Answer:
1. What was the diameter of the hailstone in inches_5.09_ and in cm___12.92___ and in feet ____0.42___?
2. What was the total volume of the hailstone in cubic inches___68.64___ and cubic feet____0.03____?
3. What was the fall velocity of this hailstone in m/s_____2.584____ and in mph___5.78_____?
Explanation:
1. If the circumference (L) of the stone is 16 inches, then from the following equation
1 inch = 2.54 cm = 0.08 ft, so 5.09 inches = 12..92 cm = 0.42 ft
2. The total volume is
3. The fall velocity is V = kd, where k = 20 if d is in cm. Let's calculate the fall velocity in cm.
2.584 m/s = 5.78 mph
Answer:
Fp = 36 N
Fq = 58 N
Explanation:
Let the left end be the reference end with string p closest to it.
Let CCW moments be positive
Sum moments about p to zero
1(9.8)[2 - 1] + Fq[6 - 2] - 5(9.8)[8/2 - 2] - 1.5(9.8)[5 - 2] - 2(9.8)[7 - 2] = 0
Fq[4] = 23.5(9.8)
Fq = 57.575 ≈ 58 N
Sum moments about q to zero
1(9.8)[6 - 1] - Fp[6 - 2] + 5(9.8)[6 - 8/2] + 1.5(9.8)[6 - 5] - 2(9.8)[7 - 6] = 0
Fp = 35.525 N
or
Sum vertical forces to zero
Fp + 57.575 - (9.8)(1 + 1.5 + 2 + 5) = 0
Fp = 35.525 ≈ 36 N
To calculate the horizontal distance traveled by the ball, we first calculate the total time it takes to reach the ground as follows:
t = √[2y/g] = <span>√[2(0.55) / 9.81]
t = 0.33 s
The horizontal distance would be
</span><span>X = Vx*t = 1.2*√[2*.55/9.8] = 0.4 m
</span>
Hope this helps.
The object will move if the forces are unbalanced.
Newtons second tells you that when a net force (the unbalanced force) is applied to and object it will produce an acceleration (movement) in direct proportion to the force and in inverse proportion to the mass of the object.
