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IgorC [24]
3 years ago
10

If block D weighs 300 lb and block B weighs 275 lb determine the required weight of block C and the angle theta for equilibrium?

Physics
1 answer:
adoni [48]3 years ago
5 0
Set up a free body diagram. 

<span>and by reason, Tcd = Tbd </span>

<span>Tbd y = 275 - 300*sinθ </span>
<span>Tcd y = Tc - 300*sin30 </span>

<span>Tbd x = 300*cosθ </span>
<span>Tcdx = 300 * cos30 </span>

<span>Tbd^2 = (275 - 300*sinθ)^2 + (300*cosθ)^2 </span>
<span>Tcd^2 = (300*sin30)^2 + (300 * cos30)^2 </span>

<span>(275 - 300*sinθ)^2 + (300*cosθ)^2 = (300*sin30)^2 + (300 * cos30)^2 </span>
<span>etc.</span>
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Explanation:

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3 years ago
Elliot jumps up and down on a pogo stick. He weighs 600.N, and his pogo stick has a spring with spring constant 1100N/m. What is
tia_tia [17]

From conservation of energy, the height he will reach when he has gravitational potential energy 250J is 0.42 meters approximately

The given weight of Elliot is 600 N

From conservation of energy, the total mechanical energy of Elliot must have been converted to elastic potential energy. Then, the elastic potential energy from the spring was later converted to maximum potential energy P.E of Elliot.

P.E = mgh

where mg = Weight = 600

To find the height Elliot will reach, substitute all necessary parameters into the equation above.

250 = 600h

Make h the subject of the formula

h = 250/600

h = 0.4167 meters

Therefore, the height he will reach when he has gravitational potential energy 250J is 0.42 meters approximately

Learn more about energy here: brainly.com/question/24116470

4 0
2 years ago
A 25 kg bear slides, from rest, 12 m down a lodgepole pine tree, moving with a speed of 5.6 m/s just before hitting the ground.
Anuta_ua [19.1K]

Answer:

(A) -2940 J

(B) 392 J

(C) 212.33 N

Explanation:

mass of bear (m) = 25 kg

height of the pole (h) = 12 m

speed (v) = 5.6 m/s

acceleration due to gravity (g) = 9.8 m/s

(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)

height at the bottom = 0

         = 25 x 9.8 x (0-12) = -2940 J

(B) kinetic energy of the Bear (KE) = 0.5mv^{2}

           = 0.5 x 25 x 5.6^{2}  = 392 J

(C) average frictional force = \frac{change in thermal energy}{height} = \frac{-(ΔKE+ΔU)}{h}

  • change in KE (ΔKE) = initial KE - final KE
  • ΔKE = 0.5mv^{2} - 0.5mvf^{2}            
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 \frac{-(ΔKE+ΔU)}{h}[/tex] = \frac{-(392 + (-2940))}{12}

=  \frac{(-392 + 2940)}{12} = 212.33 N

5 0
3 years ago
Explain the following observations:
Deffense [45]

Answer and Explanation:

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4 0
3 years ago
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