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MrMuchimi
3 years ago
10

The lift accelerates upwards at 2.5m/s?.

Physics
1 answer:
ipn [44]3 years ago
8 0

The magnitude of force exerted on the woman by the floor when the lift is accelerating is 150 Newton.

<u>Given the following data:</u>

  • Mass of woman = 60 kg.
  • Acceleration = 2.5 m/s^2

To calculate the force exerted on the woman by the floor when the lift is accelerating, we would apply Newton's Second Law of Motion:

<h3>Newton's Second Law of Motion.</h3>

Mathematically, Newton's Second Law of Motion is given by this formula;

Force = mass \times acceleration

Substituting the given parameters into the formula, we have;

Force = 60 \times 2.5

Force = 150 Newton.

Read more on force here: brainly.com/question/1121817

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Did you think about this over Christmas? I did (-: Before Christmas a 65kg student consumes 2500 Cal each day and stays at the s
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Answer:

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

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1. Bone has a Young’s modulus of about
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#1

As we know that

Y = \frac{stress}{strain}

now plug in all data into this

1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}

strain = 9.33 \times 10^{-3}

now from the formula of strain

strain = \frac{\Delta L}{L}

9.33 \times 10^{-3} = \frac{\Delta L}{0.54}

\Delta L = 5.04 \times 10^{-3} m

\Delta L = 5.04 mm

#2

As we know that

pressure * area = Force

here we know that

Area = 3.53 \times 11.6 = 40.95 m^2

P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa

now force is given as

F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N

#3

As we know that density of water will vary with the height as given below

\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}

here we know that

\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa

B = 2.3 \times 10^9 N/m^2

now density is given as

\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}

\rho = 1185.3 kg/m^3

#4

as we know that pressure changes with depth as per following equation

P = P_o + \rho g h

here we know that

P = 3 P_0

now we will have

3P_0 = P_0 + \rho g h

2P_0 = \rho g h

2(1.01 \times 10^5) = 1025 (9.81)(h)

here we will have

h = 20.1 m

so it is 20.1 m below the surface

#5

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

mg = \rho_1V_1g + \rho_2V_2g

A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)

46.6 = 43.89 - 922x + 1000x

x = 3.48 cm

so it is 3.48 cm below the interface

5 0
4 years ago
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