Answer:
Explanation:
First of all, we need to find the volume of the room, which is given by
Now we can find the mass of the air by using
where
is the density of the air
is the volume of the room
Substituting,
The position of an object is given by x = at3 - bt2 + ct,where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s, and x and t are in SI un
1 answer:
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Answer:
T=1022.42 N
Explanation:
Given that
l = 32 cm ,μ = 1.5 g/cm
L =2 m ,V= 344 m/s
The pipe is closed so n= 3 ,for first over tone
f= 129 Hz
The tension in the string given as
T = f²(4l²) μ
Now by putting the values
T = f²(4l²) μ
T = 129² x (4 x 0.32²) x 1.5 x 10⁻³ x 100
T=1022.42 N
Answer:
Explanation:
Given:
As we know that
Force=Mass * Acceleration
Putting the value of mass and Acceleration we get ,
Therefore Net force is :
Answer:
a) The maximum contact pressure is 274.58 MPa and the width of contact is 0.058 mm
b) The maximum shear stress is 82.37 MPa at a distance of 0.023 mm
Explanation:
Given data:
L = 20 mm
F = 250 N
r₁ = 10 mm
r₂ = 15 mm
v = 0.3
E = 2.07x10⁵ MPa
a) The maximum contact pressure is:
The width of contact is:
b) According the graph elastic stresses below the surface, for v = 0.3, the maximum shear stress is
T = 0.3*P = 0.3 * 274.58 = 82.37 MPa
At a distance of
0.8*b = 0.8*0.029 = 0.023 mm