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.
Answer:
Newtons third law
Explanation:
Because it states that every force has an equal and opposite force
Answer:
A hypothesis. In science, a hypothesis is an idea or explanation that you then test through study and experimentation. Outside science, a theory or guess can also be called a hypothesis. A hypothesis is something more than a wild guess but less than a well-established theory. My hypothesis was right because although it was n educated guess it was still correct
The viscosity value decreased with the increase of temperature. The effect caused by temperature in the range studied was more important in the low range of temperature, whereas at high temperature, the viscosity showed less variation.The studies carried out in the present work indicated that the effect of temperature upon the viscosity of Galician honeys was prominent, and it is necessary to take into account this variable when the operation temperature is under 25°C, since below this value a small variation in temperature produces high changes in the viscosity value. On the other hand, different models to fit experimental values of viscosity/temperature have been proved. The glass transition temperature was determined necessarily since several models include this parameter in the equation, because it provides important information related to the viscosity value. All the models show a good behaviour but the best results were obtained using the Arrhenius one.
Answer:
V = 411.43 V
Explanation:
The two forces as a result of each of the 2 charges are;
F1 = kq1•q/r
F2 = kq2.q/r
Where r = r/2 since we are dealing with potential difference at a point midway between the charges.
q1 = 5 nC = 5 × 10^(-9) C
q2 = 3 nC = 3 × 10^(-9) C
k = 9 × 10^(9) N.m²/C²
r = 35 cm = 0.35m
r/2 = 0.35/2
Thus;
F1 = (9 × 10^(9) × 5 × 10^(-9) × q)/(0.35/2)²
F1 = 1469.39q
F2 = (9 × 10^(9) × 3 × 10^(-9) × q)/(0.35/2)²
F2 = 881.63q
Net force acting midway is;
F_net = F1 + F2
F_net = 1469.39q + 881.63q
F_net = 2351.02q
Now, we know that formula for electric potential is;
V = kq/r
Thus ;
V = Fr/q derived from the earlier equation for force we used.
Where F is F_net.
V = 2351.02q × r/q
V = 2351.02r
Recall that we are dealing with midpoint and r = r/2
Thus;
V = 2351.02 × 0.35/2
V = 411.43 V