A gamma ray with an energy of 3.40 × 10-14 joules strikes a photographic plate. We know that Planck's constant is 6.63 × 10-34 j
2 answers:
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Answer:
The length of the stick is 0.28 m.
The time the stick take to move is 0.97 ns.
Explanation:
Given that,
Relative speed of stick v= 0.96 c
Speed of light
Proper length of stick = 1 m
We need to calculate the length of the stick
Using formula of length
Put the value into the formula
We need to calculate the time the stick take to move
Using formula of time
Put the value into the formula
Hence, The length of the stick is 0.28 m.
The time the stick take to move is 0.97 ns.
Answer:
F_ab = 260.17 N
Explanation:
Given:
- Moment of force F about CD, (M)_cd = 57 Nm
Find:
- First we will write down the position vectors of points A, B , C , D:
- We will take left and bottom most corner of cube to be the origin.
- The unit vectors i , j , k are along vertical planes and outside the plane, respectively.
- The position vectors wrt to the origin are:
Point A = 0.2 k
Point B = 0.4 i + 0.2 j
Point C = 0.2 j + 0.4 k
Point D = 0.4 i + 0.4 k
- Now we will determine the Force vector F_ab along vector AB.
vec (AB) = B - A = 0.4 i + 0.2 j - 0.2 k
unit (AB) = 0.4 i + 0.2 j - 0.2 k / sqrt ( 0.4^2 + 2*0.2^2)
= [5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )
Hence,
vec(F_ab) = Fab*[5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )
- Now, form a unit vector along the line CD:
vec(CD) = D - C = 0.4 i - 0.2 j
unit (CD) = 0.4 i - 0.2 j / sqrt ( 0.4^2 + 0.2^2)
= [sqrt(5)]*(0.4 i - 0.2 j)
- Now select a point on line CD, lets say C. Find the moment arm from line of action of force along AB and line CD. Hence, vector AC is:
vec(AC) =r_ac = C - A = 0.2 j + 0.2 k
- Now the moment about a line CD due to force is:
(M)_cd = unit(CD) . ( r_ac x vec(F_ab) )
The cross product of r_ac and vec(F_ab) is as follows:
(M)_c = ( r_ac x vec(F_ab) ) :
(M)_c = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k]
The dot product of (M)_c and unit (CD) is as follows:
(M)_cd = unit(CD) . (M)_c :
(M)_cd = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k] . [sqrt(5)]*(0.4 i - 0.2 j)
(M)_cd = F_ab*(sqrt(30) / 25)
- The given magnitude of the moment is (M)_cd. Calculate F_ab:
57 = F_ab*(sqrt(30) / 25)
F_ab = 260.17 N