Answer:
Explanation:
This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is
Filling in:
which simplifies to
5400 + 0 = 3300v
so v = 1.6 m/s to the east, choice B
a given box of 2 kg is in contact with the floor experiences an applied force of 150 form somebody pushing it.while in motion,it
1 answer:
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The work done is 400 J
Explanation:
The work done by you in pushing the box along the slope is given by
where
F is the magnitude of the force applied
d is the distance covered by the box along the slope
Here we have the following:
F = 200 N is the magnitude of the force applied
is the distance covered, where
h = 1 m is the vertical rise
is the slope of the plane
Substituting and solving, we find
Learn more about work:
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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.
Force applied on the car due to engine is given as
towards right
Also there is a force on the car towards left due to air drag
towards left
now the net force on the car will be given as
now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.
So we can say
So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.