Answer:
the ans will be because it has 1.672
Explanation:
Though the diagram that is mentioned in the questions is not given. I have given general information related to the Earth's magnetic axis with respect to geographic/rotational axis.
Axis is an imaginary line around which a body rotates. The rotational axis of the Earth enters into and exits from the Earth at two points namely: North and South pole.
We know that the Earth behaves like a huge bar magnet so just like the bar magnet it must also have magnetic axis and poles. Earth has north magnetic pole and south magnetic pole. It has a strong magnetic field as well known as magnetosphere.
The interesting point about magnetic axis is that it is not same as rotational axis. In fact it makes an angle with the rotational axis. This is known as magnetic inclination. This inclination varies at different points on Earth.
Another interesting point is that the geographic and magnetic poles are opposite. That means near the geographic north pole we have the magnetic south pole and vice versa.
Action reaction when you push on the ground the ground pushes you back
Explanation:
Let us first calculate long does it take to go 12m at 30m/s( assumed speed)
12/30 = 0.4 seconds
horizontal distance the ball drop in that time
H= (0)(0.4)+1/2(-9.8)(0.4)2
H= -0.78m
negative sign shows that the height of the ball at the net from the top.
Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m
As 1.62m>0.9m so the ball will clear the net.
H_1= V0y t’ + ½ g t’^2
-2.4= (0)t’ + ½ (-9.8) t’^2
t’= 0.69s
X’=V0x t’
X’=(30)(0.96)
X’= 20.7m
Answer:
The distance will be x = 41.7 [m]
Explanation:
We must first find the components in the x & y axes of the initial velocity.
![(v_{o})_{x} = 15*cos(20)= 14.09[m/s]\\(v_{o})_{y} = 15*sin(20)= 5.13[m/s]](https://tex.z-dn.net/?f=%28v_%7Bo%7D%29_%7Bx%7D%20%3D%2015%2Acos%2820%29%3D%2014.09%5Bm%2Fs%5D%5C%5C%28v_%7Bo%7D%29_%7By%7D%20%3D%2015%2Asin%2820%29%3D%205.13%5Bm%2Fs%5D)
The acceleration is the gravity acceleration therefore.
g = 9.81 [m/s^2]
Now we can calculate how long it takes to fall.
![y=(v_{o})_{y}*t-0.5*g*t^2\\-28 = 5.13*t-0.5*9.81*t^2\\-28=-4.905*t^2+5.13*t\\4.905*t^2-5.13*t=28\\t = 2.96[s]](https://tex.z-dn.net/?f=y%3D%28v_%7Bo%7D%29_%7By%7D%2At-0.5%2Ag%2At%5E2%5C%5C-28%20%3D%205.13%2At-0.5%2A9.81%2At%5E2%5C%5C-28%3D-4.905%2At%5E2%2B5.13%2At%5C%5C4.905%2At%5E2-5.13%2At%3D28%5C%5Ct%20%3D%202.96%5Bs%5D)
With this time we can find the horizontal distance that runs the projectile.
![x=(v_{o})_{x}*t\\x=14.09*2.96\\x=41.7[m]](https://tex.z-dn.net/?f=x%3D%28v_%7Bo%7D%29_%7Bx%7D%2At%5C%5Cx%3D14.09%2A2.96%5C%5Cx%3D41.7%5Bm%5D)
