Answer:
So, the evaporating pressure of the R410A = 118 psig
Explanation:
Solution:
For R410A system:
Data Given:
Evaporator Outlet Temperature = 50°F
Evaporator Superheat = 10°F
Required:
Evaporating Pressure in the system = ?
For this, first of all, we need to calculate inlet temperature on R410A system from the given value of outlet temperature.
Evaporator inlet temperature is the difference of outlet temperature and evaporator superheat.
Evaporator inlet temperature = Outlet Temperature - Evaporator Superheat
Evaporator inlet Temperature = 50°F - 10°F
Evaporator inlet Temperature = 40°F
Now, as we have the inlet temperature and the R410A system. We can consult the pressure temperature chart or PT chart, which I have attached and highlighted the value of evaporating pressure for 40°F inlet temperature.
So, the evaporating pressure of the R410A = 118 psig
Answer:
Part a)

Part b)

Part c)

Explanation:
Part a)
As we know that the friction force on two boxes is given as



Now we know by Newton's II law

so we have




Part b)
For block B we know that net force on it will push it forward with same acceleration so we have




Part c)
If Alex push from other side then also the acceleration will be same
So for box B we can say that Net force is given as




Answer:
The speed of the banana just before it hits the water is:
√(2 · g · h) = v
Explanation:
Hi there!
Before Emily throws the banana, its potential energy is:
PE = m · g · h
Where:
PE = potential energy.
m = mass of the banana.
g = acceleration of the banana due to gravity.
h = height of the bridge (distance from the bridge to the ground).
When the banana reaches the water, all its potential energy will have converted to kinetic energy. The equation for kinetic energy is as follows:
KE = 1/2 · m · v²
Where:
KE = kinetic energy.
m = mass of the banana.
v = speed.
Then, when the banana hits the water:
m · g · h = 1/2 · m · v²
multiply by 2 and divide by m both sides of the equation:
2 · g · h = v²
√(2 · g · h) = v
<h2>
Answer:20.97g N,32.63g N</h2>
Explanation:
We consider the forces at the knot.
The vertical forces are
is the vertical component of tension
at the knot.
is the weight of the mass
acting downwards.
The horizontal forces are
is the tension in the rope acting left.
is the horizontal component of tension
acting towards right.
Since the knot has no mass,it is always in equilibrium.
So,the sum of forces acting on it will be zero.
Balancing vertical forces gives,


=
Balancing horizontal forces gives,



Answer:
(D) any of these objects can be used as long as all points on the object lie farther away from the lens than the focal length of the lens.
Points lying between the focal point and the lens will be erect and will not form an object visible on a screen.