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Valentin [98]
3 years ago
11

*8–52. Beam AB has a negligible mass and thickness, and supports the 200-kg uniform block. It is pinned at A and rests on the to

p of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to P=300 N, the post slips at both B and C simultaneously.

Physics
1 answer:
denis23 [38]3 years ago
8 0

Answer:

μ₁ = 0.1048

μ₂ = 0.1375

Explanation:

Using  static equation can find in both point the moment and the forces so:

∑ M = F *d  , ∑ F = 0

∑ M A = 0

N₁ * 3 - 200 * 9.81 * 1.5 = 0

N₁ = 981  

∑ M y = 0

N₂ + 300 * ³/₅ - 981 - 20 * 9.81 = 0

N₂ = 997.2 N

∑ M C = 0

F₁ * 1.75  - 300 * ⁴/₅  * 0.75 = 0

F₁ = 102.86

∑ M B = 0

300 * ⁴/₅ * 1 - F₂ * 1.75 = 0

F₂ = 137.14 N

The Force F1 and F2 related the coefficients of static friction

F₁ = μ₁ * N₁   ⇒  102.86 N = μ₁ * 981 ⇒ μ₁ = 0.1048

F₂= μ₂ * N₂  ⇒  137.14 N = μ₂ * 997  ⇒ μ₂ = 0.1375

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Answer:

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A light-year is the distance light travels in one year (at speed = 2.998 × 108 m/s). (a) how many meters are there in 11.0 light
larisa [96]
<span>The answers are as follows:

(a) how many meters are there in 11.0 light-years?

11.0 light years ( 365 days / 1 year ) ( 24 h / 1 day ) ( 60 min / 1 h ) ( 60 s / 1 min ) ( 2.998x10^8 m/s ) = 1.04x10^17 m

(b) an astronomical unit (au) is the average distance from the sun to earth, 1.50 × 108 km. how many au are there in 11.0 light-years?

1.04x10^17 m ( 1 au / </span>1.50 × 10^8 km <span>) ( 1 km / 1000 m) = 693329.472 au

(c) what is the speed of light in au/h? au/h

</span>2.998 × 10^8 m/s ( 1 au / 1.50 × 10^8 km ) ( 1 km / 1000 m) ( 3600 s / 1 h ) = 7.1952 au/h

8 0
3 years ago
A 25 kh inductor carries a current of 60 ma. how much energy is stored in this system
IgorLugansk [536]
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8 0
3 years ago
A long wire carrying a 5.0 A current perpendicular to the xy-plane intersects the x-axis at x= - 2.0 cm . A second, parallel wir
mario62 [17]

Answer:

a . 0.35cm

b.  11.33cm

Explanation:

a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:

\frac{\mu_oi_1}{2\pi x}=\frac{\mu_oi_2}{2\pi(4-x)}\\\\5/x=\frac{3.5}{4-x}\\\\x=2.35cm\\\\N=2.35-2=0.35cm

Hence, for currents in same direction, the point is 0.35cm

b. Given both currents flow in opposite directions, the null point lies on the other side.

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:

Let x be distance of N from first wire, then distance from 2nd wire is 4+x:

\frac{\mu_oi_1}{2\pi(4+ x)}=\frac{\mu_oi_2}{2\pi x}\\\\5/(4+x)=\frac{3.5}{x}\\\\x=9.33cm\\\\N=9.33+2=11.33cm

Hence, if currents are in opposite directions the point on x-axis is 11.33cm

8 0
3 years ago
k) A stream of warm water is produced in a steady-flow mixing process by combining 1.0 kg s-1of cool water at 25°C with 0.8 kg s
Irina18 [472]

Answer:

T_ww = 43,23°C

Explanation:

To solve this question, we use energy balance and we state that the energy that enters the systems equals the energy that leaves the system plus losses. Mathematically, we will have that:

E_in=E_out+E_loss

The energy associated to a current of fluid can be defined as:

E=m*C_p*T_f

So, applying the energy balance to the system described:

m_CW*C_p*T_CW+m_HW*C_p*T_HW=m_WW*C_p*T_WW+E_loss

Replacing the values given on the statement, we have:

1.0 kg/s*4,18 kJ/(kg°C)*25°C+0.8  kg/s*4,18 kJ/(kg°C)*75°C=1.8 kg/s*4,18 kJ/(kg°C)*T_WW+30  kJ/s

Solving for the temperature Tww, we have:

(1.0 kg/s*4,18 kJ/(kg°C)*25°C+0.8 kg/s*4,18 kJ/(kg°C)*75°C-30 kJ/s)/(1.8 kg/s*4,18 kJ/(kg°C))=T_WW

T_WW=43,23 °C

Have a nice day! :D

6 0
3 years ago
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