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Alex73 [517]
2 years ago
6

A 1500 kg car has an applied forward force of 5000 N and experiences an air resistance of 1250 N. What is the car's acceleration

?
Physics
1 answer:
marshall27 [118]2 years ago
8 0

Answer:

2.33\ m/s^2

Explanation:

<u>Net Force </u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

Fn = m.a

Where a is the acceleration of the object.  The net force is the sum of the individual vector forces applied to the object.

The m=1500 Kg car has two horizontal forces applied: the forward force of 5000 N that causes the movement and the air resistance force of 1250 N that opposes motion.

The net force is Fn = 5000 N - 1500 N = 3500 N

To find the acceleration, we solve the equation for a:

\displaystyle a=\frac{Fn}{m}

\displaystyle a=\frac{3500}{1500}

\boxed{a = 2.33\ m/s^2}

The car's acceleration is a = 2.33\ m/s^2

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8 0
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Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton
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Hello!

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.

Data:

Hooke represented mathematically his theory with the equation:

F = K * Δx  

On what:

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F = K * \Delta{x}

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4\:N/cm*\Delta{x} = 2\:N

\Delta{x} = \dfrac{2\:\diagup\!\!\!\!\!N}{4\:\diagup\!\!\!\!\!N/cm}

simplify by 2

\Delta{x} = \dfrac{2}{4}\frac{\div2}{\div2}

\boxed{\boxed{\Delta{x} = \dfrac{1}{2}\:cm}}\Longleftarrow(distance)\end{array}}\qquad\checkmark

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_______________________

I Hope this helps, greetings ... Dexteright02! =)

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