When an object is moving, we use the following coefficient for friction calculations a)-μk b)-μs c)-γk d)- γs
1 answer:
Answer:
Explanation:
We use kinetic friction when a body is moving i.e. for calculations.
Static friction is used when a body is in rest while kinetic friction is used when a body is moving and its value is quite low as compared to static friction .
Static friction value increases as we apply more force while kinetic friction occurs when there is relative motion between bodies.
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The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :
J = 207.6
So the shear stress at point A is :
Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
Note: This question is incomplete and lacks necessary data to solve. But I have found the similar question on the internet. So, I will be using the data from that question to solve this question for the sack of concept and understanding.
Data Given:
x = 27 , 44 , 32 , 47, 23 , 40, 34, 52
y = 30, 19, 24, 13 , 29, 19, 21, 14
It is given that,
∑x = 299
∑y = 167
∑ = 11887
∑ = 3773
We are asked to verify the above values manually in this question.
So,
1. ∑x = 299
Let's verify it:
∑x = 27 + 44 + 32 + 47 + 23 + 40 + 34 + 52
∑x = 299
Yes, it is equal to the given value. Hence, verified.
2. ∑y = 167
Let's verify it:
∑y = 30 + 19 + 24 + 13 + 29 + 19 + 21 + 14
∑y = 169
No, it is not equal to the given value.
3. ∑ = 11887
Let's verify it:
For this to find, first we need to square all the value of x individually and then add them together to verify.
∑ =
+
+
+
+
+
+
+
∑ = 11,887
Yes, it is equal to the given value. Hence, verified.
4. ∑ = 3773
Let's verify it:
Again, for this we need to find the squares of all the y values and then add them together to verify it.
∑ =
+
+
+
+
+
+
+
∑ = 3,845
No, it is not equal to the given value.
Answer:
T₂ =93.77 °C
Explanation:
Initial temperature ,T₁ =27°C= 273 +27 = 300 K
We know that
Absolute pressure = Gauge pressure + Atmospheric pressure
Initial pressure ,P₁ = 300+1=301 kPa
Final pressure ,P₂= 367+1 = 368 kPa
Lets take temperature=T₂
We know that ,If the volume of the gas is constant ,then we can say that
Now by putting the values in the above equation we get
The temperature in °C
T₂ = 366.77 - 273 °C
T₂ =93.77 °C