Answer:
Explanation:
We are given that
Electric field,E=
We have to find the value of maximum charge density that can be placed on the surface of the plane before dielectric breakdown of the surrounding air occurs.
We know that
Where
Using the formula
Answer:
Explanation:
Using second law of motion
where m1 is mass of block, m2 is mass of flywheel, g is acceleration due to gravity whose value is taken as
, T is torque and r is radius
Substituting 5.5 Kg for m1, 13 Kg for m2, 0.33 m for r, 2.5 Nm for T we obtain
Answer:
(a) Ra = 9.25 kN; Rb = 5.75 kN
(b) 26.7 kN
Explanation:
(a) Draw a free-body diagram of the crane. There are four forces:
Reaction Ra pushing up at A,
Reaction Rb pushing up at B,
Weight force 12.5 kN pulling down at G,
and weight force 2.5 kN pulling down at F.
Sum of moments about B in the counterclockwise direction:
∑τ = Iα
-Ra (0.66 m + 0.42 m + 2.52 m) + 12.5 kN (2.52 m + 0.42 m) − 2.5 kN ((3.6 m + 0.9 m) cos 30° − 2.52 m) = 0
-Ra (3.6 m) + 12.5 kN (2.94 m) − 2.5 kN (1.38 m) = 0
Ra = 9.25 kN
Sum of moments about A in the counterclockwise direction:
∑τ = Iα
Rb (0.66 m + 0.42 m + 2.52 m) − 12.5 kN (0.66 m) − 2.5 kN ((3.6 m + 0.9 m) cos 30° + 0.66 m + 0.42 m) = 0
Rb (3.6 m) − 12.5 kN (0.66 m) − 2.5 kN (4.98 m) = 0
Rb = 5.75 kN
Alternatively, you can use sum of the forces in the y direction as your second equation.
∑F = ma
Ra + Rb − 12.5 kN − 2.5 kN = 0
Ra + Rb = 15 kN
9.25 kN + Rb = 15 kN
Rb = 5.75 kN
However, you must be careful. If you make a mistake in the first equation, it will carry over to this equation.
(b) At the maximum weight, Ra = 0.
Sum of the moments about B in the counterclockwise direction:
∑τ = Iα
12.5 kN (2.52 m + 0.42 m) − F ((3.6 m + 0.9 m) cos 30° − 2.52 m) = 0
12.5 kN (2.94 m) − F (1.38 m) = 0
F = 26.7 kN
Answer:
Explanation:
Modulus of elasticity of brick, Y = 28 x 10^9 Pa
mass of each, m = 40 kg
Area of each, A = 0.5 m²
number of bricks, n = 4
height, h = 5 m
Let the change in height is Δh.
Use the formula of Modulus of elasticity
E = stress/ strain
stress = n x m x g/A = 4 x 40 x 9.8 / 0.5 = 3136 Pa
So,
Δh = 5.6 x 10^-7 m
Answer:
F = 316.22 N
Explanation:
Given that,
The wind blows a jay bird south with a force of 300 Newtons.
The jay bird flies north, against the wind, with a force of 100 newtons.
Both the forces are acting perpendicular to each other. The net force is given by the resultant of forces as follows :
Hence, the net force on the jay bird is 316.22 N.