Ask question

Ask question

All categories

4 days ago

5

Three pendulums with strings of the same length and bobs of the same mass are pulled out to angles θ1, θ2, and θ3, respectively,

and released. The approximation sin θ = θ holds for all three angles, with θ3 > θ2 > θ1. How do the angular frequencies of the three pendulums compare?

1 answer:

Sav [2.2K]4 days ago

6 0

Answer:

All three pendulums will have the same angular frequencies.

Explanation:

For a simple pendulum, the time period using the approximation $sin(\theta )\approx \theta$ is expressed as:

$T=2\pi\sqrt{\frac{l}{g}}$

The angular frequency $\omega$ is defined as

$\omega =\frac{2\pi }{T}\\\\\omega =\frac{2\pi}{2\pi \sqrt{\frac{l}{g}}}\\\\\therefore \omega =\sqrt{\frac{g}{l}}$

Since the angular frequency remains unaffected by the initial angle (valid strictly for small angle approximations), we deduce that the angular frequencies of the three pendulums are identical.

You might be interested in

A positive charge moves in the direction of an electric field. Which of the following statements are true?

kicyunya [2264]

Answer:

The potential energy tied to the charge diminishes.

The electric field performs negative work on the charge.

Explanation:

3 0

1 month ago

Read 2 more answers

You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acce

serg [2593]

The duration required for the seventh car to pass amounts to 13.2 seconds. The train's movement is characterized by uniform acceleration, enabling the application of suvat equations. Initially, we analyze the movement of the first car, utilizing the equation for distance s covered in time t, which corresponds to the length of one car, with u = 0 as the initial velocity and a representing acceleration, over t = 5.0 s. We can rearrange the equation reflecting L as the length of one car. This is similarly applicable for the initial seven cars, accounting for the distance of 7L and the required time t'. With constant acceleration retained, we can derive t' through substitution in the equation, leading to fundamental conclusions regarding the relationship exhibited in the graph of distance against time in uniformly accelerated motion.

8 0

8 days ago

Astronomers determine that a certain square region in interstellar space has an area of approximately 2.4 \times 10^72.4×10 7

Sav [2226]

Answer:

1.5 × 10³⁶ light-years

Explanation:

A particular square area in interstellar space measures roughly 2.4 × 10⁷² (light-years)². To find the area of a square, the following formula is utilized:

A = l²

where,

A represents the area of the square

l denotes the length of one side of the square

Thus, l = √A = √2.4 × 10⁷² (light-years)² = 1.5 × 10³⁶ light-years

5 0

26 days ago

For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of Italia

serg [2593]

Answer:

Part a)

A = 0.0581 m

Part b)

T = 0.37 s

Explanation:

A slice is dropped onto the plate from a height of 0.250 m,

therefore the speed of the slice upon impact is calculated as

$v = \sqrt{2gh}$

We know that

$v = \sqrt{2(9.81)(0.250)}$

$v = 2.21 m/s$

Now applying the conservation of momentum:

$mv = (m + M)v_f$

$m = 0.300 kg$

$M = 0.400 kg$

From this equation, we find:

$0.300 (2.21) = (0.300 + 0.400) v_f$

$v_f = 0.95 m/s$

$0.400 (9.81) = 200 x_1$

When the slice rests on the plate, the new mean position can be expressed as

$x_1 = 0.01962 m$

$(0.300 + 0.400)9.81 = 200 x_2$

We also determine that the speed of SHM is represented as

$x_2 = 0.0343 m$

Here, we derive values from

$v = \omega\sqrt{A^2 - x^2}$

$\omega = \sqrt{\frac{k}{m + M}}$

$\omega = \sqrt{\frac{200}{0.300 + 0.400}}$

$\omega = 16.9 rad/s$

$a = x_2 - x_1 = 0.0343 - 0.01962 = 0.0147 m$

Using the previous formula gives:

$0.95 = 16.9\sqrt{A^2 - 0.0147^2}$

$A = 0.0581 m$

Part b)

The time period for the scale is computed as

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{16.9}$

$T = 0.37 s$

8 0

19 days ago

You are working on a laboratory device that includes a small sphere with a large electric charge Q. Because of this charged sphe

Yuliya22 [2420]

Response:

The primary consequence is an increase in induced charge at the nearest points. However, the overall net charge remains zero, meaning it does not influence the flow.

We can utilize Gauss's law to solve this problem

Ф = ∫ e. dA = $q_{int}$ / ε₀

The flow of the field is directly correlated to the charge within it. Consequently, placing a Gaussian surface beyond the non-conductive spherical shell means the flow will be zero since the sphere’s charge equals the charge induced in the shell, resulting in a net charge of zero. This evaluation shows that the shell effectively obstructs the electric field.

According to Gauss's law, if the sphere is offset, the only effect it generates is an increment in induced charge at the nearest points. Nevertheless, the net charge remains zero, so it does not impact the flow; irrespective of the sphere's position, the total induced charge is consistently equal to the charge on the sphere.

5 0

22 days ago