Incomplete question.The complete question is here
Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.
Answer:
Torque=0.51 Btu
Explanation:
Given Data
Power=225 hp
Revolutions =3000 rpm
To find
T( torque )=?
Solution
As

As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.
So

The force of F=10 N produces an extension of

on the string, so the spring constant is equal to

Then the string is stretched by

. The work done to stretch the string by this distance is equal to the variation of elastic potential energy of the string with respect to its equilibrium position:
Answer:
we have to find out the critical resolved shear stress. As it it given in the question
Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.
a) Slip will occur in the direction where cosФ cosλ are maximum. Cosine for all possible λ values are given as follows.
cos(62.4°) = 0.46
cos(72.0°) = 0.31
cos(81.1°) = 0.15
Thus, the slip direction is at the angle of 62.4° along the tensile axis.
b) now the critical resolved shear stress can be find out by the following equation.
τ
= σ
( cosФ cosλ)
now by putting values,
= (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23
Answer:
<h2>a)
Nathan's acceleration is 5 m/s²
</h2><h2>b)
Nathan's displacement during this time interval is 15.625 m</h2><h2>
c) Nathan's average velocity during this time interval is 6.25 m/s</h2>
Explanation:
a) We have equation of motion v = u + at
Initial velocity, u = 0 m/s
Final velocity, v = 12.5 m/s
Time, t = 2.5 s
Substituting
v = u + at
1.25 = 0 + a x 2.5
a = 5 m/s²
Nathan's acceleration is 5 m/s²
b) We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 5 m/s²
Time, t = 2.5 s
Substituting
s = ut + 0.5 at²
s = 0 x 2.5 + 0.5 x 5 x 2.5²
s = 15.625 m
Nathan's displacement during this time interval is 15.625 m
c) Displacement = 15.625 m
Time = 2.5 s
We have
Displacement = Time x Average velocity
15.625 = 2.5 x Average velocity
Average velocity = 6.25 m/s
Nathan's average velocity during this time interval is 6.25 m/s
Complete Question:
Check the file attached to get the complete question
Answer:
In the film Ice word Revenge, vehicle 2 did not fall of the cliff because,
but in Claire's test, vehicle 2 off the cliff because 
Explanation:
In Claire's test, the weight of vehicle 1 is either equal to or greater than the weight of vehicle 2, so it was sufficient to push it down the cliff. In the film Ice word revenge, the weight of vehicle 1 is less than the weight of vehicle 2, it is not sufficient to make it fall off the cliff ( Note: Looking exactly the same in the movie, as Claire claimed, does not mean they have the same mass). Therefore if Claire wants a collision that will not make the vehicle 2 fall off the cliff, he should collide it with a vehicle of lesser mass/weight.