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Ivenika [448]
3 years ago
13

Newton's second law problem. The spring constant can be determined from the balance of forces in equilibrium. Suppose the spring

extends a distance LaTeX: \Delta Δ x when additional mass LaTeX: \Delta Δ m is added. Write down Newton's second law for this case and solve for k.[Answer: K=delta m g/delta x.] Make sure you understand how this formula was derived. If delta m=41 g and delta x=6cm,what is k?
Physics
1 answer:
mario62 [17]3 years ago
6 0

Answer:

K =6.697 Kg/s²

Explanation:

Given:

delta m =41 g = 0.041 kg

delta x = 6cm = 0.06m

g = 9.8 m/s²

according to the given formula

K = delta m g /delta x

K = (0.041 kg × 9.8 m/s²) / 0.06m

K =6.697 Kg/s²

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If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
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The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

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As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

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Equation of motion in horizontal direction is given as

s_x=v_x_0 t

t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s

Now for the vertical distance

vy_o=0

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s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2

Now using this acceleration the value of electric field is calculated as

E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

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3 years ago
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