you expect if 4.40-MHz ultrasou

dusya [7]

Answer:

The beat frequency is 0.0019 MHz.

Explanation:

Given that,

Velocity = 0.32 m/s

Frequency = 4.40 MHz

Speed of wave = 1540 m/s

We need to calculate the frequency

Case (I),

Observer is moving away from the source

Using Doppler's effect

$f'=\dfrac{v-v'}{v}f$

Where, v' = speed of observer

Put the value into the formula

$f'=\dfrac{1540-0.32}{1540}\times4.40$

$f'=4.399\ MHz$

Case (II),

Cell is as the source of sound of frequency f' and it moving away from the observer.

Using formula of frequency

$f''=\dfrac{v-v_{s}}{v+v_{s}}\times f$

$f''=\dfrac{1540-0.32}{1540+0.32}\times4.399$

$f''=4.3971\ MHz$

We need to calculate the beat frequency

$\Delta f= f'-f''$

$\Delta f=4.399-4.3971=0.0019\ MHz$

Hence, The beat frequency is 0.0019 MHz.

4 0

3 years ago

A bus decreases its speed

eduard

Answer:

-6 km/h per minute

Explanation:

7 0

3 years ago

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

3 years ago

Describe what happens to the electric field lines when two objects with unlike charges are brought near each other.

harina [27]

Hello.

The answer is:

It creates a spark.

Then thespark can sometimes start a fire or other serious problems.

Have a nice day

4 0

3 years ago

Read 2 more answers