To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.
According to Newton's second law we have to
where,
m= mass
g = gravitational acceleration
For the balance to break, there must be a mass M located at the right end.
We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.
In this way, applying the static equilibrium equations, we have to sum up torques at point B,
Regarding the forces we have,
Re-arrange to find M,
Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg
The answer is a because if you look really close
Answer:
Is this for zoommmmmmmmmm?
Explanation:
It had to be 20 letters I had to do something
Answer:
(a) 1.85 m/s
(b) 4.1 m/s
Explanation:
Data
- initial bullet velocity, Vbi = 837 m/s
- wooden block mass, Mw = 820 g
- initial wooden block velocity, Vwi = 0 m/s
- final bullet velocity, Vbf = 467 m/s
(a) From the conservation of momentum:
Mb*Vbi + Mw*Vwi = Mb*Vbf + Mw*Vwf
Mb*(Vbi - Vbf)/Mw = Vwf
4.1*(837 - 467)/820 = Vwf
Vwf = 1.85 m/s
(b) The speed of the center of mass speed is calculated as follows:
V = Mb/(Mb + Mw) * Vbi
V = 4.1/(4.1 + 820) * 837
V = 4.1 m/s
Answer:
In D: 3J
Explanation:
Potential energy: Ep=mgh where m is the mass, h altitude.
In point A: h=20cm=0.2m
Epa=12=0.2×mg. Thus mg=12/0.2=60N
For point D: hd=5cm=0.05m
Epd=mg×0.05=60×0.05=3J