We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
Answer:
The kinetic energy is ![KE = 7.59 *10^{10} \ J](https://tex.z-dn.net/?f=KE%20%20%3D%20%207.59%20%20%2A10%5E%7B10%7D%20%5C%20%20J)
Explanation:
From the question we are told that
The radius of the orbit is ![r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m](https://tex.z-dn.net/?f=r%20%3D%20%202.3%20%2A10%5E%7B4%7D%20%5C%20km%20%20%3D%202.3%20%20%2A10%5E%7B7%7D%20%5C%20m)
The gravitational force is ![F_g = 6600 \ N](https://tex.z-dn.net/?f=F_g%20%20%3D%206600%20%5C%20N)
The kinetic energy of the satellite is mathematically represented as
![KE = \frac{1}{2} * mv^2](https://tex.z-dn.net/?f=KE%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20mv%5E2)
where v is the speed of the satellite which is mathematically represented as
![v = \sqrt{\frac{G M}{r^2} }](https://tex.z-dn.net/?f=v%20%20%3D%20%5Csqrt%7B%5Cfrac%7BG%20%20M%7D%7Br%5E2%7D%20%7D)
=> ![v^2 = \frac{GM }{r}](https://tex.z-dn.net/?f=v%5E2%20%20%3D%20%20%5Cfrac%7BGM%20%7D%7Br%7D)
substituting this into the equation
![KE = \frac{ 1}{2} *\frac{GMm}{r}](https://tex.z-dn.net/?f=KE%20%20%3D%20%20%5Cfrac%7B%201%7D%7B2%7D%20%2A%5Cfrac%7BGMm%7D%7Br%7D)
Now the gravitational force of the planet is mathematically represented as
![F_g = \frac{GMm}{r^2}](https://tex.z-dn.net/?f=F_g%20%20%3D%20%5Cfrac%7BGMm%7D%7Br%5E2%7D)
Where M is the mass of the planet and m is the mass of the satellite
Now looking at the formula for KE we see that we can represent it as
![KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r](https://tex.z-dn.net/?f=KE%20%20%3D%20%20%5Cfrac%7B%201%7D%7B2%7D%20%2A%5B%5Cfrac%7BGMm%7D%7Br%5E2%7D%5D%20%2A%20r)
=> ![KE = \frac{ 1}{2} *F_g * r](https://tex.z-dn.net/?f=KE%20%20%3D%20%20%5Cfrac%7B%201%7D%7B2%7D%20%2AF_g%20%2A%20r)
substituting values
![KE = \frac{ 1}{2} *6600 * 2.3*10^{7}](https://tex.z-dn.net/?f=KE%20%20%3D%20%20%5Cfrac%7B%201%7D%7B2%7D%20%2A6600%20%2A%202.3%2A10%5E%7B7%7D)
![KE = 7.59 *10^{10} \ J](https://tex.z-dn.net/?f=KE%20%20%3D%20%207.59%20%20%2A10%5E%7B10%7D%20%5C%20%20J)
Twin type has less to do with what twins look like and more to do with how they formed.
Identical, or monozygotic, twins form when a single fertilized egg splits and develop as two babies in the uterus. Identical twins originate from the same combination of cells and have the same genetic origin. They are ALWAYS the same sex, two girls/two boys. They may look very similar and it may be difficult to tell them apart.
Fraternal, or dizygotic, twins are two individuals from the same pregnancy who from TWO SEPARATE eggs fertilized by TWO SEPARATE SPERM. The genetic similarity between fraternal twins is the same as any two siblings, about 50 percent. They can be boys, girls, or one of each.
Answer: D. wealthy
Explanation: on the e2020 test its right
Answer:
thick wire and cold temperatures