Answer:
0.88 cm
Explanation:
Initial length at 10°c, L = 23 m
Rise of temperature, Δt = 42 - 10 = 32°c
Coefficient of linear expansion of concrete, α = 12 x 10^(-6) per°c
Minimum spacing to be left = α L Δt
![=12\times10^{-6}\times23\times32](https://tex.z-dn.net/?f=%3D12%5Ctimes10%5E%7B-6%7D%5Ctimes23%5Ctimes32)
![=8.832\times10^{-3}m](https://tex.z-dn.net/?f=%3D8.832%5Ctimes10%5E%7B-3%7Dm)
= 8.832 x 10^{-1} cm
= 0.88 cm
Note: I have a chosen a general value of 12 x 10^-6 per deg c for coefficient of expansion of concrete. However, please refer to the value given in your text book and substitute it for an accurate answer.
Answer:
![\mu =0.75](https://tex.z-dn.net/?f=%5Cmu%20%3D0.75)
Explanation:
<u>Frictional Force
</u>
When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:
![F_c=m.a_c](https://tex.z-dn.net/?f=F_c%3Dm.a_c)
The centripetal acceleration a_c is computed as
![\displaystyle a_c=\frac{v^2}{r}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a_c%3D%5Cfrac%7Bv%5E2%7D%7Br%7D)
Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one
![F_c=m.\frac{v^2}{r}](https://tex.z-dn.net/?f=F_c%3Dm.%5Cfrac%7Bv%5E2%7D%7Br%7D)
For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as
![F_r=\mu N](https://tex.z-dn.net/?f=F_r%3D%5Cmu%20N)
The normal force N is equal to the weight of the car, thus
![F_r=\mu .m.g](https://tex.z-dn.net/?f=F_r%3D%5Cmu%20.m.g)
Equating both forces
![\displaystyle \mu .m.g=m.\frac{v^2}{r}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cmu%20.m.g%3Dm.%5Cfrac%7Bv%5E2%7D%7Br%7D)
Simplifying
![\displaystyle \mu =\frac{v^2}{rg}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cmu%20%3D%5Cfrac%7Bv%5E2%7D%7Brg%7D)
Substituting the values
![\displaystyle \mu =\frac{19^2}{(49)(9.8)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cmu%20%3D%5Cfrac%7B19%5E2%7D%7B%2849%29%289.8%29%7D)
![\boxed{\mu =0.75}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cmu%20%3D0.75%7D)
Answer:
Plane mirror causes regular reflection of light. When a parallel beam of light is incident on a plane mirror it is reflected back parallely. Due to this property a plane mirror is used to see our image and not a rough surface
A) 300cm/h
B)1 hr=60 min
300/60=5
5cm/min
C)1m=100cm
300/100=3
3m/h