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3 years ago

Extra credit calculate the total mechanical energy of a pendulum as it swings from its highest point to its lowest point

1 answer:

4 0

Is there any numbers to your question?
Keep in mind, the energy is conserved in a pendulum.
Here’s more information:
https://blogs.bu.edu/ggarber/interlace/pendulum/energy-in-a-pendulum/

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Un movil de masa 12Kg sobre el cual estan actuando varias fuerzas F_1=48N, F_2=60N y F_3=30N Calcular la aceleracion con la cual

Nikitich [7]

Answer:

Lamentablemente el problema está incompleto, pues no sabemos la dirección en la que se aplican las fuerzas. Por ello, voy a resolver el problema asumiendo dos casos. (abajo se puede ver una imagen donde se describe cada caso)

1) Todas las fuerzas están en la misma dirección.

Entonces la fuerza neta será la suma de las 3 fuerzas, entonces:

F = 48N + 60N + 30N = 138N

Y por la segunda ley de Newton sabemos que:

F = m*a

fuerza igual a masa por aceleración.

Entonces la aceleración está dada por:

a = F/m = 138N/12kg = 11.5 m/s^2

2) Segundo caso, suponemos que F1 es opuesta a F2 y F3

En este caso, la fuerza neta será:

F = F2 + F3 - F1 = 60N + 30N - 48N = 42N

En este caso, la aceleración será:

a = 42N/12kg = 3.5 m/s^2

7 0

3 years ago

A grating with 250 grooves mm is used with an incandescent light source. Assume the visible spectrum to range in wavelength from

NeTakaya

The number of orders that one can see the entire visible speed is 5 orders.

<h3>How to calculate the orders?</h3>

From the information given, it should be noted that the grating spacing will be:

d = (1.00 × 10^-4) / 250

d = 4000nm

Therefore, the number of times that are needed to complete the order will be the same as the number of orders which the long wavelength time will be visible. This will be:

= (4000 × sin 90°)/700

= 5.71

Therefore, the maximum orders will be 5.

Learn more about speed on:

brainly.com/question/13943409

#SPJ1

7 0

1 year ago

When listening to tuning forks of frequency 256 Hz and 260 Hz, one hears the following number of beats per second. (A) 0 (B) 2 (

Degger [83]

Answer:

Explanation:

As we know that the no of beats can be calculated as.

No. of beats is equal to difference in the tuning forks frequencies.

So,

$n= \nu _{1}- \nu _{2}$ .

Substitute the values of frequencies of 2 tuning forks in the above equation.

$n=(260 Hz-256 Hz)\\n=4$

Therefore the number of beats per second will be hear by the observer is 4 beats per second.

3 0

3 years ago

Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value o

Anna35 [415]

Answer:

the correct one is 2. the equipotential lines must be closer together where the field has more intensity

Explanation:

The equipotential line concept is a line or surface where a test charge can move without doing work, therefore the potential in this line is constant and they are perpendicular to the electric field lines.

In this exercise we have a charge and a series of equipotential lines, if this is a point charge the lines are circles around the charge, where the potential is given by

V = k q / r

also the electric field and the electuary potential are related

E = $- \frac{dV}{dr}$

therefore the equipotential lines must be closer together where the field has more intensity

When checking the answers, the correct one is 2

3 0

3 years ago

A 12-V battery is connected across a 100-Ω resistor. How many electrons flow through the wire in 1.0 min?

Vanyuwa [196]

Answer:

The quantity of charge or electron flowing the wire in the given time is 4.5 x 10¹⁹ electrons.

Explanation:

Given;

emf of the battery, V = 12 V

resistance of the resistor, R = 100-Ω

time of current flow, t = 1 min

charge of 1 electron = 1.602 x 10¹⁹ C

The current through this circuit is given by;

I = V / R

I = (12) / (100)

I = 0.12 A

The quantity of charge or electron flowing the wire in the given time is calculated as;

Q =It

where;

I is the current flowing through the wire

t is the time of current flow = 1 x 60s = 60 s

Q = 0.12 x 60

Q = 7.2 C

1.602 x 10⁻¹⁹ C --------------- 1 electron

7.2 C -----------------------------? electron

$= \frac{7.2 }{1.602*10^{-19}} \\\\= 4.5*10^{19} \ electrons$

Therefore, the quantity of charge or electron flowing the wire in the given time is 4.5 x 10¹⁹ electrons.

4 0

3 years ago