Answer:
114.44 J
Explanation:
From Hook's Law,
F = ke................. Equation 1
Where F = Force required to stretch the spring, k = spring constant, e = extension.
make k the subject of the equation
k = F/e.............. Equation 2
Given: F = 10 lb = (10×4.45) N = 44.5 N, e = 4 in = (4×0.254) = 1.016 m.
Substitute into equation 2
k = 44.5/1.016
k = 43.799 N/m
Work done in stretching the 9 in beyond its natural length
W = 1/2ke²................. Equation 3
Given: e = 9 in = (9×0.254) = 2.286 m, k = 43.799 N/m
Substitute into equation 3
W = 1/2×43.799×2.286²
W = 114.44 J
Both countries will have the same advantage in producing goods.
<h3>What does the theory of comparative advantage state?</h3>
According to the theory of comparative advantage, a person should participate in an economic activity in which they have a comparative advantage, i.e., in which they are more effective than in the other activity. In terms of economies, they should engage in the production of those commodities and services at which they are better placed than the other.
<h3>Which country will have the comparative advantage in producing capital goods?</h3>
The output potential, resource level, and technological level of two nations A and B are identical. This suggests that both countries A and B are producing at the same level and that neither one is doing better than the other in terms of production.
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Answer:
It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.
V^2=U^2+2*a*x
V=0 (as it is at max height)
U=30ms^-1 (initial speed)
a=-g /-9.8ms^-2 (as it is moving against gravity)
x is the variable you want to calculate (height)
0=30^2+2*(-9.8)*x
x=-30^2/2*-9.8
x=45.92m
Distance traveled=(80 km/1hr)*(2.5 hrs/1) or 200 km.
Does This Help?
Answer:
![E = 1.85*10^{12}\frac{N}{C}](https://tex.z-dn.net/?f=%20E%20%3D%201.85%2A10%5E%7B12%7D%5Cfrac%7BN%7D%7BC%7D%20)
Explanation:
Hi!
The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.
The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction, and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:
![E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%20%5Cepsilon_0%20r%7D%5C%5C%5Clambda%3D%5Ctext%7Bcharge%20per%20unit%20length%7D%3D%5Cfrac%7B4.95%20%5Cmu%20C%7D%7B2%20m%7D%20%3D%202.475%20%5Cfrac%7BC%7D%7Bm%7D%5C%5Cr%3D%5Ctext%7Bperpendicular%20distance%20to%20wire%7D%5C%5C%5Cepsilon_0%3D8.85%2A10%5E%7B-12%7D%5Cfrac%7BC%5E2%7D%7BNm%5E2%7D)
Then the electric field at the point of interest is estimated as:
![E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B%5C22.475%7D%7B2%5Cpi%2A%28%208.85%2A10%5E%7B-12%7D%29%2A%282.4%2A10%5E%7B-2%7D%29%7D%5Cfrac%7BN%7D%7BC%7D%3D1.85%2A10%5E%7B12%7D%5Cfrac%7BN%7D%7BC%7D)