Answer:
<em>6.77m/s</em>
Explanation:
Using the law of conservation of momentum
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses of the object
u1 and u2 are the velocities before collision
v is the final collision
Given
m1 = 300g = 0.3kg
u1 = 6.0m/s
m2 = 10g = 0.01kg
u2 = 30m/s
Required
The bird's speed immediately after swallowing v
Substitute the given values into the formula
m1u1 + m2u2 = (m1+m2)v
0.3(6) + 0.01(30) = (0.3+0.01)v
1.8+0.3 = 0.31v
2.1 = 0.31v
v = 2.1/0.31
<em>v = 6.77m/s</em>
<em>Hence the bird's speed immediately after swallowing is 6.77m/s</em>
Answer: I showed you all calculation . You did not attach any graph to question .
Explanation:
Lets first find Velocity
Vr=o m/s
Ve=?
a=1m/s²
t=2s
----------
a=(Vr-V)/t
1m/s²=Vr-0m/s/2s
2m/s=Vr
Lets find the time neeeded to stop :
a=1m/s²
Vs=2m/s
Vf=0m/s
a=(Vf-Vs)/t
t*1m/s²=2m/s
t=2 s
