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Softa [21]
3 years ago
7

What is the maximum value the string tension can have before the can slips? The coefficient of static friction between the can a

nd the ground is 0.76.

Physics
1 answer:
Naya [18.7K]3 years ago
8 0

Answer:

T= 38.38 N

Explanation:

Here

mass of can = m = 3 kg

g= 9.8 m/sec2

angle θ = 40°

From figure we see the vertical and horizontal component of tension force T

If the can is to slip - then horizontal component of tension force should become equal to force of friction.

First we find force of friction

Fs= μ R

where

μ = 0.76

R = weight of can = mg = 3 × 9.8 = 29.4 N

Now horizontal component of tension

Tx= T cos 40 = T× 0.7660  N

==>T× 0.7660 = 29.4

==> T= 38.38 N

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Answer:

Explanation:

Initial angular velocity ω₀ = 151 x 2π / 60

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Angular deceleration α = 2.23 rad / s

ω² = ω₀² -  2 α θ

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Q11) If you were standing at the top of a building and you dropped a rock.
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Answer:

Part A

The distance travel by the rock is approximately 132.496 m

Part B

The speed when the rock hits the ground is approximately 50.96 m/s

Explanation:

Part A

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The given parameter is the time it takes the rock to hit the ground, t = 5.2 s

For an object in free fall, we have;

h = 1/2·g·t²

Where;

h = The height from which the object is dropped

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∴ h = 1/2 × 9.8 m/s² × (5.2 s)² ≈ 132.496 m

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Part B

The speed, 'v', when the rock hits the ground, is given by the following kinematic equation,

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