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FinnZ [79.3K]
3 years ago
11

You would like a pendulum that swings back and forth once every 2 seconds, but the one you have swings once every 1.9 seconds. W

hich of the following should you do to adjust it so that it has the desired period?Remove some mass from the pendulum.Make the pendulum slightly shorter.Add more mass to the pendulum.Make the pendulum slightly longer.
Physics
1 answer:
patriot [66]3 years ago
6 0

Your pendulum does a complete swing in 1.9 seconds.  You want to SLOW IT DOWN so it takes 2.0 seconds.

Longer pendulums swing slower.

You need to <em>make your pendulum slightly longer</em>.

If your pendulum is hanging by a thread or a thin string, then its speed doesn't depend at all on the weight at the bottom.  You can add weight or cut some off, and it won't change the speed a bit.  

You might be interested in
Two boxers are fighting. Boxer 1 throws his 5 kg fist at boxer 2 with a speed of 9 m/s.
Sladkaya [172]

Answer:

0.001 s

Explanation:

The force applied on an object is equal to the rate of change of momentum of the object:

F=\frac{\Delta p}{\Delta t}

where

F is the force applied

\Delta p is the change in momentum

\Delta t is the time interval

The change in momentum can be written as

\Delta p=m(v-u)

where

m is the mass

v is the final velocity

u is the initial velocity

So the original equation can be written as

F=\frac{m(v-u)}{\Delta t}

In this problem:

m = 5 kg is the mass of the fist

u = 9 m/s is the initial velocity

v = 0 is the final velocity

F = -45,000 N is the force applied (negative because its direction is opposite to the motion)

Therefore, we can re-arrange the equation to solve for the time:

\Delta t=\frac{m(v-u)}{F}=\frac{(5)(0-9)}{-45,000}=0.001 s

4 0
3 years ago
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
konstantin123 [22]

Answer:

I=2.71\times 10^{-5}\ A

Explanation:

A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.  

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

Let given is,

The diameter of a parallel plate capacitor is 6 cm or 0.06 m

Separation between plates, d = 0.046 mm

The potential difference across the capacitor is increasing at 500,000 V/s

We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :

C=\dfrac{A\epsilon_o}{d}\\\\C=\dfrac{\pi r^2\epsilon_o}{d}, r is radius

Let I is the displacement current. It is given by :

I=C\dfrac{dV}{dt}

Here, \dfrac{dV}{dt} is rate of increasing potential difference

So

I=\dfrac{\pi r^2\epsilon_o}{d}\times \dfrac{dV}{dt}\\\\I=\dfrac{\pi (0.03)^2\times 8.85\times 10^{-12}}{0.46\times 10^{-3}}\times 500000\\\\I=2.71\times 10^{-5}\ A

So, the value of displacement current is 2.71\times 10^{-5}\ A.

4 0
3 years ago
Please I need HELP WITH THIS QUESTION!!!
JulsSmile [24]
Oil, grease and dry lubricants
8 0
3 years ago
An umbrella tends to move upward on a windy day because _____.
masha68 [24]
E. all of the above

An umbrella tends to move upward on a windy day because _<span>A. buoyancy increases with increasing wind speed </span>
<span>B. air gets trapped under the umbrella and pushes it up </span>
<span>C. the wind pushes it up </span>
<span>D. a low-pressure area is created on top of the umbrella </span>

3 0
3 years ago
Se apunta un rifle horizontalmente con mira a un blanco pequeño que está a 200m en el suelo. La velocidad inicial de la bala es
vfiekz [6]

Answer:

Lo importante a tener en cuenta sobre esta pregunta es que la velocidad horizontal de la bala no hace ninguna diferencia en cuanto al tiempo que tarda en caer al suelo.

Debido a que el arma no ha aplicado ninguna fuerza vertical a la bala, la única fuerza que afecta la bala es la gravedad. Esto significa que la bala tarda tanto en caer al suelo como lo haría si se cayera, a pesar de que ahora viaja una gran distancia horizontal en la duración.

Para encontrar el tiempo de viaje antes de tocar el suelo, tenemos 3 valores:

-El desplazamiento desde el suelo que la bala debe viajar, s = 1.5m

-La aceleración que experimenta la bala. Como la gravedad está acelerando la bala hacia abajo, a = g = ~ 9.81m / s ^ 2

-La velocidad inicial de la bala verticalmente. Como la bala es estacionaria verticalmente (solo viaja horizontalmente al inicio), u = 0m

Examinamos nuestras ecuaciones de movimiento, comúnmente conocidas como ecuaciones SUVAT. Es posible que necesite aprender estos para su examen, pero algunas tablas de examen los proporcionan.

Debido a que tenemos s, u y a, y estamos buscando el tiempo t, la ecuación relevante es

s = ut + 0.5 (en ^ 2)

Completando nuestros valores tenemos:

1.5 = 0t + 0.5 (9.81 x t ^ 2)

1.5 = 4.905 x t ^ 2

Divide 1.5 entre 4.905 para encontrar t ^ 2

t ^ 2 = 0.3058 ...

Simplemente encontramos la raíz cuadrada de t ^ 2 para encontrar t, el tiempo que tarda la bala en llegar al suelo:

t = 0.553s (3 cifras significativas)

Para encontrar la distancia horizontal, d, que la bala ha viajado antes de tocar el suelo, podemos usar la ecuación que vincula el desplazamiento s con cierta velocidad v durante un tiempo t:

s = vt

La velocidad horizontal de la bala, v = 430

El tiempo antes de que la bala toque el suelo, t = 0.553

Entonces d = vt = 430 * 0.553 = 238m (3 cifras significativas)

3 0
3 years ago
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