Answer:
Coefficient of friction will be 0.296
Explanation:
We have given initial speed of the stone u = 8 m /sec
It comes to rest so final speed v = 0 m /sec
Distance traveled before coming to rest s = 11 m
According to third equation of motion
![v^2=u^2+2as](https://tex.z-dn.net/?f=v%5E2%3Du%5E2%2B2as)
So ![0^2=8^2+2\times a\times 11](https://tex.z-dn.net/?f=0%5E2%3D8%5E2%2B2%5Ctimes%20a%5Ctimes%2011)
![a=\frac{-64}{22}=-2.90m/sec^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B-64%7D%7B22%7D%3D-2.90m%2Fsec%5E2)
Acceleration due to gravity ![g=9.8m/sec^2](https://tex.z-dn.net/?f=g%3D9.8m%2Fsec%5E2)
We know that acceleration is given by
![a=\mu g](https://tex.z-dn.net/?f=a%3D%5Cmu%20g)
So ![2.90=9.8\times \mu \\](https://tex.z-dn.net/?f=2.90%3D9.8%5Ctimes%20%5Cmu%20%5C%5C)
![\mu =\frac{2.9}{9.8}=0.296](https://tex.z-dn.net/?f=%5Cmu%20%3D%5Cfrac%7B2.9%7D%7B9.8%7D%3D0.296)
So coefficient of friction will be 0.296
Explanation:
Complete the first and second sentences, choosing the correct answer from the given ones.
1. T = 100 K
![^{\circ}C=K-273](https://tex.z-dn.net/?f=%5E%7B%5Ccirc%7DC%3DK-273)
Put T = 100 K
![T=100-273=-173^{\circ} C](https://tex.z-dn.net/?f=T%3D100-273%3D-173%5E%7B%5Ccirc%7D%20C)
A temperature of 100 K corresponds on a Celsius scale to (-173 °C)
2. T = 50 °C
![K=^{\circ}C+273\\\\K=50+273\\\\T=323\ K](https://tex.z-dn.net/?f=K%3D%5E%7B%5Ccirc%7DC%2B273%5C%5C%5C%5CK%3D50%2B273%5C%5C%5C%5CT%3D323%5C%20K)
So, At 50 °C, it corresponds to a Kelvin scale of 323 K.
The Nucleus contains Protons and Neutrons.
The Neutrons does not have a charge.
The Protons are positively charge.
Hence the charge on the Nucleus, would be the charge of the proton, which is positive.
Hence Nucleus is Positively Charged.
Answer:
5.8 kg
Explanation:
We are given that
Force,F=40 N
![\theta=30^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D30%5E%7B%5Ccirc%7D)
Acceleration,a=![2 m/s^2](https://tex.z-dn.net/?f=2%20m%2Fs%5E2)
We have to find the mass of block.
![F_{net}=ma](https://tex.z-dn.net/?f=F_%7Bnet%7D%3Dma)
![F-mgsin\theta=ma](https://tex.z-dn.net/?f=F-mgsin%5Ctheta%3Dma)
![F=mgsin\theta+ma=m(gsin\theta+a)](https://tex.z-dn.net/?f=F%3Dmgsin%5Ctheta%2Bma%3Dm%28gsin%5Ctheta%2Ba%29)
![40=m(9.8sin30+2)](https://tex.z-dn.net/?f=40%3Dm%289.8sin30%2B2%29)
Where
![g=9.8m/s^2](https://tex.z-dn.net/?f=g%3D9.8m%2Fs%5E2)
![40=6.9m](https://tex.z-dn.net/?f=40%3D6.9m)
![m=\frac{40}{6.9}=5.8 kg](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B40%7D%7B6.9%7D%3D5.8%20kg)
Answer:
8.7 m/s
Explanation:
Given data
Mass= 0.6kg
radius= 0.9m
Force= 50N
A. Please see attached the FBD of the stone
B. Apply the formula
F= mv^2/r
Substitute
50= 0.6*v^2/0.9
cross multiply
50*0.9= 0.6v^2
45= 0.6v^2
v^2= 45/0.6
v^2= 75
v= √75
v= 8.7 m/s
Hence the maximum speed attained without breaking the string is 8.7 m/s