Hoosier Manufacturing operates a production shop that is designed to have the lowest unit production cost at an output rate of 1
1 answer:
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Answer:
the work done by gravity on the boy is 604.62 J
Explanation:
Given;
distance the boy slides, d = 3 m
angle of inclination of the playground, θ = 40⁰
mass of the boy, m = 32 kg
The vertical height, h, above the ground through which the boy falls represents the height of the triangle which is the opposite side.
The distance through which the boy slides, d, represents the hypotenuse side of the right triangle.
The work done by gravity on the boy is calculated as;
W = P.E = mgh
= 32kg x 9.8m/s² x 1.928m
= 604.62 J
Therefore, the work done by gravity on the boy is 604.62 J
To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through
Where,
Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is
Where
Angular acceleration
Angular velocity
t = Time
Our values are
Replacing at the previous equation we have that the angular velocity is
Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be
Therefore the angular acceleration of a point on the outer edge of the tires is
Answer:
2.36 x 10^6 J
Explanation:
Tc = 0°C = 273 K
TH = 22.5°C = 295.5 K
Qc = heat used to melt the ice
mass of ice, m = 85.7 Kg
Latent heat of fusion, L = 3.34 x 10^5 J/kg
Let Energy supplied is E which is equal to the work done
Qc = m x L = 85.7 x 3.34 x 10^5 = 286.24 x 10^5 J
Use the Carnot's equation
QH = 309.8 x 10^5 J
W = QH - Qc
W = (309.8 - 286.24) x 10^5
W = 23.56 x 10^5 J
W = 2.36 x 10^6 J
Thus, the energy supplied is 2.36 x 10^6 J.