A skydiver reaches terminal velocity. Then he opens his parachute.
2 answers:
Answer:
Explanation:
Remark
He's pulled upwards by the parachute. The formula is given below. Details are found searching for terminal velocity.
Whatever that calculation turns out to be, a person has to be going a whole lot slower before he can safely hit the ground. The parachute is designed so that the person will be slowed very quickly to a safe velocity so when he hits the ground the impulse will not be as great as it could be. The force that the parachute delivers acts upward and opposes the force of gravity which is a downward force.
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Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:
where F is the force applied, and
is the stretch of the spring with respect to its equilibrium position. Using the data, we find
Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
where F is the force applied, and
Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
Answer:
The mass m is 0.332 kg or 332 gm
Explanation:
Given
The platform is rotating with angular speed ,
Mass m is moving on platform in a circle with radius ,
Force sensor reading to which spring is attached ,
Now for the mass m to move in circle the required centripetal force is given by
=>
Thus the mass m is 0.332 kg or 332 gm