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3 years ago

I need help with answering the given question.

Physics

1 answer:

Viktor [21]3 years ago

8 0

The two hydrogen gas molecules split and the two oxygen gas molecules split and made two water molecules by having two hydrogen gas molecules and one oxygen gas molecule join twice...
Therefore, there would be 3 or 6
if i am not mistaken
hope it helps

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A speaker generates a continuous tone of 440 Hz. In the drawing, sound travels into a tube that splits into two segments, one lo

chubhunter [2.5K]

Answer:

The minimum difference between the lengths of the two tubes should be 0.385 meters.

Explanation:

As we known that for any two waves to arrive in phase at any point the difference in the path traveled by the waves should be an integral multiple of the wavelength of the wave.

Mathematically we can write:

$\Delta x=n\frac{\lambda }{2}$

For the given wave we have

$\lambda =\frac{v}{\nu }$

Applying values we get

$\lambda =\frac{339}{440 }=0.77m$

Thus the minimum difference in the lengths of the tubes can be obtained by putting the value of n = 1

$\therefore \Delta x=1\times \frac{0.77}{2}=0.385m$

7 0

4 years ago

In simple harmonic motion, the acceleration is proportional to In simple harmonic motion, the acceleration is proportional to:__

erik [133]

Answer:

e. all of the above

Explanation:

In simple harmonic motion, the acceleration is given as;

a = -ω²x = -(2πf)²x

$a_{max} = -\omega ^2A$

where;

ω is the angular velocity

f is the frequency

x is the displacement

A is the amplitude

Thus, In simple harmonic motion, the acceleration is proportional to the amplitude, velocity, frequency, and displacement.

The correct option is E.

"all of the above"

5 0

3 years ago

At the presentation ceremony, a championship bowler is presented a 1.64-kg trophy which he holds at arm's length, a distance of

solong [7]

To solve the problem it is necessary to take into account the concepts of the kinetic equations for the description of the torque at the rate of force and distance.

By definition the torque is given by,

$\tau = F*d$

where,

$F= Force$

$d = Distance$

For the problem in question the mass of the trophy is 1.64Kg and the distance of the tropeo to the board (the shoulder) is 0.655m

PART A) For part A, the torque with the given mass and the stipulated torque in the horizontal plane must be calculated as well,

$\tau = F*d$

For Newton's second law

$\tau = mg*d$

$\tau = 1.64*9.81*0.655$

$\tau = 10.5Nm$

PART B) For part B there is an angle of 26 degrees with respect to the horizontal, therefore to know the net torque it is necessary to know the horizontal component to the formed angle, that is,

$\tau = F*dcos\theta$

$\tau = mgdcos\theta$

$\tau = 1.64*9.81*0.655*cos26$

$\tau = 9.471Nm$

3 0

4 years ago

A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.65 m/s2 for 17.0 s. 2

nlexa [21]

Hello

Let's solve the problem in the three different steps

1) Uniformly accelerated motion, with acceleration $a_1 = 2.65~m/s^2$ and for a total time of $t_1=17~s$ . The body is initially at rest, so the distance covered is given by
$S= \frac{1}{2}a_1t_1^2=382.9~m$
Calling $v_f$ and $v_i$ the final and initial velocity, and since the $v_i=0~m/s$ because the body starts from rest, we can use
$a= \frac{v_f-v_i}{t}$
to find the final velocity after this first leg:
$v_{f}=v_i+a_1t_1=45~m/s$
And the average velocity in this first leg is
$v_1= \frac{v_f+v_i}{2}=22.5~m/s$

2) Uniform motion. The velocity is constant and it is equal to the final velocity of the first leg: $v_2=45~m/s$ . This is also the average velocity of the second leg.
The total time of this second leg is $t_2=1.60~min = 96~s$ . The distance covered is given by
$S_2=v_2t_2=45~m/s \cdot 96~s=4320~m$

3) Uniformly decelerated motion, with constant deceleration of $a_3=-9.39~m/s^2$ and for a total time of $t_3=4.8~s$ . Here, the initial velocity of the body is the final velocity of the previous leg, i.e. $v_i=45~m/s$ . Therefore, the distance covered in this leg is given by
$S_3=v_i t_3 + \frac{1}{2} a_3 t^2 =107.8~m$
The final velocity in this leg is given by
$v_f=v_i+at=45~m/s-9.39~m/s^2 \cdot 4.8~s = -0.07~m/s$
The negative sign means that after decelerating, the body has started to go in the opposite direction. Similarly to step 1, the average velocity in this leg is given by
$v_3 = \frac{1}{2}(v_f+v_i)= \frac{1}{2}(-0.07~m/s+45~m/s)= 22.5~m/s$

4) Finally, the total distance covered in the motion is
$S=S_1+S_2+S_3=382.9~m+4320~m+107.8~m=4810.7~m$
To find the average velocity, we must "weigh" the average velocity of each leg for the correspondent time of that leg:
$v_{ave}= \frac{v_1t_1+v_2t_2+v_3t_3}{t_1+t_2+t_3}=40.8~m/s$

8 0

3 years ago

A —————— error is the perceived shift in a objective position as it is viewed from

Citrus2011 [14]

Answer:

Hi student

Your name ???

3 0

3 years ago