Ask question

Ask question

All categories

3 years ago

9

Explain why the amplitude of the wave did not change when you increased the frequency of the stimulation. how well did the resul

ts compare with your prediction?

1 answer:

barxatty [35]3 years ago

5 0

The amplitude did not change when the recurrence was expanded on the grounds that the long headstrong time of the heart forestalls adjustment. It is the most extreme removal or separation moved by a point on a vibrating body or wave measured from its balance position. It is equivalent to the one-a large portion of the length of the vibration way.

You might be interested in

The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff

stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet m = 45 kg

coefficient of static friction μ = 0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

$\sum f_y =0$

The upward force and the downward force is :

$N_A+N_B = mg$

where;

$\mathbf { N_A \ and \ N_B}$ are the normal contact force at center point A and B respectively .

$N_A+N_B = 45*9.8$

$N_A+N_B = 441$ ------- equation (1)

Considering the forces on the horizontal axis:

$\sum f_x = 0$

$F_A +F_B = P$

where ;

$\mathbf{ F_A \ and \ F_B }$ are the static friction at center point A and B respectively.

which can be written also as:

$\mu_s N_A + \mu_s N_B = P$

$\mu_s( N_A + N_B) = P$

replacing our value from equation (1)

P = 0.30 ( 441)

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for the cabinet is not to tip over is calculated by determining the limiting condition of the unbalanced torque whose effect is canceled by the normal reaction at $N_A$ and it is shifted to $N_B$ :

Then:

$\sum M _B = 0$

$P*h = mg*0.24$

$h =\frac{45*9.8*0.24}{132.3}$

h = 0.8 m

Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

6 0

2 years ago

The specific heat of Aluminum is 0.9 J/g K. The specific heat of Copper is 0.39 J/g K. If samples of equal mass of both Aluminum

AleksandrR [38]

<span> <span><span> <span> Answer:The Aluminum loses a little more than twice the heat of the Copper.Explanation:<span>
Since specific heat is part of the equation. A smaller specific heat will create a smaller heat gain or loss. </span>
<span>Hope this helped!!!!</span></span> </span> </span></span>

6 0

2 years ago

Read 2 more answers

A projectile falls beneath the straight-line path it would follow if there were no gravity. How many meters does it fall below t

maria [59]

Answer:

Explanation:

Suppose v is the initial velocity and $\theta$ is the angle of inclination

distance traveled in vertical direction in t=1 s

When gravity is present

$y=vt+\frac{1}{2}at^2$

where $y=vertical\ distance$

$a=acceleration$

$t=time$

$v=initial\ velocity$

here initial velocity is v\sin \theta [/tex] so

$y=v\sin \theta \times 1-\frac{1}{2}gt^2$

$y=v\sin \theta -0.5g$

In absence of gravity

$y_2=v\sin \theta \times t$

$y_2=v\sin \theta \times 1$

$\Delta y=y_2-y=v\sin \theta -v\sin \theta +0.5 g=4.9\ m$

4 0

2 years ago

You notice that you always get sleepy right after lunch. What part of the scientific process are you performing by noticing this

masya89 [10]

Always getting sleepy after lunch is part of data collection.

7 0

3 years ago

Read 2 more answers

How much work is required to accelerate a proton from rest up to a speed of 0.993 c? Express your answer with the appropriate un

Keith_Richards [23]

Answer:

(a). The work done is 7001 MeV.

(b). The momentum of this proton is $4.20\times10^{8}\ kg-m/s$ .

Explanation:

Given that,

Speed = 0.993 c

We need to calculate the work done

Using work energy theorem

The work done is equal to the kinetic energy relative to the proton

$W=K.E$

$W=\dfrac{m_{p}c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}-m_{p}c^2$

Put the value into the formula

$W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(\dfrac{0.993c}{c})^2}}-1.67\times10^{-27}\times(3\times10^{8})^2$

$W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(0.993)^2}}-1.67\times10^{-27}\times(3\times10^{8})^2$

$W=1.122\times10^{-9}\ J$

$W=7001\ MeV$

(b). We need to calculate the momentum of this proton

Using formula of momentum

$p=\dfrac{m_{0}v}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Put the value into the formula

$p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(\dfrac{0.993c}{c})^2}}$

$p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(0.993)^2}}$

$p=1.404\times10^{-26}c$

$p=4.20\times10^{8}\ kg-m/s$

Hence, (a). The work done is 7001 MeV.

(b). The momentum of this proton is $4.20\times10^{8}\ kg-m/s$ .

4 0

3 years ago