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3 years ago

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If a proton and an electron are released when they are 7.00×10−10 m apart (typical atomic distances), find the initial accelerat

ion of each of them. aelectron = nothing m/s2

1 answer:

Ugo [173]3 years ago

7 0

Answer:

The acceleration of the proton is 2.823 x 10¹⁷ m/s²

The acceleration of the electron is 5.175 x 10²⁰ m/s²

Explanation:

Given;

distance between the electron and proton, r = 7 x 10⁻¹⁰ m

mass of proton, $m_p$ = 1.67 x 10⁻²⁷ kg

mass of electron, $m_e$ = 9.11 x 10⁻³¹ kg

The attractive force between the two charges is given by Coulomb's law;

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

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/c²

$F = \frac{k(q_p)(q_e)}{r^2} \\\\F = \frac{(9*10^9)(1.602*10^{-19})(1.602*10^{-19})}{(7*10^{-10})^2} \\\\F = 4.714 *10^{-10} \ N$

Acceleration of proton is given by;

F = ma

$F = m_pa_p\\\\a_p = \frac{F}{m_p}\\\\a_p = \frac{4.714*10^{-10}}{1.67*10^{-27}}\\\\a_p = 2.823 *10^{17} \ m/s^2$

Acceleration of the electron is given by;

$F = m_ea_e\\\\a_e = \frac{F}{m_e}\\\\a_e = \frac{4.714*10^{-10}}{9.11*10^{-31}}\\\\a_e = 5.175 *10^{20} \ m/s^2$

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Given that the concentration of bovine carbonic anhydrase is 3.3 pmol ⋅ L − 1 and R max ( V max ) = 222 μmol ⋅ L − 1 ⋅ s − 1 , d

LuckyWell [14K]

Answer:

The turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1.

Explanation:

Given:

The concentration of bovine carbonic anhydrase = total enzyme concentration = Et = 3.3 pmol⋅L^–1 = 3.3 × 10^–12 mol.L^–1

The maximum rate of reaction = Rmax (Vmax) = 222 μmol⋅L^–1⋅s^–1 = 222 × 10^–6 mol.L^–1⋅s^–1

The formula for the turnover number of an enzyme (kcat, or catalytic rate constant) = Rmax ÷ Et = 222 × 10^–6 mol.L^–1⋅s^–1 ÷ 3.3 × 10^–12 mol.L^–1 = 67,272,727.27 s^–1

Therefore, the turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1

3 0

3 years ago

The electric field between two parallel plates is uniform, with magnitude 628 N/C. A proton is held stationary at the positive p

aliina [53]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

$F_{p}$ = $q_{p}$ E

and, we know that, F = ma

So,

$m_{p}$ a = $q_{p}$ E

a = $\frac{q_{p}.E }{m_{p} }$ Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2 $at^{2}$

Here, u = 0.

S = 1/2 $at^{2}$

Put equation 1 into the above equation:

S = 1/2 x ( $\frac{q_{p}.E }{m_{p} }$ ) $t^{2}$ Equation 2

So,

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x ( $\frac{q_{e}.E }{m_{e} }$ ) $t^{2}$ Equation 3

We know that the charge of electron is equal to the charge of proton so,

$q_{p}$ = $q_{e}$ = q

By dividing the equation 2 by equation 3, we get:

$\frac{S}{D-S}$ = $\frac{m_{e} }{m_{p} }$

Solve the above equation for S,

S $m_{p}$ = $m_{e}$ D - $m_{e}$ S

So,

S = $\frac{m_{e}.D }{(m_{e} + m_{p}) }$

Plugging in the values,

As we know the mass of electron is 9.1 x $10^{-31}$ and the mass of proton is 1.67 x $10^{-27}$

S = $\frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27} }$

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = $\frac{m_{Cl}.D }{(m_{Cl} + m_{Na}) }$

As we know the mass of chlorine is 35.5 and of sodium is 23

S = $\frac{35.5 . 4.22}{(35.5 + 23)}$

S = 2.56 cm

7 0

2 years ago

Which statements are true about ionic compounds?A. An ionic compound can be named using prefixes.B. An ionic compound is made of

Damm [24]

The correct answer is d.

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2 years ago

A large insulating sheet has a surface charge density σ. what is the electric field strength near the insulating sheet?

MakcuM [25]

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2 years ago

A wheel decelerates from 13.5 rad s−1 to 6.0 rad s−1 in 7 s. Calculate the angular displacement

Igoryamba

Answer:

<em>Angular displacement=68.25 rad</em>

Explanation:

<u>Circular Motion</u>

If the angular speed varies from ωo to ωf in a time t, then the angular acceleration is given by:

$\displaystyle \alpha=\frac{\omega_f-\omega_o}{t}$

The angular displacement is given by:

$\displaystyle \theta=\omega_o.t+\frac{\alpha.t^2}{2}$

The wheel decelerates from ωo=13.5 rad/s to ωf=6 rad/s in t=7 s, thus:

$\displaystyle \alpha=\frac{6-13.5}{7}$

$\displaystyle \alpha=\frac{-7.5}{7}$

$\displaystyle \alpha=-1.071 \ rad/s^2$

Thus, the angular displacement is:

$\displaystyle \theta=13.5*7+\frac{-1.071*7^2}{2}$

$\displaystyle \theta=94.5-26.25$

$\boxed{\displaystyle \theta=68.25\ rad}$

Angular displacement=68.25 rad

3 0

3 years ago