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Sergio039 [100]
3 years ago
10

I need help finding the solution

Physics
1 answer:
omeli [17]3 years ago
6 0

Answer:

objects with greater mass should have greater inertia

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A small block with mass 0.0350 kgkg slides in a vertical circle of radius 0.550 mm on the inside of a circular track. During one
Lisa [10]

Answer:

 W_net = μ 5.58,   μ = 0.1     W_net = 0.558 J

Explanation:

The work is defined by the related

          W = F. d ​​= F d cos θ

where bold indicates vectors.

In the case, the work of the friction force on a circular surface is requested.

The expression for the friction force is

              fr = μ N

the friction force opposes the movement, therefore the angle is 180º and the cos 180 = -1

               W = - fr d

the path traveled half the length of the circle

               L = 2 π R

               d = L / 2

               d = π R

we substitute

                 W = - μ N d

Total work is initial to

                W_neto = - μ π R (N_b - N_a)

let's calculate

                W_net = - μ π 0.550 (0.670 - 3.90)

                W_net = μ 5.58

for the complete calculation it is necessary to know the friction coefficient, if we assume that μ = 0.1

                W_net = 0.1 5.58

                W_net = 0.558 J

7 0
3 years ago
a man checked himself into seattle hospital. he didnt know who he was and had no idea how he got to seattle. the hospital staff
8090 [49]

Amnesia would be the answer

7 0
3 years ago
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If you were trying to cross a river with the shortest possible time, would you aim your boat slightly upstream, directly across
Katyanochek1 [597]
I would say downstream since the stream can push your boat, then you would have momentum and would just have to row towards the land.
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What is the region where the magnetic force will affect other magnets?
bulgar [2K]
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3 years ago
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Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const
kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is  L=  20cm = 0.2m

Thermal conductivity of the wall is  K = 2.79 W/m·K

Temperature at the left side surface is T₁ =  50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is  h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

Q_{conduction} = Q_{convection}

Expression for the heat conduction process is

Q_{conduction} = \frac{K(T_1 - T)}{L}

Expression for the heat convection process is

Q_{convection} = h(T_2 - T)

Substitute the expressions of conduction and convection in equation above

Q_{conduction} = Q_{convection}

\frac{K(T_1 - T_2)}{L} = h(T_2 - T)

Substitute the values in above equation

\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC

Now heat flux through the wall can be calculated as

q_{flux} = Q_{conduction} \\\\q_{flux}  = \frac{K(T_1 - T_2)}{L}\\\\q_{flux}  = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

6 0
3 years ago
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