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Tamiku [17]
3 years ago
15

A disoriented physics professor drives 3.25 km north, then 2.20 km west, and then 1.50 km south. Find the magnitude and directio

n of the resultant displacement, using the method of components. In a vector-addition diagram (roughly to scale), show that the resultant displacement found from your diagram is in qualitative agreement with the result you obtained by using the method of components

Physics
1 answer:
Nana76 [90]3 years ago
4 0

Answer: magnitude of resultant = 2.811km, direction of resultant = 209.6°

Explanation: The vector ( in this case displacement) that lies on the y axis is 3.25km north and 1.50km due south.

Their resultant is gotten below

Magnitude of resultant = 3.25 - 1.50 = 1.75km

Direction of resultant = north (direction of the bigger vector)

2.20km is the only vector acting on the x axis due east, combining this vector with the resultant of the vectors above, we realize that 2.2km west is perpendicular to 1.75km due north. Since 1.75km us due north, it implies that it is the vector on the positive y axis (vy) and 2.20km due west implies that it is the vector on the negative x axis(vx), thus their resultant is gotten using phythagoras theorem.

R = √ vx² + vy²

R = √ 2.20² + 1.75²

R = √ 7.9025

R= 2.1811km.

θ = tan^-1 (vy/vx)

θ = tan ^-1 (1.25/2.20)

θ = tan ^-1 (0.5618)

θ = 29.6°.

The direction of the vector is south west which implies the third quadrant of the trigonometric quadrant which implies 180 + θ

Thus the direction of the vector is 180 + 29.60 = 209.6°

From the picture attached to this answer we can see roughly that the magnitude of resultant is longer than it component and the vector is placed on the 3rd quadrant which verifies our quantitative claim.

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The moment of inertia is 24.8 kg m^2

Explanation:

The total moment of inertia of the system is the sum of the moment of inertia of the rod + the moment of inertia of the two balls.

The moment of inertia of the rod about its centre is given by

I_r = \frac{1}{12}ML^2

where

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L = 0.96 m is the length of the rod

Substituting,

I_r = \frac{1}{12}(24)(0.96)^2=1.84 kg m^2

The moment of inertia of one ball is given by

I_b = mr^2

where

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So we have

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An elevator cab and its load have a combined mass of 1600 kg. Find the tension in the supporting cable when the cab, originally
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The tension in the supporting cable when the cab originally moves downward is 18422.4 N

What is tension?

Tension is described as the pulling force by the means of a three-dimensional object.

Tension might also be described as the action-reaction pair of forces acting at each end of said elements.

Here,

m =combined mass = 1600 kg

s = Displacement of the elevator = 42 m

g = Acceleration due to gravity = 9.81 m/s²

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v = Final velocity = 0

According to the equation of motion:

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Now let's write the equation of the forces acting on the elevator. Taking upward as positive direction:

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Hence,

The tension in the supporting cable when the cab, originally moving downward is 18422.4 N

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