Answer:
A measured force of (46.5 0.8 N ) would not be in agreement with a theoretically calculated force of (48.4 0.6 N )
Explanation:
From the question we are told that
Measured force is ![F_M = [46.5 \pm 0.8 \ N ]](https://tex.z-dn.net/?f=F_M%20%20%3D%20%20%5B46.5%20%5Cpm%200.8%20%5C%20%20N%20%5D)
Calculated force is ![F_c = [48.4 \pm 0.6 \ N ]](https://tex.z-dn.net/?f=F_c%20%3D%20%20%5B48.4%20%5Cpm%200.6%20%5C%20%20N%20%5D)
Generally the measured force in interval form is
=> 
Generally the calculated force in interval form is
=> 
Generally looking both interval we see that they do not intersect at any point Hence
A measured force of (46.5 0.8 N ) would not be in agreement with a theoretically calculated force of (48.4 0.6 N )
Answer:
(C) 40m/s
Explanation:
Given;
spring constant of the catapult, k = 10,000 N/m
compression of the spring, x = 0.5 m
mass of the launched object, m = 1.56 kg
Apply the principle of conservation of energy;
Elastic potential energy of the catapult = kinetic energy of the target launched.
¹/₂kx² = ¹/₂mv²
where;
v is the target's velocity as it leaves the catapult
kx² = mv²
v² = kx² / m
v² = (10000 x 0.5²) / (1.56)
v² = 1602.56
v = √1602.56
v = 40.03 m/s
v ≅ 40 m/s
Therefore, the target's velocity as it leaves the spring is 40 m/s
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