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Furkat [3]
3 years ago
5

question why if you keep dieing in zombies on cold war u think im gunna rev you. i just got the key and the d.i.e remote and am

geting the wonder wepon. but soon as that door explode you so quick to grab it you die. played me like a fool. why the heck you keep tryna exfil on round 15 BOI IF YOU DONT WQBFUFHJEWFHFNQC! packapunched knife to tier 1 head aS BOI!!!
Physics
1 answer:
tino4ka555 [31]3 years ago
6 0

Answer:

??????? umm what r u saying??? kinda confusing..............what does question why if you keep dieing in zombies on cold war u think im gunna rev you. i just got the key and the d.i.e remote and am geting the wonder wepon. but soon as that door explode you so quick to grab it you die. played me like a fool. why the heck you keep tryna exfil on round 15 BOI IF YOU DONT WQBFUFHJEWFHFNQC! packapunched knife to tier 1 head aS BOI!!! mean????????

very very very confusing

Explanation:

plz mark brainlyist :()

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What could be the possible answer to the question ?<br><br>thankyou ~​
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The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

  • The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

T_1 = \mathbf{\dfrac{M \cdot g}{2}}

Which gives;

M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}

T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})}  = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}

F₀ = T₂·sin(37°)

Which gives;

F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2}  \approx  \mathbf{0.377  \cdot M \cdot g}

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Learn more about equilibrium of forces here:

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Determine the magnitude of the resultant force acting on a 1.5 −kg particle at the instant t=2 s, if the particle is moving alon
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Answer:

F = 63N

Explanation:

M= 1.5kg , t= 2s, r = (2t + 10)m and

Θ = (1.5t² - 6t).

magnitude of the resultant force acting on 1.5kg = ?

Force acting on the mass =

∑Fr =MAr

Fr = m(∇r² - rθ²) ..........equation (i)

∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)

The horizontal path is defined as

r = (2t + 10)

dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

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at t = 2

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but dr/dt = 2m/s and d²r/dt² = 0m/s

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dθ/dt =3(2) - 6 = 0rads

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substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

F = √(Fr² +FΘ²) = √(0² + 63²) = 63N

7 0
3 years ago
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