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3 years ago

1. Distinguish between amplitude and frequency and give an example.

Physics

1 answer:

Svetlanka [38]3 years ago

5 0

Answer:

The difference between frequency and amplitude is that frequency is a measurement of cycles per second, and amplitude is a measurement of how large a wave is. Amplitude represents the wave's energy. Large waves contain more energy than small waves. For example, a sound wave with a high amplitude is perceived as loud.

Explanation:

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A physical proprety of gold is its ?

nydimaria [60]

Density because it can measured and felt.

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3 years ago

Two volumes of William Shakespeare stand on a bookshelf next to each other: volume one, then volume two. Each volume is 4 cm thi

Setler79 [48]

Answer:

L = 7 [cm]

Explanation:

To solve this problem we must analyze each of the distances mentioned and take into account the number of covers and thicknesses of these.

The worm crosses through the sheets of the first book, this distance can be determined by the following length analysis.

4 = P + 2*C

Where:

P = thicknesses of the pages [cm]

C = thicknesses of each cover [cm]

P = 4 - 2*(0.5)

P = 3 [cm]

The distance crossed was:

L = P + 2C + P "the pages of the first book + 2 covers + the pages of the second book"

L = 3 + (2*0.5) + 3

L = 7 [cm]

7 0

3 years ago

An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the o

Sliva [168]

Answer:

t = 0.37 seconds

Explanation:

t = (1/4)T

Maximum acceleration is;

a_max = Aω²

In simple harmonic motion, we know that v_max = Aω

Thus, a_max = v_max•ω

ω = a_max/v_max

We know that Period is given by;

T = 2π/ω

From initially, t = (1/4)T so, T = 4t

Thus, 4t = 2π/(a_max/v_max)

t = (2π/4)(v_max/a_max)

We are given;

Maximum velocity;v_max = 1.47 m/s

Max acceleration;a_max =6.24 m/s²

Thus,

t = (2π/4)(1.47/6.24)

t = 0.37 seconds

8 0

3 years ago

I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

The period of a simple pendulum is given by

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

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

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

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

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5 0

3 years ago

What is kinematics?<br>explain!!~<br><br>thankyou ~

artcher [175]

Answer:

Kinematics is the branch of mechanics concerned with the motion of objects without reference to the forces which cause the motion.

-the features or properties of motion in an object.

Explanation:

Hope this helps <3

Have a great day!

7 0

2 years ago

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