Answer:
F = 85696.5 N = 85.69 KN
Explanation:
In this scenario, we apply Newton's Second Law:
![Unbalanced\ Force = ma\\Upthrust - Weight\ of\ Space\ Craft = ma\\F - W = ma\\F - mg = ma\\F = m(g + a)\\](https://tex.z-dn.net/?f=Unbalanced%5C%20Force%20%3D%20ma%5C%5CUpthrust%20-%20Weight%5C%20of%5C%20Space%5C%20Craft%20%3D%20ma%5C%5CF%20-%20W%20%3D%20ma%5C%5CF%20-%20mg%20%3D%20ma%5C%5CF%20%3D%20m%28g%20%2B%20a%29%5C%5C)
where,
F = Upthrust = ?
m = mass of space craft = 5000 kg
g = acceleration due to gravity on surface of Kepler-10b = (1.53)(9.81 m/s²)
g = 15.0093 m/s²
a = acceleration required = 2.13 m/s²
Therefore,
![F = (5000\ kg)(15.0093\ m/s^{2} + 2.13\ m/s^{2})\\](https://tex.z-dn.net/?f=F%20%3D%20%285000%5C%20kg%29%2815.0093%5C%20m%2Fs%5E%7B2%7D%20%2B%202.13%5C%20m%2Fs%5E%7B2%7D%29%5C%5C)
<u>F = 85696.5 N = 85.69 KN</u>
Answer:112.82 m/s
Explanation:
Given
range of arrow=68 m
![Angle=3^{\circ}](https://tex.z-dn.net/?f=Angle%3D3%5E%7B%5Ccirc%7D)
as the arrow travels it acquire a vertical velocity ![v_y](https://tex.z-dn.net/?f=v_y)
![v_y=u+at](https://tex.z-dn.net/?f=v_y%3Du%2Bat)
-------1
Range is given by
R=ut
where u=initial velocity
![68=u\times t](https://tex.z-dn.net/?f=68%3Du%5Ctimes%20t)
![t=\frac{68}{u}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B68%7D%7Bu%7D)
substitute the value of t in eqn 1
![v_y=9.81\times \frac{68}{u}](https://tex.z-dn.net/?f=v_y%3D9.81%5Ctimes%20%5Cfrac%7B68%7D%7Bu%7D)
--------2
and ![tan(3)=\frac{v_y}{u}](https://tex.z-dn.net/?f=tan%283%29%3D%5Cfrac%7Bv_y%7D%7Bu%7D)
![v_y=utan(3)=0.0524u](https://tex.z-dn.net/?f=v_y%3Dutan%283%29%3D0.0524u)
substitute it in 2
![0.0524 u^2=667.08](https://tex.z-dn.net/?f=0.0524%20u%5E2%3D667.08)
![u^2=12,728.644](https://tex.z-dn.net/?f=u%5E2%3D12%2C728.644)
u=112.82 m/s
The quantum mechanical model describes the allowed energies an electron can have. It also describes how likely it is to find the electrons in various locations around an atom's nucleus.
If l and m both are doubled then the period becomes √2*T
what is a simple pendulum?
It is the one which can be considered to be a point mass suspended from a string or rod of negligible mass.
A pendulum is a weight suspended from a pivot so that it can swing freely.
Here,
A certain frictionless simple pendulum having a length l and mass m
mass of pendulum = m
length of the pendulum = l
The period of simple pendulum is:
![T = 2\pi \sqrt{\frac{l}{g} }](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7Bl%7D%7Bg%7D%20%7D)
Where k is the constant.
Now the length and mass are doubled,
m' = 2m
l' = 2l
![T' = 2\pi \sqrt{\frac{2l}{g} }](https://tex.z-dn.net/?f=T%27%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7B2l%7D%7Bg%7D%20%7D)
![T' = \sqrt{2}* 2\pi \sqrt{\frac{l}{g} }](https://tex.z-dn.net/?f=T%27%20%3D%20%5Csqrt%7B2%7D%2A%202%5Cpi%20%5Csqrt%7B%5Cfrac%7Bl%7D%7Bg%7D%20%7D)
![T' = \sqrt{2} * T](https://tex.z-dn.net/?f=T%27%20%3D%20%5Csqrt%7B2%7D%20%2A%20T)
Hence,
If l and m both are doubled then the period becomes √2*T
Learn more about Simple Harmonic Motion here:
<u>brainly.com/question/17315536</u>
#SPJ4
(-5)/3 - 6/(-5)
You can solve it now :)