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Tresset [83]
2 years ago
15

The process by which land remove surface materials is called

Physics
2 answers:
yuradex [85]2 years ago
4 0
The answer is Deflation
Citrus2011 [14]2 years ago
3 0

Answer:

Deflation

Explanation:

The process by which wind removes surface materials is called deflation.

Deflation is a process by which wind erodes the Earth's exterior and the regions which experience severe erosion are called deflation zones.

deflation originates by the erosive force of a wind that removes the loosened area and this process is facilitated by a dry climate and a loss of vegetative cover that helps to lose the sediment.

deflation is common in the arid regions.

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Which phase of matter contains particles that split into ions and electrons?
Lemur [1.5K]

Answer:

Plasma

Explanation:

7 0
2 years ago
A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa
Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

Q=P\tan\phi...(I)

We need to calculate the \phi

Using formula of \phi

\phi=\cos^{-1}(Power\ factor)

Put the value into the formula

\phi=\cos^{-1}(0.6)

\phi=53.13^{\circ}

Put the value of Φ in equation (I)

Q=600\tan(53.13)

Q=799.99\ kVAR

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

Q'=600\tan(0.95)

Q'=9.94\ KVAR

We need to calculate the difference between Q and Q'

Q''=Q-Q'

Put the value into the formula

Q''=799.99-9.94

Q''=790.05\ KVAR

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0
3 years ago
A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline
balandron [24]

Answer:

<em>2.78m/s²</em>

Explanation:

Complete question:

<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>

According to Newton's second law of motion:

\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x\\

Where:

\mu is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

gsin\theta - \mu g cos\theta = a_x\\

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>

5 0
3 years ago
An ant clings to the outside edge of the tire of an exercise bicycle. When you start pedaling, the ant's speed increases from ze
konstantin123 [22]

Answer:

1- 3.64 m/s

2- 0.28 m

3- 70.8 rad

4- 19.8 m

Explanation:

see attachment

3 0
3 years ago
newton's second law state that a force on a drag racer is equal to the mass of the car times it's acceleration. one of the most
kaheart [24]
I believe the answer you are looking for is the friction of the tires on the race track
7 0
3 years ago
Read 2 more answers
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