Answer:
M = 1433.5 kg
Explanation:
This exercise is solved using the Archimedean principle, which states that the hydrostatic thrust is equal to the weight of the desalinated liquid,
B = ρ g V
with the weight of the truck it is in equilibrium with the push, we use Newton's equilibrium condition
Σ F = 0
B-W = 0
B = W
body weight
W = M g
the volume is
V = l to h
rho_liquid g (l to h) = M g
M = rho_liquid l a h
we calculate
M = 1000 4.7 6.10 0.05
M = 1433.5 kg
In order for a hypothesis to be tested, and experiment needs to be designed.
Many trials need to be runned in order to get accurate results and to form a valid conclusion that can prove or disprove the hypothesis
Plug in the corresponding values into y = mx + b
8.18 in for y
1.31 in for m
17.2 in for b
8.18 = 1.31x + 17.2
Now bring 17.2 to the left side by subtracting 17.2 to both sides (what you do on one side you must do to the other). Since 17.2 is being added on the right side, subtraction (the opposite of addition) will cancel it out (make it zero) from the right side and bring it over to the left side.
8.18 - 17.2 = 1.31x
-9.02 = 1.31x
Then divide 1.31 to both sides to isolate x. Since 1.31 is being multiplied by x, division (the opposite of multiplication) will cancel 1.31 out (in this case it will make 1.31 one) from the right side and bring it over to the left side.
-9.02/1.31 = 1.31x/1.31
x ≈ -6.8855
x is roughly -6.89
Hope this helped!
~Just a girl in love with Shawn Mendes
Answer:
a) 
b) 
Explanation:
Given:
height of water in one arm of the u-tube, 
a)
Gauge pressure at the water-mercury interface,:
we've the density of the water 
b)
Now the same pressure is balanced by the mercury column in the other arm of the tube:
<u>Now the difference in the column is :</u>
Answer:
a) The Energy added should be 484.438 MJ
b) The Kinetic Energy change is -484.438 MJ
c) The Potential Energy change is 968.907 MJ
Explanation:
Let 'm' be the mass of the satellite , 'M'(6×
be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×
N/m) be the universal constant of gravitation.
We know that the orbital velocity(v) for a satellite -
v=
[(R+h) is the distance of the satellite from the center of the earth ]
Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)
For initial conditions ,
h = 
= 98 km = 98000 m
∴Initial Energy (
) = 
m
+ 
Substituting v=
in the above equation and simplifying we get,
= 
Similarly for final condition,
h=
= 198km = 198000 m
∴Final Energy(
) = 
a) The energy that should be added should be the difference in the energy of initial and final states -
∴ ΔE = -
= 
(
- 
)
Substituting ,
M = 6 × kg
m = 1036 kg
G = 6.67 ×
R = 6400000 m
= 98000 m
= 198000 m
We get ,
ΔE = 484.438 MJ
b) Change in Kinetic Energy (ΔKE) = m[
- 
]
= [
- 
]
= -ΔE
= - 484.438 MJ
c) Change in Potential Energy (ΔPE) = GMm[ - ]
= 2ΔE
= 968.907 MJ
