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3 years ago

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You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 1515 lb and was trav

eling eastward. Car B weighs 1125 lb and was traveling westward at 43.0 mph. The cars locked bumpers and slid eastward with their wheels locked for 17.5 ft before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.750. How fast (in miles per hour) was car A traveling just before the collision? (This problem uses English units because they would be used in a U.S. legal proceeding.)

1 answer:

serious [3.7K]3 years ago

6 0

what was the speed limit, and it was 40 mph

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For a point charge, how does the potential vary with distance from the point charge, r?

mr_godi [17]

For a point charge, how does the potential vary with distance from the point charge, r?

a constant

b. r.

c. 1/r.

d. $1/r^2$ .

e. $r^2$ .

Answer:

The correct option is C

Explanation:

Generally for a point charge the electric potential is mathematically represented as

$V = \frac{k Q }{r }$

Here we can deduce that the electric potential varies inversely with the distance i.e

$V \ \alpha \ \frac{1}{r}$

So

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2 years ago

imagine you are working as a rollercoaster designer. you are building a ride the top speed of a 65m/s at the bottom of the first

polet [3.4K]

Rollercoasters generate speed by converting gravitational potential energy into kinetic energy by taking the cart to the top of a large hill and letting it go. The conversion of a perfectly efficient system would be like so:

PE = KE

and using the formulas for potential and kinetic energy:

mgh = 1/2mv^2.

However, the efficiency of this system is 50%, meaning that the kinetic energy obtained from this conversion would appear as so:

PE=0.5 KE

mgh=0.5(1/2mv^2)

mgh=1/4mv^2.

The masses cancel out, leaving:

gh=1/4v^2

The goal is to achieve 65 m/s, and with Earth’s innate gravity of 9.806 m/s^2, we have:

gh=1/4v^2

(9.806)h=1/4(65)^2

h=107.71 meters

The height of the first hill must be 107.7 meters to generate a speed of 65 m/s with a conversion efficiency of 50%.

Hope this helps!

5 0

3 years ago

If the kinetic and potential energy in a system are equal, then the potential energy increases. What happens as a resul

Kobotan [32]

-- If the system is 'closed', then nothing ... including energy ... can get in or out, and the total energy inside has to be constant.

If half of the energy in the system starts out as potential energy and the rest starts out as kinetic, and then the potential energy increases, there's only one place the increase could have come from ... it could only have been converted from kinetic energy. So the <em>kinetic energy</em> in the system <em>must</em> <em>decrease</em>.

In fact, this isn't even a "result". The kinetic energy has to decrease <em><u>before</u></em> the potential energy can increase, because that's where the increase has to come from.

If the system is 'open', then energy can come in and go out. If the potential energy inside suddenly increases, we don't know where it came from, so we can't say anything about what happens to the system.

7 0

2 years ago

The force of attraction between 2 objects is called gravity. Why do you think that an object, like your notebook, pulled towards

Gekata [30.6K]

This question is asking: Why does gravity occur? why will your notebook get pulled towards you?

6 0

2 years ago

Read 2 more answers

The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

MA_775_DIABLO [31]

Answer:

$q=49Q/64$

and

$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

so : $x =16R/15$

7 0

2 years ago