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pantera1 [17]
2 years ago
15

Compared to Earth's moon, the moons of Mars are

Physics
1 answer:
Mice21 [21]2 years ago
7 0
I think is b I think
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(1- name 1 of Faouzia’s songs from her stripped album
Mnenie [13.5K]

Answer:

I thought  "Born Without a Heart" was pretty good.

Hi

Explanation:

5 0
3 years ago
According to the text, there is no energy shortage now, nor will there ever be. what reason (s) is given to support this stateme
Bas_tet [7]
The reason why there is no energy shortage nor will there ever be is because energy is being preserved and conserved and only changes form. It never gets lost or increased.
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2 years ago
In the following reaction, which substance is the precipitate? (NH4)2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2NH4NO3(aq) (NH4)2SO4 Ba(
Mariana [72]
I think it's Barium sulfate, the soild and percipitate
4 0
3 years ago
Read 2 more answers
A conical container of radius 6 ft and height 24 ft is filled to a height of 19 ft of a liquid weighing 64.4 lb divided by ft cu
kobusy [5.1K]

Answer:

Part (i) work required to pump the contents to the​ rim is 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5ft above the​ cone's rim is 426,484.878 lb.ft

Explanation:

The center of mass of a uniform solid right circular cone of height h lies on the axis of symmetry at a distance of h/4 from the base and 3h/4 from the top.

Center mass of the liquid Z = (24-19)ft + 19/4 = 5ft + 4.75ft = 9.75 ft

Mass of liquid in the cone = volume × density (ρ) =  ¹/₃.π.r².h.ρ

where;

r is the radius of the liquid surface = [6*(19/24)]ft = 4.75ft

ρ is the density of liquid = 64.4 lb/ft³

h is the height of the liquid = 19 ft

Mass of liquid in the cone = ¹/₃ × π × (4.75)² × 19 × 64.4 = 28,914.229 lbs

Part (i)  work required to pump the contents to the​ rim

Work required = 28,914.229 lbs × 9.75 ft = 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5 ft above the​ cone's rim

Extra work required = 28,914.229 lb ×  5ft = 144571.145 lb.ft

Total work required = (281,913.733 +  144571.145) lb.ft

                                 = 426,484.878 lb.ft

5 0
3 years ago
The charge that passes a cross-sectional area A=10-4 m2 varies with time according
Anarel [89]

Answer:

Current = dQ/dt

or I = dQ/dt

Where I represents current.

Which is the rate of flow of charge.

Q=4 + 2t + t²

dQ/dt = 2 + 2t --- This is the relation that gives the instantaneous current.

At time t=2sec

dQ/dt = I = 2 + 2t

= 2 + 2(2)

=2 + 4

= 6A.

6 0
3 years ago
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