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vladimir1956 [14]
2 years ago
11

A treasure chest full of silver and gold coins is being lifted from a pirate ship to the shore using two ropes as shown in the f

igure. The mass of the treasure chest is 75.6 kg.
If the tension in rope A is 7.42x10^2 N and the tension rope B carries is 7.52x10^2 N, what is the tension in rope C?

Physics
1 answer:
stiks02 [169]2 years ago
7 0

The tension in rope C is determined as 122.23 N.

<h3>Tension in rope C</h3>

The tension in rope C is calculated as follows;

B² = A² + C²

C² = B² - A²

C² = (752²) - (742²)

C² = 14,940

C = √14,940

C = 122.23 N

Thus, the tension in rope C is determined as 122.23 N.

Learn more about tension here: brainly.com/question/24994188

#SPJ1

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6 0
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the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
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Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

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R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

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