A1.0-cm-diameter pipe widens to 2.0, then narrows to 0.5. liquid flows through the first segment at a speed of 4.0. what is the volume flow rate through the pipe?

3.14 ×  10⁻⁴  m³  /s

Explanation:

The flow rate (Q) of a fluid is passing through different cross-sections remains of pipe always remains the same.

Q = Area x velocity

Given:

Diameters of 3 sections of the pipe are given as  

d1  =  1.0  cm,  d2  =  2.0  cm  and  d3  =  0.5  cm.

Speed in the first segment of the pipe is  

v1  =  4  m/s.

From the equation of continuity the flow rate through different cross-sections remains the same.

Flow  rate  =  Q  =  A1  v1  =  A2  v2  =  A3  v3.

Q = A1v1

   =π/4  d²1  v1  =  π/4  * 0.01² ×4.0 m³/s  =  3.14 ×  10⁻⁴  m³  /s

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards