<span>Energy = h nu, where nu is the frequency
h = 6.63 x 10^-34 J-s, Planck's constant
So nu = E/h = 1 x 10^5 J /h = 0.15 x 10^29 / s
nu lambda = c, the speed of light.
lambda = wavelength = c / nu =3 x 10^8 / 0.15 x 10^29 = 20 x 10^-21 m.
this can possibly be a gamma ray. Gamma rays are very penetrating. It's both matter and an energy. They are electromagnetic radiation that results from a radioactive material.
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Answer:0.3
Explanation:
Given
velocity of car=15 m/s
truck brought to halt in a distance of 38 m
We know
Final velocity (v)=0
(deceleration)
Therefore minimum coefficient of friction \mu will be
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!!
The main units for acceleration are <span>the meter per squared second as told by Galileo Galilei. Although there can be more than one example, I consider this one to be correct. Hope it will help you in some measure.</span>
Momentum is Mass × Velocity
= 1875×22 = 41250 kg m/sec
