Answer:
The COP of the system is = 4.6
Explanation:
Given data
Higher pressure = 1.8 M pa
Lower pressure = 0.12 M pa
Now we have to find out high & ow temperatures at these pressure limits.
Higher temperature corresponding to pressure 1.8 M pa
°c = 335.9 K
Lower temperature corresponding to pressure 0.2 M pa
°c = 262.9 K
COP of the system is given by


COP = 4.6
Therefore the COP of the system is = 4.6
You should come off like would you like to hang out sometime
Answer:

Explanation:
Our values are,

We have all the values to apply the law of linear momentum, however, it is necessary to define the two lines in which the study will be carried out. Being an intersection the vehicle of mass m_1 approaches through the X axis, while the vehicle of mass m_2 approaches by the y axis. In the collision equation on the X axis, we despise the velocity of object 2, since it does not come in this direction.

For the particular case on the Y axis, we do the same with the speed of object 1.

By taking a final velocity as a component, we can obtain the angle between the two by relating the equations through the tangent

Replacing in any of the two functions, given above, we will find the final speed after the collision,



Answer:
C. Impulse = F*t=(m*a)*t= m*(a*t) = m*Dv= D(Momentum) (“D” here’s mean Delta so change in)
Explanation:
In fact, the impulse is equal to the change in momentum of an object.
Impulse is defined as the product between the force (F) and the time (t):

however, the force is defined as the product between mass (m) and acceleration (a):

But the product a (acceleration) times t (time) is equal to the change in velocity of the object:

And this is exactly the definition of change in momentum:
