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GaryK [48]
3 years ago
8

Se deja caer una moneda desde cierta altura. Si se desprecian los efectos del aire, ¿cómo varía la fuerza neta sobre la moneda a

medida que cae? Explique. Máximo en 3 líneas
Physics
1 answer:
forsale [732]3 years ago
3 0

Answer:

Ok, primero pensemos en una situación normal.

La moneda comienza a caer, pero la moneda esta inmersa en una sustancia, el aire. El aire comienza a aplicar una resistencia al movimiento de la moneda, y esta resistencia incremente a medida que la velocidad de la moneda incremente. Llega un punto en el que esta nueva fuerza es igual a la fuerza gravitatoria, y en sentido opuesto, lo que causa que la fuerza neta sea 0, y que la moneda caiga a velocidad constante hasta que esta impacta con el suelo.

Ahora, en este caso tenemos que ignorar los efectos del aire, entonces no hay ninguna fuerza que se oponga a la fuerza gravitatoria, entonces la fuerza neta no cambia a medida que cae (La fuerza neta cambia cuando la moneda impacta el suelo).

También se puede analizar el caso en el que, como la fuerza gravitatoria decrece con el radio al cuadrado, a medida que la moneda cae, la fuerza gravitatoria incrementa. El tema es que en para estas dimensiones, ese cambio en la fuerza gravitacional es imperceptible,

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viktelen [127]

Answer:

Vi = 32 [m/s]

Explanation:

In order to solve this problem we must use the following the two following kinematics equations.

v_{f} =v_{i} - (a*t)\\

The negative sign of the second term of the equation means that the velocity decreases, as indicated in the problem.

where:

Vf = final velocity = 8[m/s]

Vi = initial velocity [m/s]

a = acceleration = [m/s^2]

t = time = 5 [s]

Now replacing:

8 = Vi - 5*a

Vi = (8 + 5*a)

As we can see we have two unknowns the initial velocity and the acceleration, so we must use a second kinematics equation.

v_{f}^{2} = v_{i}^{2} - (2*a*d)

where:

d = distance = 100[m]

(8^2) = (8 + 5*a)^2 - (2*a*100)

64 = (64 + 80*a + 25*a^2) - 200*a

0 = 80*a - 200*a + 25*a^2

0 = - 120*a + 25*a^2

0 = 25*a(a - 4.8)

therefore:

a = 0 or a = 4.8 [m/s^2]

We choose the value of 4.8 as the acceleration value, since the zero value would not apply.

Returning to the first equation:

8 = Vi - (4.8*5)

Vi = 32 [m/s]

6 0
3 years ago
an electric current 0.75 a passes through a circuit that has a resistance of 175. according to ohm's law, what is the voltage of
Aleksandr [31]

Answer:

131.25

Explanation:

i worked it out on a diffrent sheet so its hard to explain

8 0
3 years ago
A car travels along a highway with a velocity of 24 m/s, west. The car exits the highway; and 4.0 s later, its instantaneous vel
zloy xaker [14]

Answer:

4.25 m/s^{2}

Explanation:

Change in velocity considering the x component will be

Final velocity-Initial velocity

\triangle v_x= 16cos 45^{\circ}-24=-12.6862915 m/s

Change in velocity considering the y component will be

Final velocity-Initial velocity

\triangle v_y= 16sin 45^{\circ}-0=11.3137085 m/s

Resultant change in velocity=\sqrt {(-12.6862915 m/s)^{2}+(11.3137085 m/s)^{2}}=16.9982938 m/s

Acceleration= change in velocity per unit time hence

a= \frac {16.9982938}{4}=4.24957345\approx 4.25 m/s^{2}

5 0
3 years ago
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Vsevolod [243]

Answer:

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Explanation:

Given that,

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We have to find the distance covered.

Firstly, let's convert time in seconds.

→ 1 minute = 60 seconds

→ 4 minutes = (4 × 60) seconds

→ 4 minutes = 240 seconds

Now, we know that,

→ Distance = Speed × Time

→ Distance = (4 × 240) m

→ Distance = 960 m

Therefore, distance covered is 960 m.

7 0
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Leno4ka [110]
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