Answer:
306 m/s
Explanation:
Law of conservation of momentum
m1v1 + m2v2 = (m1+m2)vf
m1 is the bullet's mass so it is 0.1 kg
v1 is what we're trying to solve
m2 is the target's mass so it is 5.0 kg
v2 is the targets velocity, and since it was stationary, its velocity is zero
vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s
plugging in, we get
(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)
(0.1)(v1) + 0 = 30.6
(0.1)(v1) = 30.6
v1 = 306 m/s
As long as all the waves stay in the same medium, the intensity of
any waves ... electromagnetic or mechanical ... decrease in proportion
to the square of the distance.
If the distance increases to 3 x the original distance, then the intensity
changes to 1/3² or 1/9 of the original intensity.
I suppose choice-'d' is the correct one, but I have to tell you that
the phrase "nine times as low" is mathematically meaningless,
and it really grinds my gears.
1) According to the law of conservation of momentum ..
<span>Horiz recoil mom of gun (M x v) = horiz. mon acquired by shell (m x Vh) </span>
<span>1.22^6kg x 5.0 m/s = 7502kg x Vh </span>
<span>Vh = 1.22^6 x 5 / 7502 .. .. Vh = 813 m/s </span>
<span>Barrel velocity V .. .. cos20 = Vh / V .. ..V = 813 /cos20 .. .. ►V = 865 m/s </span>
<span>2) Using the standard range equation .. R = u² sin2θ /g </span>
<span>R = 865² x sin40 / 9.80 .. .. ►R = 49077 m .. (49 km)</span>
Answer:
6667.9 J
Explanation:
We are given that
Mass of cannon.m=14 kg
Time,t=3.15 s
Initial velocity,u=0
We have to find the work done by gravity on the falling cannon ball.


Substitute the values

Work done,W=mgh
Using the formula
