T = 3.5 secs
Velocity (v) = g * t = 10 m/s^2 * 3.5 sec = 35 m/s
Explanation:
Given that,
Size of object, h = 0.066 m
Object distance from the lens, u = 0.210 m (negative)
Focal length of the converging lens, f = 0.14 m
If v is the image distance from the lens, we can find it using lens formula as follows :
(a) Magnification,

(b) Magnification, 
h' is image height

Hence, this is the required solution.
Answer:3.87*10^-4
Explanation:
What is the decrease in mass, delta mass Xe , of the xenon nucleus as a result of this deca
We have been given the wavelength of the gamma ray, find the frequency using c = freq*wavelength.
C=f*lambda
3*10^8=f*3.44*10^-12
F=0.87*10^20 hz
Then with the frequency, find the energy emitted using equation
E=hf E = freq*Plank's constant
E=.87*10^20*6.62*10^-34
E=575.94*10^(-16)
With this energy, convert into MeV from joules.
With the energy in MeV, use E=mc^2 using c^2 = 931.5 MeV/u.
Plugging and computing all necessary numbers gives you
3.87*10^-4 u.
Answer:
110.87 dB
Explanation:
(I got it right on Acellus)
I= P/4(pi)r^2 = 60/4(pi)6.25^2
60/4(pi)6.25^2=0.12223
B=10log(I/Io)
B=10log(0.12223/1*10^-12) = 110.87 dB
111 in sigfigs