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3 years ago

4. What is the consequence of atoms being a singular closed system?

Physics

1 answer:

kaheart [24]3 years ago

5 0

Answer:

*Atomos of a certain element contain a fixed amount of energy.

Explanation:

The atoms are said to be closed systems because the number of protons, neutrons and electrons is fixed, so these elements each have an energy given by the electrical interaction with the other elements.

Consequently, the correct answer is: Atomos of a certain element contain a fixed amount of energy.

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A particle of charge 3.53×10 −8 C experiences a force of magnitude 6.03×10 −6 N when it is placed in a particular point

Cloud [144]

<h2>Electric field at the location of the charge is 169.97 N/C</h2>

Explanation:

Electric field is the ratio of force and charge.

Force, F = 6 x 10⁻⁶ N

Charge, q = 3.53 x 10⁻⁸ C

We have

$E=\frac{F}{q}\\\\E=\frac{6\times 10^{-6}}{3.53\times 10^{-8}}\\\\E=169.97N/C$

Electric field at the location of the charge is 169.97 N/C

6 0

3 years ago

Please answer of this question

Bogdan [553]

Answer:

$\frac{\pi }{15}$ m

Explanation:

At 10am, the minute hand and hour hand are ' 2 hours apart', since the minute hand is at 12pm and hour hand is at 10am.

Angle between the two hands = 2/12 * 360

= 60°

Arc Length = $2\pi r(\frac{60}{360} )$

= $2\pi (0.2)(\frac{1}{6} )\\= \frac{\pi }{15} m$

8 0

2 years ago

2. Kevin works as a janitor, and he is pushing a fully-

dybincka [34]

The time taken for him to move the bin 6.5 m is 2.30 s.

The given parameters;

weight of the load, w = 557 N
force applied , F = 410 N
angle of force, = 15°
coefficient of kinetic friction = 0.46
distance moved, d = 6.5 m

The net horizontal force on the recycling bin is calculated as follows;

$Fcos\theta - F_k = ma$

where;

m is the mass of the recycling bin
$F_k$ is the frictional force

W = mg

$557 = 9.8m\\\\m = \frac{557}{9.8} \\\\m = 56.84 \ kg$

The net horizontal force on the recycling bin is calculated as;

$Fcos \theta - F_k = ma\\\\Fcos\theta - \mu_kF_n = ma\\\\410\times cos(15) \ - \ 0.46(557) = 56.84 a\\\\139.8 = 56.84a\\\\a = \frac{139.8}{56.84} \\\\a = 2.46 \ m/s^2$

The time taken for him to move the bin 6.5 m is calculated as follows;

$s = v_0t + \frac{1}{2} at^2\\\\6.5 = 0 + \frac{1}{2} \times 2.46\times t^2\\\\6.5 = 1.23 t^2\\\\t^2 = \frac{6.5 }{1.23} \\\\t^2 = 5.285\\\\t = \sqrt{5.285} \\\\t = 2.30 \ s$