Answer:
60 m
Explanation:
After 3 seconds of travel at 20 m/s, the projectile is 3·20 = 60 meters horizontally from the cannon.
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The vertical height after 3 seconds is 0.9 m, so the straight-line distance from cannon to target is √(60^2 +0.9^2) ≈ 60.007 meters.
Answer:
<h3> b. 1.18</h3>
Explanation:
The fundamental frequency in string is expressed as;
F1 = 1/2L√T/m .... 1
L is the length of the string
T is the tension
m is the mass per unit length
If the tension is increased by 40%, the new tension will be;
T2 = T + 40%T
T2 = T + 0.4T
T2 = 1.4T
The new fundamental frequency will be;
F2 = 1/2L√1.4T/m ..... 2
Divide 1 by 2;
F2/F = (1/2L√1.4T/m)/1/2L√T/m)+
F2/F = √1.4T/m ÷ √T/m
F2/F = √1.4T/√m ×√m/√T
F2/F = √1.4T/√T
F2/F = 1.18√T/√T
F2/F = 1.18
F2 = 1.18F
Hence the fundamental frequency of vibration changes by a factor of 1.18
When an object is falling and reaches a constant velocity, the net force on the object is <em>zero</em> (it's not accelerating), and the weight of the object is equal to <em>the force of air resistance against the object</em>. (choice-D)
