Answer:
Explanation:
We are given that
Initial velocity=u=18ft/s
Final velocity,v=38ft/s
Time=t=3 s
We have to find the average acceleration over that 3 s period.
We know that
Average acceleration,a=
Using the formula
Average acceleration,a=
Average acceleration,a=
Average acceleration,a=
Hence, the average acceleration=
By definition, the kinetic energy is given by:
K = (1/2) * m * v ^ 2
where
m = mass
v = speed
We must then find the speed of both objects:
blue puck
v = root ((0) ^ 2 + (- 3) ^ 2) = 3
gold puck
v = root ((12) ^ 2 + (- 5) ^ 2) = 13
Then, the kinetic energy of the system will be:
K = (1/2) * m1 * v1 ^ 2 + (1/2) * m2 * v2 ^ 2
K = (1/2) * (4) * (3 ^ 2) + (1/2) * (6) * (13 ^ 2)
K = <span>
525</span> J
answer
The kinetic energy of the system is<span>
<span>525 </span></span>J
Answer: the correct answer is 7.8026035971 x 10^(-13) joule
Explanation:
Use Energy Conservation. By ``alpha decay converts'', we mean that the parent particle turns into an alpha particle and daughter particles. Adding the mass of the alpha and daughter radon, we get
m = 4.00260 u + 222.01757 u = 226.02017 u .
The parent had a mass of 226.02540 u, so clearly some mass has gone somewhere. The amount of the missing mass is
Delta m = 226.02540 u - 226.02017 u = 0.00523 u ,
which is equivalent to an energy change of
Delta E = (0.00523 u)*(931.5MeV/1u)
Delta E = 4.87 MeV
Converting 4.87 MeV to Joules
1 joule [J] = 6241506363094 mega-electrón voltio [MeV]
4 mega-electrón voltio = 6.40870932 x 10^(-13) joule
4.87 mega-electrón voltio = 7.8026035971 x 10^(-13) joule
We want to know the amount of force that stretches the spring 0.22 m.
That force is the WEIGHT of the mass hung from it.
The weight of the mass is (mass) times (gravity).
To do that calculation, we need to know the value of gravity, but
gravity has different values on every planet. I shall assume that
this whole springy question is taking place on Earth, so that the
value of gravity is 9.8 m/s² .
The weight of the mass is (0.4 kg) x (9.8 m/s²) = 3.92 Newtons.
The spring constant is
(force/length of the stretch)
= (3.92 Newtons) / (0.22 meters)
= (3.92 / 0.22) Newtons/meter
= 17.82 N/m .
The question for this problem would be the minimum headphone delay, in ms, that will cancel this noise.
The 200 Hz. period = (1/200) = 0.005 sec. It will need to be delayed by 1/2, so 0.005/2, that is = 0.0025 sec. So converting sec to ms, will give us the delay of:Delay = 2.5 ms.
