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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
Answer:
The answer is a," their receptors are sensitive to chemical molecules."
Given : Time taken to reach the maximum height t=3 s a=−g=−10m/s
2
The initial velocity of the ball can be calculated by,
Using v=u+at
∴ 0=u−10×3 ⟹u=30 m/s
Using S=ut+
2
1
at
2
∴ S=30×3+
2
1
×(−10)×3
2
=45m
