Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        A flat plate moves normally with a speed v` towards a horizontal jet of water of uniform area of cross section . The jet discharges water at the rate of volume V per second at a speed of v° . The density of water is ¶ .assume that water splashes along the surface of the plate at right angles to the original motion . The magnitude of the force acting on the plate due to the jet of water is `
2 months ago

Vikas TU
9468 Points
```							Length of liquid jet hitting wall / sec = v1 + v2 Vol of liquid cong out / sec V = Area of pipe * v2 Vol of liquid hitting wall /sec = V/v2 *(v1 +v2)Momentum of liquid hitting wall per second is P => V/v2 *(v1 +v2)¶ * (v1+v2) = (v1+v2)^2 Vp /v2 Finaal momentum perpendicular to wall =0 Force is the change in momentum / sec F = (v+v2)^2 Vp /v2
```
2 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on General Physics

View all Questions »

### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 18 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions

Post Question