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:
13.33 seconds
Explanation:
I = Q/t
t = Q/I = 4/0.3 = 13.33 seconds
The Cosmic Ray is a natural way for nuclear fission and nucleosynthesis to occur. It refers to the formation of chemical elements from the impact of cosmic rays on an object.
Answer:
Efficiency = StartFraction T Subscript h Baseline minus T Subscript C Baseline over T Subscript h Baseline EndFraction times 100. Efficiency equals T Subscript c Baseline minus T Subscript h Baseline over T Subscript h Baseline all times 100.
Answer:
• 36.4 kg of coal.
• 80 pounds of coal.
Explanation:
Using proportionality constant,
Mass of coal = 1,000,000/27,500,000 btus/metric ton
= 0.0364 metric tons of coal
Mass of coal = 1,000,000/25,000,000 btus/ton
= 0.04 tons of coal.
Converting metric tons to kilogram,
1 metric ton = 1000kg,
0.0364 metric ton;
= 36.4 kg of coal.
Converting tons to pounds,
1 ton = 2000 pounds,
0.04 metric ton;
= 80 pounds of coal.
