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3 years ago

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Two pans of a balance are 24.1 cm apart. The fulcrum of the balance has been shifted 1.33 cm away from the center by a dishonest

shopkeeper.
By what percentage is the true weight of the goods being marked up by the shopkeeper? Assume the balance has negligible mass.
Answer in units of %.

1 answer:

Citrus2011 [14]3 years ago

8 0

As we know that fulcrum is shifted to one side by 1.33 cm

so here let say left side mass is m1 while on other side mass is m2

so we will have

$m_1g (24.1 - 1.33) = m_2 g(24.1 + 1.33)$

$m_2 = 0.895 m_1$

now in order to find the percentage we have

$percentage = \frac{m_1 - m_2}{m_1}\times 100$

$percentage = (1 - 0.895)\times 100$

$percentage = 10.5%$

so it is 10.5 % less mass

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A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

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Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

So

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P_f = k * \frac{q_1 * q_2 }{R_f } - 0$

So

$\frac{1}{2} * m * \frac{1}{8} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R_f }$

=> $\frac{1}{2} * m *v_o^2 [-\frac{15}{16} ] = k * \frac{q_1 * q_2 }{R_f }$

=> $- \frac{15}{32} * m *v_o^2 = k * \frac{q_1 * q_2 }{R_f } ---(2)$

Divide equation 2 by equation 1

$\frac{- \frac{15}{32} * m *v_o^2 }{- \frac{3}{8} * m *v_0^2 } } = \frac{k * \frac{q_1 * q_2 }{R_f } }{k * \frac{q_1 * q_2 }{R } }}$

=> $-\frac{15}{32 } * -\frac{8}{3} = \frac{R}{R_f}$

=> $\frac{5}{4} = \frac{R}{R_f}$

=> $R_f = \frac{4}{5} R$

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Answer:

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Kinetic energy than parked

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A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa

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Answer:

$V_p = 267.258 m/s$

$V_n = 38.375 m/s$

Explanation:

using the law of the conservation of the linear momentum:

$P_i = P_f$

where $P_i$ is the inicial momemtum and $P_f$ is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

$P_i = m(270 m/s)$

$P_f = mV_P + M_nV_n$

where m is the mass of the proton and $V_p$ is the velocity of the proton after the collision, $M_n$ is the mass of the nucleus and $V_n$ is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = m $V_p$ + 14m $V_n$

then, m is cancelated and we have:

270 = $V_p$ + $14V_n$

This is a elastic collision, so the kinetic energy K is conservated. Then:

$K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2$

and

Kf = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

then,

$\frac{1}{2}m(270)^2$ = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

here we can cancel the m and get:

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

now, we have two equations and two incognites:

270 = $V_p$ + $14V_n$ (eq. 1)

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

in the second equation, we have:

36450 = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$ (eq. 2)

from this last equation we solve for $V_n$ as:

$V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

and replace in the other equation as:

270 = $V_p +$ 14 $\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

so,

$V_p = -267.258 m/s$

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

$V_n = \frac{270+267.258}{14}$

$V_n = 38.375 m/s$

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