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2 years ago

You work for NASA and an engineer. Your supervisor is thinking of building a space station instead of a

Physics

1 answer:

expeople1 [14]2 years ago

3 0

Heosndıxnzjdndkdnfmfmfjfmcbvdkfmdj ı dont under stand but ı am dont write

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Xelga [282]

13.6) When any boat displaces a weight of water equal to its own weight, it floats.

7 0

2 years ago

Temperature which is solid becomes a liquid

Brums [2.3K]

Answer:

The answer would be melting point.

Explanation:

Hope this helps. Can you please mark me brainliest

5 0

2 years ago

Read 2 more answers

A confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m a

AlekseyPX

Answer:

Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day

Part (b) The specific discharge of the flow is 0.0363 m/day

Part (c) The average linear velocity of the flow is 0.242 m/day

Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours

Explanation:

Part (a) the flow rate per unit width of the aquifer

From Darcy's law;

$q = -Kb\frac{dh}{dl}$

where;

q is the flow rate

K is the permeability or conductivity of the aquifer = 25 m/day

b is the aquifer thickness

dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m

dl is the change in the horizontal hight = 1000 m

q = -(25*30)*(-1.45/1000)

q = 1.0875 m³/day

Part (b) the specific discharge of the flow

$V = \frac{Q}{A} = \frac{q}{b} = -K\frac{dh}{dl}\\\\V = -(25 m/d).(\frac{-1.45 m}{1000 m}) = 0.0363 m/day$

V = 0.0363 m/day

Part (c) the average linear velocity of the flow assuming steady unidirectional flow

Va = V/Φ

Φ is the porosity = 0.15

Va = 0.0363 / 0.15

Va = 0.242 m/day

Part (d) the time taken for a tracer to travel the distance between the observation wells

The distance between the two wells = 1000 m

average linear velocity = 0.242 m/day

Time = distance / speed

Time = (1000 m) / (0.242 m/day)

Time = 4132.23 days

$= 4132.23 days *\frac{24 .hrs}{1.day} = 99173.52, hours$

4 0

2 years ago

The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?

Yuki888 [10]

Answer:

-.457 m/s^2

Explanation:

Actual weight = 60 .3 (9.81) = 591.54 N

Accel of lift changes this to 60.3 ( 9.81 - L) where L - accel of lift

60.3 ( 9.81 - L ) = 564

solve for L = .457 m/s^2 DOWNWARD

so L = - .457 m/s^2

4 0

2 years ago

Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R.From the express

AlladinOne [14]

Answer:

Vb = k Q / r r <R

Vb = k q / R³ (R² - r²) r >R

Explanation:

The electic potential is defined by

ΔV = - ∫ E .ds

We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product

VB - VA = - ∫ E dr

Let's substitute every equation they give us and we find out

r> R

Va = - ∫ (k Q / r²) dr

-Va = - k Q (- 1 / r)

We evaluate with it Va = 0 for r = infinity

Vb = k Q / r r <R

We perform the calculation of the power with the expression of the electric field that they give us

Vb = - int (kQ / R3 r) dr

We integrate and evaluate from the starting point r = R to the final point r <R

Vb = ∫kq / R³ r dr

Vb = k q / R³ (R² - r²)

This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity

8 0

3 years ago