Answer:
So the minimum force is
32.2Newton
Explanation:
To solve for the minimum force, let us assume it to be F (N)
So
F=mgsinA
But
=>>>> coefficient of static friction x (F + mgcosA
=>3 x 9.8 x sin35 = 0.3 x (F + 3 x 9.8 x cos35)
So making F subject of formula
F + 24.0 = 56.2
F = 32.2N
Explanation:
this is a physical property
Based on the length of the Ethernet cable and the mass, the tension in the cable can be found to be 80 N.
<h3>How much tension is in the cable?</h3>
The tension in the cable can be found as:
= 4 x mass x length x frequency
Solving for the frequency is:
= 1 / (0.800 / 4)
= 1 / 0.20
= 5.0 Hz
The tension is therefore:
= 4 x 0.20 x 4.00 x 5
= 80N
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