Answer:

Explanation:
Given that,
Initial angular velocity, 
Acceleration of the wheel, 
Rotation, 
Let t is the time. Using second equation of kinematics can be calculated using time.

Let
is the final angular velocity and a is the radial component of acceleration.

Radial component of acceleration,

So, the required acceleration on the edge of the wheel is
.
5.6 g/ml. That is the density.
Answer:
Vf=3
Explanation:
you must first write your data
data before impact
M1=1000 M2=5000
V1=0 m/s V2 =0m/s
data after impact
M1=1000 M2=5000
V1=15m/s V2=?
M1V1 +M2V2=M1V1 +M2V2f
(1000)(0)+(5000)(0)=(1000)(15)+(5000)Vf
0=15000+5000Vf
- 15000÷5000=5000Vf÷5000
Vf= -3
Vf =3
Answer:
Block A will have a final charge of 3.5nC.
Explanation:
This is because at the point of contact with Block B, which is electrically positive, the electrons in Block A will be attracted to the excess 'unpaired' protons in block B. Hence, the electrons will flow into Block B causing unpaired protons to remain in Block A.
This process is called Charging by Conduction.
This charging process will continue until the charges are evenly distributed between both objects.
In case you're wondering, "<em>how's all this possible within a few seconds</em>?", remember that electrons travel very fast and so, this process is a rather rapid one.
Answer:
Acceleration = 4 m/s²
Explanation:
Given the following data;
Force = 8 N
Mass = 2 kg
To find the acceleration of the block;
Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.
Mathematically, it is given by the formula;
Substituting into the formula, we have;
Acceleration = 4 m/s²