A (b) would be 42 (c) from x to Y mark me as brainlist thanks
It is the branch of science, in which we study different phenomena of atmosphere including climate and weather.
Answer:
a) α=7.9x10^-4 rad
b) θ=1.12x10^-4 rad
c) The Earth and the Moon cannot be seen without a telescope.
Explanation:
In this exercise we will use the concepts of angular resolution, which depends on both the wavelength of the rays and the diameter of the eye or lens on the meter. Its unit of measure is the radian. The attached image shows the solution step by step.
"<em>F = dP/dt. </em> The net force acting on an object is equal to the rate at which its momentum changes."
These days, we break up "the rate at which momentum changes" into its units, and then re-combine them in a slightly different way. So the way WE express and use the 2nd law of motion is
"<em>F = m·A.</em> The net force on an object is equal to the product of the object's mass and its acceleration."
The two statements say exactly the same thing. You can take either one and work out the other one from it, just by working with the units.
Hello!
a) Assuming this is asking for the minimum speed for the rock to make the full circle, we must find the minimum speed necessary for the rock to continue moving in a circular path when it's at the top of the circle.
At the top of the circle, we have:
- Force of gravity (downward)
*Although the rock is still connected to the string, if the rock is swinging at the minimum speed required, there will be no tension in the string.
Therefore, only the force of gravity produces the net centripetal force:
![\Sigma F = F_g\\\\F_c = F_g\\\\\frac{mv^2}{r} = mg](https://tex.z-dn.net/?f=%5CSigma%20F%20%3D%20F_g%5C%5C%5C%5CF_c%20%3D%20F_g%5C%5C%5C%5C%5Cfrac%7Bmv%5E2%7D%7Br%7D%20%3D%20mg)
We can simplify and rearrange the equation to solve for 'v'.
![\frac{v^2}{r} = g\\\\v^2 = gr\\\\v = \sqrt{gr}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%5E2%7D%7Br%7D%20%3D%20g%5C%5C%5C%5Cv%5E2%20%3D%20gr%5C%5C%5C%5Cv%20%3D%20%5Csqrt%7Bgr%7D)
Plugging in values:
![v = \sqrt{9.8 * 0.75} = \boxed{2.711 m/s^}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B9.8%20%2A%200.75%7D%20%3D%20%5Cboxed%7B2.711%20m%2Fs%5E%7D)
b)
Let's do a summation of forces at the bottom of the swing. We have:
- Force due to gravity (downward, -)
- Tension force (upward, +)
The sum of these forces produces a centripetal force, upward (+).
![\Sigma F = T - F_g\\\\F_c = T - F_g\\\\\frac{mv^2}{r} = T - mg](https://tex.z-dn.net/?f=%5CSigma%20F%20%3D%20T%20-%20F_g%5C%5C%5C%5CF_c%20%3D%20T%20-%20F_g%5C%5C%5C%5C%5Cfrac%7Bmv%5E2%7D%7Br%7D%20%3D%20T%20-%20mg)
Rearranging for 'T":
![T = \frac{mv^2}{r} + mg\\\\](https://tex.z-dn.net/?f=T%20%3D%20%20%20%5Cfrac%7Bmv%5E2%7D%7Br%7D%20%2B%20%20mg%5C%5C%5C%5C)
Plugging in the appropriate values:
![T = \frac{(0.2)(2.711^2)}{(0.75)} + 0.2(9.8) = \boxed{3.92 N}](https://tex.z-dn.net/?f=T%20%3D%20%20%5Cfrac%7B%280.2%29%282.711%5E2%29%7D%7B%280.75%29%7D%20%2B%200.2%289.8%29%20%3D%20%5Cboxed%7B3.92%20N%7D)