Answer:
Fy = 14.3 [N]
Explanation:
To be able to solve this problem we must know that the force is a vector and has magnitude and direction, therefore it can be decomposed into the force in the X & y components:
When we have the components on the horizontal and vertical axes we must use the Pythagorean theorem.
where:
F = 15 [N]
Fx = horizontal component = 4.5 [N]
Fy = vertical component [N]
![15=\sqrt{4.5^{2}+F_{y}^{2}}\\ 15^{2}= (\sqrt{4.5^{2}+F_{y}^{2}})^{2} \\225 = 4.5^{2}+F_{y} ^{2}\\ F_{y}^{2} =225 -4.5^{2}\\ F_{y}^{2}=204.75\\F_{y}=\sqrt{204.75}\\ F_{y}=14.3 [N]](https://tex.z-dn.net/?f=15%3D%5Csqrt%7B4.5%5E%7B2%7D%2BF_%7By%7D%5E%7B2%7D%7D%5C%5C%2015%5E%7B2%7D%3D%20%28%5Csqrt%7B4.5%5E%7B2%7D%2BF_%7By%7D%5E%7B2%7D%7D%29%5E%7B2%7D%20%5C%5C225%20%3D%204.5%5E%7B2%7D%2BF_%7By%7D%20%5E%7B2%7D%5C%5C%20%20F_%7By%7D%5E%7B2%7D%20%3D225%20-4.5%5E%7B2%7D%5C%5C%20F_%7By%7D%5E%7B2%7D%3D204.75%5C%5CF_%7By%7D%3D%5Csqrt%7B204.75%7D%5C%5C%20%20F_%7By%7D%3D14.3%20%5BN%5D)
Answer: mass x height x gravitational field strength (g)
note: gravitational field strength (g) = 10 N/Kg
55 x 15 x 10 = 8250
gpe = 8250j
Explanation:
Answer:
i. Cv =3R/2
ii. Cp = 5R/2
Explanation:
i. Cv = Molar heat capacity at constant volume
Since the internal energy of the ideal monoatomic gas is U = 3/2RT and Cv = dU/dT
Differentiating U with respect to T, we have
= d(3/2RT)/dT
= 3R/2
ii. Cp - Molar heat capacity at constant pressure
Cp = Cv + R
substituting Cv into the equation, we have
Cp = 3R/2 + R
taking L.C.M
Cp = (3R + 2R)/2
Cp = 5R/2
Answer:
Explanation:
Given:
- time taken by the sun to complete one revolution,
- radial distance of the sunspot,
<u>Therefore, angular speed of rotation of sun:</u>
<u>Now the tangential velocity of the sunspot can be given by:</u>
Answer:
If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface.
Explanation:
Option A is incorrect because, given this case, it is easier to calculate the field.
Option B is incorrect because, in a situation where the surface is placed inside a uniform field, option B is violated
Option C is also incorrect because it is possible to be a field from outside charges, but there will be an absence of net flux through the surface from these.
Hence, option D is the correct answer. "If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface."
