Which particle is a negatively charged ion?

A simple pendulum has a period of 3.45 second, when the length of the pendulum is shortened by 1.0m, the period is 2.81 second c

den301095 [7]

Answer:

Original length = 2.97 m

Explanation:

Let the original length of the pendulum be 'L' m

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Original time period of the pendulum (T) = 3.45 s

Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, $L_1=L-1$

New time period of the pendulum is, $T_1=2.81\ s$

We know that, the time period of a simple pendulum of length 'L' is given as:

$T=2\pi\sqrt{\frac{L}{g}}$ -------------- (1)

So, for the new length, the time period is given as:

$T_1=2\pi\sqrt{\frac{L_1}{g}}$ ------------ (2)

Squaring both the equations and then dividing them, we get:

$\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1$

Now, plug in the given values and calculate 'L'. This gives,

$L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m$

Therefore, the original length of the simple pendulum is 2.97 m

4 0

3 years ago

Can someone answer and explain these questions?

Andru [333]

Answer:

oh that looks hard but sorry im not good at math ether so sorry i couldnt help

Explanation:

7 0

2 years ago

9.46 Air enters the compressor of an ideal cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. T

Colt1911 [192]

Answer:

a) 67pa

Explanation:

8 0

3 years ago

What metal is used to make hard magnets?

Lelu [443]

The most common metals used for permanent magnets are iron, nickel, cobalt and some alloys of rare earth metals

3 0

4 years ago

Figure 21.62 shows a box on whose surfaces the electric field is measured to be horizontal and to the right. On the left face (3

natima [27]

(a) The electric flux on left face is 0.24 Vm and on the right face is 1 Vm.

(b) The total flux is 1.24 Vm.

(c) The total amount of charge that is inside the box is 1.1 x 10⁻¹¹ C.

The area of the left face is calculated as follows;

A1 = 0.03 m x 0.02 m = 0.0006 m²

<h3>Electric flux on the left face</h3>

Ф1 = EA1

Ф1 = (400 V/m)( 0.0006 m²) = 0.24 Vm

Let the dimension of the right face = 5 cm by 2 cm

<h3>Area of the right face</h3>

A2 = 0.05 m x 0.02 m = 0.001 m²

<h3>Electric flux on the right face</h3>

Ф2 = EA2

Ф2 = (1000 V/m)( 0.001 m²) = 1 Vm

<h3>Total flux</h3>

Ф = Ф1 + Ф2

Ф = 0.24 Vm + 1 Vm = 1.24 Vm

<h3>Total charge inside the box</h3>

Ф = Q/ε

Q = εФ

Q = (8.85 x 10⁻¹²)(1.24)

Q = 1.1 x 10⁻¹¹ C

Thus, the electric flux on left face is 0.24 Vm and on the right face is 1 Vm.

The total flux is 1.24 Vm and the total amount of charge that is inside the box is 1.1 x 10⁻¹¹ C.

Learn more about total flux here: brainly.com/question/26289097

#SPJ1

7 0

2 years ago