A light year is a unit of distance. It is a distance that light can travel in a years time which is six trillion miles. It is used to measure the distances in space. To take one example, the distance to the next nearest big galaxy, the Andromeda Galaxy, from earth is 21,000,000,000,000,000,000 km.
Do you understand it? <span />
Answer:
t=67.7s
Explanation:
From this question we know that:
Vo = 6m/s
a = 1.8 m/s2
D = 1500m
And we also know that:
Replacing the known values:
Solving for t we get 2 possible answers:
t1 = -44.3s and t2 = 67.7s Since negative time represents an instant before the beginning of the movement, t1 is discarded. So, the final answer is:
t = 67.7s
A) average acceleration = final velocity - initial velocity / time
= 7700 - 0 / 11
= 700ms^-2
B) force = mass x acceleration
= (3.05 x 105) x 700
= 320.25 x 700
= 224,175N
Answer:
V=14.9 m/s
Explanation:
In order to solve this problem, we are going to use the formulas of parabolic motion.
The velocity X-component of the ball is given by:
![Vx=V*cos(\alpha)\\Vx=15.7*cos(31^o)=13.5m/s](https://tex.z-dn.net/?f=Vx%3DV%2Acos%28%5Calpha%29%5C%5CVx%3D15.7%2Acos%2831%5Eo%29%3D13.5m%2Fs)
The motion on the X axis is a constant velocity motion so:
![t=\frac{d}{Vx}\\t=\frac{20.0}{13.5}=1.48s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bd%7D%7BVx%7D%5C%5Ct%3D%5Cfrac%7B20.0%7D%7B13.5%7D%3D1.48s)
The whole trajectory of the ball takes 1.48 seconds
We know that:
![Vy=Voy+(a)*t\\Vy=15.7*sin(31^o)+(-9.8)*(1.48)=-6.42m/s](https://tex.z-dn.net/?f=Vy%3DVoy%2B%28a%29%2At%5C%5CVy%3D15.7%2Asin%2831%5Eo%29%2B%28-9.8%29%2A%281.48%29%3D-6.42m%2Fs)
Knowing the X and Y components of the velocity, we can calculate its magnitude by:
![V=\sqrt{Vx^2+Vy^2} \\V=\sqrt{(13.5)^2+(-6.42)^2}=14.9m/s](https://tex.z-dn.net/?f=V%3D%5Csqrt%7BVx%5E2%2BVy%5E2%7D%20%5C%5CV%3D%5Csqrt%7B%2813.5%29%5E2%2B%28-6.42%29%5E2%7D%3D14.9m%2Fs)
south = -(north)
Displacement = (4 km north) + (2 km south) + (5 km north) + (5 km south)
Displacement = (4 km north) - (2 km north) + (5 km north) - (5 km north)
Displacement = (4 - 2 + 5 - 5) km north
<u>Displacement = 2 km north </u>