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Elena L [17]
2 years ago
5

A tank is full of oil weighing 40 lb/ft^3. The tank is an inverted right circular cone (with the base at the top) with a height

of 5 feet and a radius of 2 feet. Find the work required to pump the water to a height of 1 feet above the top of the tank
Physics
1 answer:
krok68 [10]2 years ago
8 0

Answer:

26945.6 ft⋅lbf

Explanation:

Volume of Right Circular Cone = pi*(radius^2)*(height/3)

Pi*(4)*(5/3) = 20.94 ft^3

Density = Mass / Volume

Mass = Density*Volume

Mass = (40)*(20.94)

Mass = 837.6 lb

Work = Force*Height

Force = Mass*Acceleration

Acceleration will be gravitational acceleration

Work = (837.6)*(32.17)*(1)

Work = 26945.6 ft⋅lbf

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120 J of work was done to lift an object 6 m above
Slav-nsk [51]

Answer:

The last option, 20 N and 2.04 kg

Explanation:

work = (force)(distance)

work = 120 joules

distance: 6 m

rearrange to find force:

120=(6)F

F= 120/6 = 20 Newtons.

Assuming its lifted from Earth's surface, the force of gravity will be 9.81 m/s^2. Let's find mass:

F=mg

m=F/g

m=(20)/(9.81)= 2.038 kg

6 0
2 years ago
What is relationship between latent heat and water ?
Wittaler [7]
Latent heat, energy absorbed or released by a substance during a change in its physical state (phase) that occurs without changing its temperature.
5 0
3 years ago
Find the acceleration if a 32.5 N force is<br> used on an object that has a mass of<br> 128.6 kg.
Sholpan [36]

Answer:

Acceleration=3.95

Explanation:Use the formula a=m/f

a=128.6/32.5

a=3.95

3 0
2 years ago
PLEASE HELP ASAP
DaniilM [7]

Your answers:

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I'm sorry I cannot answer all your questions because there's a typing limit. :)

Thank you for your patience, and I hope I helped!

Would appreciate Brainliest ;)

4 0
3 years ago
Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude o
Anna [14]

The magnitude of the tension in the string marked A is 52.5N

Generally, the equation for is mathematically given as

Let's take θ be an angle at A

So, tanθ = 3/8

Let's take α be an angle at B (Below X)

tanα = 5/4

Let's take β be an angle at C (Below x)

tanβ = 1/6

First we take the Horizontal Components

74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)

By solving the equation, we get

A = 78.9 - 0.668B … (1)

Now, we take the vertical components

74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)

By solving the equation, we get

40.07 = 1.015B

B = 39.5N

By substituting the value of B in equation (1)

A = 78.9 - 0.6668× 39.5

A = 52.5N

Hence, the magnitude of the tension in the string marked A is 52.5N

Learn more about Tension here brainly.com/question/2287912

#SPJ1

8 0
2 years ago
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