applied forces would be push for example.
normal forces would seem to be a force such as gravity.
friction for example when you try to slide on carpet but the fabric or whatever its made of stops you.
Answer:
tension in rope = 25.0 N
Explanation:
- Two forces act on the suspended weight. The force coming down is the gravitational force and the upward force by the tension in the rope.
- Since the suspended weight is not accelerating so that the net force will be zero. Therefore the tension in the rope should be 25 N.
∑F = F - W = 0
so
F = W
so tension in rope = F = T = 25 N
Since Astronaut and wrench system is isolated in the space and there is no external force on it
So here momentum of the system will remain conserved
so here we can say
![m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}](https://tex.z-dn.net/?f=m_1v_%7B1i%7D%20%2B%20m_2v_%7B2i%7D%20%3D%20m_1v_%7B1f%7D%20%2B%20m_2v_%7B2f%7D)
initially both are at rest
so here plug in all values
![0 = 100 v_{1f} + 2\times 10](https://tex.z-dn.net/?f=0%20%3D%20100%20v_%7B1f%7D%20%2B%202%5Ctimes%2010)
![v_{1f} = -0.20 m/s](https://tex.z-dn.net/?f=v_%7B1f%7D%20%3D%20-0.20%20m%2Fs)
so here the astronaut will move in opposite direction and its speed will be equal to 0.20 m/s
Answer:
the change in momentum = Force x change in time
Answer:
Pressure,P=6×10^3Pa
Explanation:
The gas has an ideal gas behaviour and ideal gas equation
PV=NKT
T= V/N p/K ...eq1
Average transitional kinetic energy Ktr=1.8×10-23J
Ktr=3/2KT
T=2/3Ktr/K....eq2
Equating eq1 and 2
V/N p/K = 2/3Ktr/K
Cancelling K on both sides
P= 2/3N/V( Ktr)
Substituting the value of N/V and dividing by 10^-6 to convert cm^3 to m^3
P = 2/3 (5.0×10^20)/10^-6 × 1.8×10^-23
P= 6 ×10^3Pa