Answer:
Fx1 (6 m) sin 60 = 300 (3 m) cos 60 balancing torques about floor
Fx1 = 900 * 1/2 / 5.20 = 86.6 N this is the horizontal force that must be supplied by the wall to balance torques about the floor
This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.
Fx1 = Fx2 = 86.6 N
Answer:
P = 0.0644 atm
Explanation:
Given that,
The pressure of a sample of gas is measured as 49 torr.
We need to convert this temperature to atmosphere.
The relation between torr and atmosphere is as follow :
1 atm = 760 torr
1 torr = (1/760) atm
49 torr = (49/760) atm
= 0.0644 atm
Hence, the presssure of the sample of gas is equal to 0.0644 atm.
Answer:
Explanation:
Given that,
The compression in the spring, x = 0.0647 m
Speed of the object, v = 2.08 m/s
To find,
Angular frequency of the object.
Solution,
We know that the elation between the amplitude and the angular frequency in SHM is given by :
A is the amplitude
In case of spring the compression in the spring is equal to its amplitude
So, the angular frequency of the spring is 32.14 rad/s.
Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa
<u>Explanation:</u>
Given -
Stress Direction, A = [1 0 0 ]
Slip plane = [ 1 1 1]
Normal to slip plane, B = [ 1 1 1 ]
Critical stress, Sc = 2.92 MPa
Let the direction of slip on = [ 1 1 0 ]
Let Ф be the angle between A and B
cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3
cos Ф = 1/√3
σ = Sc / cosФ cosλ
For slip along [ 1 1 0 ]
cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1
cos λ = 1/√2
Therefore,
σ = 2.92 / 1/√3 1/√2
σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa
Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa
