Answer:
Explanation:
⁵⁷Co₂₇ + e⁻¹ = ²⁷Fe₂₆
mass defect = 56.936296 + .00055 - 56.935399
= .001447 u
equivalent energy
= 931.5 x .001447 MeV
= 1.3479 MeV .
= 1.35 MeV
energy of gamma ray photons = .14 + .017
= .157 MeV .
Rest of the energy goes to neutrino .
energy going to neutrino .
= 1.35 - .157
= 1.193 MeV.
Answer:
According to the data given in the question, experiment on table two pulling and falling masses are arranged in the fig. 250 g is pulling right side and 100 g pulling down. The gravitational force is common to both the masses, so we cannot say that the block moves towards heavier mass, also the block does not move towards the lighter mass.
Obviously, the effect of heavier mass of 250 g is more on the block, so the block moves towards right bottom corner. i.e., diagonally between two masses
please find the attachment.
Answer:
aaksj
Explanation:
a) the capacitance is given of a plate capacitor is given by:
C = \epsilon_0*(A/d)
Where \epsilon_0 is a constant that represents the insulator between the plates (in this case air, \epsilon_0 = 8.84*10^(-12) F/m), A is the plate's area and d is the distance between the plates. So we have:
The plates are squares so their area is given by:
A = L^2 = 0.19^2 = 0.0361 m^2
C = 8.84*10^(-12)*(0.0361/0.0077) = 8.84*10^(-12) * 4.6883 = 41.444*10^(-12) F
b) The charge on the plates is given by the product of the capacitance by the voltage applied to it:
Q = C*V = 41.444*10^(-12)*120 = 4973.361 * 10^(-12) C = 4.973 * 10^(-9) C
c) The electric field on a capacitor is given by:
E = Q/(A*\epsilon_0) = [4.973*10^(-9)]/[0.0361*8.84*10^(-12)]
E = [4.973*10^(-9)]/[0.3191*10^(-12)] = 15.58*10^(3) V/m
d) The energy stored on the capacitor is given by:
W = 0.5*(C*V^2) = 0.5*[41.444*10^(-12) * (120)^2] = 298396.8*10^(-12) = 0.298 * 10 ^6 J
Answer:
3.16 × 
W/
Explanation:
β(dB)=10 × 
=
W/
β=55 dB
Therefore plugging into the equation the values,
55=10 ![log_{10}(\frac{I}{ [tex]10^{-12}](https://tex.z-dn.net/?f=log_%7B10%7D%28%5Cfrac%7BI%7D%7B%20%5Btex%5D10%5E%7B-12%7D)
})[/tex]
5.5= })[/tex]
= 
316227.76×= I
I= 3.16 × W/
Answer:A student shoots a spitball with a perfectly horizontal velocity of 9.7 m/s from a height of 1.8 meters. How long will it take for the spitball to hit the ground?
(ignore air resistance) (include units and correct number of significant figures)
Explanation:La respuesta es porque esa es la respuesta, la respuesta al número es 9.7 1.8 Divide =53.888
