Jogger moves in three displacements
d1 = 10 blocks East
d2 = 5 blocks South
d3 = 2 blocks East
now we can say
total displacement towards East direction will be

Total displacement towards South

now to find the net displacement we can use vector addition



<em>so magnitude of net displacement will be equal to 13 blocks</em>
Answer:
The algebraic equation is:

Explanation:
Given information:
mb = book's mass
vb = tangential speed
R = radius of the path
Question: Derive an algebraic equation for the vertical force, Fv = ?
To derive the equation, we need to draw a force diagram for this case, please, see the attached diagram. As you can see, there are three types of forces acting on the system. Two up and one of the weight acting down. Therefore, the algebraic equation is as follows:

The variables were defined above and g is the gravity.
The answer for this problem is clarified through this, the
system is absorbing (+). And now see that it uses that the SURROUNDINGS are
doing 84 KJ of work. Any time a system is overshadowing work done on it by the
surroundings the sign will be +. So it's just 12.4 KJ + 4.2 = 16.6 KJ.
when the apple moves in a horizontal circle, the tension force in the string provides the necessary centripetal force to move in circle. the tension in the string is given as
T=mv²/r
where T = tension force in the string , m = mass of the apple
v = speed of apple , r = radius of circle.
clearly , tension force depends on the square of the speed. hence greater the speed, greater will be the tension force.
at some point , the speed becomes large enough that it makes the tension force in the string becomes greater than the tensile strength of the string. at that point , the string breaks
Answer:
75.6J
Explanation:
Hi!
To solve this problem we must use the first law of thermodynamics that states that the heat required to heat the air is the difference between the energy levels of the air when it enters and when it leaves the body,
Given the above we have the following equation.
Q=(m)(h2)-(m)(h1)
where
m=mass=1.3×10−3kg.
h2= entalpy at 37C
h1= entalpy at -20C
Q=m(h2-h1)
remember that the enthalpy differences for the air can approximate the specific heat multiplied by the temperature difference
Q=mCp(T2-T1)
Cp= specific heat of air = 1020 J/kg⋅K
Q=(1.3×10−3)(1020)(37-(-20))=75.6J