Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
<h2>Solution :</h2>
Here ,
• Height of sign post = 30 m
• Distance between signpost and truck = 24 m
Let the
• Top of signpost = A
• Bottom of signpost = B
• The end of truck facing sign post be = C
Now as we can clearly imagine that the ladder will act as an hypotenuse to the Triangle ABC .
Where
• AB = Height of signpost = 30 m
• BC = distance between both = 24 m
• AC = Minimum length of ladder
→ AC² = AB² + BC² ( As we can see AB is perpendicular to BC )
→ AC² = (30)² + (24)²
→ AC² = 900 + 576
→ AC² = 1476
→ AC = 38.41875
or AC apx = 38.42
So minimum height of ladder = 38.42
Answer:
0.00001266 m
Explanation:
D = Distance from source to screen
m = Order
d = Slit separation
The distance from a point on the screen to the center line
At m = 0
At m = 1
The slit separation is 0.00001266 m
Answer:
a) 2250 J
b) 0 J
c) 2250 J
Explanation:
a) Since, the process is isochoric
the change in internal energy
Here, n = 0.2 moles
Cv = 12.5 J/mole.K
We have to find T_f so we can use gas equation as
![\frac{P_1V_1}{P_2V_2} =\frac{T_i}{T_f}\\Since, V_1=V_2 [isochoric/process]\\\Rightarrow \frac{P_{atm}}{4P_{atm}} = \frac{300}{T_f} \\\Rightarrow T_f = 1200 K](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BP_2V_2%7D%20%3D%5Cfrac%7BT_i%7D%7BT_f%7D%5C%5CSince%2C%20V_1%3DV_2%20%20%20%20%5Bisochoric%2Fprocess%5D%5C%5C%5CRightarrow%20%5Cfrac%7BP_%7Batm%7D%7D%7B4P_%7Batm%7D%7D%20%3D%20%5Cfrac%7B300%7D%7BT_f%7D%20%5C%5C%5CRightarrow%20T_f%20%3D%201200%20K)
So, 
b) Since, the process is isochoric no work shall be done.
c) By first law of thermodynamics we have
Since, Q is positive 2250 J of heat will flow into the system.
We can first calculate the net force using the given information.
By Newton's second law, F(net) = ma:
F(net) = 25 * 4.3 = 107.5
We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):
f = F(net) - F(app) = 150 - 107.5 = 42.5 N
Now we can calculate the coefficient of friction, u, using the normal force, F(N):
f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17
