Answer:
Id say C
Explanation:
The hypotenuse for c is up and to the right so the unit vectors show that
B4 the tackle:
<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>
<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>
<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>
<span>The vector triangle is right angled: </span>
<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>
<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>
<span>v(f) = 5.6 m/s (to 2 sig figs) </span>
<span>direction of v(f) is the same as the direction of the final momentum </span>
<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>
<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>
<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
A) We differentiate the expression for velocity to obtain an expression for acceleration:
v(t) = 1 - sin(2πt)
dv/dt = -2πcos(2πt)
a = -2πcos(2πt)
b) Any value of t can be plugged in as long as it is greater than or equal to 0.
c) we integrate the expression of velocity to find an expression for displacement:
∫v(t) dt = ∫ 1 - sin(2πt) dt
x(t) = t + cos(2πt)/2π + c
x(0) = 0
0 = = + cos(0)/2π + c
c = -1/2π
x(t) = t + cos(2πt)/2π -1/2π
Answer:
Energy of Photon = 4.091 MeV
Explanation:
From the conservation of energy principle, we know that total energy of the system must remain conserved. So, the energy or particles before collision must be equal to the energy of photons after collision.
K.E OF electron + Rest Energy of electron + K.E of positron + Rest Energy of positron = 2(Energy of Photon)
where,
K.E OF electron = 3.58 MeV
Rest Energy of electron = 0.511 MeV
Rest Energy of positron = 0.511 MeV
K.E OF positron = 3.58 MeV
Energy of Photon = ?
Therefore,
3.58 MeV + 0.511 MeV + 3.58 MeV + 0.511 MeV = 2(Energy of Photon)
Energy of Photon = 8.182 MeV/2
<u>Energy of Photon = 4.091 MeV</u>
