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GaryK [48]
4 years ago
7

On a horizontal frictionless floor, a worker of weight 0.900 kN pushes horizontally with a force of 0.200 kN on a box weighing 1

.8 kN. As a result of this push, which statement could be true?
A) The worker will accelerate at 1.08 m/s^2 and the box will accelerate at 2.17 m/s^2, but in opposite directions.
B) The box will not move because the push is less than the weight.
C) The worker and box will bot have an acceleration of 2.17 m/s^2, but in opposite directions.
D) The worker will accelerate at 2.17 m/s^2 and the box will accelerate at 1.08 m/s^2, but in opposite directions.
E)The worker and box will bot have an acceleration of 1.08 m/s^2, but in opposite directions.
Physics
1 answer:
Ad libitum [116K]4 years ago
5 0

Answer:

D) The worker will accelerate at 2.17  m/s²  and the box will accelerate at 1.08  m/s² , but in opposite directions.

Explanation:

Newton's third law

Newton's third law or principle of action and reaction states that when two interaction bodies appear equal forces and opposite directions. in each of them.

F₁₂= -F₂₁

F₁₂: Force of the box on the worker

F₂₁: Force of the worker on the box

Newton's second law

∑F = m*a

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Formula to calculate the mass (m)

m =  W/g

Where:

W : Weight (N)

g : acceleration due to gravity  (m/s²)

Data

W₁ =1.8 kN   : box weight

W₂ = 0.900 kN : worker weight

F₂₁ = 0.200 kN

F₁₂ = - 0.200 kN

g = 9.8 m/s²

Newton's second law for the box

∑F = m*a

F₂₁ = m₁*a₁    m₁=W₁/g

0.2 kN = (1.8kN)/(9.8 m/s² ) *a₁

a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{1.8 kN}

a₁= 1.08 m/s² : acceleration of the box

Newton's second law for the worker

∑F = m*a

F₁₂ = m₂*a₂ , m₂=W₂/g

- 0.2 kN =( (0.9 kN) /(9.8 m/s² ) )*a₂

a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{0.9 kN}

a₂=  -2.17 m/s² : acceleration of the worker

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Answer:

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Explanation:

To solve this circuit we are going to find the equivalent resistance of each branch, let's remember

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         R_{eq} = ∑ R_{i}

* For resistance in parallel

        1 / R_{eq} = ∑ 1/R_{i}

We solve the two branches of the wheatstone bridge

Series resistors

Branch B

         R_B = Rb + R4

         R_B = 2 + 18

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Branch C

         R_C5 = Rc + R5

         R_C5 = 3 + 12

         R_C5 = 15 Ω

Resistance in parallel R_B and R_C5

         1 / R_BC = 1 / R_B + 1 / R_C5

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Let's remember that the voltage in a series circuit is the sum of the voltages

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           i = 10 / (R_a + R_BC)

           i = 10 / (14.57)

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             V_C = i R_C

              V_C = 0.39213 3

              V_C = 1.176 V

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