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larisa [96]
3 years ago
8

A pendulum has a 0.35\ \text{kg}0.35 kg0, point, 35, space, start text, k, g, end text mass oscillating at a small angle from a

light string of length 1.0\ \text{m}1.0 m1, point, 0, space, start text, m, end text. What is the frequency of oscillation?
Physics
1 answer:
expeople1 [14]3 years ago
4 0

Answer:

The frequency of oscillation of the simple pendulum is 0.49 Hz.

Explanation:

Given that,

Mass of the simple pendulum, m = 0.35 kg

Length of the string to which it is attached, l = 1 m

We need to find the frequency of oscillation. The frequency of oscillation of the simple pendulum is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{9.8}{1}} \\\\f=0.49\ Hz

So, the frequency of oscillation of the simple pendulum is 0.49 Hz. Hence, this is the required solution.

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ow long must a simple pendulum be if it is to make exactly ten swings per second? (That is, one complete vibration takes exactly
Igoryamba
The period T of a pendulum is given by:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the length of the pendulum while g=9.81 m/s^2 is the gravitational acceleration.

In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is T=0.200 s. Using this data, we can solve the previous formula to find L:
L=g ( \frac{T}{2\pi} )^2=(9.81 m/s^2)( \frac{0.2 s}{2 \pi} )^2=1 \cdot 10^{-3} m=1 mm
4 0
3 years ago
There are Z protons in the nucleus of an atom, where Z is the atomic number of the element. An α particle carries a charge of +2
qaws [65]

Answer:

r = 2.84 \times 10^{-14} m

Explanation:

As per energy conservation we know that the electrostatic potential energy of the charge system is equal to the initial kinetic energy of the alpha particle

So here we can write it as

\frac{1}{2}mv^2 = \frac{k(2e)(ze)}{r}

now we know that

m = 1.67 \times 10^{-27} kg

e = 1.6 \times 10^{-19} C

z = 79

here kinetic energy of the incident alpha particle is given as

KE = 6.4 \times 10^{-13} J

now we have

6.4 \times 10^{-13} = \frac{(9\times 10^9)(1.6 \times 10^{-19})^2(79)}{r}

now we have

r = 2.84 \times 10^{-14} m

7 0
3 years ago
A spring with a force constant of 5.3 n/m has a relaxed length of 2.60 m. when a mass is attached to the end of the spring and a
olganol [36]
This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant. 
<span> U = (1/2)kx^2 
</span><span> U = (1/2)(5.3)(3.62-2.60)^2 
</span> U = <span> <span>2.75706 </span></span>J
6 0
3 years ago
On which factor potential energy depends?​
Ludmilka [50]

▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂☘️

The potential energy of the object depends on

  • the height of the object with respect to some reference points,
  • the mass of the object,
  • the gravitational field the object is in.

▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂☘️

Hope it helps ~

3 0
2 years ago
Read 2 more answers
A protein molecule in an electrophoresis gel has a negative charge.The exact charge depends on the pH of the solution, but 30 ex
Ad libitum [116K]

Answer:

The magnitude of the electric force on a protein with this charge is 7.2\times10^{-15}\ N

Explanation:

Given that,

Electric field = 1500 N/C

Charge = 30 e

We need to calculate the magnitude of the electric force on a protein with this charge

Using formula of electrostatic force

F=Eq

Where, F = force

E = electric field

q = charge

Put the value into the formula

F=1500\times30\times1.6\times10^{-19}

F=7.2\times10^{-15}\ N

Hence, The magnitude of the electric force on a protein with this charge is 7.2\times10^{-15}\ N

6 0
3 years ago
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