You have to use the specific heat equation.
Q = cmΔT where Q is the energy, c is specific heat, m is mass, and ΔT is change in temp.
So we can substitute our variables into the equation.
30000J = (390g)(3.9J*g/C)ΔT
Solving for ΔT, we get:
30000J/[(390g)*(3.9J*g/C) = ΔT
ΔT = 19.72386588C
I'm assuming the temperature is C, since it was not specified.
Hope this helps!
A concave lens can only form a virtual image. The correct option among all the options that are given in the question is the third option or option "C". Concave lenses are mostly thinner in the middle compared to its edges. I hope that this answer has come to your help.
Answer:
a= -0.83m\s^2
Explanation:
a = v \ t
a = -25 \ 30 = -0.833 m\s^2
the object is slowing down 0.83 meter every second
Steel is more dense because it’s heavy while water is light
Answer:
Explanation:
q = 2e = 3.2 x 10^-19 C
mass, m = 6.68 x 10^-27 kg
Kinetic energy, K = 22 MeV
Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A
(a) time, t = 2.8 s
Let N be the alpha particles strike the surface.
N x 2e = q
N x 3.2 x 10^-19 = i t
N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8
N = 2.36 x 10^12
(b) Length, L = 16 cm = 0.16 m
Let N be the alpha particles
K = 0.5 x mv²
22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²
v² = 1.054 x 10^15
v = 3.25 x 10^7 m/s
So, N x 2e = i x t
N x 2e = i x L / v
N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)
N = 4153.85
(c) Us ethe conservation of energy
Kinetic energy = Potential energy
K = q x V
22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V
V = 1.17 x 10^7 V