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baherus [9]
2 years ago
10

B)A man walks 95 km, East, then 55 km, north. Calculate his RESULTANT

Physics
1 answer:
Varvara68 [4.7K]2 years ago
8 0

The resultant displacement of the man is 109.77 km in the direction N60°E.

<h3>Displacement</h3>

Displacement is the distance travelled in a specified direction.

To calculate displacement, the straight line from starting point to end point of travel is taken and calculated.

<h3>Resultant displacement of the man </h3>

In the example above, a man walks 95 km, East, then 55 km, north.

The two distances form a right-angled triangle with two sides 95 and 55 units. The hypotenuse gives the resultant displacement, D.

Using Pythagoras rule:

D^2 = 95^2 + 55^2

D^2 = 12050

D = 109.77

Thus, the resultant displacement is 109.77 km

To calculate the direction:

Let the direction be y

y + x = 90°

tan x = 55/95

tanx x = 0.578

x = 30°

Then, y = 90 - 30

y = 60°

Therefore, the resultant displacement of the man is 109.77 km in the direction N60°E.

Learn more about displacement at: brainly.com/question/321442

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The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.
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The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

  • The maximum steepness of the slope where the truck can be parked without tipping over is approximately <u>54.55 %</u>.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, <em>C</em> represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}

The steepness of the slope is therefore;

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100

Where;

\overline{AM} = Half the width of the truck = \dfrac{2.4 \, m}{2} = 1.2 m

\overline{CM} = The elevation of the center of gravity above the ground = 2.2 m

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%

tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}

Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right)  \approx 28.6 ^{\circ}

The maximum steepness of the slope where the truck can be parked is <u>54.55 %</u>.

Learn more here:

brainly.com/question/20793607

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