The coefficient of static friction is 0.357 and the coefficient of kinetic friction is 0.265.
Explanation:
Coefficient of static friction is defined as the proportionality constant for the frictional force between the crate and floor for starting the motion of crate and normal force acting on the crate. As the normal force of the crate is equal to the influence of acceleration due to gravity acting on the mass of the crate, the frictional force for static friction coefficient will be the force applied to move the crate.
Thus,
Since, the static friction force is 70 N, the normal force is equal to
So normal force is 196 N and static force is 70 N, and the ratio of static friction force to the normal force will give the coefficient of static friction.
Similarly, the coefficient of kinetic friction can be determined from the ratio of kinetic friction force to normal force. Here the kinetic friction force will be equal to the force applied on the crate to keep it moving.
Thus, the coefficient of static friction is 0.357 and the coefficient of kinetic friction is 0.265.
Answer: have "cis C=C double bonds" and "liquid" at room temperature.
Explanation:
The unsaturated fatty acids have one or more C=C double bonds in the cis formation. Thus, this results in the molecules not been as stable as the saturated fats. They have weaker intermolecular bonds thus resulting in lower melting point . The consequently results in it being liquid at room temperature.
Answer:
The remaining percentage of drug concentration is about 88.7% 2 years after manufacture.
Explanation:
Recall the formula for the decay of a substance at an initial concentration at manufacture:
where k is the decay rate (in our case 0.06/year), and t is the elapsed time in years. Therefore, after 2 years since manufacture we have:
This in percent form is 88.7 %. That is, the remaining percentage of drug concentration is about 88.7% 2 years after manufacture.
Answer:
4.8 ton
Explanation:
We are given that
Mass,m=77 ton
Displacement,s=3500 ft
We have to find the force parallel to the incline would be required to hold the monolith on this causeway.
The force parallel to the incline would be required to hold the monolith on this causeway=
Using the formula
The force parallel to the incline would be required to hold the monolith on this causeway=