Answer:
Explanation: Weight of space probes on earth is given by:
W= weight of the object( in N)
m= mass of the object (in kg)
g=acceleration due to gravity(9.81 
)
Therefore,
Similarly,
Now, considering these two parts as uniform spherical objects
Also, according to Superposition principle, gravitational net force experienced by an object is sum of all individual forces on the object.
Force between these two objects is given by:
G= gravitational constant (
)
= masses of the object
R= distance between their centres (in m)(18 m)
Substituiting all these values into the above formula
This is the magnitude of force experienced by each part in the direction towards the other part, i.e the gravitational force is attractive in nature.
Answer:
1) Current decreases; 2) Inverse proportionally; 3) 1[A]
Explanation:
1)
As we can see as the resistance increases the current decreases, if we take two points as an example, when the resistance is equal to 50 [ohms] the current is equal to 1[amp] and when the resistance is equal to 200 [ohms] the current tends to have a value below 0.5 [amp]. Thus demonstrating the decrease in current.
2)
Inverse proportionally, by definition we know that the law of ohm determines the voltage according to resistance and amperage. This is the voltage will be equal to the product of the voltage by the resistance.
![V=I*R\\V = voltage [volts]\\I = current[amp]\\R = resistance [ohms]](https://tex.z-dn.net/?f=V%3DI%2AR%5C%5CV%20%3D%20voltage%20%5Bvolts%5D%5C%5CI%20%3D%20current%5Bamp%5D%5C%5CR%20%3D%20resistance%20%5Bohms%5D)
where:
And whenever we have in a fractional number the denominator the variable we are interested in, we can say that this is inversely proportional to the value we are interested in determining. In this case, we can see from the two previous expressions that both the current and the resistance appear in the denominator, therefore they are inversely proportional to each other.
3)
If we place ourselves on the graph on the resistance axis, we see that at 50 [ohm] will correspond a current value equal to 1 [A].
Answer:
a)W=8.333lbf.ft
b)W=0.0107 Btu.
Explanation:
<u>Complete question</u>
The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.
Solution
Preload = F₀=0 lbf
Spring constant k= 200 lbf/in
Initial length of spring x₁=0
Final length of spring x₂= 1 in
At any point, the force during deflection of a spring is given by;
F= F₀× kx where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.
Change to lbf.ft by dividing the value by 12 because 1ft=12 in
100/12 = 8.333 lbf.ft
work required to compress the spring, W=8.333lbf.ft
The work required to compress the spring in Btu will be;
1 Btu= 778 lbf.ft
?= 8.333 lbf.ft----------------cross multiply
(8.333*1)/ 778 =0.0107 Btu.
