Answer:
The speed of Susan is 2.37 m/s
Explanation:
To visualize better this problem, we need to draw a free body diagram.
the work is defined as:
here we have the work done by Paul and the friction force, so:
Now the change of energy is:
When you are finding work, the easiest way is to use the formula.
W = F*D
Where F is the force and D is the distance. Simply take the constant force of 209N and multiply it by the distance of 10m. Which will give you 2090J
<u>Answer</u>:
The coefficient of static friction between the tires and the road is 1.987
<u>Explanation</u>:
<u>Given</u>:
Radius of the track, r = 516 m
Tangential Acceleration = 3.89 m/s^2
Speed,v = 32.8 m/s
<u>To Find:</u>
The coefficient of static friction between the tires and the road = ?
<u>Solution</u>:
The radial Acceleration is given by,
Now the total acceleration is
=>
=>
=>
=>
The frictional force on the car will be f = ma------------(1)
And the force due to gravity is W = mg--------------------(2)
Now the coefficient of static friction is
From (1) and (2)
Substituting the values, we get
The mass number of an isotope is the sum of the numbers of protons and the numbers of the neutron. Given that the mass number is 65 and the number of neutrons is 35, the number of protons is 30. The atom is then Zinc (Zn). The charge is equal to +2, as it lacks 2 more electrons.
Answer:
the answer is that the dough has the same mass before and after it was flattened