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3 years ago

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A car is driven 110 km west and then 40 km southwest, how far is the car from the point of origin?

1 answer:

Margaret [11]3 years ago

8 0

Answer:

141.147 km

Explanation:

$sin45=\frac{BD}{BC}\\\Rightarrow BD=BCsin45\\\Rightarrow BD=40\times sin45\\\Rightarrow BD=28.28\ km$

Now, AE=BD also CD=BD as the angle B and C are equal

So, from Pythagoras theorem

AC² = AE²+(CD+DE)²

$AC=\sqrt{AE^2+(CD+DE)^2}\\\Rightarrow AC=\sqrt{(40\times sin45)^2+(110+40\times sin45)^2}\\\Rightarrow AC=141.147\ km$

Displacement of the car is 141.147 km

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The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

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Answer:

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(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

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According to figure,

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The horizontal component is

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The vertical component is

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$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

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(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

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$F_{x}=T\cos\theta$

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$F_{x}=658.33\times\dfrac{4}{5}$

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(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

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(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

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