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2 years ago

two cars with initial speed 2v and v, lock their brakes and skid to a stop. what is the ratio of the distance travelled

Physics

1 answer:

denpristay [2]2 years ago

8 0

Answer:

4:1

Explanation:

Given that the initial speed of the first car, u = 2v while the initial speed of the second car, u = v. To find the distance travelled, we are going to apply one of the equations of motion. The equation chosen is

v² = u² - 2as, where

s = the distance needed

a = acceleration due to gravity

u = initial velocity which is v & 2v

v = final velocity which is 0

For the first car with initial velocity, 2v, on substituting into the equation, we have

v² = u² - 2as(1)

0 = 4v - 2as(1)

4v = 2as(1)

2v = as(1), making s(1) subject of formula we have

s(1) = 2v/a

Taking the second car, we have u = v

v² = u² - 2as(2)

0 = v - 2as(2)

v = 2as(2), making s(2) subject of formula, we have

s(2) = v/2a

Not, ratio of s1 : s2 =

2v/a : v/2a

s1/s2 = 2v/a ÷ v/2a

s1/s2 = 2v/a * 2a/v

s1/s2 = 4av/av

s1/s2 = 4/1

Therefore, the ratio of the first car to the second car is 4:1

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Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

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Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

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According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

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The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

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$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

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The acceleration of automobiles is often given in terms of the time it takes to go from 0 mi/h to 60 mi/h. One of the fastest st

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Answer:

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