I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
Answer:
Wavelength = 0.48 m (Approx)
Explanation:
Given:
Speed of sound = 340 m/s
Frequency = 706 hz
Find:
Wavelength
Computation:
Wavelength = Speed of sound / Frequency
Wavelength = 340 / 706
Wavelength = 0.48 m (Approx)
Answer:
5.096*10^-8
Explanation:
Given that
The average value of the electromagnetic wave is 310 mW/m²
To find the maximum value of the magnetic field the wave is closest to, we say
Emax = √Erms
Emax = √[(2 * 0.310 * 3*10^8 * 4π*10^-7)]
Emax = √233.7648
Emax = 15.289
Now, with our value of maximum electromagnetic wave gotten, we divide it by speed of light to get our final answer
15.289 / (3*10^8) = 5.096*10^-8 T
Suffice to say, The maximum value of the magnetic field in the wave is closest to 5.096*10^-8
