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12 days ago

The acceleration of segment D is m/s2. Rank segments A, B , C from least accelerations to greatest acceleration. Least

Physics

2 answers:

Ostrovityanka [2.2K]12 days ago

8 0

Answer:

D, C, B, A

Explanation:

The derivative from a velocity-time graph provides the acceleration value.

Segment A

$\frac{dy}{dx} = \frac{15m/s}{1s} = 15m/s^2$

Segment B

$\frac{dy}{dx} = \frac{5m/s}{1s} = 5m/s^2$

Segment C

$\frac{dy}{dx} = \frac{0m/s}{2s} = 0m/s^2$

Segment D

$\frac{dy}{dx} = \frac{-20m/s}{1s} = -20m/s^2$

Sorted from the lowest to the highest acceleration:

D, C, B, A

Keith_Richards [2.2K]12 days ago

7 0

Segment D is -20 m/s squared

I concluded this after completing the calculations.

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10 days ago

A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele

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Answer:

a)n= 3.125 x $10^{19$ electrones.

b)J= 1.515 x $10^{6$ A/m²

c) $V_{d$ =1.114 x $10^{4$ m/s

d) ver explicación

Explanation:

La corriente 'I' = 5A =>5C/s

diámetro 'd'= 2.05 x $10^{-3$ m

radio 'r' = d/2 => 1.025 x $10^{-3$ m

número de electrones 'n'= 8.5 x $10^{28}$

a) La cantidad de electrones que pasan por la bombilla cada segundo se determina mediante:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

Como sabemos que: Q= ne

donde e es la carga del electrón, es decir, 1.6 x $10^{-19$ C

n= Q/e => 5/ 1.6 x $10^{-19$

n= 3.125 x $10^{19$ electrones.

b) La densidad de corriente 'J' en el cable se calcula como

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x $10^{-3$ )²)

J= 1.515 x $10^{6$ A/m²

c) La velocidad típica ' $V_{d$ ' de un electrón se expresa como:

$V_{d$ = $\frac{J}{n|q|}$

=1.515 x $10^{6$ / 8.5 x $10^{28}$ x |-1.6 x $10^{-19$ |

$V_{d$ =1.114 x $10^{4$ m/s

d) De acuerdo con estas ecuaciones,

J= I/A

$V_{d$ = $\frac{J}{n|q|}$ = $\frac{I}{nA|q|}$

Si utilizaras un cable de doble diámetro, ¿cuáles de las respuestas anteriores cambiarían? ¿Aumentarían o disminuirían?

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1 month ago

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Answer:

Explanation:

To approach this problem, we need to understand two key concepts.

First, the gravitational force on an object in orbit equals its mass multiplied by centripetal acceleration.

Secondly, Newton's law of universal gravitation defines the force between two masses: Fg = mMG/r², where Fg denotes gravitational force, m and M signify the masses, G represents the gravitational constant, and r indicates the distance separating the two masses.

Thus:

Fg = m v²/r

mMG/r² = m v²/r

v² = MG/r

Potential energy for each planet is expressed as:

PE = mgr = m (MG/r²) r = mMG/r

Kinetic energy for each planet is computed as:

KE = 1/2 mv² = 1/2 m (MG/r) = 1/2 mMG/r

Total mechanical energy is calculated as:

ME = PE + KE = 3/2 mMG/r

Since both planets share the same mass, the only variable is their orbital radius. Consequently, Planet A, with a smaller radius, possesses greater potential, kinetic, and mechanical energy.

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1 month ago

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Answer:

Explanation:

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