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morpeh [17]
4 years ago
13

A box rests on a horizontal, frictionless surface. a girl pushes on the box with a force of 17 n to the right and a boy pushes o

n the box with a force of 13 n to the left. the box moves 3.5 m to the right. find the work done by (a) the girl, (b) the boy, and (c) the net force.

Physics
1 answer:
Komok [63]4 years ago
5 0
Refer to the diagram shown below.

The net force acting on the box is 17 - 13 = 4 N to the right.
The box moves on a friction surface by 3.5 m to the right.

By definition,
Work = Force x Distance.

(a) The work done by the girl is
     W₁ = (17 N)*(3.5 m) = 59.5 J

(b) The work done by the boy is
     W₂ = (13 N)*(-3.5 m) = - 45.5 J

(c) The work done by the net force  is
    W₃ = (4 N)*(3.5 m) = 14 J

Note that W₃ = W₁ + W₂

Answers:
(a) 59.5 J
(b) - 45.5 J
(c) 14 J

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The temperature change is \frac{dT}{dt} = 1.016 ^oC/m

Explanation:

From the question we are told that

   The velocity field with which the bird is flying is  \vec V =  (u, v, w)= 0.6x + 0.2t - 1.4 \ m/s

   The temperature of the room is  T(x, y, u) =  400 -0.4y -0.6z-0.2(5 - x)^2 \  ^o C

    The time considered is  t =  10 \  seconds

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 Given that the bird is inside the room then the temperature of the room is equal to the temperature of the bird

Generally the change in the bird temperature with time is mathematically represented as

      \frac{dT}{dt} = -0.4 \frac{dy}{dt} -0.6\frac{dz}{dt} -0.2[2 *  (5-x)] [-\frac{dx}{dt} ]

Here the negative sign in \frac{dx}{dt} is because of the negative sign that is attached to x in the equation

 So

       \frac{dT}{dt} = -0.4v_y  -0.6v_z -0.2[2 *  (5-x)][ -v_x]

From the given equation of velocity field

    v_x  =  0.6x

    v_y  =  0.2t

     v_z  =  -1.4

So

\frac{dT}{dt} = -0.4[0.2t]  -0.6[-1.4] -0.2[2 *  (5-x)][ -[0.6x]]    

substituting the given values of x and t

\frac{dT}{dt} = -0.4[0.2(10)]  -0.6[-1.4] -0.2[2 *  (5-1)][ -[0.61]]      

\frac{dT}{dt} = -0.8 +0.84 + 0.976  

\frac{dT}{dt} = 1.016 ^oC/m  

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Answer:

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Answer:

Ratio of resistance of heater A to resistance of heater B is 5.80

Explanation:

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Mass of water in heater B, m₂ = 5.80 L

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Final temperature, T₁ = 90 ⁰C

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Amount of heat required to raise the temperature of water by heaters A and B are given by:

Q₁ = m₁C(T₁ - T₀)       and  

Q₂ = m₂C(T₁ - T₀)

Ratio of power used by both the heaters A and B is:

\frac{P_{1} }{P_{2} } =\frac{Q_{1} }{t} \times\frac{t}{Q_{2} }

Since, time t, temperature difference(T₁ - T₀) and specific heat C are same for both the heaters A and B. So, the above equation becomes:

\frac{P_{1} }{P_{2} } =\frac{m_{1} }{m_{2} }    ...(1)

The relation to determine electrical power for both heaters A and B are:

P_{1}=\frac{V^{2} }{R_{1} }     and

P_{2}=\frac{V^{2} }{R_{2} }

Here V is the voltage applied to both the heaters and is equal.

So, the ratio of electrical power of heaters is:

\frac{P_{1} }{P_{2} } =\frac{R_{2} }{R_{1} }     ....(2)

But according to the problem, the electrical power is converted into the thermal power. So,equation (1) and (2) are equal. Hence,

\frac{m_{1} }{m_{2} } =\frac{R_{2} }{R_{1} }

Substitute the suitable values in the above equation.

\frac{1 }{5.80 } =\frac{R_{2} }{R_{1} }

\frac{R_{1} }{R_{2} }=5.80

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