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anzhelika [568]
3 years ago
7

A flatbed truck is carrying an 800-kg load of timber that is not tied down. The maximum friction force between the truck bed and

the load is 2400 N. What is the greatest acceleration that the truck can have without losing its load
Physics
1 answer:
ycow [4]3 years ago
4 0

Answer:

Acceleration a ≤ 3 m/s^2

the greatest acceleration that the truck can have without losing its load is 3 m/s^2

Explanation:

For the truck to accelerate without losing its load.

Acceleration force of truck must be less than or equal to the maximum friction force between the truck bed and the load.

Fa ≤ F(friction)

But;

Fa = mass × acceleration

Fa = ma

ma ≤ F(friction)

a ≤ (F(friction))/m ......1

Given;

Fa = mass × acceleration

Fa = ma

mass m = 800 kg

F(friction) = 2400 N

Substituting the given values into equation 1;

a ≤ F(friction)/m

a ≤ 2400N/800kg

a ≤ 3 m/s^2

the greatest acceleration that the truck can have without losing its load is 3 m/s^2

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A)F_0d

Explanation

If you graph the force on an object as a function of the position of that object, then the area under the curve will equal the work done on that object, so we need to find the area under the function to find the work

Step 1

find the area under the function.

so

Area:

\text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red}\begin{gathered} \text{the area of a rectangle is given by} \\ A_{rec}=lenght\cdot widht \\ \text{and} \\ \text{the area of a triangle is given by:} \\ A_{tr}=\frac{base\cdot height}{2} \end{gathered}

so

\begin{gathered} \text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red} \\ \text{replace} \\ \text{Area}=(F_0\cdot d)+\frac{(F_0\cdot d)}{2}-\frac{(F_0\cdot d)}{2} \\ \text{Area}=(F_0\cdot d) \\ Area=F_0d \end{gathered}

therefore, the answer is

A)F_0d

I hope this helps you

4 0
1 year ago
What causes atoms to bond together<br> ?!?
lisov135 [29]

Answer:

electrostatic attraction

Explanation:

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7 0
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Snezhnost [94]

1,300 or more.

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When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal speed. Once he has reache
Masteriza [31]

Answer:

b) True.    the force of air drag on him is equal to his weight.

Explanation:

Let us propose the solution of the problem in order to analyze the given statements.

The problem must be solved with Newton's second law.

When he jumps off the plane

     fr - w = ma

Where the friction force has some form of type.

     fr = G v + H v²

Let's replace

     (G v + H v²) - mg = m dv / dt

We can see that the friction force increases as the speed increases

At the equilibrium point

      fr - w = 0

      fr = mg

      (G v + H v2) = mg

For low speeds the quadratic depended is not important, so we can reduce the equation to

     G v = mg

     v = mg / G

This is the terminal speed.

Now let's analyze the claims

a) False is g between the friction force constant

b) True.

c) False. It is equal to the weight

d) False. In the terminal speed the acceleration is zero

e) False. The friction force is equal to the weight

3 0
3 years ago
A 5 kg ball is suspended on one cable. Calculate the tension in the cable
ludmilkaskok [199]

Answer:

49N

Explanation:

F=ma

We know the mass is 5kg, and since the ball is suspended on one cable, the acceleration is g, 9.8m/s^2

F=5kg*9.8m/s^2

 = 49N

Hope this helps!

7 0
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