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2 years ago

Shotguns are examples of

Physics

1 answer:

tankabanditka [31]2 years ago

6 0

The answer would be {B}

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two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel .calculate the equivalent series resist

Novosadov [1.4K]

Explanation:

Given that,

Two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel.

For series combination,

$R_{eq}=R_1+R_2$

For parallel combination,

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

When 6 ohm and 3 ohm are in series,

$R_s=6+3\\\\R_s=9\ \Omega$

When 6 ohm and 3 ohm are in paralle,

$\dfrac{1}{R_p}=\dfrac{1}{6}+\dfrac{1}{3}\\\\R_p=2\ \Omega$

So, the equivalent resistance in series combination is 9 ohms and in parallel combination it is 2 ohms.

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2 years ago

When U-235 splits, it usually emits

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Splitting atoms. 'Fission' is another word for splitting. The process of splitting a nucleus is called nuclear fission. ... For fission to happen, the uranium-235 or plutonium-239 nucleus must first absorb a neutron.

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2 years ago

Bill is learning to play tennis. He does pretty well hitting the ball back to his opponent but, many times he misses the ball wh

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Answer: Place his feet parallel to the baseline prior to tossing the ball

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2 years ago

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical ener

defon

Complete question:

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find

(a) the force constant of the spring and (b) the amplitude of the motion.

Answer:

(a) the force constant of the spring = 47 N/m

(b) the amplitude of the motion = 0.292 m

Explanation:

Given;

mass of the spring, m = 200g = 0.2 kg

period of oscillation, T = 0.410 s

total mechanical energy of the spring, E = 2 J

The angular speed is calculated as follows;

$\omega = \frac{2\pi}{T} \\\\\omega = \frac{2\pi}{0.41} \\\\\omega = 15.33 \ rad/s$

(a) the force constant of the spring

$\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\k = m \omega^2\\\\k = (0.2)(15.33)^2\\\\k = 47 \ N/m$

(b) the amplitude of the motion

E = ¹/₂kA²

2E = kA²

A² = 2E/k

$A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2\times 2}{47} }\\\\A = 0.292 \ m$

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