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egoroff_w [7]
3 years ago
11

A hockey puck is at rest on a flat surface. Then one end of the surface is lifted gradually, forming an inclined plane of increa

sing slope. By measuring the angle of the slope at which the puck begins to slide, what quantity can be calculated?
the coefficient of kinetic friction between the puck and the surface

the mass of the puck

the volume of the puck

the coefficient of static friction between the puck and the surface

Physics
2 answers:
Jlenok [28]3 years ago
6 0

The coefficient of static friction between the puck and the surface.

In fact, that coefficient describes exactly how "hard" it is to cause the puck to start moving, if it starts from an idle condition.

ikadub [295]3 years ago
6 0

Answer:

the coefficient of static friction between the puck and the surface

Explanation:

As per the FBD we can say that as we increase the angle of the inclination the component of force of gravity will counterbalance the friction force

At the condition of just equilibrium when it is just begin to slide then in that case we can say

F_f = mg sin\theta

so here for finding frictional force maximum value we can say

F_f = \mu F_n

here for finding Normal force we can write force balance normal to the inclined plane

F_n = mgcos\theta

now from above equations

\mu (mgcos\theta) = mg sin\theta

\mu = tan\theta

so angle depends on the coefficient of static friction

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3 years ago
8-14 A Cu-30% Zn alloy tensile bar has a strain-hardening coefficient of 0.50. The bar, which has an initial diameter of 1 cm an
77julia77 [94]

Answer:

The true stress at true strain 0.05cm/cm is 80MPa

Explanation:

Given that

the strength coefficient is K

true strain is ε

strain hardening exponent is n

initial diameter of bar is d = 1cm, (10mm)

tensile force is F

engineering stress(S) = 120

the engineering stress(S) = \frac{F}{\frac{\pi }{4}(d^2) }

To find force (F) =

                    120 = \frac{F}{\frac{\pi }{4}(100^{2} )}

                     F = 120 * (π/4) * (100)

                     F = 9425N

Calculate the true strain  (ε) = In (l₀ / l₁)

where

l₀ =  initial length of the metallic bar = 3cm

l₁ = final length of metallic bar = 3.5cm

ε = In (3.5 / 3)

  = In 1.1667

  = 0.154cm/cm

Calculate the true stress (σ) at fracture point

          = \frac{F}{\frac{\pi }{4}(d^2) }}

tensile force is F and final diameter of bar is d₁ (d in the eqn)

Substitute 9425 N for F and 0.926 cm (9.26mm) for d₁ (d in the eqn)

σ = \frac{9425}{\frac{\pi }{4}(9.26^2) }}

   = 140MPa

To find the strength coefficient (K) of the material bar

K = \frac{140}{\sqrt{0.154} }

K = \frac{140}{0.3925}

   = 356.75MPa

To calculate the true stress σ true strain of 0.05cm/cm

K  = 356.75MPa

σ = 356.75(0.05)^0^.^5

  = 356.75 ( 0.2236)

  = 80MPa

The true stress at true strain 0.05cm/cm is 80MPa

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Answer:

The third choice: 2.81*10^{16} \: \text{protons}

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therefore, the number n that makes up a charge of +4.5mC is

n = \dfrac{+4.5mC}{+1.6*10^{-19}C}  = \dfrac{+4.5*10^{-3}C}{+1.6*10^{-19}C}

\boxed{n =2.81*10^{16} \: \text{protons}}

which is the 3rd choice.

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