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muminat
3 years ago
5

If a car is moving on a road at 70 km/hr going due north, and then changes direction and starts traveling north-east staying at

70 km/hr, what happens to its speed and velocity?
answers are in the link guys

Physics
2 answers:
alex41 [277]3 years ago
4 0

Question: If a car is moving on a road at 70 km/hr going due north, and then changes direction and starts traveling north-east staying at 70 km/hr, what happens to its speed and velocity?

Answer: the velocity of the car changes, but the speed stays the same.

Explanation: basically velocity is speed with a direction and speed is the absolute value or magnitude of velocity.

question answered by

(jacemorris04)

Dmitrij [34]3 years ago
4 0

Answer:

Velocity of the car changes while the speed will remain the same

Explanation:

As we know that the speed is the magnitude of rate of change in the position of the object

So here we have

speed = \frac{distance}{time}

while for finding velocity we know that it is rate of change in the position along with the direction of the motion

So it is given as

velocity = \frac{displacement}{time}

so here we can say that magnitude of the velocity in both is same while the direction is different

so correct answer will be

Velocity of the car changes while the speed will remain the same

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A pulled tablecloth exerts a frictional force of 0.6 n on a plate with a mass of 0.4 kg. what is the acceleration of the plate?
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3 years ago
Hello, I wanted an answer from a mathematician. The number 1.04 is closer to the number 1, 2, 1.25 or 1.5.
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Answer:

Explanation:

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2 years ago
A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and
Alexxandr [17]

Answer:

a

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b

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Explanation:

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          The mass of the rock is  m_r  =  200 \ kg

          The  length of the small object from the rock is  d  =  2 \ m

          The  length of the small object from the branch l  =  12 \ m

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The  force exerted by the weight of the rock is mathematically evaluated as

      W =  m_r *  g

substituting values

     W =   200 *  9.8

     W =   1960 \ N

 So  at  equilibrium the sum  of the moment about the fulcrum is mathematically represented as

         \sum  M_f  =  F * cos \theta *  l  -  W cos\theta  *  d =  0

Here  \theta is very small so  cos\theta  *  l  =  l

                               and  cos\theta  *  d  =  d

Hence

       F *   l  -  W  * d =  0

=>    F  = \frac{W * d}{l}

substituting values

        F  = \frac{1960 *  2}{12}

       F  =326.7 \ N

The  mechanical advantage is mathematically evaluated as

          M  = \frac{W}{F}

substituting values

        M  = \frac{1960}{326.7}

       M  = 6

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