Physics, 23.04.2021 14:00 minileiva211

30. For the figure below, the vector product T x W will be calculated as: 22 N
W
500
12 N
20 N
A. 242.42 N
B. 196.24 N
C. 185.52 N
D. 169.69 N

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initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

$a = 32 ft/s^2$

so as we know that

$a = \frac{v^2}{r}$

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

$a_c = 32 ft/s^2$

$a_t = 0$

now radius of curvature is given as

$r = \frac{v^2}{a_c}$

$r = \frac{150^2}{32}$

$r = 703.125ft$

so the radius of curvature just after it is dropped is 703.125 ft

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30. For the figure below, the vector product T x W will be calculated as: 22 N
W
500

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