Physics, 03.01.2020 22:31 nicolew647

Did you think about this over christmas? i did (-: before christmas a 65kg student consumes 2500 cal each day and stays at the same weight. for three days in a row while visiting her parents she eats 3500 cal and, wanting to keep from gaining weight decides to "work off" the excess by jumping up and down at the christmas tree. with each jump she accelerates to a speed of 3.2 m/s before leaving the ground. a) how high will she jump each jump? b) how many jumps must she do to keep her weight? assume that the efficiency of the body in using energy is 25%. c) do you suggest that is a reasonable way for the student not to gain weight over christmas? d) possible enhancement: what other way/ways would you suggest for the student to keep her weight?

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a) Em = 332.8 J , b) # jump = 13, c) It is reasonable since there are not too many jumps , d) lower the calories consumed

Explanation:

a) Let's use energy conservation

Initial. On the floor

Em₀ = K = ½ m v²

Final. The highest point

Emf = U = m g h

Energy is conserved

Em₀ = Emf

½ m v² = m g h

h = ½ v² / g

h = ½ 3.2² /9.8

h = 0.52 m

b) When he was at home he maintained his weight with 2500 cal / day. In his parents' house he consumes 3500 cal / day, the excess of calories is

Q = 3500 -2500 = 1000cal / day

Let's reduce this value to the SI system

Q = 1000 cal (4,184 J / 1 cal) = 4186 J / day

Now the energy in each jump is

Em = K = ½ m v²

Em = ½ 65 3.2²

Em = 332.8 J

They indicate that the body can only use 25% of this energy

Em effec = 0.25 332.8 J

Em effec = 83.2 J

This is the energy that burns the body

Let's use a Proportion Rule (rule of three), if a jump spends 83.2J how much jump it needs to spend 1046 J

# jump = 1046 J (1 jump / 83.2 J)

# jump = 12.6 jumps / day

# jump = 13

c) It is reasonable since there are not too many jumps

d) That some days consume more vegetables to lower the calories consumed

Answer from: Quest

amplitude

hope this

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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