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ryzh [129]
3 years ago
6

A chain 72 meters long whose mass is 29 kilograms is hanging over the edge of a tall building and does not touch the ground. How

much work is required to lift the top 14 meters of the chain to the top of the building?
Use that the acceleration due to gravity is 9.8 meters per second squared. Your answer must include the correct units.
Physics
1 answer:
dangina [55]3 years ago
5 0

Answer:

Work done required is 3567.2 J

Explanation:

Given :

Length of chain, l = 72 m

Mass of chain, M = 29 kg

Linear mass density of chain, μ = \frac{Mass\ of\ chain }{Length\ of\ chain} = \frac{29}{72}  = 0.40 kg/m

Let x be the length of the chain which lift to the top of the building.

Work done required to lift the chain is equal to the potential energy of the chain.

W = ∫μg (72 - x ) dx

Here g is acceleration due to gravity.

The limit of integration is from 0 to 14.

W = μg ( 72x - x²/2)

Substitute 0.40 kg/m for μ, 9.8 m/s² for g and 14 m for x in the above equation.

W = 0.40\times9.8\times(72\times14\ - \frac{14^{2} }{2})

W = 3567.2 J

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PIT_PIT [208]

Answer:

Fluid fricton goes to Static friction and sliding friction goes to rolling friction

Explanation:

6 0
2 years ago
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In a physics lab, Asha is given a 10.7 kg uniform rectangular plate with edge lengths 67.3 cm by 53.5 cm . Her lab instructor re
Leya [2.2K]

Answer:

I=2.6363\ kg.m^2

Explanation:

Given:

dimension of uniform plate, (0.673\times 0.535)\ m^2

mass of plate, m=10.7\ kg

Now we find the moment of inertia about the center of mass of the rectangular plate is given as:

I_{cm}=\frac{1}{12} \times m(L^2+B^2)

where:

L= length of the plate

B= breadth of the plate

I_{cm}=\frac{1}{12} \times 10.7\times(0.673^2+0.535^2)

I_{cm}=0.6591\ kg.m^2

We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .

Now we find the distance between the center of mass and the corner:

s=\frac{\sqrt{ (0.673^2+0.535^2)}}{2}

s=0.4299\ m

Now using parallel axis theorem:

I=I_{cm}+m.s^2

I=0.6591+10.7\times 0.4299^2

I=2.6363\ kg.m^2

6 0
3 years ago
n deep space, sphere A of mass 47 kg is located at the origin of an x axis and sphere B of mass 110 kg is located on the axis at
Volgvan

Answer:

a)-1.014x 10^{-7J

b)3.296 x  10^{-7J

Explanation:

For Sphere A:

mass 'Ma'= 47kg

xa= 0

For sphere B:

mass 'Mb'= 110kg

xb=3.4m

a)the gravitational potential energy is given by

U_{i = -GMaMb/ d

U_{i= - 6.67 x 10^{-11} x 47 x 110/ 3.4 => -1.014x 10^{-7J

b) at d= 0.8m (3.4-2.6) and U_{i=-1.014x 10^{-7J

The sum of potential and kinetic energies must be conserved as the energy is conserved.

K_{i + U_{i= K_{f + U_{f

As sphere starts from rest and sphere A is fixed at its place, therefore K_{i is zero

U_{i= K_{f + U_{f

The final potential energy is

U_{f= - GMaMb/d

Solving for 'K_{f '

K_{f = U_{i + GMaMb/d => -1.014x 10^{-7 + 6.67 x 10^{-11} x 47 x 110/ 0.8

K_{f = 3.296 x  10^{-7J

6 0
3 years ago
.
lesya692 [45]

Answer:

John Dalton

Explanation:

Dalton's atomic theory was the foundation for a new understanding of chemical structures. He proposed that matter was constituted by indivisible and indestructible particles "atoms." He theorized that all atoms of a particular substance were equal, and the atoms of different substances had atoms of different sizes and masses.

He also proposed that all compounds of elements were combinations of elements but in a very precise ratio.

7 0
3 years ago
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A box is at rest on a ramp at an incline of 22°. The normal force on the box is 538 N.
fomenos

Answer: 580 N

Refer to attached figure.

The angle of inclination is 22 degrees

weight (gravitational force) acts downwards.

Normal force is a contact force which acts perpendicular to the point of contact.

The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.

Gravitational force on an object = mg

The normal force N= mg cos 22

\Rightarrow mg =\frac{N}{cos22}=\frac{538 N}{0.927}=580 N





8 0
3 years ago
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