A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb
1 answer:
To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
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Answer:
minimum length of a surface crack is 18.3 mm
Explanation:
Given data
plane strain fracture toughness K = 82.4 MPa m1/2
stress σ = 345 MPa
Y = 1
to find out
the minimum length of a surface crack
solution
we will calculate length by this formula
length = 1/π ( K / σ Y)²
put all value
length = 1/π ( K / σ Y)²
length = 1/π ( 82.4 / 345× 1)²
length = 18.3 mm
minimum length of a surface crack is 18.3 mm
Answer:
(a). Index of refraction are = 1.344 &
= 1.406
(b). The velocity of red light in the glass 2.23 ×
The velocity of violet light in the glass 2.13 ×
Explanation:
We know that
Law of reflection is
Here
= angle of incidence
= angle of refraction
(a). For red light
1 × =
×
= 1.344
For violet light
1 × =
×
= 1.406
(b). Index of refraction is given by
= 1.344
2.23 ×
This is the velocity of red light in the glass.
The velocity of violet light in the glass is given by
2.13 ×
This is the velocity of violet light in the glass.