Answer:
See the answers below.
Explanation:
to solve this problem we must make a free body diagram, with the forces acting on the metal rod.
i)
The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.
We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.
For the summation of forces we will take the forces upwards as positive and the negative forces downwards.
ΣF = 0
Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.
ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.
ΣM = 0
![(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]](https://tex.z-dn.net/?f=%2815%2A9%29%20-%20%2818%2AW%29%20%3D%200%5C%5C135%20%3D%2018%2AW%5C%5CW%20%3D%207.5%20%5BN%5D)
Given:
Amount of heat produced = 100 kcal per hour
Let's find the rate of energy production in joules.
We know that:
1 calorie = 4.184 Joules
1 kcal = 4.184 Joules
To find the rate of energy production in Joules, we have:
Therefore, the rate of energy production in joules is 418.4 kJ/h which is equivalent to 418400 Joules
ANSWER:
418.4 kJ/h
Answer:
θ=142.9°
Explanation:
d=1 *r
angle ϕ= 37.1°
the line connecting pebble and target should be tangent to a circle so
cos(180-ϕ-θ)=
=
∴ θ=180-ϕ-
θ= 180-37.1-0
θ=142.9°
To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of 
(Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as
Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing
![W = (14*7*62)\big [15y-\frac{y^2}{2}\big ]^{15}_0](https://tex.z-dn.net/?f=W%20%3D%20%2814%2A7%2A62%29%5Cbig%20%5B15y-%5Cfrac%7By%5E2%7D%7B2%7D%5Cbig%20%5D%5E%7B15%7D_0)
![W = (14*7*62)[15(15)-\frac{(15)^2}{2}]](https://tex.z-dn.net/?f=W%20%3D%20%2814%2A7%2A62%29%5B15%2815%29-%5Cfrac%7B%2815%29%5E2%7D%7B2%7D%5D)
Therefore the total work in the system is 
