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nikitadnepr [17]
3 years ago
8

. An iron cube (ρ = 7860 kg/m3) having sides of 1 cm is submerged in open fresh water (ρ = 1000kg/m3) with its top surface at a

depth of 2.5 m. Considering that its top surface is held horizontal with respect to the water surface, determine the difference in pressure between the top and bottom surfaces of the cube?
Physics
1 answer:
Korolek [52]3 years ago
5 0

Answer:

\Delta P = 98.07\,Pa

Explanation:

The difference between top and bottom surfaces is computed by the following hydrostatic equivalence:

\Delta P = \rho_{w}\cdot g \cdot \Delta h

\Delta P = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.01\,m)

\Delta P = 98.07\,Pa

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Answer:

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Explanation:

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A hungry lion is looking for food. It went 5 miles (mi) before it spotted a zebra to stalk. It took this lion 0.2 hours to get t
siniylev [52]
The lion covered 5 miles in 0.2 hours. With a quick division you can find the speed per hour: 5 / 0.2 = 25 mi/h
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A balloon drifts 140 m toward the west in 45 s; then the wind suddenly changes and the balloon flies 90 m toward the east in the
rusak2 [61]

Answer:

Explanation:

Given

1 ) 140 m west in 45 s .

2 ) 90 m east in 25 s .

a )

distance travelled in first 45 s = 140 m

b ) distance travelled in next 25 s = 90 m

c ) Total distance travelled = 140 m + 90 m

= 230 m

d ) average speed in first 45 s

= distance in 45 s  45

= 140 / 45 = 3.11 m /s

e ) average speed in next 25 s

distance in 25 s / 25

= 90 / 25 = 3.6 m /s

f ) average in entire trip

=  total distance / total time

= (140 + 90) / ( 25 + 45 )

= 3.28 m /s

g )

displacement in first 45 s = 140 m towards west

h )

displacement in next 25 s = 90 m towards east

i )

total displacement = 140 - 90

= 50 m towards west .

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3 years ago
If an object increases speed from 4 meters per second to 10 meters per second in 3 seconds, what is the acceleration of that obj
kirill [66]

Answer:

Explanation:

Acceleration=(v-u)/t

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4 0
3 years ago
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Hi, so i have to find T1, can some1 help?
iragen [17]

30.1 N

Explanation:

Given:

W_1 = 16\:\text{N}

W_2 = 8\:\text{N}

Let's write the components of the net forces at the intersections. Note that the system is equilibrium so all the net forces are zero.

<u>Forces</u><u> </u><u>involving</u><u> </u><u>W1</u><u>:</u>

x:\:\:\:-T_1 + T_3\cos \alpha = 0\:\: \\ \text{or}\:\:T_2 = T_3\cos \alpha\:\:\:\:\:(1)

y:\:\:\:T_3\sin \alpha - W_1 = 0\:\:\: \\ \text{or}\:\:\:T_3\sin \alpha = W_1\:\:\:\:\:\:(2)

<u>Forces</u><u> </u><u>involving</u><u> </u><u>W2</u><u>:</u>

x:\:\:\:T_1\sin 53 - T_3\sin \alpha = W_2\:\:\:\:\:\:\:(3)

y:\:\:\:T_4 - T_1\cos 53 - T_3\cos \alpha = 0\:\:\:\;(4)

Substitute (2) into (3) and we get

T_1\sin 53 - W_1 = W_2

Solving for T_1,

T_1 = \dfrac{W_1 + W_2}{\sin 53} = 30.1\:\text{N}

7 0
3 years ago
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