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aleksandr82 [10.1K]
3 years ago
8

The drawing shows a top view of a door that is free to rotate about an axis of rotation that is perpendicular to the screen. Fin

d the net torque (magnitude and direction) produced by the forces F1 and F2 about the axis.
Physics
1 answer:
dangina [55]3 years ago
4 0

Answer:

τ = 23.34 Nm

Direction = Anti-Clockwise (Positive)

Explanation:

The image of the question is attached in the answer.

First, we will calculate the torque due to force F₁.

τ₁ = F₁ d₁ Sin θ₁

where,

τ₁ = Torque due to F₁ = ?

F₁ = 20 N

d₁ = moment arm of F₁ = 0.5 m

θ₁ = Angle between F₁ and d₁ = 90°

Therefore,

τ₁ = (20 N)(0.5 m)Sin 90°

τ₁ = - 10 Nm

Negative sign due to clockwise direction.

Now, we will calculate the torque due to force F₂.

τ₂ = F₂ d₂ Sin θ₂

where,

τ₂ = Torque due to F₂ = ?

F₂ = 35 N

d₂ = moment arm of F₂ = 1.10 m

θ₂ = Angle between F₂ and d₂ = 120°

Therefore,

τ₁ = (35 N)(1.1 m)Sin 120°

τ₁ = 33.34 Nm

Positive sign due to anti-clockwise direction.

Now, we find net torque on door:

τ = τ₁ + τ₂

τ = - 10 Nm + 33.34 Nm

<u>τ = 23.34 Nm</u>

<u>Direction = Anti-Clockwise (Positive)</u>

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Katena32 [7]

Answer:

1

 t_a  =  13.49 \  s

2

 The distance is   D = 55.2 \  m    

Explanation:

From the question we are told that

   The maximum speed of the cheetah is  v =  103 \  km/h =  28.61 \  m/s

    The maximum of  gazelle is  u =  76.5 \  km/h =  21.25 \  m/s

     The distance ahead is d =  99.3 \  m

Let  t_a denote the time which the cheetah catches the gazelle

Gnerally the equation representing the distance the cheetah needs to move in order to catch the gazelle is

           v* t_a = d +  u* t_a

=>          28.61 t_a = 99.3 +  21.25t_a

=>          7.36 t_a = 99.3

=>         t_a  =  13.49 \  s

Now at t =  7.5 s  

             7.5 v = D+  7.5u

=>          28.61 * 7.5  = D +  21.25* 7.5

=>          7.36 *  7.5 =D

=>         D = 55.2 \  m        

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5 0
2 years ago
An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10
s2008m [1.1K]

Answer:

<em>1.228 x </em>10^{-6}<em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm  

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x 10^{3} N

modulus of elasticity E = 85 GN/m^2  = 85 x 10^{9} Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = \frac{\pi D^{2} }{4} =  \frac{3.142* 40^{2} }{4} = 1256.8 mm^2

area of hole a = \frac{\pi(D^{2} - d^{2}) }{4} = \frac{3.142*(40^{2} - 30^{2})}{4} = 549.85 mm^2

Total contraction of the bar = \frac{F*L}{AE} + \frac{Fl}{aE}

total contraction = \frac{F}{E} * (\frac{L}{A} +\frac{l}{a})

==> \frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85}) = <em>1.228 x </em>10^{-6}<em> mm </em>

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