Answer: There is only one Sun in the galaxy … that is the thing that rises in the morning and sets at night. However, there is a use of “sun” to signify any old star … nobody knows exactly there might be trillions out there
Explanation:
I cant see the paragraph so i cant see. It srry
Answer:
c.Law
Explanation:
i think you must learn and write however good luck
A) ![2.03 m/s^2](https://tex.z-dn.net/?f=2.03%20m%2Fs%5E2)
Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:
Parallel:
(1)
where
m is the mass
g = 9.8 m/s^2 the acceleration of gravity
![\theta=22.5^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D22.5%5E%7B%5Ccirc%7D)
is the coefficient of friction
R is the normal reaction
a is the acceleration
Perpendicular:
(2)
From (2) we find
![R=mg cos \theta](https://tex.z-dn.net/?f=R%3Dmg%20cos%20%5Ctheta)
And substituting into (1)
![mg sin \theta - \mu_k mg cos \theta = ma](https://tex.z-dn.net/?f=mg%20sin%20%5Ctheta%20-%20%5Cmu_k%20mg%20cos%20%5Ctheta%20%3D%20ma)
Solving for a,
![a=g sin \theta - \mu_k g cos \theta = (9.8)(sin 22.5)-(0.19)(9.8)(cos 22.5)=2.03 m/s^2](https://tex.z-dn.net/?f=a%3Dg%20sin%20%5Ctheta%20-%20%5Cmu_k%20g%20cos%20%5Ctheta%20%3D%20%289.8%29%28sin%2022.5%29-%280.19%29%289.8%29%28cos%2022.5%29%3D2.03%20m%2Fs%5E2)
B) 5.94 m/s
We can solve this part by using the suvat equation
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
Here we have
v = ?
u = 0 (it starts from rest)
![a=2.03 m/s^2](https://tex.z-dn.net/?f=a%3D2.03%20m%2Fs%5E2)
s = 8.70 m
Solving for v,
![v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bu%5E2%2B2as%7D%3D%5Csqrt%7B0%2B2%282.03%29%288.70%29%7D%3D5.94%20m%2Fs)