By US Fg u u high I’ll fucouoyog you go y DC oh dad dc Sc um c o CBC cyo FCC ztdA
The speed of the target with the bullet lodged in it is 5.6 m/s.
<h3>What is speed?</h3>
This is the rate of change of distance.
To calculate the speed of the target and the bullet lodged, we use the formula below.
Formula:
- V = mu/(m+M)............. Equation 1
Where:
- V = Speed of the target and the bullet
- m = mass of the bullet
- M = mass of the target
- u = Initial speed of the target
From the question,
Given:
- m = 0.04 kg
- u = 1400 m/s
- M = 9.96 kg
Substitute these values into equation 1
- V = (0.04×1400)/(0.04+9.96)
- V = 56/10
- V = 5.6 m/s
Hence, the speed of the target with the bullet lodged in it is 5.6 m/s.
Learn more about speed here: brainly.com/question/6504879
#SPJ1
"<span>H-C=N:" is the one answer among the choices given in the question that shows the correct dot diagram. The correct option among all the options that are given in the question is the fourth option or option "D". The other choices can be neglected. I hope that this is the answer that has come to your help.</span>
Answer:
Explanation:
For answer this we will use the law of the conservation of the angular momentum.
so:
where 
is the moment of inertia of the merry-go-round, 
is the initial angular velocity of the merry-go-round, 
is the moment of inertia of the merry-go-round and the child together and 
is the final angular velocity.
First, we will find the moment of inertia of the merry-go-round using:
I = 
I = 
I = 359.375 kg*m^2
Where 
is the mass and R is the radio of the merry-go-round
Second, we will change the initial angular velocity to rad/s as:
W = 0.520*2
rad/s
W = 3.2672 rad/s
Third, we will find the moment of inertia of both after the collision:
Finally we replace all the data:
Solving for :
