Answer:
Along path BC of the Otto cycle, heat transfer Qh into the gas occurs at constant volume, causing a further increase in pressure and temperature. This process corresponds to burning fuel in an internal combustion engine, and takes place so rapidly that the volume is nearly constant.
Answer:
19.6N
Explanation:
Given parameters:
Mass of rock = 2kg
Speed = 30m/s
Unknown:
Net force on the rock = ?
Solution:
The net force acting on this rock is a function of the acceleration due to gravity acting upon it.
Net force = weight = mass x acceleration due to gravity
Net force = 2 x 9.8 = 19.6N downward
Answer:
a) 
b) 
c) 
Explanation:
Given masses:


Velocity of mass 1, 
Velocity of mass 2, 
a)
Initial momentum:



b)
magnitude of initial momentum:


From the conservation of momentum:



is the magnitude of final velocity.
Direction of final velocity will be in the direction of momentum:




c)
Vertical component of final velocity:


<span> y=y0 + vt +1/2gt^2
(solve for t here) cause you know y,y0,v,g
you will do quad formula here
then:
v=v0 +at solve for v
(remember the direction of the ball too (signs))
The main thing to remember here is that when the ball passes exactly (height) where it was launched it will travel the speed at which it was launched. *its almost like the ball was thrown in the downward direction. </span>
Here is your answer
C. towards the floor
REASON:
Using Fleming's Left hand rule we can determine the direction of force applied on a moving charged particle placed in a magnetic field.
The direction of current will be just opposite to the direction of electron(negative charge) because current moves from positive to negative terminal whereas electron moves from negative to positive terminal.
So, direction of current- North to South
Now applying Fleming's Left hand rule we get the direction of force in downward direction, i.e. towards the floor.
HOPE IT IS USEFUL