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Sphinxa [80]
3 years ago
14

The spring-mounted 0.84-kg collar A oscillates along the horizontal rod, which is rotating at the constant angular rate rad/s. A

t a certain instant, r is increasing at the rate of 610 mm/s.
If the coefficient of kinetic friction between the collar and the rod is 0.61, calculate the friction force F exerted by the rod on the collar at this instant.
Physics
1 answer:
vfiekz [6]3 years ago
8 0

Answer:

Friction = 5.027 N

Explanation:

Kinetic friction does not change with speed or acceleration. The only thing that is required is to state that the object is moving.

Solving the force equation for friction we get:

Friction = coefficient of friction* normal force exerted by the object

Friction = u * R

here R is = 0.84 * Gravity = 0.84 * 9.81 = 8.2404 N

u = 0.61 (as stated in the question)

So frictional force will be:

Friction = 0.61 * 8.2404

Friction = 5.027 N

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1. draught

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A student on an amusement park ride moves in circular path with a radius of 3.5 m once every 4.5 s. What is the tangential veloc
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Answer:

the tangential velocity of the student is 4.89 m/s.

Explanation:

Given;

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let the student's tangential velocity = v

The tangential velocity of the student is calculated as follows;

v = \frac{2\pi r}{t} \\\\v = \frac{2 \pi \times 3.5}{4.5} \\\\v = 4.89 \ m/s

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A centrifuge is a device used to separate materials by their masses. A sample in a centrifuge is rotated at high speeds along a
german

Answer:

 F = 1.047 10⁻² N

Explanation:

Let's use kinematics to find the angular acceleration

             w = w₀ + α t

as for rest w₀ = 0

             w = α t

             α = w / t

let's reduce the magnitudes to the SI system

              w = 1000 rev / min (2π rad/ 1 rev) (1 min/ 60s) = 104.72 rad / s

              m = 1.00 g (1 kg / 1000 g) = 1,000 10⁻³ kg

              r = 10.0 cm (1 m / 100 cm) = 0.100 m

let's calculate

              α = 104.72 / 1

              α = 104.72 rad / s²

angular and linear variables are related

               a = α  r

               a = 104.72 0.100

               a = 10.47 m / s²

finally we substitute in Newton's second law

               F = 1 10⁻³ 10.47

               F = 1.047 10⁻² N

8 0
3 years ago
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