A meter stick is balanced at the 50 cm mark. You tie a 20 N weight at the 20 cm mark. Where should a 30 N weight be placed so th
1 answer:
Answer:
20 cm to the right of the center or 20+50 = 70 cm from the left side.
Explanation:
The length of meter stick is 1 m = 100 cm
Balance point on 50 cm
From the center the 20 N weight is 50-20 = 30 cm
Torque is obtained when force is multiplied with the distance
As the force is conserved we have
The distance will be 20 cm to the right of the center or 20+50 = 70 cm from the left side.
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Answer:
Explanation:
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Given that,
Time = 0.5 s
Acceleration = 10 m/s²
(I). We need to calculate the speed of apple
Using equation of motion
Where, v = speed
u = initial speed
a = acceleration
t = time
Put the value into the formula
(III). We need to calculate the height of the branch of the tree from the ground
Using equation of motion
Put the value into the formula
(II). We need to calculate the average velocity during 0.5 sec
Using formula of average velocity
Where, = final position
= initial position
Put the value into the formula
Hence, (I). The speed of apple is 5 m/s.
(II). The average velocity during 0.5 sec is 2.5 m/s
(III). The height of the branch of the tree from the ground is 1.25 m.