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geniusboy [140]
3 years ago
10

A book is moved once around the edge of a tabletop with dimensions 1.75 m à 2.25 m. If the book ends up at its initial position,

what is its displacement? If it completes its motion in 23 s, what is its average velocity? What is its average speed?
Physics
1 answer:
uysha [10]3 years ago
6 0

Explanation:

The dimension of the book is 1.75 m × 2.25 m. If the book ends up at its initial position. The displacement of the book is equal to zero as the object reaches to its initial position.

If it completes its motion in 23 s, t = 23 s

Total displacement of the book is equal to its perimeter. It is given by :

d=2(1.75+2.25)=8\ m

The net displacement divided by total time taken is called the average velocity of an object. Here, the displacement is 0. So, average velocity is 0.

The average speed of an object is given by :

v=\dfrac{d}{t}

v=\dfrac{8\ m}{23\ s}

v = 0.347 m/s

So, the average speed of the book is 0.347 m/s. Hence, this is the required solution.

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What are the average velocity and average acceleration of the tip of a 2.4 cm long hour hand of a clock?
ra1l [238]
A circle has a revolution of 360°. Since there are 12 hour markings, each hour interval has an angle of 30°. In radians, that would be equal to π/6 radians. So, in every 1 hour that passes, it covers π/6 of an angle. So, the angular velocity denoted as ω is π/6 ÷ 1 hour = π/6 rad/h. We can compute the average linear velocity, v, from the relationship:

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Therefore, the average velocity is 1.257 cm per hour.

For the average acceleration, it is equal to zero. The hands of the clock move at a constant velocity. Since acceleration is the change of velocity per unit time, there is no change of velocity because it's constant. That's why it is zero.

8 0
3 years ago
A point P1 is located by the vector A = (3.74)î + (1.64)ĵ and a point P2 is located by the vector B = (1.60)î + (3.66)ĵ. The vec
seropon [69]

Answer:

\vec{C} = ( \ - 2.14 \ , \ 2.02 \ )

Explanation:

The vector that point from point P1 to point P2 its found simply by taking the vector at which point P2 its located and subtracting the vector at which point P1 its located:

\vec{C} = \vec{B} - \vec{A}

So:

\vec{C} = ( \ 1.60 \ , \ 3.66 \ ) - ( \ 3.74 \ , \ 1.64 \ )

\vec{C} = ( \ 1.60 \ - \ 3.74 \ , \ 3.66 \ - \ 1.64 \ )

\vec{C} = ( \ - 2.14 \ , \ 2.02 \ )

4 0
3 years ago
You throw a bouncy rubber ball and a wet lump of clay, both of mass m, at a wall. Both strike the wall at speed v, but while the
lana [24]

Answer:

<em>The fifth option is the correct answer: mv; 2 mv</em>

Explanation:

<u>Change of Momentum</u>

Assume an object has a momentum p1 and after some interaction it now has a momentum p2, the change of momentum is

\Delta p=p_2-p_1

The momentum is computed as

p=mv

Where m is the mass of the object and v its speed. Now let's analyze the situation of both the ball and the clay.

The clay has an initial speed v and a mass m, thus its initial momentum is

p_1=mv

When it hits the wall, it sticks, thus its final speed is 0 and

p_2=0

The change of momentum is

\Delta p=0-mv=-mv

The absolute change is mv

Now for the ball, the initial condition is the same as it was for the clay, but the ball hits back at the same speed, thus its final momentum is

p_2=-mv

The change of momentum is

\Delta p=-mv-mv=-2mv

The absolute change is 2mv

The fifth option is the correct answer: mv; 2 mv

3 0
3 years ago
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