A _______ is a repeating disturbance or vibration that transfers or moves energy from place to place without transporting mass.

Which expression correctly describes energy using SI units

mote1985 [20]

Answer:

Joule

Explanation:

energy, work, quantity of heat

m2·kg·s-2

5 0

3 years ago

A transverse, sinusoidal wave travels in a string and can be described by the function: y(x,t)=0.87 sin(21x−4.9t). What is the s

pshichka [43]

Answer:

2.61m/s

Explanation:

Given the wave function;

y(x,t)=0.87 sin(21x−4.9t).

The general wave equation is expressed as;

$y = Asin(2\pi ft + 2\pi x /\lambda)$

f is the frequency of the wave

t is the time

$\lambda\\$ is the wavelength

On comparing;

2πft = 4.9t

2πf= 4.9

f = 4.9/2π

f = 4.9/2(3.14)

f = 4.9/6.28

f = 0.78Hz

Get the wavelength;

2πx/ $\lambda$ = 21x

2π/ $\lambda$ = 21

2π = 21 $\lambda$

$\lambda$ = 21/2π

$\lambda$ = 21/2(3.14)

$\lambda$ = 21/6.28

$\lambda$ = 3.34m

Speed = frequency * wavelength

Speed of the wave = 0.78 * 3.34

Speed of the wave = 2.61m/s

Hence the speed of the wave is 2.61m/s

5 0

3 years ago

Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note

mote1985 [20]

Answer:

Frequency = 658 Hz

Explanation:

The third harmonics of A is,

$f_{A}=(3)(440Hz)=1320Hz$

The second harmonics of E is,

$f_{E}=(2)(659Hz)=1318Hz$

The difference in the two frequencies is,

delta_f = 1320 Hz - 1318 Hz = 2 Hz

The beat frequency between the third harmonic of A and the second harmonic of E is,

delta_f = $3f_{A}-2f_{E}$

$f_{E}=\frac{3f_{A}-delta_f}{2}$

We have calculate the frequency of the E string when she hears four beats per second, then

delat_f = 4 Hz

$f_{E}=\frac{3(440Hz)-4Hz}{2}$

$f_{E}=658Hz$

Hope this helps!

5 0

3 years ago

A 150.0-kg crate rests in the bed of a truck that slows from 50.0 km/h to a stop in 12.0 s. The coefficient of static friction b

Leokris [45]

By Newton's second law,

• the net force acting vertically on the crate is 0, and

∑ F = n - mg = 0 ==> n = mg = 1470 N

where n is the magnitude of the normal force; and

• the net force acting in the horizontal direction on the crate is also 0, with

∑ F = f - b = 0 ==> b = f = µn = 0.645 (1470 N) = 948.15 N

where b is the magnitude of the braking force, f is (the maximum) static friction, and µ is the coefficient of static friction. This is to say that static friction has a maximum magnitude of 948.15 N. If the brakes apply a larger force than this, then the crate will begin to slide.

Note that we are taking the direction of the truck's motion as it slows down to be the positive horizontal direction. The brakes apply a force in the negative direction to slow down the truck-crate system, and static friction keeps the crate from sliding off the truck bed so that the frictional force points in the positive direction.

Let a be the acceleration felt by the crate due to either the brakes or friction. Use Newton's second law again to solve for a :

f = ma ==> a = (948.15 N) / (150.0 kg) = 6.321 m/s²

With this acceleration, the truck will come to a stop after time t such that

0 = 50.0 km/h - (6.321 m/s²) t ==> t ≈ (13.9 m/s) / (6.321 m/s²) ≈ 2.197 s

and this is the smallest stopping time possible.

8 0

3 years ago

Describing Acceleration

FrozenT [24]

I’m going to order the list by ABCs
The ones I put down are correct

A
B
C
D

4 0

3 years ago

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