A block moving at an initial velocity of 18 m/s accelerates at a rate of -9m/s2. What is the velocity of the block after 6
seconds?

-36 m/s

Explanation:

u = 18m/s

a = -9m/s²

v = ?

t = 6s

from,

v = u + at

v = 18 + (-9)(6)

v = 18 - 54

v = -36m/s

heat the gas and measure the pressure

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

so as we know that

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

so

now radius of curvature is given as

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