A 2.0-kg ball is at rest when a horizontal force of 5.0 N is applied. In the absence of friction, what is the speed of the ball
1 answer:
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Answer:
The gravitational potential energy between two particles, if the distance between them is halved, is multiplied by 4 (option c).
Explanation:
The gravitational force is the force of mutual attraction that two objects with mass experience.
The Law of Universal Gravitation enunciated by Newton says that every material particle attracts any other material particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance that separates them. Mathematically this is expressed as:
where m1 and m2 are the masses of the objects, r the distance between them and G a universal constant that receives the name of constant of gravitation.
If the distance between two particles is reduced by half, then, where F' is the new value of the gravitational force:
F'=4*F
<u><em> The gravitational potential energy between two particles, if the distance between them is halved, is multiplied by 4 (option c).</em></u>
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.
Answer:
(A) The period of its rotation is 0.5 s (2) The frequency of its rotation is 2 Hz.
Explanation:
Given that,
a ball is spun around in circular motion such that it completes 50 rotations in 25 s.
(1). Let T be the period of its rotation. It can be calculated as follows :
(2). Let f be the frequency of its rotation. It can be defined as the number of rotations per unit time. So,
Hence, this is the required solution.