<span>Assuming pulley is frictionless. Let the tension be ‘T’. See equation below.</span>
<span> </span>
Answer:
a) The sign of the charge is positive.
b) The magnetic force on the particle is 0.050 newtons.
Explanation:
The magnetic force F on a moving charge with velocity v passing through a magnetic field B is:
(1)
a)
Because it is a cross product, we can find the direction of the force using the right-hand rule, that is too the direction of the movement. We have two possibilities here because the velocity vector and magnetic field are perpendicular: the particle deflects towards east or toward west, which depends on the charge of the particle. Note that if you put your right hand fingers, except thumb, pointing towards north (direction of velocity) and later close them in the direction of the magnetic field, if you maintain your thumb perpendicular to this movement it will point towards east (See figure), so that will be de direction of the force if the charge is positive, but if the charge is negative, the direction will be opposite (towards west). So the charge has to be positive to deflects towards east.
b)
Now by 1:

Answer:
x2 = 0.99
Explanation:
from superheated water table
at pressure p1 = 0.6MPa and temperature 200 degree celcius
h1 = 2850.6 kJ/kg
From energy equation we have following relation



![2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]](https://tex.z-dn.net/?f=2850.6%20%2B%20%5B%5Cfrac%7B50%5E2%7D%7B2%7D%20%2A%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2FS%5E2%7D%5D%20%3D%20h2%20%2B%5B%20%5Cfrac%7B600%5E2%7D%7B2%7D%20%2A%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2FS%5E2%7D%5D)
h2 = 2671.85 kJ/kg
from superheated water table
at pressure p2 = 0.15MPa
specific enthalpy of fluid hf = 467.13 kJ/kg
enthalpy change hfg = 2226.0 kJ/kg
specific enthalpy of the saturated gas hg = 2693.1 kJ/kg
as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have
quality of steam x2
h2 = hf + x2(hfg)
2671.85 = 467.13 +x2*2226.0
x2 = 0.99
Answer:
The velocity of the truck after the collision is 20.93 m/s
Explanation:
It is given that,
Mass of car, m₁ = 1200 kg
Initial velocity of the car, 
Mass of truck, m₂ = 9000 kg
Initial velocity of the truck, 
After the collision, velocity of the car, 
Let
is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.




So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.
I’m not completely sure but most likely is is the 10 mile bike ride, I hope I can help! (: