22. a - (vf^2 - vi^2)/(2d)
a = (0 - 23^2)/(170)
a = -3.1 m/s^2
23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3
33 = 3t
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long.
24. The sprinter starts from rest. The average acceleration is found from:
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s
25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m
26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s
acceleration, a = -7.00m/s^2
displacement, s - 92m
Using v^2 = u^2 - 2as
0^2 - u^2 + 2 (-7.00) (92)
initial velocity, u = sqrt (1288) = 35.9 m/s
This is the speed pf the car just bore braking.
I hope this helps!!
Answer:
11.6532 x 10⁻¹¹ J or 7.3 MeV is given off
Explanation:
Mass of an alpha particle = 4.0026u, ∴ mass of three = 12.0078u
Find the difference in mass.
Mass of three alpha - Mass of Carbon nucleus
12.0078u - 12u = 0.0078u
Since 1u = 1.66 x 10⁻²⁷ kg
Therefore, 0.0078u = 1.2948 x 10⁻²⁷
Now that we know Mass(m) = 1.2948 x 10⁻²⁷ and Speed (c) 3 x 10⁸ m²s⁻²
Formular for Energy ==> E₀ = mc²
E = (1.2948 x 10⁻²⁷) (3 x 10⁸ m²s⁻²)²
E = (1.2948 x 10⁻²⁷) (9 x 10¹⁶) J
E = 11.6532 x 10⁻¹¹ J
Or, if you need your energy in MeV
1 MeV = 1.60x10⁻¹³ J
Just do the conversion by dividing 11.6532 x 10⁻¹¹ J by 1.60x10⁻¹³ J
It will give you 7.3 MeV
