Given :
Thin hoop with a mass of 5.0 kg rotates about a perpendicular axis through its center.
A force F is exerted tangentially to the hoop. If the hoop’s radius is 2.0 m and it is rotating with an angular acceleration of 2.5 rad/s².
To Find :
The magnitude of F.
Solution :
Torque on hoop is given by :
( Moment of Inertia of hoop is MR² )
Putting value of M, R and α in above equation, we get :
Therefore, the magnitude of force F is 25 N.
Hence, this is the required solution.
Answer:
23.1 N/C
Explanation:
OP = 3 m , OQ = 4 m
q = - 8 nC, Q = 75 nC
Electric field at P due to the charge Q is
Electric field at P due to the charge q is
According to the diagram, tanθ = 3/4
Resolve the components of E1 along x axis and along y axis.
So, Electric field along X axis, Ex = - E1 Cos θ
Ex = - 27 x 4 / 5 = - 21.6 N/C
Electric field along y axis, Ey = E1 Sinθ - E2
Ey = 27 x 3 /5 - 8 = 8.2 N/C
The resultant electric field at P is given by
Answer:
The airspeed must be 7.78 m/s for the rectangular plate kept at 30°.
Explanation:
By looking at the images below wee see that the airspeed on one side of the rectangular plate decreases the statical pressure over this side. Since over the downside, the pressure still bein the atmospheric pressure. This difference in pressure produces a lift force in the plate. The list force is the net force obtained between the difference of the forces that produce the pressure over the upside and the downside:
Where up and down relate to what movement the forces produce. And p and V are the respective air density and velocity.
When the plate is kept horizontal the lift force balance the moment due to the weight of the plate and considering that both forces act at the same point:
By replacing the known values it is possible to find the plate's weight:
When the plate kept to 30° from the vertical the moment equation balance is written as:
The sine of 30° is due to the weight is 30° oriented, therefore the new value for the airspeed is:
