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yaroslaw [1]
3 years ago
15

Beginning 145 miles directly south of the city of Hartville, a car travels due west. If the car is travelling at a speed of 42 m

iles per hour, determine the rate of change of the distance between Hartville and the car when the car has been travelling for 55 miles. (Do not include units in your answer, and round to the nearest hundredth.)

Physics
1 answer:
ziro4ka [17]3 years ago
8 0

Answer:

The rate of change of the distance is 14.89.

Explanation:

Given that,

Distance = 145 miles

Speed of car = 42 miles/hr

Distance covered by car = 55 miles

We need to calculate the the rate of change of the distance

According to figure,

Let OA is x, and AB is y.

Now, using Pythagorean theorem

x^2=y^2+145^2

On differentiating

2x\dfrac{dx}{dt}=2y\dfrac{dy}{dt}

\dfrac{dx}{dt}=\dfrac{y}{x}\dfrac{dy}{dt}

\dfrac{dx}{dt}=\dfrac{55\times42}{\sqrt{55^2+145^2}}

\dfrac{dx}{dt}=14.89\ miles/hr

Hence, The rate of change of the distance is 14.89.

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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

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