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3 years ago

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A small car of mass m and a large car of mass 4m drive along a highway at constant speed. they approach a curve of radius r. bot

h cars maintain the same acceleration a as they travel around the curve. how does the speed of the small car vs compare to the speed of the large car vl as they round the curve? hints

2 answers:

matrenka [14]3 years ago

7 0

Hints?????????????????????????

natali 33 [55]3 years ago

7 0

Answer:

Both speeds are equals ⇒ $V_{S}=V_{L}$

Explanation:

We know that the small car has mass '' $m$ '' and the large car has mass '' $4m$ ''.

We also know that they approach a curve of radius '' $r$ ''.

Both cars maintain the same acceleration as they travel around the curve.

The centripetal acceleration in a curve has the following equation :

$a_{c}=\frac{V^{2}}{r}$ (I)

Where '' $a_{c}$ '' is centripetal acceleration

Where '' $V$ '' is the tangential speed and where '' $r$ '' is the radius of the curve.

If we use the equation (I) to find $V$ , we will find that the speed depends of the centripetal acceleration and the radius of the curve.

It doesn't depend of the mass. Therefore, the small car and the large car will have the same speed while they are travelling around the curve ⇒

$V_{S}=V_{L}$

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Natali5045456 [20]

The question is incomplete. The complete question is :

A plate of uniform areal density $\rho = 2 \ kg/m^2$ is bounded by the four curves:

$y = -x^2+4x-5m$

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Solution :

Given :

$y = -x^2+4x-5$

$y = x^2+4x+6$

$x=1$

$x=2$

and $\rho = 2 \ kg/m^2$ , $P_x=1 \$ , $P_y=-2 \$ .

So,

$dI = dmr^2$

$dI = \rho \ dA \ r^2$ , $r=\sqrt{(x-1)^2+(y+2)^2}$

$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \ dy \ dx$

$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$

$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$

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3 years ago

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Molodets [167]

Considering the equivalence between mass and energy given by the expression of Einstein's theory of relativity, the correct answer is the last option: the energy equivalent of an object with a mass of 1.05 kg is 9.45×10¹⁶ J.

The equivalence between mass and energy is given by the expression of Einstein's theory of relativity, where the energy of a body at rest (E) is equal to its mass (m) multiplied by the speed of light (c) squared:

E=m×c²

This indicates that an increase or decrease in energy in a system correspondingly increases or decreases its mass, and an increase or decrease in mass corresponds to an increase or decrease in energy.

In other words, a change in the amount of energy E, of an object is directly proportional to a change in its mass m.

In this case, you know:

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Replacing:

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Solving:

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Finally, the correct answer is the last option: the energy equivalent of an object with a mass of 1.05 kg is 9.45×10¹⁶ J.

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