Answer:
ok i don't know the answer but I send you tomorrow
Answer: 97.33 N
Explanation: In order to solve this problem we have to use the expression for the velocity of the transversal wave in wire. This is given by:
v=
whete T and μ are the Tension and linear density.
then
T=v^2*μ ; μ can be calculated as 0.166Kg/0.74m=0.22 Kg/m
v is obtained by λ*ν = 0.0336 m*626 /s=21.03 m/s
thus
T=442.4 *0.22=97.33 N
"<span>During radioactive decay, atoms break down, releasing, particles or energy" is the one statement about radioactive decay among the following choices given in the question that is true. The correct option is option "b".
"H</span>alf-life" is the term among the following that <span>refers to the time it takes for one-half of the radioactive atoms in a sample of a radioactive element to decay. The correct option is option "d".</span>
It slows the object down so it cannot move well and evetually the object cannot be pushed and farther
Answer:
The wavelengths of these two fringes is 589.61 x 10∧9
Explanation:
d = slit width = 0.001/1000 = 1 x 10-6 m
X1 = Position of first fringe = 0.7288 m
X2 = Position of second fringe = 0.73 m
D = distance of the screen
tan\theta1 = X1/D \Rightarrow\theta1 = tan-1(X1/D ) = tan-1(0.7288/1 ) = 36.085
tan\theta2 = X2/D \Rightarrow\theta2 = tan-1(X2/D ) = tan-1(0.73/1 ) = 36.129
position of first bright fringe is given as
d Sin\theta1 = n\lambda1
for bright fringe , n = 1
d Sin\theta1 = \lambda1
\lambda1 = (1 x 10-6 ) Sin36.085 = 589 x 10-9
position of second bright fringe is given as
d Sin\theta2 = n\lambda2
for bright fringe , n = 1
d Sin\theta2 = \lambda2
\lambda2 = (1 x 10-6 ) Sin36.129 = 589.61 x 10-9
