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2 years ago

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A circus tightrope walker weighing 800 N is standing in the middle of a 15 meter long cable stretched between two posts. The cab

le was originally horizontal. The lowest point of the cable is now at his feet and is 30 cm below the horizontal. Assume the cable is massless. What is the tension in the cable

1 answer:

Lorico [155]2 years ago

4 0

Answer:

T = 10010 N

Explanation:

To solve this problem we must use the translational equilibrium relation, let's set a reference frame

X axis

Fₓ-Fₓ = 0

Fₓ = Fₓ

whereby the horizontal components of the tension in the cable cancel

Y Axis

$F_{y} + F_{y} - W =0$

2 $F_{y}$ = W

let's use trigonometry to find the angles

tan θ = y / x

θ = tan⁻¹ (0.30 / 0.50 L)

θ = tan⁻¹ (0.30 / 0.50 15)

θ = 2.29º

the components of stress are

F_{y} = T sin θ

we substitute

2 T sin θ = W

T = W / 2sin θ

T = $\frac{ 800}{ 2sin 2.29}$

T = 10010 N

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irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

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Where:

$x$ is the horizontal distance between the cannon and the ball

$V_{o}=23 m/s$ is the cannonball initial velocity

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$y_{o}=1.96 m$ is the initial height of the cannonball

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Substituting (5) in (1):

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Finally:

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2 years ago

MRU(movimiento rectilineo uniforme) un movil viaja en linea recta con una velocidad media de 12metros /segundos durante 9segundo

liubo4ka [24]

Answer:

a) 141.6m

b) 8.4m/s

Explanation:

a) to find the total displacement you use the following formula for each trajectory. Next you sum the results:

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3 years ago

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nika2105 [10]

Answer:

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Explanation:

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If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area(A2) of the piston.

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2 years ago

-. A 2kg cart moving to the right at 5m/s collides with an 8kg cart at rest. As a

bulgar [2K]

Answer:

<em>The velocity of the carts after the event is 1 m/s</em>

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If we have a system of bodies, then the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2+...+m_nv_n$

If a collision occurs and the velocities change to v', the final momentum is:

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2 years ago

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nata0808 [166]

Answer:

A. 12,240.

Explanation:

1,530 times 8.0 = 12,240.

hope this helped :)

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2 years ago