Awater tank is filled to a depth of 10 m, and the bottom of the tank is 20 m above ground. a water-filled hose that is 2.0 cm in diameter extends from the bottom of the tank to the ground, but no water is flowing in this hose. the water pressure at ground level in the hose is closest to the density of water is 1000 kg/m3 .

3.95 x 10^5 Pa

Explanation:

total height, h = 10 m + 20 m = 30 m

density of water, d = 1000 kg/m^3

Atmospheric pressure, Po = 1.01 x 10^5 Pa

Pressure, P = Po + h d g

P = 1.01 x 10^5 + 30 x 1000 x 9.8

P = (1.01 + 2.94) x 10^5

P = 3.95 x 10^5 Pa

Thus, the total pressure at the ground is 3.95 x 10^5 Pa.

3.95 x 10⁵ Pa

Explanation:

given,

depth of water tank = 10 m

bottom tank height = 20 m

density of water = 1000 Kg/m³

diameter of hose = 2 cm

Pressure at ground level = ?

net head of water from the ground

H = 20 + 10

H = 30 m

pressure exerted at the ground

P = ρ g h

P = 1000 x 9.8 x 30

P = 2.94 x 10⁵ Pa

atmospheric pressure = 1.01 x 10⁵ Pa

total pressure on the ground

 = atmospheric pressure + pressure due to water tank

 = 1.01 x 10⁵ Pa +  2.94 x 10⁵ Pa

 =  3.95 x 10⁵ Pa

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

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