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Iteru [2.4K]
3 years ago
9

9. A pendulum bob is made with a ball filled with water. What would happen to the frequency of vibration of this pendulum if a h

ole in the ball allowed water to slowly leak out? (Treat the pendulum as a simple pendulum.)
Physics
1 answer:
Ilya [14]3 years ago
6 0

Answer:

The frequency of vibration will be unchanged.

Explanation:

The frequency of oscillation of a simple pendulum is given by

f = \dfrac{1}{2 \pi}\sqrt{\dfrac{g}{L}}

where, g is the acceleration due to gravity and L is the length of the pendulum.

It can be seen from the formula that the frequency of the pendulum is independent of the mass.

Thus, the frequency of vibration will be same.

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África tiene más de 1000 lenguas, pero solo 50 son habladas por más de 500 mil personas, explica si esto es un aspecto positivo
Misha Larkins [42]

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7 0
3 years ago
A certain nuclear power plant is capable of producing 1.2×10^9 W of electric power. During operation of the reactor, mass is con
alukav5142 [94]

Answer:

0.00016 kg

Explanation:  

Given:

Power = P = 1.2 × 10⁹ Watts

Power =  work done / Time

efficiency = 0.30

Input power = 1.2 × 10⁹ / 0.30 =  4  × 10⁹ W

Energy =  4  × 10⁹ x 60 x 60 = 1.44 x 10¹³ joules

E = m c² , where c is the speed of light and m is the mass.

⇒ mass = m = E / c²  = (1.44 x 10¹³) / (3 × 10⁸ )²

                                   = 0.00016 kg

6 0
4 years ago
(8c7p26) During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with lar
likoan [24]

Answer: 1175 J

Explanation:

Hooke's Law states that "the strain in a solid is proportional to the applied stress within the elastic limit of that solid."

Given

Spring constant, k = 102 N/m

Extension of the hose, x = 4.8 m

from the question, x(f) = 0 and x(i) = maximum elongation = 4.8 m

Work done =

W = 1/2 k [x(i)² - x(f)²]

Since x(f) = 0, then

W = 1/2 k x(i)²

W = 1/2 * 102 * 4.8²

W = 1/2 * 102 * 23.04

W = 1/2 * 2350.08

W = 1175.04

W = 1175 J

Therefore, the hose does a work of exactly 1175 J on the balloon

7 0
3 years ago
An airplane flying at an altitude of 6 miles passes directly over a radar antenna. When the airplane is 10 miles away (s = 10),
Novosadov [1.4K]

Answer:

Explanation:

Given

altitude of the Plane h=6\ miles

When Airplane is s=10\ miles away

Distance is changing at the rate of \frac{\mathrm{d} s}{\mathrm{d} t}=290\ mph

From diagram we can write as

h^2+x^2=s^2

differentiate above equation w.r.t time

2h\frac{\mathrm{d} h}{\mathrm{d} t}+2x\frac{\mathrm{d} x}{\mathrm{d} t}=2s\frac{\mathrm{d} s}{\mathrm{d} t}

as altitude is not changing therefore \frac{\mathrm{d} h}{\mathrm{d} t}=0

0+x\frac{\mathrm{d} x}{\mathrm{d} t}=s\frac{\mathrm{d} s}{\mathrm{d} t}

at s=10\ miles\ and\ h=6\ miles

substitute the value we get x=\sqrt{10^2-6^2}=8\ miles

8\times \frac{\mathrm{d} x}{\mathrm{d} t}=10\times 290

\frac{\mathrm{d} x}{\mathrm{d} t}=362.5\ mph

5 0
4 years ago
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