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Harrizon [31]
3 years ago
13

As the pendulum swings through the origin, it has a speed of v0 = 1.1 m/s. If the mass on the end of the pendulum is 15 g, and t

he pendulum is 30 cm long, how high does the pendulum swing before coming back down? What is the angle corresponding to that height?

Physics
1 answer:
serg [7]3 years ago
4 0

Answer:

1: 6.18 cm

2: 52.5609 degrees

Explanation:

We have the pendulum speed at the origin, and in that moment, all energy is kinetic, so we can calculate the pendulum energy by:

Ec = 0.5*m*v^2 = 0.5*0.015*1.1^2 = 0.0091 J

Now with that energy, we can calculate the height the pendulum will reach, as in that moment, the kinetic energy is totally converted to gravitational potencial energy:

Eg = m*g*h = 0.0091

0.015 * 9.81 * h = 0.0091

h = 0.0091 / (0.015 * 9.81 ) = 0.0618 m = 6.18 cm

Looking at the image attached, we can see that the pendulum will form a triangle, and one of the cathetus will be the length of the pendulum minus the height it went up, and the hypotenusa will be the pendulum length.

So, we know that the sine of the angle will be the division between the opposite cathetus and the hypotenusa:

sin(angle) = (30-6.18)/30 = 23.82/30 = 0.794 -> angle = 52.5609 degrees

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Calculate the magnitude of the force between two point charges where q1 = +5.30 ?C and Q2 = +11.2 11C where the separation betwe
MA_775_DIABLO [31]

Answer:

Magnitude of force, F = 11 Newtons

Explanation:

Charge 1, q_1=5.3\ \mu C=5.3\times 10^{-6}\ C

Charge 2, q_2=11.2\ \mu C=11.2\times 10^{-6}\ C

The separation between the charges is, r = 0.22 m

We have to find the magnitude of the force between two point charges. It can be calculated using the formula as :

F=k\dfrac{q_1q_2}{r^2}

F=9\times 10^9\times \dfrac{5.3\times 10^{-6}\ C\times 11.2\times 10^{-6}\ C}{(0.22\ m)^2}

F = 11.03 N

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F = 11 Newtons

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A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit
likoan [24]

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

Length of wire l = 84 cm = 0.84 m

So mass density \mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m

Tension in the wire T = 25 N

Frequency f = 120 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 120=753.6rad/sec

And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

Power is given by P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt

From the relation we can see that power P\ \propto\ A^2

So if amplitude is halved then power will be \frac{1}{4} times

So power will be equal to \frac{0.2023}{2}=0.0505watt

4 0
3 years ago
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