We need the frequency of the photon, it is v = c/ λ
Where c is 3 x 10^8 ms^-1 and λ
is the wave length
We also need the expression of
connecting frequency to energy of photon
which is E = hv where h is Planck’s
constant
Combining the two equations
will give us:
E = h x c/λ
Inserting the values, we will
have:
E = 6.626 x 10^-34 x 3 x 10^8 /
0.126
E = 1.578 x 10^ -24 J
#1
so mass number = 238
mass number = protons + neutrons
given that
neutrons = 146
238 = protons + 146
protons = 92
#2

so mass number = 241
mass number = protons + neutrons
given that
Protons = 94
241 = 94 + neutrons
neutrons = 147
#3

A = mass number
Protons = 90
Neutrons = 137
A = protons + Neutrons
A = 90 + 137 = 227
Answer:
Explanation:
i = Imax sin2πft
given i = 180 , Imax = 200 , f = 50 , t = ?
Put the give values in the equation above
180 = 200 sin 2πft
sin 2πft = .9
sin2π x 50t = .9
sin 360 x 50 t = sin ( 360n + 64 )
360 x 50 t = 360n + 64
360 x 50 t = 64 , ( putting n = 0 for least value of t )
18000 t = 64
t = 3.55 ms .
Answer:
T = 0.03 Nm.
Explanation:
d = 1.5 in = 0.04 m
r = d/2 = 0.02 m
P = 56 kips = 56 x 6.89 = 386.11 MPa
σ = 42-ksi = 42 x 6.89 = 289.58 MPa
Torque = T =?
<u>Solution:</u>
σ = (P x r) / T
T = (P x r) / σ
T = (386.11 x 0.02) / 289.58
T = 0.03 Nm.
Answer:
Check the explanation
Explanation:
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
The image is virtual
The image is upright
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
Kindly check the diagram in the attached image below.