Products must equal which in a chemical reaction?

Physics

1 answer:

Evgesh-ka [11]2 years ago

3 0

I think it is reactants

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In each part calculate the kinetic energy of the given objects in joules a) an automobile whose mass is 1260kg moving at a speed

Contact [7]

Kinetic energy (KE) = 1/2 m v^2

Where:

m = mass = 1260 kg

v = speed = 66.3 km/h = 18.42 m/s

Replacing:

KE = 1/2 (1260 kg ) (18.42 m/s )^2 = 213,756.7 J

5 0

1 year ago

My Notes An electron is released from rest on the axis of a uniform positively charged ring, 0.500 m from the ring's center. If

dsp73

Answer:

Velocity of the electron = v = 1.2\times 10^8\ m/s.

Explanation:

Given,

Mass of the electron = $m_e\ =\ 9\times 10^{-31}\ kg$
Charge on the electron = $q_e\ =\ 1.62\times 10^{-19}\ C$
Charge density of the ring = $\rho\ =\ +1.00\times 10^{-6}\ C/m$
Radius of the ring = R = 0.70 m
Distance between the electron ant the center or the ring = x = 0.5 m

Now total charge on the ring = $Q\ =\ \rho\times 2\pi R$

Potential energy due to the charged ring to the point on the x-axis is

$P.E.\ =\ \dfrac{KQq_e}{\sqrt{R^2\ +\ x^2}}\\$

Let v be the velocity of the electron at the center of the ring.

Total kinetic energy of the electron = $\dfrac{1}{2}m_ev^2\\$

Now, From the conservation of energy,

the total potential energy of the electron at initially is converted to the total kinetic energy of the electron at the center of the ring,

$\therefore P.E.\ =\ K.E.\\\Rightarrow \dfrac{KQ}{\sqrt{R^2\ +\ x^2}}\ =\ \dfrac{1}{2}m_ev^2\\\Rightarrow v\ =\ sqrt{\dfrac{2 KQq_e}{m_e\sqrt{R^2\ +\ x^2}}}\\\Rightarrow v\ =\ \sqrt{\dfrac{2Kq_e\rho \times 2\pi R}{m_e\sqrt{R^2\ +\ x^2}}}\\\Rightarrow v\ =\ \sqrt{\dfrac{2\times 9\times 10^9\times 1.0\times 10^{-6}\times 2\times 3.14\times 0.7\times 1.6\times 10^{-19}}{9\times 10^{-31}\times \sqrt{0.7^2\ +\ 0.5^2}}}\\\Rightarrow v\ =\ 1.2\times 10^8\ m/s.$