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emmasim [6.3K]
3 years ago
14

A flatbed truck is carrying a 20.0 kg crate along a level road. the coefficient of static friction between the crate and the bed

is 0.400, and the coefficient of kinetic friction is 0.300. what is the maximum acceleration that the truck can have if the crate is to stay in place relative to the truck?
Physics
1 answer:
Dahasolnce [82]3 years ago
6 0
<span>3.92 m/s^2 Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so 0.4 * 9.8 m/s^2 = 3.92 m/s^2 If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so 20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N Multiply by the coefficient of static friction, giving 196 N * 0.4 = 78.4 N So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass 78.4 N / 20.0 kg = 78.4 kg*m/s^2 / 20.0 kg = 3.92 m/s^2 And you get the same result.</span>
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When we jump from the truck and accelerate towards the earth surface, the earth also accelerates towards us but it's acceleration is very negligible.

To find the answer, we need to know about the acceleration of earth due to the gravitational attraction.

<h3>What's the gravitational force between the earth and a person?</h3>
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m= mass of the person

r= separation between them.

<h3>What's the acceleration of the earth towards the person when he jumps from a truck?</h3>
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so acceleration of earth= Gm/r²

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Thus, we can conclude that the earth also accelerates towards us.

Learn more about the gravitational force here:

brainly.com/question/72250

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7 0
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What is the magnitude of the force a charge 25uc exerts on a charge 3mc 35 cm away?
Misha Larkins [42]
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3 years ago
In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for
4vir4ik [10]

Answer:

The  fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is  

      k  = 0.903

Explanation:

From the question we are told that

     The time  constant  \tau  =  3

The potential across the capacitor can be mathematically represented as

     V  =  V_o  (1 -  e^{- \tau})

Where V_o is the voltage of the capacitor when it is fully charged

    So   at  \tau  =  3

     V  =  V_o  (1 -  e^{- 3})

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   Generally energy stored in a capacitor is mathematically represented as

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In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now  since capacitance is  constant  at  \tau  =  3

        The  energy stored can be evaluated at as

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       V^2 =  0.903  V_o ^2

Hence the fraction of the energy stored in an initially uncharged capacitor is  

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