Answer:
-6.8 m/s²
Explanation:
Given:
v₀ = 30.5 m/s
v = 0 m/s
t = 4.5 s
Find: a
a = (v − v₀) / t
a = (0 m/s − 30.5 m/s) / 4.5 s
a = -6.8 m/s²
Answer:
v2 = 27.3m/s
Explanation:
Assuming forward as positive.
Mass = m1 = 64kg
Let v be the common velocity of the student and the skateboard.
mass of skateboard = m2 = 5.94kg
v = 1.4m/s
Since the skateboard and the student are initially moving together at the same velocity their momentum together is
(m1 + m2)v
Let the final velocity of the student be v1 and the final velocity of the skateboard be v2
v1 = – 1.0m/s (falls backwards that's why the velocity is negative since we are assuming forward as positive)
Then from conservation of momentum, momentum before is equal to momentum after.
(m1 + m2)v = m1v1 + m2v2
m2v2= (m1 + m2)v – m1v1
v2 = ( (m1 + m2)v – m1v1)/m2
v2 = ( (64 + 5.94)×1.4 – 64×(-1.0))/5.94
v2 = ( (64 + 5.94)×1.4 + 64×1.0)/5.94
v2 = 27.3m/s
Answer:
When her hands extends, her momen of inertia is
.
Explanation:
Given that,
Initial angular speed, 
Initial moment of inertia, 
Final angular speed, 
Initially, a skater rotates with her arms crossed and finally she extends her arms. The momentum remains conserved. Using the conservation of momentum as :

is final moment of inertia

So, when her hands extends, her momen of inertia is
. Hence, this is the required solution.