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3 years ago

A empty truck whose mass is

Physics

1 answer:

trapecia [35]3 years ago

8 0

Answer:

$0.67 m/s^2$

Explanation:

We can solve the problem by using Newton's second law:

$F=ma$

where

F is the net force on the truck

m is the mass of the truck+load

a is the acceration

At the beginning, the mass of the truck is

m = 2000 kg

and its acceleration is

$a=1 m/s^2$

So, the maximum force that the engine of the truck can produce is

$F=ma=(2000)(1)=2000 N$

Later, a load of 1000 kg is put on the truck, so the new mass is

m = 2000 + 1000 = 3000 kg

The maximum force that the engine can produce is the same, so the new maximum acceleration of the truck will be:

$a=\frac{F}{m}=\frac{2000 N}{3000 kg}=0.67 m/s^2$

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3 years ago

The brakes on a 15,680 N car exert a stopping force of 640 N.

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3 years ago

A race car moves along a circular track at a speed of 0.512 m/s. If the car's centripetal acceleration is 15.4 m/s2, what is the

tiny-mole [99]

Answer:

The radius is $r = 0.0170 \ m$

Explanation:

From the question we are told that

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3 years ago

At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constan

Jet001 [13]

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At time $t_1$ , the acceleration vector has direction $\theta_1$ such that

$\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ$

which indicates the particle is situated at a point on the lower left half of the circle, while at time $t_2$ the acceleration has direction $\theta_2$ such that

$\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ$

which indicates the particle lies on the upper left half of the circle.

Notice that $\theta_1-\theta_2=90^\circ$ . That is, the measure of the major arc between the particle's positions at $t_1$ and $t_2$ is 270 degrees, which means that $t_2-t_1$ is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

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$t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s$

and the magnitude of the particle's acceleration toward the center of the circle is

$\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}$

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3 years ago

This model shows a ball being shot from the cannon. Which of these statements accurately describes the energy of the cannonball?

Fudgin [204]

The potential energy of the model will be the greatest at point C while it will be the least at point E. Option B and D are correct.

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The potential energy of an object is due to its position. It can be given as the formula,

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Where,

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From the formula since the mass and gravitation acceleration is constant, the potential energy of the model will depend upon the height only.

The potential energy will be greatest at point C because the relative position of the ball is the highest from the earth.

The potential energy will be lowest at point E because the height of the ball is the least in this position.

Therefore, the potential energy of the model will be the greatest at point C while it will be the least at point E.

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