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riadik2000 [5.3K]
1 year ago
15

another way, speed is a scalar value, while reloolg isa Which ONE of the following represents a toy car that moves at a higher a

verage speed? (12) Show your calculations. A: A toy car that travels 1,50 km in 3 min B: A toy car that travels 800 m in 1,50 min C: A toy car that travels 250 km in 8 hours CAR A: CAR B: CAR C: ​
Physics
1 answer:
erica [24]1 year ago
4 0

Car  A will have highest speed is 83.3m/s .

<h3>What is speed ? </h3>

The rate of change of position of an object in any direction.

The S.I unit is m/s . Speed is a scalar quantity it defines only magnitude not direction

.

speed = distance /time

In case of Car A ,

We have given distance 150Km in 3 min ,

First we have convert the distance km to m

150×1000m

then conversion of min to sec

 38×60sec

speed = 15000/180

speed = 83.3m/sec

In case of Car B

we have given 800m in 150 min

lets convert the time into second

150×60

Speed = 800/150×60

speed = 0.88m/ s

In case of Car C

We have given here distance 250 Km and time in 8 hours

convert km to m

25000

and time into sec

88×60×60

speed = 0.86m/ s



Hence ,Car A has highest speed amongst them .

To learn more about speed click here

brainly.com/question/7359669

#SPJ9

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Explanation:

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Answer: 4.8 s

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We have to find the time t

Well, according to Newton's second law of motion we have:

F=m.a (1)

Where a is the acceleration, which can be expressed as:

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2 years ago
Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,
Marrrta [24]

Answer:

a

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b

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   Stranded  Wire  R_1  = 0.0189 \ \Omega

Explanation:

Considering the first question

From the question we are told that

  The  radius of the first wire is  r_1  = 1.53 mm = 0.0015 \  m

  The radius of  each strand is  r_0 =  0.306 \ mm =  0.000306 \ m

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Considering the first wire

     The  cross-sectional area of the first wire is

      A   = \pi  r^2

= >  A   = 3.142 *  (0.0015)^2

= >  A   = 7.0695 *10^{-6} \  m^2

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=>  I  =  1750*7.0695 *10^{-6}

=>  I  =   0.01237 \  A

Considering the second wire  wire

The  cross-sectional area of the second wire is

     A_1  =  19 *  \pi r^2

=>     A_1  =  19 *3.142 *  (0.000306)^2

=>  A_1  =  5.5899 *10^{-6} \  m^2

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     Resistivity is  \rho  =  1.69* 10^{-8} \Omega \cdot m

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Generally the resistance of the first wire is mathematically represented as

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Generally the resistance of the first wire is mathematically represented as

    R_1  =  \frac{\rho *  l  }{A_1}

=> R_1  =  \frac{  1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }

=> R_1  = 0.0189 \ \Omega

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