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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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Frictional force increases with the increase in the _______________ of the surface.

Sedaia [141]

Answer:

Frictional force increases with the increase in the roughness of the surface.

Explanation:

You will see that the rougher the surface, the greater the wear and tear.

7 0

3 years ago

A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =

castortr0y [4]

Answer:

a) 43.20V

b) 2.71W/s

c) 40.25s

d) 7.77Nm

Explanation:

(a) The emf of a rotating coil with N turns is given by:

$emf=NBA\omega sin(\omega t)$

N: turns

B: magnitude of the magnetic field

A: area

w: angular velocity

the emf max is given by:

$emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V$

(b) the maximum rate of change of the magnetic flux is given by:

$\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}$

(c) $emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V$

(d) The torque is given by:

$\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm$

3 0

3 years ago

Because communication involves two or more people acting in both sender and receiver roles and because their messages are depend

djyliett [7]

Answer:

Transactional process

Explanation:

Because communication involves two or more people acting in both sender and receiver roles and because their messages are dependent on and influenced by those of their partner means that communication is a transactional process i.e. a give and take process.

If the sender relays a message and the receiver does not reply or reply not taking cognizance of the sender's message, communication has not taken place.

For example, if consider this.

Sender: How old are you?

Reciever: My Cat is Red.

The response of the receiver is not in tune with the sender and so communication has not taken place.

4 0

3 years ago

Two metal spheres of identical mass m = 4.20 g are suspended by light strings 0.500 m in length. The left-hand sphere carries a

Yanka [14]

Answer:

Explanation:

Answer:

0.632 m

Explanation:

let the equilibrium separation between the charges is d and the angle made by string with the vertical is θ.

According to the diagram,

d = L Sinθ + L Sinθ = 2 L Sinθ .....(1)

Let T be the tension in the string.

Resolve the components of T.

T Sinθ = k q1 q2 / d^2

T Sinθ = k q1 q2 / (2LSinθ)² .....(2)

T Cosθ = mg .....(3)

Dividing equation (2) by equation (3), we get

tanθ = k q1 q2 / (4 L² Sin²θ x mg)

tan θ Sin²θ = k q1 q2 / (4 L² m g)

For small value of θ, tan θ = Sin θ

So,

Sin³θ = k q1 q2 / (4 L² m g)

Sin³θ = (9 x 10^9 x 0.785 x 10^-6 x 1.47 x 10^-6) / (4 x 0.5 x 0.5 x 4.20 x 10^-3 x 9.8)

Sin³θ = 0.2523

Sinθ = 0.632

θ = 39.2 degree

So, the separation between the two charges, d = 2 x L x Sin θ

d = 2 x 0.5 x 0.632 = 0.632 m

6 0

3 years ago

the power rating of an electric lawn mower is 2200 watts. if the lawn mower was used for 30 mins, and 650 Newtons of force was u

Tasya [4]

Answer:

Calculate the work done by a 47 N force pushing a 0.025 kg pencil 0.25 m ... A boy on a bicycle drags a wagon full of newspapers at 0.80 m/s for 30 min ... A power mower does 9.00 x 105 J of work in 0.500 h. ... p: W 2200ch: w will320,000 T/ ... How much electrical energy (in kilowatt hours) would a 60.0 W light bulb use in ..

Explanation:

4 0

3 years ago