Momentum before collision must be equal to momentum after collision
(m1<span> + m</span>2)v = m1v1<span> + m</span>2v<span>2</span>
3.62g*270 m/s=2.30kg* (x)
convert the units to be the same that is convert the kg mass to grams
1kg=1000g
2.30kg=y
230/100*1000=2300g
3.62*270=2300X
977.4g/m/s=2300X
977.4/2300=0.4250M/S
Finding a velocity after the gun is embedded on the block of wood
3.62*270+2300g*0.425=(3.62+2300)*V
977.5+977.5=2303V
1955=2303V
V=1955/2303
V=0.8489M/S
This question involves the concepts of the equations of motion, kinetic energy, and potential energy.
a. The kinetic energy of the rocket at launch is "3.6 J".
b. maximum gravitational potential energy of the rocket is "3.6 J".
<h3>a. KINETIC ENERGY AT LAUNCH</h3>
The kinetic energy of the rocket at launch is given by the following formula:

where,
- K.E = initial kinetic energy = ?
- m = mass of rocket = 0.05 kg
= initial speed = 12 m/s
Therefore,

K.E = 3.6 J
<h3>
b. MAXIMUM GRAVITATIONAL POTENTIAL ENERGY</h3>
First, we will use the third equation of motion to find the maximum height reached by rocket:

where,
- g = -9.81 m/s²
- h = maximum height = ?
- vf = final speed = 0 m/s
Therefore,
2(-9.81 m/s²)h = (0 m/s)² - (12 m/s)²
h = 7.34 m
Hence, the maximum gravitational potential energy will be:
P.E = mgh
P.E = (0.05 kg)(9.81 m/s²)(7.34 m)
P.E = 3.6 J
Learn more about the equations of motion here:
brainly.com/question/5955789
Answer:
A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s)
Explanation:
<u>To rank the angular speed (ω) of the objects, we need first calculate its value for every object:</u>
A bowling ball of radius 12.3cm rotating at 8.21 radians per second:
ω = 8.21 rad/s
A tire of radius 0.321m rotating at 75.8 rpm:
A 6.84cm diameter top spinning at 375 degrees per second:
A square with sides (b) 0.458m long, whose corners are moving with tangential speed (v) 2.51 m/s as it rotates about its center:

A rock on a string, being swung in a circle of radius 0.521 m with a centripetal acceleration (a) of 25.4 m/s²:
<u>Now, the rank of the angular speed of the objects, from highest to lowest is: </u>
A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s)
I hope it helps you!