Answer: 6m/s
Explanation:
Using the law of conservation of momentum, the change in momentum of the bodies before collision is equal to the change in momentum after collision.
After collision, the two objects will move at the same velocity (v).
Let mA and mB be the mass of the two objects
uA and uB be their velocities before collision.
v be their velocity after collision
Since the two objects has the same mass, mA= mB= m
Also since object A is at rest, its velocity = 0m/s
Velocity of object B = 12m/s
Mathematically,
mAuA + mBuB = (mA+mB )v
m(0) + m(12) = (m+m)v
0+12m = (2m)v
12m = 2mv
12 = 2v
v = 6m/s
Therefore the speed of the composite body (A B) after the collision is 6m/s
The lithosphere is the solid outer section of Earth, which includes Earth's crust (the "skin" of rock on the outer layer of planet Earth), as well as the underlying cool, dense, and rigid upper part of the upper mantle. The lithosphere extends from the surface of Earth to a depth of about 44–62 mi (70–100 km).
Firstly, let's convert the velocities in km/hr to m/s
32*1000/3600=8.89m/s
54*1000/3600=15m/s
From the formula, acceleration=V-U/t
15-8.89/8=0.76m/s²
hope this helps.
I cant see the paragraph so i cant see. It srry
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
