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:
The man is on the verge of having a heart attack or a stroke.
Explanation:
If he has a family history of coronary (heart) disease, it means it could normally affect. Normally here means without anything aggravating it. It's already in his lineage so he could have it.
Now, he's past middle age - he's 45. He's past the growing stages of life. His organs are fully developed herefore.
Now also, he suffers from Type 2 diabetes. Although this is sometimes milder than Type 1 diabetes, it increases the risk of having a heart disease or a stroke!
Soda, especially sweetened one, is not to be taken too often because it can cause Diabetes Mellitus. For a diabetes patient, this should be a "no-go-area". Taking this constantly (everyday at work) will now put this 45-year-old man in harm's way.
He is no more at risk of having complications but already on the path to a heart disease or a stroke.
Answer:
4.5sec
Explanation:
From the question above, the following are the parameters that are given
u= 30m/s
v= 50m/s
s= 180m
First of all we have to find the acceleration by using the third equation of motion
V^2= U^2 + 2as
50^2= 30^2 + 2(a)(180)
2500= 900 + 360a
Collect the like terms
2500-900= 360a
1600=360a
Divide both sides by the coefficient of a which is 360
1600/360=360a/360
a= 4.44m/s
The next step is to find the time. To do this we will have to use the first equation of motion
v= u + at
50= 30 + 4.44t
Collect the like terms
50-30= 4.44t
20= 4.44t
Divide both sides by the coefficient of t which is 4.44
20/4.44= 4.44t/4.44
t= 4.5sec
Hence 4.5secs elapses while the auto moves at a distance of 180m
Answer:
Explanation:
The question relates to motion on a circular path .
Let the radius of the circular path be R .
The centripetal force for circular motion is provided by frictional force
frictional force is equal to μmg , where μ is coefficient of friction and mg is weight
Equating cenrtipetal force and frictionl force in the case of car A
mv² / R = μmg
R = v² /μg
= 26.8 x 26.8 / .335 x 9.8
= 218.77 m
In case of moton of car B
mv² / R = μmg
v² = μRg
= .683 x 218.77x 9.8
= 1464.35
v = 38.26 m /s .
Answer:
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Explanation:
