 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
`        the coefficint of friction is 0.75.if sin 37’=0.6,the angle of friction is?`
5 months ago

```							sin37 = 3/5 so cos37 = root(1 - (3/5)^2) = root( 1- 9/25) = root(16/25) cos37 = 4/5 so tan37 = 3/4 = 0.75 angle of friction t = arc[tan(coefficient of friction)] t = arc[tan(3/4)] t = 37 degree
```
5 months ago
```							sin37 = 3/5 so cos37 = root(1 - (3/5)^2) = root( 1- 9/25) = root(16/25) cos37 = 4/5 so tan37 = 3/4 = 0.75 angle of friction t = arc[tan(coefficient of friction)] t = arc[tan(3/4)] t = 37 degree
```
5 months ago
```							Dear Student, Suppose the compression is x.The kinetic energy before colliding with spring is 12mv2=12×100×1032=15000 JThe potential energy after compression is 12kx2=250x2Frictional force is 0.75×100×10=750 N (taking g=10 ms-2)So energy loss due to friction is Ef=F.x=75xEquating total energy we have750x+150x2=15000⇒x=-12.8, 7.8Here negative sign means expansion. So the answer is 7.8 m. t = arc[tan(3/4)] ,t = 37 degree
```
5 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »  ### Course Features

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

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