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Temka [501]
3 years ago
5

A meter stick is supported by a pivot at its center of mass. Assume that the meter stick is uniform and that the center of mass

is at the 50 cm mark.a)If a massm1= 80 g is suspended at the 30 cm mark, at which cm mark would a mass m2= 110 g need to be suspended for the system to be in equilibrium?
Physics
1 answer:
iragen [17]3 years ago
5 0

Answer:

64.5 cm

Explanation:

30 * 80 + x * 110 = 50* 190 => x = 64.5

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What is the least possible initial kinetic energy kmin the oxygen atom could have and still excite the cesium atom?
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If perfectly elastic there would be no energy left over for exciting the atom. if the collision were partially elastic, then some of the initial kinetic energy would be converted into internal energy,

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3 years ago
When two 10-Ohm resistors are in parallel in a circuit, what is the net resistor value
Elza [17]

Answer:

Net capacitance

=

let net capacitance be R

1/R = 1/10+1/10

1/R=2/10

1/R=1/5

cross mutiply

R=5

the net capacitance is 5ohms

6 0
3 years ago
Container A and container B hold samples of the same ideal gas. The volume and the pressure of container A is equal to the volum
SVETLANKA909090 [29]

Answer:

A. TA = TB/2.

Explanation:

Since container A has half as many molecules of the ideal gas in it as container B. Therefore, container A will have half the volume of gas as in container B:

V_A = \frac{1}{2}V_B

Now, from Charle's Law:

\frac{V_A}{T_A}=\frac{V_B}{T_B}\\\\\frac{1}{2}\frac{V_B}{T_A}=\frac{V_B}{T_B}\\\\T_A = \frac{T_B}{2}

Hence, the correct option is:

<u>A. TA = TB/2.</u>

6 0
2 years ago
The Sun appears to move across the sky during the day. The best explanation for this apparent motion is that Earth is
Harrizon [31]

Answer:

C

Explanation:

5 0
3 years ago
Read 2 more answers
Si el coeficiente de fricción cinética entre los neumáticos y el pavimento seco es de 0.80. ¿Cuál es la distancia mínima para de
vovangra [49]

Answer: 52.9 metros.

Explanation:

Podemos escribir la fuerza de fricción cinética como

F = μ*N

donde N es la fuerza normal entre el coche y el suelo, cuya magnitud es igual al peso en esta situación.

F = μ*m*g

donde m es la masa del coche y g es 9.8m/s^2

y sabemos que μ = 0.8

Por la segunda ley de Newton, sabemos que:

F = m*a

fuerza es igual a masa por aceleración.

a = F/m

entonces la aceleración causada por la fuerza de rozamiento es:

F = 0.8*m*g

a = F/m = (0.8*m*g)/m = 0.8*g.

Entonces ya encontramos la aceleración, hay que recordar que esta aceleración es en sentido opuesto a la sentido de movimiento, entonces podemos escribir la aceleración como:

a(t) = -0.8*g

Para la velocidad, podemos integrar sobre el tiempo para obtener.

v(t) = -0.8*g*t + v0

donde v0 es la velocidad inicial del auto = 28.7m/s

v(t) = -0.8*g*t + 28.8m/s

Ahora podemos encontrar el tiempo necesario para que la velocidad del coche sea cero, en ese momento, como deja de moverse, ya no tendremos rozamiento cinético, entonces no habrá aceleración y el coche se detendrá completamente.

v(t) = 0m/s = -0.8*9.8m/s^2*t + 28.8m/s

7.84m/s^2*t = 28.8m/s

                 t   = (28.8m/s)/(7.84m/s^2) = 3.63 segundos.

Ahora vamos a la ecuación de movimiento, donde asumimos que la posición inicial del coche es 0m, así que no tendremos constante de integración.

p(t) = -(1/2)*(0.8*9.8m/s^2)*t^2 + 28.8m/s*t

Ahora podemos evaluar la posición en t = 3.63 segundos, y esto nos dara la distancia que el coche se movio mientras frenaba.

p(3.63s) = -(1/2)*(0.8*9.8m/s^2)*(3.63s)^2 + 28.8m/s*(3.63s) = 52.9 metros.

6 0
2 years ago
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