Answer:
(a) 81.54 N
(b) 570.75 J
(c) - 570.75 J
(d) 0 J, 0 J
(e) 0 J
Explanation:
mass of crate, m = 32 kg
distance, s = 7 m
coefficient of friction = 0.26
(a) As it is moving with constant velocity so the force applied is equal to the friction force.
F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N
(b) The work done on the crate
W = F x s = 81.54 x 7 = 570.75 J
(c) Work done by the friction
W' = - W = - 570.75 J
(d) Work done by the normal force
W'' = m g cos 90 = 0 J
Work done by the gravity
Wg = m g cos 90 = 0 J
(e) The total work done is
Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J
I am 11 year old and u just needed to login in so I can't answer ur questions
Answer:
I think the answer is a
Explanation:
for it to be accurate has be to exactly 0.9 rad
it is not precise because the answer she is getting is different everytime and not even close. For instance,
It would have been precise if she had gotten 0.37 rad in every attempt. or 0.89 every attempt...
Explanation:
Given that,
Angle by the normal to the slip α= 60°
Angle by the slip direction with the tensile axis β= 35°
Shear stress = 6.2 MPa
Applied stress = 12 MPa
We need to calculate the shear stress applied at the slip plane
Using formula of shear stress
Put the value into the formula
Since, the shear stress applied at the slip plane is less than the critical resolved shear stress
So, The crystal will not yield.
Now, We need to calculate the applied stress necessary for the crystal to yield
Using formula of stress
Put the value into the formula
Hence, This is the required solution.
It depends how you want it to work do you want to take a picture oh yea and to do that you must create a account btw
