Answer:
A. 26.17 B. 1.17 C. 30.86 D. 5.86
Explanation:
Explanation:
Value of the cross-sectional area is as follows.
A =
= 3.45 
The given data is as follows.
Allowable stress = 14,500 psi
Shear stress = 7100 psi
Now, we will calculate maximum load from allowable stress as follows.

= 
= 50025 lb
Now, maximum load from shear stress is as follows.

= 
= 48990 lb
Hence,
will be calculated as follows.

= 48990 lb
Thus, we can conclude that the maximum permissible load
is 48990 lb.
Answer:
Approximately
.
Explanation:
The formula for the kinetic energy
of an object is:
,
where
is the mass of that object, and
is the speed of that object.
Important: Joule (
) is the standard unit for energy. The formula for
requires two inputs: mass and speed. The standard unit of mass is
while the standard unit for speed is
. If both inputs are in standard units, then the output (kinetic energy) will also be in the standard unit (that is: joules,
Convert the unit of the arrow's mass to standard unit:
.
Initial
of this arrow:
.
That's the same as the energy output of this bow. Hence, the efficiency of energy transfer will be:
.
Answer: 
Explanation:
Given
Wavelength of light 
Screen is
away
Distance between two adjacent bright fringe is 
When same experiment done in water, wavelength reduce to 
So, the distance between the two adjacent bright fringe is 
Keeping other factor same, distance becomes
