Answer:
technically yes
Explanation:
with a gun depending on how fast it shoots so when you fire at something you shoot in front of it a little bit so you hit it but a plane that fast you shoot like 100 feet infront of it...
The position vector can be
transcribed as:
A<span> = 6 i + y j
</span>
i <span>points in the x-direction and j points
in the y-direction.</span>
The magnitude of the
vector is its dot product with itself:
<span>|A|2 = A·A</span>
<span>102 = (6 i +
y j)•(6 i+ y j)
Note that i•j = 0, and i•i = j•j =
1 </span>
<span>100 = 36 + y2
</span>
<span>64 = y2</span>
<span>get the square root of 64 = 8</span>
<span>The vertical component of the vector is 8 cm.</span>
Answer:
Explanation:
Given that,
At one instant,
Center of mass is at 2m
Xcm = 2m
And velocity =5•i m/s
One of the particle is at the origin
M1=? X1 =0
The other has a mass M2=0.1kg
And it is at rest at position X2= 8m
a. Center of mass is given as
Xcm = (M1•X1 + M2•X2) / (M1+M2)
2 = (M1×0 + 0.1×8) /(M1 + 0.1)
2 = (0+ 0.8) /(M1 + 0.1)
Cross multiply
2(M1+0.1) = 0.8
2M1 + 0.2 =0.8
2M1 = 0.8-0.2
2M1 = 0.6
M1 = 0.6/2
M1 = 0.3kg
b. Total momentum, this is an inelastic collision and it momentum after collision is given as
P= (M1+M2)V
P = (0.3+0.1)×5•i
P = 0.4 × 5•i
P = 2 •i kgm/s
c. Velocity of particle at origin
Using conversation of momentum
Momentum before collision is equal to momentum after collision
P(before) = M1 • V1 + M2 • V2
We are told that M2 is initially at rest, then, V2=0
So, P(before) = 0.3V1
We already got P(after) = 2 •i kgm/s in part b of the question
Then,
P(before) = P(after)
0.3V1 = 2 •i
V1 = 2/0.3 •i
V1 = 6 ⅔ •i m/s
V1 = 6.667 •i m/s
B. It's an example of velocity due to the fact that it has a measurement of speed, divided by time, and has a specific direction. Acceleration doesn't have any direction on it, but has speed divided by time. C and D have a different mode of measurement despite of the fact that it still needs meters/miles/km.
