<h2>Given that,</h2>
Mass of two bumper cars, m₁ = m₂ = 125 kg
Initial speed of car X is, u₁ = 10 m/s
Initial speed of car Z is, u₂ = -12 m/s
Final speed of car Z, v₂ = 10 m/s
We need to find the final speed of car X after the collision. Let v₁ is its final speed. Using the conservation of momentum to find it as follows :
v₁ is the final speed of car X.
So, car X will move with a velocity of -12 m/s.
Answer:
Magnetic force, 
Explanation:
Given that,
A beryllium-9 ion has a positive charge that is double the charge of a proton, 
Speed of the ion in the magnetic field, 
Its velocity makes an angle of 61° with the direction of the magnetic field at the ion's location.
The magnitude of the field is 0.220 T.
We need to find the magnitude of the magnetic force on the ion. It is given by :
So, the magnitude of magnetic force on the ion is 
.
Ionic bonds are formed between a cation (metal) and an anion (nonmetal)
NO musical instrument produces a 'pure' tone with only a
single frequency in it.
EVERY instrument produces more or less harmonics (multiples)
in addition to the basic frequency it's playing.
The percussion instruments (drums etc) are the richest producers
of bunches of different frequencies.
Fuzzy electric guitars are next richest.
The strings and brass instruments are moderate producers of
harmonics ... I can't remember which is greater than the other.
Then come the woodwinds ... clarinet, oboe, etc.
The closest to 'pure' tones of single frequency are the sounds
made by the flute and piccolo, but even these are far from 'pure'.
The only way to get a true single-frequency sound is from an
electronic 'sine wave' generator.
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
