The coefficient of static friction is 0.234.
Answer:
Explanation:
Frictional force is equal to the product of coefficient of friction and normal force acting on any object.
So here the mass of the object is given as 2 kg, so the normal force will be acting under the influence of acceleration due to gravity.
Normal force = mass * acceleration due to gravity
Normal force = 2 * 9.8 = 19.6 N.
And the frictional force is given as 4.6 N, then
![Coefficient of static friction = Frictional force/Normal force](https://tex.z-dn.net/?f=Coefficient%20of%20static%20friction%20%3D%20Frictional%20force%2FNormal%20force)
Coefficient of static friction = 4.6 N / 19.6 N = 0.234
So the coefficient of static friction is 0.234.
It causes a movement of convection in the water resulting in a pulling current.
Hope this helpss
The macromolecule that is made up of subunits that are called amino acids are the protein. The answer is therefore letter D. The carbohydrates are made up of subunits called monosaccarides. The lipids are mades up of subunits referred to as the fatty acid while nucleic acid are made up of nucleotides.
Change in displacement per unit of time (Ex. 25m/s East)
Answer:
42.99°
Explanation:
= Kinetic friction force
= Pulling force at angle ![\theta](https://tex.z-dn.net/?f=%5Ctheta)
= Weight of the box = 150 N
Kinetic friction force
![F_h=\muN_h](https://tex.z-dn.net/?f=F_h%3D%5CmuN_h)
Pulling force at angle ![\theta](https://tex.z-dn.net/?f=%5Ctheta)
![F_{\theta}=\muN_{\theta}](https://tex.z-dn.net/?f=F_%7B%5Ctheta%7D%3D%5CmuN_%7B%5Ctheta%7D)
N = Pulling force
According to question
![\frac{F_h}{F_{\theta}}=\frac{2}{1}\\\Rightarrow \frac{\muN_h}{\muN_{\theta}}=2\\\Rightarrow \frac{N_h}{N_{\theta}}=2\\\Rightarrow N_{\theta}=\frac{N_h}{2}\\\Rightarrow N_{\theta}=\frac{150}{2}\\\Rightarrow N_{\theta}=75\ N](https://tex.z-dn.net/?f=%5Cfrac%7BF_h%7D%7BF_%7B%5Ctheta%7D%7D%3D%5Cfrac%7B2%7D%7B1%7D%5C%5C%5CRightarrow%20%5Cfrac%7B%5CmuN_h%7D%7B%5CmuN_%7B%5Ctheta%7D%7D%3D2%5C%5C%5CRightarrow%20%5Cfrac%7BN_h%7D%7BN_%7B%5Ctheta%7D%7D%3D2%5C%5C%5CRightarrow%20N_%7B%5Ctheta%7D%3D%5Cfrac%7BN_h%7D%7B2%7D%5C%5C%5CRightarrow%20N_%7B%5Ctheta%7D%3D%5Cfrac%7B150%7D%7B2%7D%5C%5C%5CRightarrow%20N_%7B%5Ctheta%7D%3D75%5C%20N)
Applying Newton's second law in the vertical direction we get
![N_h-Nsin\theta=N_{\theta}\\\Rightarrow 150-110sin\theta=75\\\Rightarrow \theta=sin^{-1}\frac{75}{110}\\\Rightarrow \theta=42.99\ ^{\circ}](https://tex.z-dn.net/?f=N_h-Nsin%5Ctheta%3DN_%7B%5Ctheta%7D%5C%5C%5CRightarrow%20150-110sin%5Ctheta%3D75%5C%5C%5CRightarrow%20%5Ctheta%3Dsin%5E%7B-1%7D%5Cfrac%7B75%7D%7B110%7D%5C%5C%5CRightarrow%20%5Ctheta%3D42.99%5C%20%5E%7B%5Ccirc%7D)
The angle is 42.99°