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3 years ago

If the period of a spring is 5 seconds, what is the frequency?

Physics

2 answers:

masha68 [24]3 years ago

8 0

Frequency(f) = 1/Period(T)
f = 1/5 = 0.2 Hz

Anastasy [175]3 years ago

3 0

Answer:0.2Hz

Explanation:

Frequency is defined as the number of cycles completed by a wave in one seconds. Frequency is therefore measured in cycles/second or Hertz(Hz).

Period (T) is the time taken to complete an oscillation. The relationship between both is that frequency (F) is the reciprocal of the period. Mathematically, F = 1/T

From the question, Period (T) is 5seconds

Frequency = 1/5

F = 0.2Hz

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a student in your class says "electromagnetic waves can travel through outer space but mechanical waves can’t.” is it this stude

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Answer:

Right

Explanation:

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3 years ago

Lab: newton's laws of motion assignment: lab report

poizon [28]

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2 years ago

What is the kinetic energy of a 200 kg satellite as it follows a circular orbit of radius 8x106m around the earth?

motikmotik

Given the equation for the Speed of a Satellite

v = SqRt{Gravitational Constant}{Mass of Earth} divided by the radius given in your problem

we have:

(square root whole term on right side)

v = G Me
———
r

so. (6.67x10^-11)(5.97x10^24)
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(8.0x10^6)

v = 7055 m/s (which is reasonable)

so utilize the Kinetic Energy Formula

KE = 1/2mv^2

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4 0

3 years ago

Read 2 more answers

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inysia [295]

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3 years ago

A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr

rjkz [21]

Answer:

Part A

I = 1.4 mW/m²

Part B

β = 91.46 dB

Explanation:

Part A

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^{2}}$

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

$I = \frac{34}{4\pi (44)^{2}}$

I = 0.0013975 W/m²

≈ I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

Part B

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value. It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;

β $= 10log_{10}\frac{I}{I_{0}}$ dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

∴ β $= 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}$

β $= 10log_{10} (1.4 * 10^{9})$

β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.

4 0

3 years ago