Answer:
The velocity is 19.39 m/s
Solution:
As per the question:
Mass, m = 75 kg
Radius, R = 19.2 m
Now,
When the mass is at the top position in the loop, then the necessary centrifugal force is to keep the mass on the path is provided by the gravitational force acting downwards.
where
v = velocity
g = acceleration due to gravity
The charged particles are often deflated in a magnetic field.
<h3>What is a magnetic field?</h3>The term charge refers to a positive or negative entity. The can be created when a charge is made to pass through a conductor in a magnetic field.
A magnetic field is created when we have a north pole and a south pole. The charged particles could be made to pass through the electric field and when that happens, we can see a pattern a shown in the image attached.
Thus, we can see that the charged particles are often deflated in a magnetic field.
Learn more about magnetic field:brainly.com/question/23096032
#SPJ1
Answer:
W = 5701 KW
Explanation:
From the question let inlet be labelled as point 1 and exit as point 2, for the fluid steam, we can get the following;
Inlet (1): P1 = 40 bar ; T1 = 500°C and V1 = 200 m/s
Exit(2) : At saturated vapour; P2 = 0.8 bar and V2 = 150 m/s
Volumetric flow rate = 15 m^(3)/s
Now, to solve this question, we assume constant average values, steeady flow and adiabatic flow.
Specific volume for steam at P2 = 0.8 bar in the saturated vapour state can be gotten from saturated steam tables(find a sample of the table attached to this answer).
So from the table,
v2 = 2.087 m^(3)/kg
Now, mass flow rate (m) = (AV) /v
Where AV is the volumetric flow rate.
Thus, the mass flow rate at exit could be calculated as;
m = 15/(2.087) = 7.17 kg/s
We also know energy equation could be defined as;
Q-W = m[(h1 - h2) + {(V2(^2) - (V1(^2)} /2)} + g(Z2 - Z1)]
Since the flow is adiabatic, potential energy can be taken to be zero. Therefore, we get;
-W = m[(h2 - h1) + {(V2(^2) - (V1(^2)} /2)}
From, table 2, i attached , at P1 = 40 bar and T1 = 500°C; specific enthalpy was calculated to be h1 = 3445.3 KJ/Kg
Likewise, at P2 = 0.8 bar; from the table, we get specific enthalpy as;
h2 = 2665.8 KJ/Kg
So we now calculate power developed;
W = - 7.17 [(2665.8 - 3445.3) + {(150^(2) - 200^(2))/2000 = 5701KW
Since the sign is not negative but positive, it means that the power is developed from the system.
