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3 years ago

Which process is a from of mechanical weathering

1 answer:

7 0

Ice. The formation of ice in the myriad of tiny cracks and joints in a rock's surface slowly pries it apart over thousands of years. Frost wedging results when the formation of ice widens and deepens the cracks, breaking off pieces and slabs. Frost wedging is most effective in those climates that have many cycles of freezing and thawing. Frost heaving is the process by which rocks are lifted vertically from soil by the formation of ice. Water freezes first under rock fragments and boulders in the soil; the repeated freezing and thawing of ice gradually pushes the rocks to the surface.

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An automobile having a mass of 1100 kg initially moves along a level highway at 110 km/h relative to the highway. It then climbs

Greeley [361]

Answer:

Change in kinetic energy=-513.652 KJ

Change in potential energy=431.64KJ

Explanation:

We are given that

Mass of an automobile , m=1100 kg

Initial speed, u=110 km/h= $110\times \frac{5}{18}=30.56 m/s$

Where $1 km/h=5/18 m/s$

Height , $h_2=40 m$

$g=9.81 m/s^2$

Final speed, v=0

Change in kinetic energy, $\Delta K.E=\frac{1}{2}m(v^2-u^2)$

$\Delta K.E=\frac{1}{2}(1100)(0-(30.56)^2)=-513652.48 J$

$\Delta K.E=-\frac{513652.48}{1000}=-513.652 KJ$

Where 1 KJ=1000 J

Change in potential energy, $\Delta P.E=mgh(h_2-h_1)$

Initially height, h1=0

Using the formula

$\Delta P.E=1100\times 9.81(40-0)$

$\Delta P.E=431640J$

$\Delta P.E=431.64KJ$

6 0

3 years ago

What are the horizontal structures beneath a slab that help transfer the load from the slab to the columns?

Delicious77 [7]

Answer: D) Beams

Explanation:

3 0

2 years ago

Read 2 more answers

The coolant heat storage system:

Monica [59]

Answer:

Explanation:

This is because many things, such as pcs, over heat

8 0

3 years ago

Please read and answer each question carefully.

Klio2033 [76]

the answer is (c)

After the vehicle is involved in a car accident or fire

5 0

3 years ago

Find the time-domain sinusoid for the following phasors:_________

sattari [20]

Answer:

a. r(t) = 6.40 cos (ωt + 38.66°) units

b. r(t) = 6.40 cos (ωt - 38.66°) units

c. r(t) = 6.40 cos (ωt - 38.66°) units

d. r(t) = 6.40 cos (ωt + 38.66°) units

Explanation:

To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:

(i) Convert the phasor to polar form. The polar form is written as;

r∠Ф

Where;

r = magnitude of the phasor = $\sqrt{a^2 + b^2}$

Ф = direction = tan⁻¹ ( $\frac{b}{a}$ )

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:

r(t) = r cos (ωt + Φ)

Where;

ω = angular frequency of the sinusoid

Φ = phase angle of the sinusoid

(a) 5 + j4

(i) convert to polar form

r = $\sqrt{5^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

5 + j4 = 6.40∠38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt + 38.66°)

(b) 5 - j4

(i) convert to polar form

r = $\sqrt{5^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

5 - j4 = 6.40∠-38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt - 38.66°)

(c) -5 + j4

(i) convert to polar form

r = $\sqrt{(-5)^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{-5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

-5 + j4 = 6.40∠-38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt - 38.66°)

(d) -5 - j4

(i) convert to polar form

r = $\sqrt{(-5)^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{-5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

-5 - j4 = 6.40∠38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt + 38.66°)

3 0

3 years ago