First, torque is equal to force times the distance. for the first force that is applied, the torque is zero because is applied at the hinge. so the net torque:
t = ( 12 N ) ( 0 m ) ( cos 30 ) + ( 12 N ) ( 1.68 m ) cos 45
t = 14.26 Nm is the torque with respect to the hinge
Answer:

Explanation:
Given that

We know that acceleration a given as




We know that



So the magnitude of force F

Your basically breaking the sound beerier <span />
Answer:
<em>0.45 mm</em>
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A = 
where d is the diameter of the wire
1.613 x 10^-7 = 
6.448 x 10^-7 = 3.142 x 
=
d = 4.5 x 10^-4 m = <em>0.45 mm</em>
Answer:
Explanation:
Potential due to a charged metallic sphere having charge Q and radius r on its surface will be
v = k Q / r . On the surface and inside the metallic sphere , potential is the same . Outside the sphere , at a distance R from the centre potential is
v = k Q / R
a ) On the surface of the shell , potential due to positive charge is
V₁ = 
On the surface of the shell , potential due to negative charge is
V₁ = 
Total potential will be zero . they will cancel each other.
b ) On the surface of the sphere potential
= 
= 22.5 x 10⁵ V
On the surface of the sphere potential due to outer shell
= 
= -9 x 10⁵
Total potential
=( 22.5 - 9 ) x 10⁵
= 13.5 x 10⁵ V
c ) In the space between the two , potential will depend upon the distance of the point from the common centre .
d ) Inside the sphere , potential will be same as that on the surface that is
13.5 x 10⁵ V.
e ) Outside the shell , potential due to both positive and negative charge will cancel each other so it will be zero.