Ask question

Ask question

All categories

1 month ago

12

A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a

function of time according to the equation y = 0.049 cos(7t) where y is the displacement in meters and the time t is in seconds. show answer No Attempt 33% Part (a) What is the amplitude of the wave, in meters?

1 answer:

kicyunya [3.2K]1 month ago

3 0

Answer:

Amplitude, A = 0.049 meters

Explanation:

Given that

A harmonic wave propagates in the positive x direction at 6 m/s along a tight string. A fixed point along this string oscillates over time according to the equation:

$y = 0.049 \cos(7t)$ .......(1)

The general wave equation is expressed as:

$y=A\cos(\omega t)$ .......(2)

A denotes the wave's amplitude

When we compare equation (1) with (2), we find:

A = 0.049 meters

Thus, the amplitude of the wave is 0.049 meters.

You might be interested in

Which model is used to describe the interaction of external forces that affect an organization's strategy and ability to compete

ValentinkaMS [3465]

Answer:

Competitive forces model

Explanation:

The Competitive forces model is a crucial instrument in strategic analysis aiming to assess an organization’s competitiveness. Commonly referred to as the "Five Force Model of Porter", this framework includes five key factors: the intensity of rivalry among existing competitors, the negotiating power of buyers, the threat posed by potential new entrants, the bargaining strength of suppliers, and the risk of substitute products or services.

These elements significantly influence an organization's competitive strategy and its likelihood of success.

5 0

1 month ago

An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass

kicyunya [3294]

Answer:

A) The updated amplitude = 0.048 m

B) Period T = 0.6 seconds

Explanation: Please refer to the attached documents for the solution.

4 0

1 month ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Sav [3153]

Complete Question

An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.

At time 5 seconds, I = 1.2 A.

We need to find the charge Q traveling through a cross-section of the conductor from time 0 to time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

The question indicates that

The wire’s diameter is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

Aluminum's resistivity is $2.75*10^{-8} \ ohm-meters.$

The electric field variation is described as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

The charge is effectively given by the equation

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area expressed as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

Thus,

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

By substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

The question states that t = 5 seconds

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

1 month ago

"A block of metal weighs 40 N in air and 30 N in water. What is the buoyant force on the block due to the water? The density of

Sav [3153]

Answer:

The buoyant force acting on the block from the water is 10 N

Explanation:

The buoyant force (F_B) experienced by a block is defined as the difference between its actual weight in air and its weight when submerged in water.

The data provided indicates:

A metal block weighs 40 N in air and 30 N in water.

Thus, F_B = 40 - 30 = 10 N

therefore, the buoyant force acting on the block from the water amounts to 10 N

6 0

22 days ago

Read 2 more answers

A sphere of radius 0.03m has a point charge of q= 7.6 micro C located at it’s centre. Find the electric flux through it?

Sav [3153]

Answer:

The electric flux going through the sphere is $8.58 *10^{5} \frac{Nm^2}{C}$

Explanation:

Given

$Radius,\ r = 0.03m\\Charge,\ q =7.6\µC$

Required

Calculate the electric flux

Electric flux can be computed using the formula;

Ф = q/ε

Where ε stands for the electric constant permittivity

ε $= 8.8542 * 10^{-12}$

Substituting ε $= 8.8542 * 10^{-12}$ and $q =7.6\µC$ ; the formula simplifies to

Ф = $\frac{7.6\µC}{8.8542 * 10^{-12}}$

Ф = $\frac{7.6 * 10^{-6}}{8.8542 * 10^{-12}}$

Ф = $\frac{7.6}{8.8542} *\frac{10^{-6}}{10^{-12}}$

Ф = $\frac{7.6}{8.8542} *10^{12-6}}$

Ф = $0.85834970974 *10^{12-6}}$

Ф = $0.85834970974 *10^{6}}$

Ф = $8.5834970974 *10^{5}}$

Ф = $8.58 *10^{5} \frac{Nm^2}{C}$

Thus, the electric flux through the sphere is $8.58 *10^{5} \frac{Nm^2}{C}$

7 0

2 months ago