Answer:
Incomplete question
Check attachment for the given diagram
Explanation:
Given that,
Initial Velocity of drum
u=3m/s
Distance travelled before coming to rest is 6m
Since it comes to rest, then, the final velocity is 0m/s
v=3m/s
Using equation of motion to calculate the linear acceleration or tangential acceleration
v²=u²+2as
0²=3²+2×a×6
0=9+12a
12a=-9
Then, a=-9/12
a=-0.75m/s²
The negative sign shows that the cylinder is decelerating.
Then, a=0.75m/s²
So, using the relationship between linear acceleration and angular acceleration.
a=αr
Where
a is linear acceleration
α is angular acceleration
And r is radius
α=a/r
From the diagram r=250mm=0.25m
Then,
α=0.75/0.25
α =3rad/sec²
The angular acceleration is =3rad/s²
b. Time take to come to rest
Using equation of motion
v=u+at
0=3-0.75t
0.75t=3
Then, t=3/0.75
t=4 secs
The time take to come to rest is 4s
Answer:
Explanation:
the force of the rocket engine pushing it up, the force of gravity pulling it down, maybe some force of air resistance as the rocket goes fast, hmmm Free Body Diagrams (FBD) should have any and all forces on the model, unless they are negligible . or so slight they really make little difference in the total outcome.
Hi, thank you for posting your question here at Brainly.
To compute for the change in potential energy, the equation would be:
delta PE = mg*delta h
delta PE = 0.5*9.81*(2-1.8)
delta Pe = 0.98 J
The potential energy is converted to kinetic energy.
Answer:
kg m/s
Explanation:
e = Charge = C
V = Voltage = 
c = Speed of light = m/s
Momentum is given by
The unit of MeV/c in SI fundamental units is kg m/s
Answer:
vB' = 0.075[m/s]
Explanation:
We can solve this problem using the principle of linear momentum conservation, which tells us that momentum is preserved before and after the collision.
Now we have to come up with an equation that involves both bodies, before and after the collision. To the left of the equal sign are taken the bodies before the collision and to the right after the collision.
where:
mA = 0.355 [kg]
vA = 0.095 [m/s] before the collision
mB = 0.710 [kg]
vB = 0.045 [m/s] before the collision
vA' = 0.035 [m/s] after the collision
vB' [m/s] after the collison.
The signs in the equation remain positive since before and after the collision, both bodies continue to move in the same direction.
![(0.355*0.095)+(0.710*0.045)=(0.355*0.035)+(0.710*v_{B'})\\v_{B'}=0.075[m/s]](https://tex.z-dn.net/?f=%280.355%2A0.095%29%2B%280.710%2A0.045%29%3D%280.355%2A0.035%29%2B%280.710%2Av_%7BB%27%7D%29%5C%5Cv_%7BB%27%7D%3D0.075%5Bm%2Fs%5D)
