The farther away an object is from the Sun the slower it orbits around it. The closer an object is from the Sun the faster it orbits around it.
Answer:
Balanced forces: When a number of forces acting on a body do not cause any change in its state of rest or of uniform motion along a straight line then the forces are said to be balanced forces. In other words, a body is said to be underbalanced forced when the resulting force acting on the body is zero.
The balanced forces:
⋅ Cannot set any stationary body into motion.
⋅ May change the shape and size of soft objects.
⋅ Cannot change the speed/velocity of a moving body.
Unbalanced forces:
When the resultant of all the forces acting on a body is not zero, then forces are called unbalanced forces.
Example:
⋅ Game of tug of war: When the forces exerted by both the teams are equal, then the rope does not move. But, if the force applied by team A is greater than team B, then the rope, as well as members of the weaker team, i.e., B, will be pulled towards A. The unbalanced force can (a) Set a stationary body in motion.
⋅ Set a moving body at rest.
⋅ Change the direction of motion.
Explanation:
give me an one thanks please
Answer:
<h3>They eat meat and vegetation. </h3>
Answer:
100 ÷ 9.58 = 10.44 (approximate answer)
Answer:
x = 17.88[m]
Explanation:
We can find the components of the initial velocity:
![(v_{x})_{o} = 13.3*cos(41.5)=9.96[m/s]\\(v_{y})_{o} = 13.3*sin(41.5)=8.81[m/s]](https://tex.z-dn.net/?f=%28v_%7Bx%7D%29_%7Bo%7D%20%20%3D%2013.3%2Acos%2841.5%29%3D9.96%5Bm%2Fs%5D%5C%5C%28v_%7By%7D%29_%7Bo%7D%20%20%3D%2013.3%2Asin%2841.5%29%3D8.81%5Bm%2Fs%5D)
We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.
g = - 9.81[m/s^2]
Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.
![y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]](https://tex.z-dn.net/?f=y%3Dy_%7Bo%7D%20%2B%28v_%7By%7D%20%29_%7Bo%7D%20%2At-0.5%2Ag%2A%28t%29%5E%7B2%7D%20%5C%5C0%3D1.9%2B%288.81%2At%29-%284.905%2At%5E%7B2%7D%29%5C%5C-1.9%3D8.81%2At%2A%281-0.5567%2At%29%5C%5Ct%3D0%5C%5Ct%3D1.796%5Bs%5D)
With this time we can calculate the horizontal distance:
![x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]](https://tex.z-dn.net/?f=x%3D%28v_%7Bx%7D%29_%7Bo%7D%20%2At%5C%5Cx%3D9.96%2A1.796%5C%5Cx%3D17.88%5Bm%5D)
