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Romashka [77]
3 years ago
12

A robot starts from a certain point and moves east for a distance of 5.0 meters, then goes north for 3.0 meters, and then turns

west for 2.0 meters. a. What are the x-y coordinates for the resultant vector? b. What is the magnitude of the resultant vector for the robot?
Physics
1 answer:
pashok25 [27]3 years ago
4 0

Explanation:

It is given that, a robot starts from a certain point and moves east for a distance of 5.0 meters, then goes north for 3.0 meters, and then turns west for 2.0 meters.

Let east is +x, west is -x north is +y.

When it moves east for a distance of 5.0 meters, it means (5,0)

When it goes north for 3.0 meters, and then turns west for 2.0 meters, it means (0,3) +(-2,0)

(a) x-y coordinates for the resultant vector is (3,3)

(b) The magnitude of the resultant vector for the robot is given by :

R=\sqrt{3^2+3^2} \\\\R=4.24\ m

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Consider an elevator with a table and a book on top of the table. The mass of the table is 10kg and the mass of the book is 2kg.
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Newton's second law allows us to find the force of the block on the table is 126 N

Newton's second law says that the net force is proportional to the product of the mass and the acceleration of the body

                Σ F = m a

Where the bold letters indicate vectors, m is the mass and the acceleration of the body

A free body diagram is a diagram where the forces are represented without the details of the bodies, in the attached we can see a free body diagram of the system.

Let's start by finding the acceleration of the elevator with kinematics  

                 v = v₀ + a t

                 a = \frac{v-v_o}{a}  

Where v and v₀ are the current and initial velocity, respectively, at acceleration and t is the time

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                 a = 3.5 m / s²

Let's write Newton's second law for each body

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                N₂ - W₂ + N₁ = m a

               

Table

                N₁ - W₁ - W₂ = M a

                W₁ = Mg

                W₂ = mg

                N₁ = (M + m) g + M a

                N₁ = (10 + 2) 9.8 + 10 3.5

                N₁ = 152.6 N

This is the reaction of the earth to the support of the block and the table

               N₂ = ma + m g  - N₁

               N₂ = m ( a +g) - N₁  

               N₂ = 2 (3.5 + 9.8) - 152.6

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In conclusion using Newton's second law we can find the forces of the block on the table is 126 N

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