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
1 answer:
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 :
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Answer:
F = 19.1 N
Explanation:
To find the force exerted by the string on the block you use the following formula:
(1)
k: spring constant = 95.5 N/m
x: displacement of the block from its equilibrium position = 0.200 m
you replace the values of k and x in the equation (1):
Hence, the force exterted on the block is 19.1 N
Explanation:
It is given that, the height of a certain tower is 862 feet i.e to reach on the ground the object should travel, s = 862 feet
The distance traveled by a freely falling object is given by :
t = 7.34 seconds
So, the object will take 7.34 seconds to fall to the ground from the top of the building. Hence, this is the required solution.