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3 years ago

If you triple the speed from 20km/h to 60km/h, by what factor does the KE increase?

Physics

1 answer:

sertanlavr [38]3 years ago

5 0

Answer:

<em>The kinetic energy increases by a factor of 9 </em>

Explanation:

<u>Kinetic Energy </u>

It's the energy related to the speed of an object and is calculated by the formula

$\displaystyle K=\frac{mv^2}{2}$

Let's suppose the initial speed of the mass is $v_o$ , then the kinetic energy is

$\displaystyle K_o=\frac{mv_o^2}{2}$

Now the speed will triple, i.e. $v_1=3v_o$ . The new kinetic energy is

$\displaystyle K_1=\frac{mv_1^2}{2}$

$\displaystyle K_1=\frac{m(3v_o)^2}{2}$

$\displaystyle K_1=9\frac{mv_o^2}{2}$

Thus

$K_1=9K_o$

The kinetic energy increases by a factor of 9

Let's check with numbers: $v_o=20,\ v_1=60$

$\displaystyle K_o=\frac{m20^2}{2}=200m$

$\displaystyle K_1=\frac{m60^2}{2}=1800m$

The relation is

$\displaystyle \frac{K_1}{K_o}=9$

The new kinetic energy is nine times the initial kinetic energy.

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Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

Brrunno [24]

1) $+2.19\mu C$

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$ (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

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kondaur [170]

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