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Elina [12.6K]
2 years ago
13

A 12 cm radius air duct is used to replenish the air of a room 9.2 m x 5.5 m x 4.5 m every 10 min. how fast does the air flow in

the duct?
Physics
1 answer:
mote1985 [20]2 years ago
6 0
In this problem, given the dimensionns of the room and the dimension of the pipe in which air passes through, we can determine the flow of air in the duct. The area of the duct is A = pi r2 where r is equal to 12 cm. we divide this area to time which is 10 minutes, then we get the answer.
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How much work is done using a 500 W microwave for 5 minutes
frosja888 [35]

Power = work done/time

work done = power x time

                  = 500x5x60

                  = 150000J

6 0
3 years ago
Consider that 168.0 J of work is done on a system and 305.6 J of heat is extracted from the system. In the sense of the first la
Ilia_Sergeevich [38]

To solve this problem we must resort to the Work Theorem, internal energy and Heat transfer. Summarized in the first law of thermodynamics.

dQ = dU + dW

Where,

Q = Heat

U = Internal Energy

By reference system and nomenclature we know that the work done ON the system is taken negative and the heat extracted is also considered negative, therefore

W = -168J \righarrow  Work is done ON the system

Q = -305.6J \rightarrow Heat is extracted FROM the system

Therefore the value of the Work done on the system is -158.0J

3 0
3 years ago
The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to I
erma4kov [3.2K]

Answer:

Part a)

I_{end} = \frac{mL^2}{3}

Part b)

I_{edge} = \frac{2ma^2}{3}

Explanation:

As we know that by parallel axis theorem we will have

I_p = I_{cm} + Md^2

Part a)

here we know that for a stick the moment of inertia for an axis passing through its COM is given as

I = \frac{mL^2}{12}

now if we need to find the inertia from its end then we will have

I_{end} = I_{cm} + Md^2

I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2

I_{end} = \frac{mL^2}{3}

Part b)

here we know that for a cube the moment of inertia for an axis passing through its COM is given as

I = \frac{ma^2}{6}

now if we need to find the inertia about an axis passing through its edge

I_{edge} = I_{cm} + Md^2

I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2

I_{edge} = \frac{2ma^2}{3}

7 0
3 years ago
1. A hydraulic lift is to be used to lift a truck masses 5000 kg. If the diameter of the
zheka24 [161]

Answer:

a)   F = 4.9 10⁴ N,  b)   F₁ = 122.5 N

Explanation:

To solve this problem we use that the pressure is transmitted throughout the entire fluid, being the same for the same height

1) pressure is defined by the relation

           P = F / A

to lift the weight of the truck the force of the piston must be equal to the weight of the truck

          ∑F = 0

          F-W = 0

          F = W = mg

          F = 5000 9.8

          F = 4.9 10⁴ N

the area of ​​the pisto is

          A = pi r²

          A = pi d² / 4

          A = pi 1 ^ 2/4

          A = 0.7854 m²

pressure is

          P = 4.9 104 / 0.7854

          P = 3.85 104 Pa

2) Let's find a point with the same height on the two pistons, the pressure is the same

          \frac{F_1}{A_1} = \frac{F_2}{A_2}

where subscript 1 is for the small piston and subscript 2 is for the large piston

          F₁ = \frac{A_1}{A_2} \ F_2

the force applied must be equal to the weight of the truck

          F₁ = ( \frac{d_1}{d_2} )^2\  m g

          F₁ = (0.05 / 1) ² 5000 9.8

          F₁ = 122.5 N

7 0
2 years ago
A wood block, after being given a starting push, slides down a ramp at a constant speed. what is the angle of the ramp above hor
Stells [14]

The solution for the problem is:

Constant speed means Fnet = 0. 
Let m = mass of wood block and Θ = angle of ramp; then if µk = 0.35 …

The computation would be:


Fnet = 0 = mg (sin Θ) - (µk) (mg) (cos Θ) 
mg (sin Θ) = µk (mg) (cos Θ) 
µk = tan Θ 
Θ = arctan(µk)

= arctan (0.35)

≈ 19.3°

5 0
3 years ago
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