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Dimas [21]
2 years ago
6

a 90 kg architect is standing 2 meters from the center of a scaffold help up by a rope on both sides. the scaffold is 6m long an

d has a mass of 200kg. what is the tension on each rope

Physics
1 answer:
Mademuasel [1]2 years ago
8 0
We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.

1) Equilibrium of forces:
T_1 - W_p - W_s + T_2 =0
where
W_p = (90kg)(9.81 m/s^2)=883 N is the weight of the person
W_s = (200kg)(9.81 m/s^2)=1962 N is the weight of the scaffold
Re-arranging, we can write the equation as
T_1 = 2845 N-T_2 (1)

2) Equilibrium of torques:
T_1 \cdot 3 m - W_p \cdot 2 m - T_2 \cdot 3m =0
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using W_p = 883 N and replacing T1 with (1), we find
2845 N \cdot 3 m - T_2 \cdot 3 m - 833 N \cdot 2 m - T_2 \cdot 3 m=0
from which we find
T_2 = 1128 N

And then, substituting T2 into (1), we find
T_1 = 1717 N
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Answer:

1.5 u

Explanation:

The range equation is:

R = u² sin(2θ) / g

When u = v, R = 2.25 R.

2.25 R = v² sin(2θ) / g

2.25 u² sin(2θ) / g = v² sin(2θ) / g

2.25 u² = v²

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A 2.85 kg fish is attached to the lower end of a vertical spring that has negligible mass and force constant 875 N/m. The spring
Westkost [7]

Answer:

A) V = 0.324 m/s

B) V = 0.56 m/s

Explanation:

A) Change in gravitational potential energy = mass * gravity * height change

Change in G.P.E = 2.85 * 9.81 * 0.058 = 1.6215 J

Energy transferred to spring = 0.5 * k * x^2

Here k = 875 N/m

And x = 0.058 m

Thus energy transferred to spring = 1.47175 J

The G.P.E is converted to spring potential energy and kinetic energy as it moves down. So the difference in energy is accounted by kinetic energy as follows:

Kinetic energy = G.P.E - Spring P.E

Kinetic energy = 1.6215 - 1.47175 = 0.14975 J

We can now find the speed using this kinetic energy:

Kinetic energy = 0.14975

0.5 * mass * velocity^2 = 0.14975

0.5 * 2.85 * V^2 = 0.14975

V = 0.324 m/s

B) The fish accelerates because the force on it are unbalanced. These forces are the weight of the fish, and the force of the spring stopping it.

As long as the weight of the fish is more than the upward force of the spring, the fish will continue to accelerate. Using this knowledge, we can deduce that the speed is maximum when the weight and spring force are equal. Thus we set them equal and find out the displacement first:

Weight = spring force

2.85 * 9.81 = 875 * Displacement

Displacement = 0.03195 m

Similarly as (A):

Change in G.P.E = 2.85 * 9.81 * 0.03195 = 0.8933 J

Spring P.E = 0.5 * 875 * 0.03195^2 = 0.4466 J

Kinetic energy = 0.8933 - 0.4466 = 0.4467 J

0.5 * mass * V^2 = 0.4467

0.5 * 2.85 * V^2 = 0.4467

V = 0.56 m/s

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3 years ago
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Answer:

Pressure is equal in A, B, C and D

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A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s . Part A What is the can'
Reika [66]

Answer:

the can's kinetic energy is 0.42 J

Explanation:

given information:

Mass, m = 460 g = 0.46 kg

diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m

velocity, v = 1.1 m/s

the kinetic energy of the can is the total of kinetic energy of the translation and rotational.

KE = \frac{1}{2} I ω^2 + \frac{1}{2} mv^{2}

where

I = \frac{1}{2} mr^{2} and ω = \frac{v}{r}

thus,

KE = \frac{1}{2} \frac{1}{2} mr^{2} (\frac{v}{r})^2 + \frac{1}{2} mv^{2}

     = \frac{1}{2} \frac{1}{2} mr^{2} \frac{v^{2} }{r^{2}} + \frac{1}{2} mv^{2}

     = \frac{1}{4} mv^{2} + \frac{1}{2} mv^{2}

     = \frac{3}{4} mv^{2}

     = \frac{3}{4} (0.46) (1.1)^{2}

     = 0.42 J

8 0
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