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2 years ago

12

During a summer with little rainfall, your house on a hill slope experiences an interval during which your well runs dry. You ha

ve to borrow water from your neighbor who lives downslope from you. Why does your neighbor have water when you don’t?

1 answer:

irina1246 [14]2 years ago

6 0

Answer and Explanation:

You don't have water because of two possible reasons:

Because of the summer and the little rain, the underwater supply goes low.
The slope in the hill you live makes the underground water goes down by the effect of gravity. Imagine the underground water like a small tank, when the water is reduced for any reason the bottom of the tank will have the remaining water, while the top part will be "empty".

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Convertir:<br> A. 3Km a m<br> B. 250 ma Km<br> C. 1000Cm a m<br> D. 10000 mm a Cm

Katen [24]

Answer:

A. 3,000,000 m

B. 0.25 km

C. 10 m

D. 1,000 cm

Explanation:

no hablo español, así que solo ingrese esto en el traductor de G*ogle

A. One kilometer equals 1000 meters, so

3,000*1,000 = 3,000,000 m

B. One meter equals 0.001 kilometer, so

250*0.001 = 0.25 km

C. One centimeter equals 0.01 meter

1,000*0.01 = 10 m

D. One milimeter equals 0.1 centimer, so

10,000*0.1 = 1,000

4 0

2 years ago

A horizontal force, F1 = 65 N, and a force, F2 = 12.4 N acting at an angle of θ to the horizontal, are applied to a block of mas

Nezavi [6.7K]

Answer:

(a) FN = 24.18 N

(b) a = 22.87 m/s²

Explanation:

Newton's second law of the block:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the block on the surface and the y-axis in the direction perpendicular to it.

F₁ : Horizontal force

F₂ : acting at an angle of θ to the horizontal,

W: Weight of the block : In vertical direction

FN : Normal force : perpendicular to the direction the surface

fk : Friction force: parallel to the direction to the surface

Known data

m =3.1 kg : mass of the block

F₁ = 65 N, horizontal force

F₂ = 12.4 N acting at an angle of θ to the horizontal

θ = 30° angle θ of F₂ with respect to the horizontal

μk = 0.2 : coefficient of kinetic friction between the block and the surface

g = 9.8 m/s² : acceleration due to gravity

Calculated of the weight of the block

W= m*g = (3.1 kg)*(9.8 m/s²) = 30.38 N

x-y F₂ components

F₂x = F₂cos θ= (12.4)*cos(30)° = 10.74 N

F₂y = F₂sin θ= (12.4)*sin(30)° = 6.2 N

a)Calculated of the Normal force (FN)

We apply the formula (1)

∑Fy = m*ay ay = 0

FN+6.2-30.38 = 0

FN = -6.2+30.38

FN = 24.18 N

Calculated of the Friction force:

fk=μk*N= 0.2* 24.18 N = 4.836 N

b) We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax , ax= a : acceleration of the block

F₁ + F₂x -fk = ( m)*a

65 N + 10.74 -4.836 = ( 3.1)*a

70.904 = ( 3.1)*a

a = (70.904 ) / ( 3.1)

a = 22.87 m/s²

4 0

2 years ago

The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5

KiRa [710]

Answer:

0.68 s

Explanation:

We are given that

Initial velocity of box= $u=13m/s$

Final velocity of box=v=11.5 m/s

Distance=d=8.5 m

We have to find the time taken by box to slow by this amount.

We know that

$v^2-u^2=2as$

Substitute the values

$(11.5)^2-(13)^2=2a(8.5)$

$132.25-169=17a$

$-36.75=17a$

$a=\frac{-36.75}{17}=-2.2m/s^2$

We know that

Acceleration= $a=\frac{v-u}{t}$

Substitute the values

$-2.2=\frac{11.5-13}{t}$

$-2.2=\frac{-1.5}{t}$

$t=\frac{1.5}{2.2}=0.68 s$

Hence, the time taken by box to slow by this amount=0.68 s

8 0

3 years ago

The boom is supported by the winch cable that has a diameter of 0.5 in. and allowable normal stress of σallow=21 ksi. A boom ris

Andrew [12]

Explanation:

Let us assume that forces acting at point B are as follows.

$\sum F_{x} = 0$

$T + F_{AB} Sin 60$ = 0 ...... (1)

$\sum F_{y}$ = 0

$F_{AB} Cos 60 + W$ = 0 .......... (2)

Hence, formula for allowable normal stress of cable is as follows.

$\sigma_{allow} = \frac{T}{A}$

T = $(20 \times 1000) \frac{\pi}{4} \times (0.5)^{2}$

= 3925 kip

From equation (1), $F_{AB}Sin (60^{o})$ = -3925

$F_{AB} \times -0.304 8$ = -3925

$F_{AB}$ = 12877.29 kip

From equation (2), -12877.29 (Cos 60) + W = 0

$-12877.29 kip \times \frac{1}{2} + W$ = 0

W = 6438.64 kip

Thus, we can conclude that greatest weight of the crate is 6438.64 kip.

5 0

2 years ago

What is the equivalent resistance between the points A and B of the network?

Dominik [7]

Explanation:

First, simplify the circuit. Then calculate the parallel and consecutive resistances to find the answer.

3 0

2 years ago