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liberstina [14]

2 years ago

14

The brake pads for a bicycle tire are made of rubber. If a frictional force of 50 N is applied to each side of the tires, determ

ine the average shear strain in the rubber. Each pad has cross-sectional dimensions of 20 mm and 50 mm. Gr

1 answer:

mrs_skeptik [129]2 years ago

4 0

Answer:

average shear strain in the rubber is 0.25rad

Explanation:

The base has a dimension of 20mm and 50mm.

frictional force is 50 N

using the expressed shear stress,

we will determine τ = $F_t/A$

τ = 50 / 1000

= 0.05MPa

shear stress τ = γ .G

τ = shear stress

γ = shear strain

G = 0.20MPa

γ = τ / G

= 0.05 / 0.20

= 0.25rad

average shear strain in the rubber is 0.25rad

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Answer:

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we have given that

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A lead fishing weight of mass 0.2 kg is tied to a fishing line that is 0.5 m long. the weight is then whirled in a vertical circ

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In the movement of the weight in vertical circle, using momentum balance, the largest tension is at the bottom of the circle. This is represented by:

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2 years ago

Daniel is 50.0 meters away from a building. He observes that his line-of-sight to the tip of the building makes an angle of 63.0

frez [133]

Since his line of sight 63 degrees makes with the tip of the building

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2 years ago

A solid uniform sphere of mass 1.85 kg and diameter 45.0 cm spins about an axle through its center. Starting with an angular vel

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Answer:

The net torque is 0.0372 N m.

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$\frac{\omega^{2}-\omega_{0}^{2}}{2\Delta\theta}=\alpha$

Because the sphere stops the final angular velocity is zero, it's important all quantities in the SI so 2.40 rev/s = 15.1 rad/s and 18.2 rev = 114.3 rad, then:

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The negative sign indicates the sphere is slowing down as we expected.

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$\sum\overrightarrow{\tau}=I\overrightarrow{\alpha}$ (2)

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7 0

2 years ago

Read 2 more answers

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oksano4ka [1.4K]

As we know that reaction time will be

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$d = 20 \times 0.50$

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$d = 110 - 10 = 100 m$

So our distance from the intersection will be 100 m when we apply brakes

now this distance should be covered till the car will stop

so here we will have

$v_f = 0$

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now from kinematics equation we will have

$v_f^2 - v_i^2 = 2 a d$

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$a = \frac{-400}{200} = -2 m/s^2$

so the acceleration required by brakes is -2 m/s/s

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$v_f - v_i = at$

$0 - 20 = -2 (t)$

$t = 10 s$

total time to stop the car is given as

$T = 10 s + 0.5 s = 10.5 s$

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2 years ago