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3 years ago

Define amplitude and frequency as they relate to a wave?

1 answer:

3 0

<span>frequency definition. In physics, the number of crests of a wave that move past a given point in a given unit of time. The most common unit offrequency is the hertz (Hz), corresponding to one crest per second. Thefrequency of a wave can be calculated by dividing the speed of the wave by the wavelength.</span><span>amplitude, in physics, the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. It is equal to one-half the length of the vibration path.</span>

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Im Stuck In This Thing Called ''Life" Can You Help Me.

pantera1 [17]

im still trying to find someone who can help me

4 0

4 years ago

Read 2 more answers

A 3.0 kg object is loaded into a toy spring gun, and the spring has a spring constant 750N/m. The object compresses the spring b

kiruha [24]

Answer:

Explanation:

mass of object, m = 3 kg

spring constant, K = 750 n/m

compression, x = 8 cm = 0.08 m

angle of gun, θ = 30°

(a) As the ball is launched, it has some velocity due to the compression in the spring, so it has some kinetic energy.

(b) Let v be th evelocity of ball at the tim eof launch.

by using the conservation of energy

1/2 Kx² = 1/2 mv²

750 x 0.08 x 0.08 = 3 x v²

v = 1.265 m/s

By use of the formula of maximum height

$h = \frac{v^{2}Sin^{2}\theta}{2g}$

$h = \frac{1.265^{2}Sin^{2}30}{2\times 9.8}$

h = 0.02 m

h = 2 cm

4 0

3 years ago

A small glass bead has been charged to + 30.0 nC . A small metal ball bearing 2.60 cm above the bead feels a 1.80×10−2 N downwar

BlackZzzverrR [31]

Answer:

The charge on the ball bearing 4.507 × 10^-8 C

Explanation:

From Coulomb's law

F = kq1q2/r²

make q2 the subject

q2 = Fr²/kq1

q2 = (1.8×10^-2 × 0.026²) ÷ (9×10^9 × 30×10^-9)

q2 = 4.507 × 10^-8 C

8 0

3 years ago

un conductor de plata (p=1,6x10°-6ohm x m) tiene una seccion de 5x10°-6 m°2 y una longitud de 110 m . calcular la resistencia

Alexus [3.1K]

Responder:

35,2 ohm.

Explicación:

Dado:

La resistencia específica del conductor es,

La longitud del conductor es,

El área de la sección transversal del conductor es,

Sabemos que la resistencia de un conductor es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.

Por lo tanto, la resistencia se puede expresar como:

$R=\frac{\rho\times l}{A}$