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3 years ago

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You are holding one end of an elastic cord that is fastened to a wall 3.0 m away. You begin shaking the end of the cord at 2.3 H

z , creating a continuous sinusoidal wave of wavelength 1.0 m. Part A How much time will pass until a standing wave fills the entire length of the string

1 answer:

Karo-lina-s [1.5K]3 years ago

7 0

Answer:

Time take to fill the standing wave to the entire length of the string is 1.3 sec.

Explanation:

Given :

The length of the one end $x=$ $3m$ , frequency of the wave $f$ = 2.3 Hz, wavelength of the wave λ = 1 m.

Standing wave is the example of the transverse wave, standing wave doesn't transfer energy in a medium.

We know,

∴ $v = f$ λ

Where $v =$ speed of the standing wave.

also, ∴ $v=\frac{x}{t}$

where $t =$ time take to fill entire length of the string.

Compare above both equation,

⇒ $t =$ $\frac{3}{2.3}$ sec

$t = 1.3sec$

Therefore, the time taken to fill entire length 0f the string is 1.3 sec.

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Density is a physical property that relates the mass of a substance to its volume. A. Calculate the density, in g/mL , of a liqu

Angelina_Jolie [31]

Answer:

A) 0.660 g/ml

B) 1.297 ml

C) 0.272 g

Explanation:

Every substance, body or material has mass and volume, however the mass of different substances occupy different volumes. This is where density $D$ appears as a physical characteristic property of matter that establishes a relationship between the mass $m$ of a body or substance and the volume $V$ it occupies:

$D=\frac{m}{V}$ (1)

Knowing this, let's begin with the answers:

<h2 /><h2>Answer A:</h2>

Here the mass is $m=0.155g$ and th volume $V=0.000235L=0.235mL$

Solving (1) with these values:

$D=\frac{0.155g}{0.235mL}$ (2)

$D=0.660g/mL$ (3)

<h2>Answer B:</h2>

In this case the mass of a sample is $m=4.71g$ and its density is $D=3.63g/mL$ .

Isolating $V$ from (1):

$V=\frac{m}{D}$ (4)

$V=\frac{4.71g}{3.63g/mL}$ (5)

$V=1.297mL$ (5)

<h2>Answer C:</h2>

In this case the volume of a sample is $V=0.293mL$ and its density is $D=0.930g/mL$ .

Isolating $m$ from (1):

$m=D.V$ (6)

$m=(0.930g/mL)(0.293mL)$ (7)

$m=0.272g$ (8)

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3 years ago

Read 2 more answers

What is another name for the introitus? psych 230?

Vera_Pavlovna [14]

Vaginal opening. areola is the part of the breast.

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3 years ago

To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(

DerKrebs [107]

Answer:

0.0549 m

Explanation:

Given that

equation y(x,t)=Acos(kx−ωt)

speed v = 8.5 m/s

amplitude A = 5.5*10^−2 m

wavelength λ = 0.5 m

transverse displacement = ?

v = angular frequency / wave number

and

wave number = 2π/ λ

wave number = 2 * 3.142 / 0.5

wave number = 12.568

angular frequency = v k

angular frequency = 8.5 * 12.568

angular frequency = 106.828 rad/sec ~= 107 rad/sec

so

equation y(x,t)=Acos(kx−ωt)

y(x,t)= 5.5*10^−2 cos(12.568 x−107t)

when x =0 and and t = 0

maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))

maximum y(x,t)= 5.5*10^−2 m

and when x = x = 1.52 m and t = 0.150 s

y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )

y(x,t)= 5.5*10^−2 × (0.9986)

y(x,t) = 0.0549 m

so the transverse displacement is 0.0549 m

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3 years ago

A 61 kg skater is traveling at 2.5 m/s while carrying a 4.0 kg bowling ball. After he throws the bowling ball forward at twice t

gregori [183]

The final velocity of the skater is 2.34 m/s forward

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system before and after the ball is thrown must be conserved, in absence of external forces.

Before the ball is thrown, the total momentum is:

$p_i = (M+m)u$

where

M = 61 kg is the mass of the skater

m = 4.0 kg is the mass of the ball

u = 2.5 m/s (forward) is the combined velocity of the skater and the ball

After, the ball is thrown at twice the velocity, so the final total momentum is

$p_f = MV+mv$

where

V is the final velocity of the skater

v = 2(2.5) = 5.0 is the final velocity of the ball

Since the total momentum must be conserved, we can write

$p_i = p_f\\(M+m)u = MV+mv\\V=\frac{(M+m)u-mv}{M}=\frac{(61+4.0)(2.5)-(4.0)(5.0)}{61}=2.34 m/s$

So, the skater is moving at 2.34 m/s (forward) after the shot.

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3 years ago

In any problems involving circular motion, which way does the tangential speed vector point?

Anton [14]

In what may be one of the most remarkable coincidences in
all of physical science, the tangential component of circular
motion points along the tangent to the circle at every point.

The object on a circular path is moving in that exact direction
at the instant when it is located at that point in the circle. The
centripetal force ... pointing toward the center of the circle ...
is the force that bends the path of the object away from a straight
line, toward the next point on the circle. If the centripetal force
were to suddenly disappear, the object would continue moving
from that point in a straight line, along the tangent and away from
the circle.

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3 years ago