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Luden [163]
2 years ago
15

Suppose the truck that’s transporting the box In Example 6.10 (p. 150) is driving at a constant speed and then brakes and slows

at a constant acceleration. While coming to a stop, the driver looks in the rear-view mirror and notices that the box is not slipping. In what direction is the frictional force acting on the box?
Physics
1 answer:
Scorpion4ik [409]2 years ago
5 0

Answer:

Friction acts in the opposite direction to the motion of the truck and box.

Explanation:

Let's first review the problem.

A moving truck applies the brakes, and a box on it does not slip.

Now when the truck is applying brakes, only it itself is being slowed down. Since the box is slowing down with the truck, we can conclude that it is friction that slows it down.

The box in the question tries to maintains its velocity forward when the brakes are applied. We can think of this as the box exerting a positive force relative to the truck when the brakes are applied. When we imagine this, we can also figure out where the static friction will act to stop this positive force. Friction will act in the negative direction. Or in other words, friction will act in the opposite direction to the motion of the truck and box. This explains why the box slows down with the truck, as friction acts to stop its motion.

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<span>Hello!
 
We have the following data:
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Time (T) = ? (in minutes)
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Formula of the consumption of electric energy:

P =  \frac{E}{T}

Solving:

P = \frac{E}{T}

P = \frac{E}{T} \to T =  \frac{E}{P}

T =  \frac{9000\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W/s}{3\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W}

\boxed{T = 3000\:seconds}

How many minutes can it run for? (<span>Let's convert in minutes)
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1 minute --------- 60 seconds
y minute --------- 3000 seconds

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<span>Product of extremes equals product of means
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60*y = 1*3000

60y = 3000

y =  \frac{3000}{60}

\boxed{\boxed{y = 50\:minutes}}\end{array}}\qquad\quad\checkmark


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crimeas [40]

<u>Answer:</u>

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<u>Explanation:</u>

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Comparing both the equations

    We will get initial velocity = 14 m/s(already given) and \frac{1}{2} a = -1.86

     So,  Acceleration, a = -3.72 m/s^2

 Now we have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

 When time is 2 seconds we need to find final velocity.

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  So, Velocity of rock after 2 seconds = 6.56 m/s  

6 0
2 years ago
An electrical conductor is an element with __________ electrons in its outer orbit.
Setler [38]
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L_o = 89.65m

6 0
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