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ladessa [460]
3 years ago
13

After a while, the car started to go around a long bend, still maintaining its constant speed of 55 miles per hour. Is there a n

et (unbalanced) force acting on the car? 1. No; the speed of the car does not change. 2. Cannot be determined 3. No; the velocity of the car does not change. 4. Yes; its direction is not the same as the velocity. 5. Yes; its direction is the same as the velocity.
Physics
1 answer:
sukhopar [10]3 years ago
6 0

Answer:

4) True. The change of direction needs an unbalanced force

Explanation:

Let us propose the resolution of the problem using Newton's second law.

    F = m a

As the car is spinning the acceleration is centripetal

    a = v2.r

   

    F = m v2 / r

We can see that as the velocity of a vector even if its module does not change, the change of direction requires an external force.

Now we can analyze the statement if they are true or false

1) and 3) False, even when the speed changes, the direction changes

2) False with the speed change can be determined

4) True. The change of direction needs an unbalanced force

5) False are different things. the direction is where it is going and the speed is the magnitude of the vector

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2 years ago
A rock is tossed straight up from the ground with a speed of 21 m/s . When it returns, it falls into a hole 10 m deep.a.) What i
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(a) 25.2 m/s

Let's take the initial vertical position of the rock as "zero" (reference height).

According to the law of conservation of energy, the speed of the rock as it reaches again the position "zero" after being thrown upwards is equal to the initial speed of the rock, 21 m/s (in fact, if there is no air resistance, no energy can be lost during the motion; and since the kinetic energy depends only on the speed of the rock:

K=\frac{1}{2}mv^2

and the gravitational potential energy of the rock has not changed, since the rock has returned into its initial position, it means that the speed of the rock should be the same)

This means that we can only analyze the final part of the motion, the one in which the rock falls into the 10 m hole. Since it is a free fall motion, we can find the final speed by using

v^2 = u^2 + 2gd

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u = 21 m/s is the initial speed of the rock as it enters the hole

g = 9.8 m/s^2 is the acceleration due to gravity

d = 10 m is the depth of the hole

Substituting,

v=\sqrt{u^2 +2gd}=\sqrt{(21 m/s)^2+2(9.8 m/s^2)(10 m)}=25.2 m/s

(b) 4.72 s

The vertical position of the rock at time t is given by

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where

v_y = 21 m/s is the initial vertical velocity

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-10 = 21 t - \frac{1}{2}(9.8)t^2\\10+21 t -4.9t^2 = 0

which has two solutions:

t = -0.43 s --> negative, so we discard it

t = 4.72 s --> this is our solution

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