Answer:
(a) 328 Nm
(b) 79.35 Nm
Explanation:
N = =150, side = 17.5 cm = 0.175 m, i = 42 A, B = 1.7 T
A = side^2 = 0.175^2 = 0.030625 m^2
(a) Torque = N x i x A x B x Sinθ
For maximum torque, θ = 90 degree
Torque = 150 x 42 x 0.030625 x 1.7 x Sin 90
Torque = 328 Nm
(b) θ = 14 degree
Torque = 150 x 42 x 0.030625 x 1.7 x Sin 14
Torque = 79.35 Nm
The answer is B-plus.
The object is accelerating to the right, and up.
Answer:
Explanation:
By conservation of energy, the sum of the kinetic and gravitational potential energies at the surface of the Earth must be equal than their sum at infinity, so we have:
Where 
is the gravitational constant,
and 
are the mass and radius of the Earth, <em>m </em>is the mass of the particle, 
its velocity at the surface of the Earth (which would be its escape velocity) and 
and 
are the velocities and distance at infinity, which would be null and infinity respectively, so the right hand side of our equation is 0J, which leaves us with:
Also, since the force the molecule experiments is the force of gravity (disregarding drag), we can write its weight in terms of Newton's Law of Gravitation:
Which means that:
So finally putting all together we can write:
Answer:
2.1 rad(anticlockwise).
Explanation:
So, we are given the following data or parameters or information in the question above:
=> "The torsional stiffness of the spring support is k = 50 N m/rad. "
=> "If a concentrated torque of mag- nitude Ta = 500 Nm is applied in the center of the bar"
=> "L = 300 mm Assume a shear modu- lus G = 10 kN/mm2 and polar monnent of inertia J = 2000 mln"
Hence;
G × J = 10 kN/mm2 × 2000 mln = 20 Nm^2.
Also, L/2 = 300 mm /2 = 0.15 m (converted to metre).
==> 0.15/20 (V - w) + θ = 0.
==> 0.15/20 (V - w ) = -θ.
Where V = k = 50 N m/rad
w = 183.3 θ.
Therefore, w + Vθ = 500 Nm.
==> 183.3 + 50 θ = 500 Nm.
= 6.3
Anticlockwise,
θ = 2.1 rad.
